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✓ 58 results · all verified · 43 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 15 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Finite Fields and Cyclotomic Extensions

1 · Prerequisites

2 · Summary

Finite fields on this page are governed by the Frobenius endomorphism, Artin's fixed-field theorem, and the finite Galois correspondence. The page uses the published algebraic-extension and Galois pages for degree, splitting, and automorphism machinery, then adds the relative Frobenius, the finite cyclic-group lemmas it needs, Dedekind independence for normal bases, and the polynomial and unit-group facts that control cyclotomic extensions and finite-field factorisations.

The development begins with finite fields: the $q$ -power map identifies the fixed field, determines the full Galois group, describes every intermediate field, and controls Frobenius conjugates and normal bases. It then turns to roots of unity and cyclotomic extensions, defines $Φ_{n}$ by the divisor recursion, proves the primitive-root and irreducibility theorems, and uses the resulting Galois groups to study finite-field factorisation, composita and intersections over $Q$ , primes congruent to $1$ modulo $n$ , and finite abelian Galois groups over $Q$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A finite cyclic group has exactly one subgroup of each order dividing its own

Statement

Let $G = ⟨ g ⟩$ be a cyclic group of finite order $n \geq 1$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups). For every positive divisor $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ):

$⟨ g^{n / d} ⟩ is a subgroup of G of order d,$

it is the only subgroup of $G$ of order $d$ , and every subgroup of $G$ is of this form for exactly one such $d$ . Moreover, for positive divisors $d$ and $e$ of $n$ ,

$⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩ ⟺ d ∣ e .$

The two extremes are instances rather than exceptions: $d = 1$ gives the trivial subgroup ${e}$ and $d = n$ gives $G$ itself, and at $n = 1$ the only divisor is $d = 1$ , where $G$ is trivial.

Facts & Assumptions

Given: A cyclic group $G = ⟨ g ⟩$ whose underlying set is finite of order $n \geq 1$ ; divisibility of integers is that of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , and orders are those of The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity.

[A1]

$∣ G ∣ = n$ , and since $G = ⟨ g ⟩$ , [L1] gives $ord (g) = ∣ ⟨ g ⟩ ∣ = n$ .

[L1]

For an element $x$ of finite order $m$ in a group: $x^{k} = e$ if and only if $m ∣ k$ ; the powers $x^{0}, x^{1}, \dots, x^{m - 1}$ are pairwise distinct; and $⟨ x ⟩ = {x^{s} : s \in N, s < m}$ , so $⟨ x ⟩$ is finite with $∣ ⟨ x ⟩ ∣ = m = ord (x)$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L2]

$⟨ x ⟩ = {x^{k} : k \in Z}$ : the cyclic subgroup generated by $x$ is exactly the set of integer powers of $x$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L3]

For a finite group $G$ and $H \leq G$ one has $∣ G ∣ = [G : H] ∣ H ∣$ ; consequently $∣ H ∣$ divides $∣ G ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Fix a positive divisor $d$ of $n$ and put $c : = n / d$ , an integer with $c \geq 1$ and $n = c d$ .

givenalgebra

2.1

The element $g^{c}$ has order exactly $d$ : first $(g^{c})^{d} = g^{c d} = g^{n} = e$ by [A1] and [L1]; and if $1 \leq j < d$ then $1 \leq c j < c d = n$ , so $n ∤ c j$ and hence $(g^{c})^{j} = g^{c j} \neq e$ by [L1]. Therefore $∣ ⟨ g^{c} ⟩ ∣ = d$ by [L1].

step 1.1A1L1algebra

2.2

The set of $x \in G$ with $x^{d} = e$ is exactly $⟨ g^{c} ⟩$ : writing $x = g^{m}$ with $m \in Z$ by [L2] and [A1], the condition $x^{d} = g^{m d} = e$ says $n ∣ m d$ by [L1], that is $c d ∣ m d$ , that is $c ∣ m$ ; and ${g^{m} : c ∣ m} = {(g^{c})^{k} : k \in Z} = ⟨ g^{c} ⟩$ by [L2].

step 1.1A1L1L2algebra

2.3

For positive divisors $d, e$ of $n$ with $d ∣ e$ , write $e = d f$ with $f \geq 1$ ; then $n / d = (n / e) f$ , so $g^{n / d} = (g^{n / e})^{f} \in ⟨ g^{n / e} ⟩$ by [L2], and therefore $⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩$ , the latter being a subgroup containing $g^{n / d}$ .

step 1.1L2algebra

3.1

If $H \leq G$ has $∣ H ∣ = d$ , then $H = ⟨ g^{c} ⟩$ : each $h \in H$ has $ord (h) = ∣ ⟨ h ⟩ ∣$ dividing $∣ H ∣ = d$ by [L1] and [L3], so $h^{d} = e$ by [L1] and hence $h \in ⟨ g^{c} ⟩$ by step 2.2; thus $H \subseteq ⟨ g^{c} ⟩$ , and both sets have exactly $d$ elements by step 2.1, so they are equal.

step 2.1step 2.2L1L3

4.1

Every subgroup $H \leq G$ has this form for exactly one positive divisor of $n$ : $H$ is a subset of the finite set $G$ , so $d : = ∣ H ∣$ is defined and divides $n$ by [L3] and [A1], and step 3.1 gives $H = ⟨ g^{n / d} ⟩$ ; the divisor is determined by $H$ , being its order.

step 2.1step 3.1A1L3

5.1

Conversely, if $⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩$ then $d ∣ e$ , since by step 2.1 the two subgroups have orders $d$ and $e$ and [L3] applied to the subgroup $⟨ g^{n / d} ⟩$ of $⟨ g^{n / e} ⟩$ makes $d$ divide $e$ . Together with steps 2.1, 3.1, 4.1 and 2.3 this proves every clause of the lemma.

step 2.1step 3.1step 2.3step 4.1L3∎

Remarks

What the divisor lattice buys. The clause that matters downstream is not existence but uniqueness: a subgroup of a finite cyclic group is pinned down by its order alone, so the Galois correspondence turns "subgroups of $Gal (F_{q^{n}} / F_{q})$ " into "divisors of $n$ " with nothing left to choose (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).
Where cyclicity is used. Uniqueness fails without it. In the Klein four-group there are three distinct subgroups of order two, and the argument breaks at step 2.2, where the solutions of $x^{2} = e$ form the whole group rather than a single cyclic subgroup.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A cyclic group of order $n$ has exactly $φ (n)$ generators

Statement

Let $G = ⟨ g ⟩$ be a cyclic group of finite order $n \geq 1$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups). For an integer $a$ ,

$⟨ g^{a} ⟩ = G ⟺ \gcd (a, n) = 1,$

that is, $g^{a}$ generates $G$ exactly when $a$ and $n$ are coprime (Coprime integers: $\gcd (a, b) = 1$ ). Consequently $G$ has exactly $φ (n)$ generators (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

At $n = 1$ the group is trivial, every integer is coprime to $1$ , and the single element is its own generator, in agreement with $φ (1) = 1$ .

Facts & Assumptions

Given: A cyclic group $G = ⟨ g ⟩$ whose underlying set is finite of order $n \geq 1$ .

[A1]

$∣ G ∣ = n$ , and since $G = ⟨ g ⟩$ , [L1] gives $ord (g) = ∣ ⟨ g ⟩ ∣ = n$ ; in particular $g^{n} = e$ .

[L1]

For an element $x$ of finite order $m$ in a group: $x^{k} = e$ if and only if $m ∣ k$ ; the powers $x^{0}, x^{1}, \dots, x^{m - 1}$ are pairwise distinct; and $⟨ x ⟩ = {x^{s} : s \in N, s < m}$ , so $∣ ⟨ x ⟩ ∣ = m = ord (x)$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L2]

$⟨ x ⟩ = {x^{k} : k \in Z}$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian); and $⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L3]

For $a, b \in Z$ not both $0$ there are integers $x_{0}, y_{0}$ with $a x_{0} + b y_{0} = \gcd (a, b)$ (Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution).

[L4]

$\gcd (a, b)$ is a common divisor of $a$ and $b$ , and $\gcd (a, b) \geq 1$ whenever $(a, b) \neq (0, 0)$ (Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L5]

For $n \geq 1$ , every class in $Z / n$ contains exactly one integer $r$ with $0 \leq r < n$ , and $r \mapsto [r]_{n}$ is a bijection from ${0, \dots, n - 1}$ onto $Z / n$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L6]

For $n \geq 1$ and $a \in Z$ , the class $[a]_{n}$ is a unit of $Z / n$ if and only if $\gcd (a, n) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L7]

$φ (n) = ∣ (Z / n)^{\times} ∣$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Proof

technique · direct

1.1

Put $m : = \gcd (a, n)$ . Since $n \geq 1$ the pair $(a, n)$ is not $(0, 0)$ , so $m \geq 1$ , $m ∣ a$ and $m ∣ n$ by [L4].

L4given

2.1

If $m = 1$ then $g^{a}$ generates $G$ : by [L3] there are integers $u, v$ with $a u + n v = 1$ , whence $g = g^{a u + n v} = (g^{a})^{u} (g^{n})^{v} = (g^{a})^{u}$ using $g^{n} = e$ from [A1]; so $g \in ⟨ g^{a} ⟩$ by [L2], and $⟨ g^{a} ⟩$ is then a subgroup containing $g$ , so it contains $⟨ g ⟩ = G$ by [L2] and equals $G$ .

step 1.1A1L2L3

2.2

Conversely, if $g^{a}$ generates $G$ then $m = 1$ : from $m ∣ a$ and $m ∣ n$ the integer $n / m$ is at least $1$ and $(g^{a})^{n / m} = (g^{n})^{a / m} = e$ by [A1], so $ord (g^{a}) \leq n / m$ and hence $∣ ⟨ g^{a} ⟩ ∣ \leq n / m$ by [L1]; but $⟨ g^{a} ⟩ = G$ has $n$ elements, so $n \leq n / m$ and therefore $m = 1$ .

step 1.1A1L1algebra

3.1

By [A1] and [L1] the powers $g^{0}, g^{1}, \dots, g^{n - 1}$ are pairwise distinct and exhaust $G$ , so the generators of $G$ are exactly the elements $g^{r}$ with $0 \leq r < n$ and $\gcd (r, n) = 1$ , one for each such $r$ .

step 2.1step 2.2A1L1

4.1

By [L5] the map $r \mapsto [r]_{n}$ is a bijection from ${0, \dots, n - 1}$ onto $Z / n$ , and by [L6] it carries the $r$ with $\gcd (r, n) = 1$ onto the units of $Z / n$ ; so the number of such $r$ is $∣ (Z / n)^{\times} ∣ = φ (n)$ by [L7], and by step 3.1 that is the number of generators of $G$ .

step 3.1L5L6L7∎

Remarks

Why the criterion is stated and not only the count. The count answers "how many", but the development below repeatedly needs "which": that $ζ^{a}$ is again a primitive $n$ -th root of unity exactly when $a$ is coprime to $n$ is the criterion, and it is what makes the exponent of a cyclotomic automorphism a unit modulo $n$ ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ).

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields

Definition

Let $F_{q}$ be a finite field of order $q$ (Finite fields and their order) and let $E$ be a finite field having $F_{q}$ as a subfield. The relative Frobenius of the extension $E / F_{q}$ is the map

$σ_{q} ⁣ : E ⟶ E, σ_{q} (x) = x^{q} .$

It is an $F_{q}$ -automorphism of $E$ , and no separate result is needed for that. Let $p$ be the characteristic of $E$ ; the subfield $F_{q}$ has the same identity element and hence the same characteristic, so $q = p^{k}$ with $k = [F_{q} : F_{p}]$ by Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ . The Frobenius map ${Fr}_{E} ⁣ : x \mapsto x^{p}$ is an injective field endomorphism of $E$ , is an automorphism because $E$ is finite, and has $k$ -fold iterate $x \mapsto x^{p^{k}}$ (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields); that iterate is $σ_{q}$ . Every $a \in F_{q}$ satisfies $a^{q} = a$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield), so $σ_{q}$ fixes $F_{q}$ pointwise. Hence

$σ_{q} \in Aut (E / F_{q})$

(Relative field automorphisms and $Aut (K / F)$ ). Its iterates are $σ_{q}^{i} (x) = x^{q^{i}}$ for $i \in N$ , with $σ_{q}^{0}$ the identity.

Remarks

The letter. The relative Frobenius is written $σ_{q}$ rather than $φ_{q}$ because $φ$ is Euler's totient (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ) everywhere below, and the two symbols would otherwise stand side by side in the same formula.
Relative, not absolute. ${Fr}_{E}$ is intrinsic to $E$ ; $σ_{q}$ depends on the chosen base field $F_{q}$ , and it is the identity exactly when $E = F_{q}$ . Taking $F_{q}$ to be the prime field returns ${Fr}_{E}$ itself.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The elements of a finite extension fixed by the $q$ -power map are exactly the base field

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E$ be a finite field having $F_{q}$ as a subfield. Then

${x \in E : x^{q} = x} = F_{q} .$

Equivalently, the fixed field of the cyclic group generated by the relative Frobenius $σ_{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields) is the base field:

$E^{⟨ σ_{q} ⟩} = F_{q} .$

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ (Finite fields and their order), and the set $S : = {x \in E : x^{q} = x}$ .

[L1]

The relative Frobenius is $σ_{q} (x) = x^{q}$ , an $F_{q}$ -automorphism of $E$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L2]

If $F$ is a field with $q$ elements, then every $a \in F$ satisfies $a^{q} = a$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L3]

Let $D$ be an integral domain. A nonzero polynomial $f \in D [x]$ of degree $n$ has at most $n$ distinct roots in $D$ (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L4]

The fixed field of a subgroup $G$ of the automorphism group of $K$ is $K^{G} = {x \in K : σ (x) = x for every σ \in G}$ (The fixed field $K^{G}$ of a group of field automorphisms).

Proof

technique · direct

1.1

Applying [L2] to the field $F_{q}$ , which has exactly $q$ elements, every $a \in F_{q}$ satisfies $a^{q} = a$ ; hence $F_{q} \subseteq S$ .

L2given

1.2

$S$ is the set of roots in $E$ of the polynomial $t^{q} - t \in E [t]$ , which is nonzero of degree $q$ ; a field is an integral domain, so [L3] gives $∣ S ∣ \leq q$ .

L3given

2.1

Since $F_{q} \subseteq S$ , $∣ F_{q} ∣ = q$ and $∣ S ∣ \leq q$ , the finite sets $F_{q}$ and $S$ coincide: ${x \in E : x^{q} = x} = F_{q}$ .

step 1.1step 1.2given

3.1

An element of $E$ is fixed by every power of $σ_{q}$ precisely when it is fixed by $σ_{q}$ itself, so $E^{⟨ σ_{q} ⟩} = {x \in E : x^{q} = x}$ by [L1] and [L4], and step 2.1 identifies this with $F_{q}$ .

step 2.1L1L4∎

Remarks

What forces equality is a count, not an inclusion. The inclusion $F_{q} \subseteq S$ is immediate; the content is that $t^{q} - t$ cannot have more than $q$ roots, so the $q$ elements of $F_{q}$ already use them all up. The same count identifies the intermediate fields of an extension of finite fields (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E / F_{q}$ be an extension of finite fields of degree $n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). Then

$∣ E ∣ = q^{n},$

and the relative Frobenius $σ_{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields) has order exactly $n$ in $Aut (E / F_{q})$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity). At $n = 1$ this says $σ_{q}$ is the identity, of order one.

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n$ ; the prime subfield of $E$ is $F_{p}$ with $p$ the characteristic, and $F_{q}$ has the same characteristic, its identity element being that of $E$ .

[L1]

The relative Frobenius is $σ_{q} (x) = x^{q}$ , an $F_{q}$ -automorphism of $E$ , with $σ_{q}^{i} (x) = x^{q^{i}}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L2]

If $F$ is a field with $q$ elements, then every $a \in F$ satisfies $a^{q} = a$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L3]

[L4]

If $F$ is a finite field, then there is a unique prime $p$ and a unique positive integer $m$ with $∣ F ∣ = p^{m}$ ; here $p = char F$ and $m = [F : F_{p}]$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ).

[L5]

For fields $F \subseteq K \subseteq L$ with $K / F$ and $L / K$ finite, $L / F$ is finite and $[L : F] = [L : K] [K : F]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

Proof

technique · direct

1.1

By [L4] applied to $F_{q}$ , $q = p^{k}$ with $k = [F_{q} : F_{p}]$ ; by [L4] applied to $E$ , $∣ E ∣ = p^{m}$ with $m = [E : F_{p}]$ .

L4given

2.1

The tower $F_{p} \subseteq F_{q} \subseteq E$ and [L5] give $m = [E : F_{q}] [F_{q} : F_{p}] = n k$ , so $∣ E ∣ = p^{n k} = (p^{k})^{n} = q^{n}$ .

step 1.1L5algebra

3.1

Every $x \in E$ satisfies $x^{q^{n}} = x$ , by [L2] applied to $E$ , whose order is $q^{n}$ by step 2.1; by [L1] this says $σ_{q}^{n} = {i d}_{E}$ .

step 2.1L1L2

3.2

For an integer $j$ with $1 \leq j < n$ one has $σ_{q}^{j} \neq {i d}_{E}$ : otherwise every one of the $q^{n}$ elements of $E$ would be a root of the nonzero polynomial $t^{q^{j}} - t \in E [t]$ , whose degree $q^{j}$ is smaller than $q^{n}$ because $q \geq 2$ , contradicting [L3].

step 2.1L1L3algebra

4.1

The least $j \geq 1$ with $σ_{q}^{j} = {i d}_{E}$ is therefore $j = n$ , that is $ord (σ_{q}) = n$ ; for $n = 1$ steps 3.1 and 3.2 say only that $σ_{q} = {i d}_{E}$ , of order one.

step 3.1step 3.2L1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E$ be a finite field having $F_{q}$ as a subfield, with $[E : F_{q}] = n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). Then $E / F_{q}$ is a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) and

$Gal (E / F_{q}) = ⟨ σ_{q} ⟩$

is cyclic of order $n$ , generated by the relative Frobenius $σ_{q} ⁣ : x \mapsto x^{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n$ , and the subgroup $G : = ⟨ σ_{q} ⟩$ of $Aut (E / F_{q})$ .

[L1]

The relative Frobenius $σ_{q} (x) = x^{q}$ is an $F_{q}$ -automorphism of $E$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L2]

${x \in E : x^{q} = x} = F_{q}$ , that is $E^{⟨ σ_{q} ⟩} = F_{q}$ (The elements of a finite extension fixed by the $q$ -power map are exactly the base field).

[L3]

$∣ E ∣ = q^{n}$ and $σ_{q}$ has order exactly $n$ in $Aut (E / F_{q})$ (For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L4]

If $G$ is a finite group of automorphisms of $K$ , then $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

[L5]

For a finite extension $K / F$ with $G = Aut (K / F)$ , the conditions " $K / F$ is Galois", " $K$ is the splitting field over $F$ of a separable polynomial", " $∣ G ∣ = [K : F]$ " and " $K^{G} = F$ " are equivalent (Equivalent characterizations of a finite Galois extension).

[L6]

$K^{G} = {x \in K : σ (x) = x for every σ \in G}$ (The fixed field $K^{G}$ of a group of field automorphisms).

Proof

technique · direct

1.1

$G = ⟨ σ_{q} ⟩$ is a cyclic group of automorphisms of $E$ , finite of order $n$ by [L1] and [L3].

L1L3

1.2

Its fixed field is $E^{G} = F_{q}$ by [L2] and [L6].

L2L6

2.1

Applying [L4] to the finite automorphism group $G$ of $E$ and using step 1.2, $[E : F_{q}] = [E : E^{G}] = ∣ G ∣ = n$ and $Aut (E / F_{q}) = Aut (E / E^{G}) = G$ .

step 1.1step 1.2L4

3.1

Hence $∣ Aut (E / F_{q}) ∣ = n = [E : F_{q}]$ , so $E / F_{q}$ is Galois by [L5], and $Gal (E / F_{q}) = Aut (E / F_{q}) = ⟨ σ_{q} ⟩$ is cyclic of order $n$ by step 2.1 and step 1.1. At $n = 1$ the group is trivial and $E = F_{q}$ .

step 1.1step 2.1L5∎

Remarks

Separability and normality are never argued separately. The usual route checks that $E$ is a splitting field of $t^{q^{n}} - t$ and that this polynomial has no repeated root. Routing through Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ instead replaces both checks by one count: an automorphism group of order $n$ whose fixed field is $F_{q}$ already forces $∣ Aut ∣ = [E : F_{q}]$ , which is one of the equivalent Galois conditions.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$

Statement

Let $F_{q}$ be a finite field of order $q$ , let $n \geq 1$ , and let $F_{q^{n}}$ be a finite field having $F_{q}$ as a subfield with $[F_{q^{n}} : F_{q}] = n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). For a positive divisor $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ) put

$F_{q^{d}} : = {x \in F_{q^{n}} : x^{q^{d}} = x} .$

Then the fields $F_{q^{d}}$ , as $d$ runs over the positive divisors of $n$ , are exactly the intermediate fields of $F_{q^{n}} / F_{q}$ , with distinct divisors giving distinct fields;

$[F_{q^{d}} : F_{q}] = d, ∣ F_{q^{d}} ∣ = q^{d};$

and for positive divisors $d, e$ of $n$ ,

$F_{q^{d}} \subseteq F_{q^{e}} ⟺ d ∣ e .$

The two ends of the lattice are instances: $d = 1$ gives the base field $F_{q}$ and $d = n$ gives $F_{q^{n}}$ .

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E : = F_{q^{n}}$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n \geq 1$ , and the relative Frobenius $σ_{q} (x) = x^{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields), whose $i$ -th iterate is $x \mapsto x^{q^{i}}$ .

[L1]

$E / F_{q}$ is Galois and $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ is cyclic of order $n$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ ).

[L2]

In a cyclic group $⟨ g ⟩$ of finite order $n$ , for each positive divisor $c$ of $n$ the subgroup $⟨ g^{n / c} ⟩$ has order $c$ and is the unique subgroup of that order, every subgroup has this form for exactly one such $c$ , and $⟨ g^{n / c} ⟩ \subseteq ⟨ g^{n / c^{'}} ⟩$ if and only if $c ∣ c^{'}$ (A finite cyclic group has exactly one subgroup of each order dividing its own).

[L3]

For $K / F$ finite Galois with $G = Gal (K / F)$ , the assignments $H \mapsto K^{H}$ and $F^{'} \mapsto Gal (K / F^{'})$ are mutually inverse inclusion-reversing bijections between subgroups $H \leq G$ and intermediate fields $F \subseteq F^{'} \subseteq K$ , and $[K : K^{H}] = ∣ H ∣$ , $[K^{H} : F] = [G : H]$ (The fundamental theorem of finite Galois theory).

[L4]

${x \in E : x^{q} = x} = F_{q}$ (The elements of a finite extension fixed by the $q$ -power map are exactly the base field).

[L5]

For an extension of finite fields $L / F_{q}$ of degree $m$ one has $∣ L ∣ = q^{m}$ (For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L6]

$K^{H} = {x \in K : σ (x) = x for every σ \in H}$ (The fixed field $K^{G}$ of a group of field automorphisms).

[L7]

For a finite group $G$ and $H \leq G$ one has $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Write $G : = Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ , cyclic of order $n$ by [L1], and for a positive divisor $d$ of $n$ put $H_{d} : = ⟨ σ_{q}^{d} ⟩$ .

2.1

By [L2] applied to $G$ with generator $σ_{q}$ and $c = n / d$ , the subgroup $H_{d} = ⟨ σ_{q}^{n / (n / d)} ⟩$ has order $n / d$ ; every subgroup of $G$ is $H_{d}$ for exactly one positive divisor $d$ of $n$ ; and $H_{e} \subseteq H_{d}$ if and only if $d ∣ e$ , since $H_{e} \subseteq H_{d}$ reads $⟨ σ_{q}^{n / (n / e)} ⟩ \subseteq ⟨ σ_{q}^{n / (n / d)} ⟩$ , which by [L2] says $(n / e) ∣ (n / d)$ , that is $d ∣ e$ .

step 1.1L2algebra

2.2

The fixed field of $H_{d}$ is $E^{H_{d}} = {x \in E : σ_{q}^{d} (x) = x} = {x \in E : x^{q^{d}} = x} = F_{q^{d}}$ , because an element fixed by $σ_{q}^{d}$ is fixed by all its powers and conversely.

step 1.1L6given

3.1

By [L3] the map $H \mapsto E^{H}$ is a bijection from the subgroups of $G$ onto the intermediate fields of $E / F_{q}$ ; composing with the bijection of step 2.1 between positive divisors of $n$ and subgroups, the fields $F_{q^{d}} = E^{H_{d}}$ are exactly the intermediate fields, distinct divisors giving distinct fields.

step 2.1step 2.2L3

3.2

Degrees: [L3] gives $[F_{q^{d}} : F_{q}] = [E^{H_{d}} : F_{q}] = [G : H_{d}]$ , and [L7] with step 2.1 turns this into $∣ G ∣ / ∣ H_{d} ∣ = n / (n / d) = d$ ; then [L5] gives $∣ F_{q^{d}} ∣ = q^{d}$ .

step 2.1step 2.2L3L5L7algebra

4.1

Inclusions: [L3] makes the correspondence inclusion-reversing, so $F_{q^{d}} = E^{H_{d}} \subseteq E^{H_{e}} = F_{q^{e}}$ exactly when $H_{e} \subseteq H_{d}$ , which by step 2.1 holds exactly when $d ∣ e$ . At $d = 1$ one has $H_{1} = G$ and $F_{q^{1}} = E^{G} = F_{q}$ by [L4], and at $d = n$ one has $H_{n} = {i d}$ and $F_{q^{n}} = E$ ; with steps 3.1 and 3.2 this proves every clause.

step 2.1step 2.2step 3.1step 3.2L3L4∎

Remarks

Why the lattice is exactly the divisor lattice. Uniqueness of the subgroup of each order in a cyclic group is what leaves no choice: had $Gal (E / F_{q})$ been the Klein four-group, three distinct subgroups of order two would have produced three intermediate fields of the same degree, and no indexing by divisors could exist.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Galois description of the subfields of a finite field and the elementary divisibility criterion agree

Statement

Two statements in this library describe the subfields of a finite field, and they describe the same objects.

The subfields of $F_{p^{n}}$ are the unique fields $F_{p^{d}}$ for positive divisors $d$ of $n$ fixes a finite field $F$ of order $p^{m}$ with $p$ its characteristic and says that for each positive divisor $e$ of $m$ the set ${a \in F : a^{p^{e}} = a}$ is the unique subfield of $F$ of order $p^{e}$ , and that these are all of the subfields of $F$ . Its index set is therefore the divisors of $m$ , and its base point is the prime field.

The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ fixes a base field $F_{q}$ inside $F$ and says that the intermediate fields of $F / F_{q}$ are the ${x \in F : x^{q^{d}} = x}$ for the positive divisors $d$ of $n = [F : F_{q}]$ . Its index set is therefore the divisors of $n$ , and its base point is $F_{q}$ .

The dictionary. Write $q = p^{k}$ , so that $m = k n$ . For a positive divisor $d$ of $n$ the two prescriptions produce literally the same set,

${x \in F : x^{q^{d}} = x} = {x \in F : x^{p^{k d}} = x},$

which is the subfield of order $p^{k d}$ named by the first statement; and $k d$ runs exactly over the divisors $e$ of $m$ that are multiples of $k$ as $d$ runs over the divisors of $n$ . So the intermediate fields of $F / F_{q}$ are precisely those subfields of $F$ whose order is $p^{e}$ with $k ∣ e$ , which is the expected answer: a subfield of $F$ contains the unique subfield of order $p^{k}$ exactly when $k$ divides $e$ , by the divisibility clause of The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ applied over the prime field.

Neither statement is the other. The published one is elementary: it counts roots of $t^{p^{e}} - t$ and needs no Galois theory. The one proved here reads the lattice off the subgroup lattice of a cyclic Galois group, and it is that reading which the rest of this page uses, because the same correspondence also supplies the degrees and the automorphism groups of the intermediate fields. Recording their agreement here is what keeps the two vocabularies from drifting apart in later proofs.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A monic irreducible of degree $d$ over $F_{q}$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$

Statement

Let $F_{q}$ be a finite field of order $q$ , let $π \in F_{q} [t]$ be monic irreducible of degree $d \geq 1$ , and let $α$ be a root of $π$ in some extension field of $F_{q}$ . Then $F_{q} (α)$ is a finite field of order $q^{d}$ , the $d$ elements

$α, α^{q}, α^{q^{2}}, \dots, α^{q^{d - 1}}$

are pairwise distinct roots of $π$ lying in $F_{q} (α)$ ,

$π = \prod_{i = 0}^{d - 1} (t - α^{q^{i}}) in F_{q} (α) [t],$

and $F_{q} (α)$ is a splitting field of $π$ over $F_{q}$ (Polynomials that split and splitting fields of a polynomial or a family of polynomials), of degree $d$ over $F_{q}$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). In particular these $d$ elements are pairwise conjugate over $F_{q}$ (Conjugate algebraic elements over a field) and form a single orbit of the relative Frobenius. The list starts at $i = 0$ , so its first member is $α$ itself, and at $d = 1$ it is the single element $α \in F_{q}$ .

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q \geq 2$ , a monic irreducible $π \in F_{q} [t]$ of degree $d \geq 1$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree), a root $α$ of $π$ in an extension field, and the field $K : = F_{q} (α)$ .

[L1]

If $K / F$ is a field extension and $a \in K$ is algebraic, there is a unique monic irreducible $m_{a} \in F [x]$ with $\ker ({ev}_{a}) = (m_{a})$ , and for every $f \in F [x]$ one has $f (a) = 0$ if and only if $m_{a} ∣ f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L2]

If $a$ is algebraic over $F$ with minimal polynomial of degree $n$ , then $[F (a) : F] = n$ (An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite).

[L3]

If $[L : F_{q}] = m$ , then $L$ has an ordered $F_{q}$ -basis of length $m$ , and every element of $L$ has a unique coordinate list in that basis (The degree $[K : F] = \dim_{F} K$ of a finite field extension, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[L4]

An extension $E / F_{q}$ of finite fields of degree $m$ is Galois with $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ cyclic of order $m$ , where $σ_{q} (x) = x^{q}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L5]

Let $R$ be a commutative ring, $a \in R$ and $f \in R [x]$ . Then $f (a) = 0$ if and only if $x - a$ divides $f$ in $R [x]$ (Factor theorem over a commutative ring).

[L6]

A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots in that domain (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L7]

If $F$ is a field with $q$ elements, then every $a \in F$ satisfies $a^{q} = a$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L8]

Two elements algebraic over $F$ are conjugate over $F$ when they have the same minimal polynomial over $F$ (Conjugate algebraic elements over a field).

Proof

technique · direct

1.1

$π$ is the minimal polynomial of $α$ over $F_{q}$ : since $π (α) = 0$ , [L1] gives $m_{α} ∣ π$ , and $π$ is irreducible while $m_{α}$ is monic of degree at least one, so $π = m_{α}$ . Hence $[K : F_{q}] = d$ by [L2].

L1L2given

2.1

Fix the length- $d$ basis supplied by [L3]. Unique coordinates give a bijection $F_{q}^{d} \to K$ , so $∣ K ∣ = q^{d}$ and $K$ is a finite field. Now [L4] applies: $K / F_{q}$ is Galois with $Gal (K / F_{q}) = ⟨ σ_{q} ⟩$ cyclic of order $d$ , where $σ_{q} (x) = x^{q}$ .

step 1.1L3L4algebra

3.1

Each $σ_{q}^{i}$ fixes the coefficients of $π$ , which lie in $F_{q}$ , so applying the field homomorphism $σ_{q}^{i}$ to the equation $π (α) = 0$ gives $π (α^{q^{i}}) = 0$ : every $α^{q^{i}}$ is a root of $π$ lying in $K$ .

step 2.1L4given

4.1

The elements $α, α^{q}, \dots, α^{q^{d - 1}}$ are pairwise distinct. Suppose $α^{q^{i}} = α^{q^{j}}$ with $0 \leq i < j \leq d - 1$ and apply the automorphism $σ_{q}^{d - j}$ : since $x^{q^{d}} = x$ for every $x \in K$ by [L7] and step 2.1, this yields $α^{q^{r}} = α$ with $r : = i + d - j$ and $1 \leq r \leq d - 1$ . The set $S : = {x \in K : x^{q^{r}} = x}$ is the fixed set of the automorphism $σ_{q}^{r}$ , hence a subfield of $K$ ; it contains $F_{q}$ by [L7] and contains $α$ , so $K = F_{q} (α) \subseteq S$ . But $S$ is the root set in $K$ of the nonzero polynomial $t^{q^{r}} - t$ , so $∣ S ∣ \leq q^{r}$ by [L6], giving $q^{d} = ∣ K ∣ \leq q^{r} < q^{d}$ because $q \geq 2$ and $r < d$ . This is impossible.

step 2.1step 3.1L6L7algebra

5.1

The product $P : = \prod_{i = 0}^{d - 1} (t - α^{q^{i}})$ divides $π$ in $K [t]$ . Indeed, listing the distinct roots as $r_{0}, \dots, r_{d - 1}$ , [L5] writes $π = (t - r_{0}) g_{0}$ ; for $j \geq 1$ the equation $0 = π (r_{j}) = (r_{j} - r_{0}) g_{0} (r_{j})$ and $r_{j} \neq r_{0}$ in the field $K$ give $g_{0} (r_{j}) = 0$ , so the same step applies to $g_{0}$ with the remaining $d - 1$ distinct roots, and after $d$ such steps $π = P h$ for some $h \in K [t]$ .

step 3.1step 4.1L5

6.1

Both $π$ and $P$ are monic of degree $d$ , so $h$ is monic of degree $0$ , that is $h = 1$ and $π = P$ .

step 5.1givenalgebra

7.1

Consequently $π$ splits over $K$ , and $K = F_{q} (α)$ is generated over $F_{q}$ by the root $α$ ; since the subfield of $K$ generated over $F_{q}$ by all the roots of $π$ contains $α$ , it contains and hence equals $K$ , so $K$ is a splitting field of $π$ over $F_{q}$ . All $d$ roots share the minimal polynomial $π$ by step 1.1 and [L1], so they are pairwise conjugate over $F_{q}$ by [L8], and step 3.1 exhibits them as one orbit of $σ_{q}$ .

step 1.1step 3.1step 4.1step 6.1L1L8∎

Remarks

The index starts at zero. The orbit is $α^{q^{0}} = α, α^{q}, \dots, α^{q^{d - 1}}$ , so the factor $t - α$ is present in the product; dropping the term $i = 0$ would leave a polynomial of degree $d - 1$ that is not $π$ .
Where irreducibility is used. It enters twice: to identify $π$ with the minimal polynomial of $α$ in step 1.1, and through that identification to force $[K : F_{q}] = d$ , which is what makes the count in step 4.1 tight. For a reducible $π$ the conclusion fails outright, as $t^{2} - 1$ over $F_{3}$ shows: its roots $1$ and $- 1$ are not a Frobenius orbit.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$

Statement

Let $F_{q}$ be a finite field of order $q$ and, for an integer $d \geq 1$ , let $N_{q} (d)$ denote the number of monic irreducible polynomials of degree $d$ in $F_{q} [t]$ . Then each $N_{q} (d)$ is finite, and for every $n \geq 1$

$\sum_{d ∣ n} d N_{q} (d) = q^{n},$

the sum being over the positive divisors $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form). At $n = 1$ the identity reads $N_{q} (1) = q$ .

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q \geq 2$ and an integer $n \geq 1$ ; monic polynomials are as in Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree.

[L1]

In $F_{q} [t]$ one has $t^{q^{n}} - t = \prod P (t)$ , the product being over the monic irreducible $P \in F_{q} [t]$ whose degree divides $n$ , each such $P$ occurring once (Over $F_{q}$ , $x^{q^{n}} - x$ is the product of all monic irreducibles whose degrees divide $n$ ).

[L2]

If $R$ is an integral domain and $f, g \in R [x]$ are nonzero, then $f g \neq 0$ and $\deg (f g) = \deg f + \deg g$ (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L3]

If $R$ is an integral domain, then $R [x]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

Proof

technique · direct

1.1

For each $d \geq 1$ the monic polynomials of degree $d$ in $F_{q} [t]$ are the $t^{d} + a_{d - 1} t^{d - 1} + \dots + a_{0}$ with $a_{0}, \dots, a_{d - 1} \in F_{q}$ , so there are exactly $q^{d}$ of them and $N_{q} (d) \leq q^{d}$ is finite.

givenalgebra

2.1

Consequently the family of monic irreducible $P \in F_{q} [t]$ with $\deg P ∣ n$ is finite, having at most $\sum_{d ∣ n} q^{d}$ members, so the product in [L1] is a finite product of nonzero polynomials in the integral domain $F_{q} [t]$ ([L3]).

step 1.1L1L3

3.1

Taking degrees in [L1] and applying [L2] repeatedly to that finite product gives $\deg (t^{q^{n}} - t) = \sum_{\deg P ∣ n} \deg P$ , where the sum runs over the same finite family; and $\deg (t^{q^{n}} - t) = q^{n}$ because $q^{n} > 1$ .

step 2.1L1L2algebra

4.1

Splitting that sum according to the degree of $P$ : the possible degrees are exactly the positive divisors $d$ of $n$ , there are $N_{q} (d)$ monic irreducibles of degree $d$ , and each contributes $d$ ; hence $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ .

step 1.1step 3.1algebra

5.1

At $n = 1$ the only positive divisor is $d = 1$ , so the identity reads $N_{q} (1) = q$ , in agreement with the fact that the monic polynomials of degree one are the $t - a$ for $a \in F_{q}$ and each is irreducible.

step 1.1step 4.1algebra∎

Remarks

What the identity does not give. It determines $N_{q} (n)$ only once every $N_{q} (d)$ for proper divisors $d$ of $n$ is known, so it is a recursion rather than a formula. Inverting it into a closed form is a separate matter and is not carried out here.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Normal bases of a finite Galois extension

Definition

Let $K / F$ be a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) of degree $n = [K : F]$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension), and list its Galois group as

$Gal (K / F) = {σ_{1}, \dots, σ_{n}},$

which has exactly $n$ elements because $∣ Gal (K / F) ∣ = [K : F]$ for a finite Galois extension (Equivalent characterizations of a finite Galois extension). Scalar multiplication by $F$ makes $K$ an $F$ -vector space of dimension $n$ .

An element $α \in K$ is a normal basis generator for $K / F$ when the list

$(σ_{1} α, σ_{2} α, \dots, σ_{n} α)$

is an ordered basis of $K$ as an $F$ -vector space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Such a list is called a normal basis of $K$ over $F$ : a basis that is a single orbit of the Galois group.

Two conditions, not one. A list is an ordered basis when it is injective and its image is a basis, so a normal basis generator must in particular have $n$ distinct conjugates $σ_{i} α$ . Neither half implies the other: a basis of $K$ over $F$ need not be a Galois orbit, and a Galois orbit of size $n$ need not be a basis.

The list is indexed by the group, not ordered by it. Reordering $σ_{1}, \dots, σ_{n}$ permutes the list and leaves the property of being a basis unchanged, since a basis is a property of the underlying set together with injectivity of the list (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). The automorphisms are those of Relative field automorphisms and $Aut (K / F)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial

Statement

Let $S$ be an integral domain, let $R \subseteq S$ be a subring whose underlying set is infinite, let $m \geq 1$ , and let $f \in S [x_{1}, \dots, x_{m}]$ (Polynomial rings in finitely many commuting indeterminates by iteration). If

$f (a_{1}, \dots, a_{m}) = 0 for all a_{1}, \dots, a_{m} \in R,$

then $f = 0$ in $S [x_{1}, \dots, x_{m}]$ .

Here $f (a_{1}, \dots, a_{m})$ is the iterated evaluation of Evaluation and roots of a polynomial in a commutative target ring, carried out one indeterminate at a time along the construction of $S [x_{1}, \dots, x_{m}]$ .

Facts & Assumptions

Given: An integral domain $S$ , an infinite subring $R \subseteq S$ , and the polynomial rings $S [x_{1}, \dots, x_{m}]$ built by iteration (Polynomial rings in finitely many commuting indeterminates by iteration).

[L1]

A nonzero polynomial $g \in D [x]$ of degree $k$ over an integral domain $D$ has at most $k$ distinct roots in $D$ (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L2]

If $R$ is an integral domain, then $R [x_{1}, \dots, x_{n}]$ is an integral domain for every $n \in N$ , including $n = 0$ (A polynomial ring in finitely many indeterminates over an integral domain is an integral domain).

[L3]

A nonzero $g = \sum_{i} a_{i} x^{i}$ has a largest index with $a_{i} \neq 0$ , its degree; the zero polynomial has no degree (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

Proof

technique · induction

1.1

Base case $m = 1$ : let $g \in S [x]$ vanish at every element of $R$ . If $g \neq 0$ it has a degree $k$ by [L3], so by [L1] it has at most $k$ distinct roots in $S$ ; but every one of the infinitely many elements of $R \subseteq S$ is a root, and an infinite set has more than $k$ elements. Hence $g = 0$ .

L1L3base

1.2

Inductive hypothesis: fix $m \geq 1$ and assume that every $g \in S [x_{1}, \dots, x_{m}]$ vanishing at all tuples from $R$ is zero.

2.1

Let $f \in S [x_{1}, \dots, x_{m + 1}] = S [x_{1}, \dots, x_{m}] [x_{m + 1}]$ vanish at every tuple from $R$ , and write $f = \sum_{j \leq k} g_{j} x_{m + 1}^{j}$ with $g_{j} \in S [x_{1}, \dots, x_{m}]$ . Fix $a = (a_{1}, \dots, a_{m}) \in R^{m}$ ; then $\sum_{j \leq k} g_{j} (a) x_{m + 1}^{j}$ is an element of $S [x_{m + 1}]$ vanishing at every element of $R$ , so it is zero by step 1.1, and therefore $g_{j} (a) = 0$ for every $j \leq k$ .

step 1.1L2given

3.1

Since $a \in R^{m}$ was arbitrary, each $g_{j}$ vanishes at every tuple from $R$ , so $g_{j} = 0$ by step 1.2 and hence $f = 0$ . This completes the induction, and the statement holds for every $m \geq 1$ .

step 1.2step 2.1discharge-induction∎

Remarks

Infinite, not merely large. The hypothesis cannot be weakened to a finite $R$ of any size: over $R = S = F_{q}$ the nonzero polynomial $x^{q} - x$ vanishes at every element, and in $m$ indeterminates so does $x_{1}^{q} - x_{1}$ . This is exactly why the normal basis theorem needs a separate argument over a finite base field (Every finite cyclic extension has a normal basis).

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For a finite Galois extension, $(α_{j})$ is a base-field basis exactly when the matrix $(σ_{i} α_{j})$ is invertible

Statement

Let $K / F$ be a finite Galois extension of degree $n$ , list its Galois group as $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ , and let $α_{1}, \dots, α_{n} \in K$ . Let $A \in M_{n} (K)$ be the matrix with entries

$A_{i j} = σ_{i} (α_{j}) (1 \leq i, j \leq n) .$

Then $(α_{1}, \dots, α_{n})$ is an ordered basis of $K$ as an $F$ -vector space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) if and only if $A$ is invertible over $K$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ ).

Facts & Assumptions

Given: A finite Galois extension $K / F$ of degree $n$ with $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ ; elements $α_{1}, \dots, α_{n} \in K$ ; the matrix $A \in M_{n} (K)$ with $A_{i j} = σ_{i} (α_{j})$ (Finite rectangular matrices over a commutative ring, their entries, rows and columns); and the $F$ -linear map $Φ ⁣ : F^{n} \to K$ given by $Φ (a) = \sum_{j} a_{j} α_{j}$ . Each $σ_{i}$ is an $F$ -automorphism of $K$ (Relative field automorphisms and $Aut (K / F)$ ), hence additive and $F$ -linear.

[L1]

For a finite Galois extension $K / F$ with $G = Gal (K / F)$ one has $∣ G ∣ = [K : F]$ (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and $Gal (K / F)$ ); so $\dim_{F} K = n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L2]

Let $G$ be a group and $K$ a field. Every finite family of distinct group homomorphisms $G \to K^{\times}$ is linearly independent over $K$ as a family of functions (Dedekind's linear independence theorem for distinct characters).

[L3]

For a linear map $T : V \to W$ of $F$ -vector spaces with $V$ finite-dimensional, $\dim_{F} V = \dim_{F} (\ker T) + \dim_{F} (im T)$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L4]

For $A \in M_{n} (F)$ and $L_{A} (x) = A x$ on $M_{n \times 1} (F)$ : $A$ is invertible if and only if $L_{A}$ is a linear isomorphism (A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms).

[L5]

Let $R$ be a commutative ring, $n \geq 1$ , $A \in M_{n} (R)$ . Then $A$ is invertible if and only if $\det (A)$ is a unit of $R$ (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

[L6]

$\det (A^{T}) = \det (A)$ for every $A \in M_{n} (R)$ over a commutative ring (For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ ); the transpose is $(A^{T})_{j i} = A_{i j}$ (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

Proof

technique · direct

1.1

By [L1] the $F$ -vector space $K$ has dimension $n$ , and $Φ$ is a map between $F$ -vector spaces of dimension $n$ ; so by [L3] it is injective if and only if it is surjective, and $(α_{1}, \dots, α_{n})$ is an ordered basis of $K$ over $F$ exactly when $Φ$ is bijective.

L1L3

1.2

For the implication that a basis has an invertible matrix, suppose $(α_{1}, \dots, α_{n})$ is an ordered basis, and let $c \in K^{n}$ satisfy $A^{T} c = 0$ , that is $\sum_{i} c_{i} σ_{i} (α_{j}) = 0$ for every $j$ . The map $θ ⁣ : K \to K$ , $θ (x) = \sum_{i} c_{i} σ_{i} (x)$ , is $F$ -linear because each $σ_{i}$ is, and it vanishes at every $α_{j}$ , hence on their $F$ -span, which is $K$ .

given

2.1

For the implication that a non-basis has a singular matrix, suppose $(α_{1}, \dots, α_{n})$ is not an ordered basis. By step 1.1 the map $Φ$ is not injective, so there is $a \in F^{n}$ with $a \neq 0$ and $\sum_{j} a_{j} α_{j} = 0$ . Applying $σ_{i}$ and using $σ_{i} (a_{j}) = a_{j}$ for $a_{j} \in F$ gives $\sum_{j} σ_{i} (α_{j}) a_{j} = 0$ for every $i$ , that is $A a = 0$ with $a \neq 0$ in $K^{n}$ . Were $A$ invertible with inverse $B$ , this would force $a = B (A a) = 0$ ; so $A$ is not invertible.

step 1.1given

2.2

So $\sum_{i} c_{i} σ_{i}$ is the zero function on $K$ , in particular on $K^{\times}$ . The restrictions $σ_{i} ∣_{K^{\times}} ⁣ : K^{\times} \to K^{\times}$ are group homomorphisms and are pairwise distinct, since two automorphisms of $K$ agreeing on $K^{\times}$ agree on $K$ ; so [L2] forces $c_{i} = 0$ for every $i$ .

step 1.2L2

3.1

Hence the $K$ -linear map $x \mapsto A^{T} x$ on $K^{n}$ has zero kernel, so by [L3] over $K$ it is also surjective and therefore a linear isomorphism; by [L4] the matrix $A^{T}$ is invertible, so $\det (A^{T})$ is a unit of $K$ by [L5], and $\det (A) = \det (A^{T})$ by [L6] is a unit, whence $A$ is invertible by [L5].

step 2.2L3L4L5L6

4.1

Step 2.1 gives one implication, that a list which is not a basis has a matrix that is not invertible, and step 3.1 gives the other, that a list which is a basis has an invertible matrix; together they are the stated equivalence.

step 2.1step 3.1∎

Remarks

Why the transpose appears. The dependence relation among the $α_{j}$ produces a null vector on the right of $A$ , while the Dedekind relation among the $σ_{i}$ produces one on the right of $A^{T}$ . Only the determinant sees both, which is why the two halves are joined through For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ rather than by a single rank computation.
Where the Galois hypothesis is used. Twice: to know that the group has exactly $n = [K : F]$ elements, so that $A$ is square, and to know that the $σ_{i}$ are $F$ -linear, which is what lets step 2.1 pull the scalars $a_{j}$ through.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Over an infinite base field, no nonzero polynomial vanishes at the conjugate tuple of every element

Statement

Let $K / F$ be a finite Galois extension of degree $n$ whose base field $F$ is infinite, and list $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ . Then for every nonzero $f \in K [x_{1}, \dots, x_{n}]$ (Polynomial rings in finitely many commuting indeterminates by iteration) there exists $α \in K$ with

$f (σ_{1} α, \dots, σ_{n} α) \neq 0.$

Facts & Assumptions

Given: A finite Galois extension $K / F$ of degree $n$ with $F$ infinite and $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ ; by [L4] the $F$ -vector space $K$ has dimension $n$ , so an ordered $F$ -basis $(α_{1}, \dots, α_{n})$ of $K$ exists (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L1]

Let $S$ be an integral domain and $R \subseteq S$ an infinite subring. If $g \in S [x_{1}, \dots, x_{m}]$ satisfies $g (a_{1}, \dots, a_{m}) = 0$ for all $a_{1}, \dots, a_{m} \in R$ , then $g = 0$ (A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial).

[L2]

For a finite Galois extension $K / F$ of degree $n$ with $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ and $α_{1}, \dots, α_{n} \in K$ , the list $(α_{1}, \dots, α_{n})$ is an ordered $F$ -basis of $K$ if and only if the matrix $A$ with $A_{i j} = σ_{i} (α_{j})$ is invertible (For a finite Galois extension, $(α_{j})$ is a base-field basis exactly when the matrix $(σ_{i} α_{j})$ is invertible).

[L3]

For commutative rings $R, S$ , a unital ring homomorphism $φ ⁣ : R \to S$ and $s \in S$ , there is a unique unital ring homomorphism $R [x] \to S$ extending $φ$ on constants and sending $x$ to $s$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism). Iterating this along Polynomial rings in finitely many commuting indeterminates by iteration gives, for any commutative $K$ -algebra $T$ and any $t_{1}, \dots, t_{n} \in T$ , a unique $K$ -algebra homomorphism $K [x_{1}, \dots, x_{n}] \to T$ sending $x_{i}$ to $t_{i}$ .

[L4]

For a finite Galois extension $K / F$ with $G = Gal (K / F)$ one has $∣ G ∣ = [K : F]$ (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and $Gal (K / F)$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L5]

$A \in M_{n} (K)$ is invertible when some $B \in M_{n} (K)$ satisfies $A B = I_{n} = B A$ ; such a $B$ is unique and written $A^{- 1}$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ , Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

Proof

technique · contrapositive

1.1

It suffices to prove the contrapositive: if $f \in K [x_{1}, \dots, x_{n}]$ satisfies $f (σ_{1} α, \dots, σ_{n} α) = 0$ for every $α \in K$ , then $f = 0$ . Assume that hypothesis on $f$ .

contrapositive-reduceassume-hyp

1.2

Fix an ordered $F$ -basis $(α_{1}, \dots, α_{n})$ of $K$ and put $A_{i j} = σ_{i} (α_{j})$ ; by [L2] the matrix $A$ is invertible, with inverse $A^{- 1}$ as in [L5].

L2L4L5given

2.1

For $c \in F^{n}$ write $α (c) : = \sum_{j} c_{j} α_{j}$ ; then $c \mapsto α (c)$ is a bijection $F^{n} \to K$ because the $α_{j}$ form a basis, and $σ_{i} (α (c)) = \sum_{j} c_{j} σ_{i} (α_{j}) = (A c)_{i}$ since each $σ_{i}$ fixes $F$ pointwise and is additive.

step 1.2given

2.2

Let $ψ ⁣ : K [x_{1}, \dots, x_{n}] \to K [x_{1}, \dots, x_{n}]$ be the unique $K$ -algebra homomorphism with $ψ (x_{i}) = \sum_{j} A_{i j} x_{j}$ , and $ψ^{'}$ the unique one with $ψ^{'} (x_{i}) = \sum_{j} (A^{- 1})_{i j} x_{j}$ ; both exist by [L3].

step 1.2L3

3.1

For $c \in F^{n}$ , the evaluation homomorphism $K [x_{1}, \dots, x_{n}] \to K$ at $c$ composed with $ψ$ sends $x_{i}$ to $\sum_{j} A_{i j} c_{j} = (A c)_{i}$ , so by the uniqueness clause of [L3] it is evaluation at $A c$ ; hence $ψ (f)$ evaluated at $c$ equals $f (A c) = f (σ_{1} α (c), \dots, σ_{n} α (c))$ , which is $0$ by the hypothesis of step 1.1.

step 1.1step 2.1step 2.2L3

4.1

So $ψ (f)$ vanishes at every tuple from the infinite subring $F$ of the integral domain $K$ , and [L1] gives $ψ (f) = 0$ .

step 3.1L1given

5.1

The composite $ψ^{'} \circ ψ$ is a $K$ -algebra endomorphism sending $x_{i}$ to $\sum_{j} A_{i j} \sum_{k} (A^{- 1})_{j k} x_{k} = \sum_{k} (A A^{- 1})_{i k} x_{k} = x_{i}$ , so it is the identity by the uniqueness clause of [L3]; applying $ψ^{'}$ to step 4.1 therefore gives $f = ψ^{'} (ψ (f)) = ψ^{'} (0) = 0$ , which is the contrapositive.

step 2.2step 4.1L3L5discharge-contrapositive∎

Remarks

Where infiniteness of $F$ enters. Only in step 4.1, through [L1]. Over a finite base field the conclusion is false: with $∣ F ∣ = q$ and $n = [K : F]$ , the nonzero polynomial $x_{1}^{q^{n}} - x_{1}$ vanishes at every conjugate tuple, since every element of $K$ satisfies $x^{q^{n}} = x$ . The finite case of the normal basis theorem is therefore proved by a different argument (Every finite cyclic extension has a normal basis).

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every finite Galois extension of an infinite field has a normal basis

Statement

Let $K / F$ be a finite Galois extension whose base field $F$ is infinite. Then $K / F$ has a normal basis (Normal bases of a finite Galois extension): there is $α \in K$ such that $(σ_{1} α, \dots, σ_{n} α)$ is an ordered $F$ -basis of $K$ , where $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ .

Facts & Assumptions

Given: A finite Galois extension $K / F$ (Finite Galois extensions and $Gal (K / F)$ ) of degree $n$ with $F$ infinite, and $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ numbered so that $σ_{1} = {i d}_{K}$ ; the matrix $B$ over $K [x_{1}, \dots, x_{n}]$ with $B_{i j} : = x_{k}$ where $k$ is the index determined by $σ_{i} σ_{j} = σ_{k}$ ; and $D : = \det B$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L1]

For every nonzero $f \in K [x_{1}, \dots, x_{n}]$ there is $α \in K$ with $f (σ_{1} α, \dots, σ_{n} α) \neq 0$ (Over an infinite base field, no nonzero polynomial vanishes at the conjugate tuple of every element).

[L2]

$(α_{1}, \dots, α_{n})$ is an ordered $F$ -basis of $K$ if and only if the matrix $A$ with $A_{i j} = σ_{i} (α_{j})$ is invertible (For a finite Galois extension, $(α_{j})$ is a base-field basis exactly when the matrix $(σ_{i} α_{j})$ is invertible).

[L3]

$\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i} a_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix); a matrix over a commutative ring is invertible if and only if its determinant is a unit (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

[L4]

A matrix $A$ is invertible when some $B$ satisfies $A B = I_{n} = B A$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ ); matrix products, the identity and the transpose are as in Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose, and obey the arithmetic laws of Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products.

[L5]

$K [x_{1}, \dots, x_{n}]$ is an integral domain (A polynomial ring in finitely many indeterminates over an integral domain is an integral domain, Polynomial rings in finitely many commuting indeterminates by iteration), and for any commutative $K$ -algebra $T$ and $t_{1}, \dots, t_{n} \in T$ there is a unique $K$ -algebra homomorphism $K [x_{1}, \dots, x_{n}] \to T$ with $x_{i} \mapsto t_{i}$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

Proof

technique · direct

1.1

$B$ is a well-defined $n \times n$ matrix over $K [x_{1}, \dots, x_{n}]$ : for each pair $(i, j)$ the product $σ_{i} σ_{j}$ lies in the group $Gal (K / F)$ and so equals $σ_{k}$ for exactly one index $k$ .

given

2.1

Let $ε ⁣ : K [x_{1}, \dots, x_{n}] \to K$ be the $K$ -algebra homomorphism with $ε (x_{1}) = 1$ and $ε (x_{k}) = 0$ for $k \neq 1$ , supplied by [L5]. Applying $ε$ entrywise to $B$ gives the matrix $P \in M_{n} (K)$ with $P_{i j} = 1$ when $σ_{i} σ_{j} = σ_{1} = i d$ , that is when $σ_{j} = σ_{i}^{- 1}$ , and $P_{i j} = 0$ otherwise.

step 1.1L5given

3.1

$P$ is invertible, with $P^{T}$ as an inverse: the $(i, i^{'})$ entry of $P P^{T}$ is $\sum_{j} P_{i j} P_{i^{'} j}$ , and a term is nonzero exactly when $σ_{j} = σ_{i}^{- 1}$ and $σ_{j} = σ_{i^{'}}^{- 1}$ , which happens for exactly one $j$ when $i = i^{'}$ and for no $j$ otherwise; so $P P^{T} = I_{n}$ , and the same computation on $P^{T} P$ gives $I_{n}$ . Hence $\det P$ is a unit of $K$ by [L3], and in particular $\det P \neq 0$ .

step 2.1L3L4

4.1

Since $\det$ is a polynomial expression in the entries by [L3] and $ε$ is a ring homomorphism, $ε (D) = ε (\det B) = \det P \neq 0$ ; hence $D \neq 0$ in $K [x_{1}, \dots, x_{n}]$ .

step 2.1step 3.1L3L5

5.1

By [L1] there is $α \in K$ with $D (σ_{1} α, \dots, σ_{n} α) \neq 0$ . Substituting $x_{k} \mapsto σ_{k} (α)$ in $B$ replaces the entry $B_{i j} = x_{k}$ , where $σ_{i} σ_{j} = σ_{k}$ , by $σ_{k} (α) = σ_{i} (σ_{j} (α))$ ; since substitution is a ring homomorphism, it carries $D = \det B$ to the determinant of the matrix $A$ with $A_{i j} = σ_{i} (α_{j})$ for $α_{j} : = σ_{j} (α)$ . So $\det A \neq 0$ .

step 4.1L1L3L5

6.1

A nonzero element of the field $K$ is a unit, so $A$ is invertible by [L3], and [L2] makes $(α_{1}, \dots, α_{n}) = (σ_{1} α, \dots, σ_{n} α)$ an ordered $F$ -basis of $K$ ; that is a normal basis.

step 5.1L2L3∎

Remarks

What the specialisation is for. The matrix of indeterminates is a device for showing that one determinant polynomial is not the zero polynomial, and the cheapest way to see that is to send it to a permutation matrix. Nothing about the particular substitution $x_{1} \mapsto 1$ survives into the conclusion: the element $α$ produced in step 5.1 has no relation to it.
Why $σ_{1}$ is the identity. Only so that the specialised matrix is the permutation matrix of $σ \mapsto σ^{- 1}$ ; any other choice of which indeterminate to set to $1$ would give the permutation matrix of a different bijection, with the same conclusion.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every finite cyclic extension has a normal basis

Statement

Let $K / F$ be a finite Galois extension whose Galois group is cyclic, say $Gal (K / F) = ⟨ σ ⟩$ of order $n = [K : F]$ . Then $K / F$ has a normal basis (Normal bases of a finite Galois extension): there is $α \in K$ for which

$(α, σ α, \dots, σ^{n - 1} α)$

is an ordered $F$ -basis of $K$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Facts & Assumptions

Given: A finite Galois extension $K / F$ with $Gal (K / F) = ⟨ σ ⟩$ cyclic of order $n$ ; by [L6] $\dim_{F} K = [K : F] = n$ , and $σ$ is an $F$ -linear endomorphism of the $F$ -vector space $K$ , written $T$ when regarded as such. Evaluation of a polynomial at $T$ is that of Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ .

[L1]

[L2]

For every endomorphism $T$ of a finite-dimensional $F$ -vector space, $Ann (T) = {p \in F [x] : p (T) = 0}$ is a nonzero ideal with a unique monic generator $μ_{T}$ , and $p (T) = 0$ if and only if $μ_{T} ∣ p$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ , The annihilator set $Ann (T) = {p \in F [x] : p (T) = 0}$ ; once existence is proved, its unique monic generator $μ_{T}$ is the minimal polynomial).

[L3]

For $A \in M_{n} (F)$ the polynomial $χ_{A}$ is monic of degree $n$ ( $χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$ ); $χ_{T}$ is defined as $χ_{[T]_{B}}$ for any ordered basis $B$ and is independent of it (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ in dimension zero).

[L4]

$μ_{T} ∣ χ_{T}$ for every endomorphism $T$ of a finite-dimensional vector space (The minimal polynomial divides the characteristic polynomial, $μ_{T} ∣ χ_{T}$ ).

[L5]

An endomorphism $T$ of a finite-dimensional vector space has a cyclic vector if and only if $μ_{T} = χ_{T}$ (A cyclic vector exists exactly when the minimal and characteristic polynomials agree); $v$ is a cyclic vector when $Z (v; T) = {p (T) v : p \in F [x]}$ is all of $V$ (Cyclic subspaces, cyclic vectors, and vector annihilators).

[L6]

For a finite Galois extension $K / F$ with $G = Gal (K / F)$ one has $∣ G ∣ = [K : F]$ (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and $Gal (K / F)$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L7]

With $m_{T, v}$ the unique monic generator of ${Ann}_{T} (v) = {p : p (T) v = 0}$ (The vector annihilator is the unique monic generator of ${Ann}_{T} (v)$ and divides the minimal polynomial) and $d = \deg m_{T, v}$ , the list $(v, T v, \dots, T^{d - 1} v)$ is an ordered basis of $Z (v; T)$ (A vector annihilator gives a power basis and its companion matrix).

Proof

technique · direct

1.1

$T^{n} = {i d}_{K}$ , because $σ$ has order $n$ in $Gal (K / F)$ ; so the polynomial $x^{n} - 1$ lies in $Ann (T)$ and $μ_{T} ∣ x^{n} - 1$ by [L2]. In particular $\deg μ_{T} \leq n$ .

L2given

1.2

The maps $i d, σ, σ^{2}, \dots, σ^{n - 1}$ are pairwise distinct elements of $Gal (K / F)$ , and their restrictions to $K^{\times}$ are pairwise distinct group homomorphisms $K^{\times} \to K^{\times}$ , since two field automorphisms of $K$ agreeing on $K^{\times}$ agree on $K$ .

L1given

2.1

If $p = \sum_{i < n} a_{i} x^{i}$ with $a_{i} \in F$ satisfies $p (T) = 0$ , then $\sum_{i < n} a_{i} σ^{i}$ is the zero function on $K$ , hence on $K^{\times}$ , so [L1] applied to the family of step 1.2 forces every $a_{i}$ to be $0$ ; thus no nonzero polynomial of degree less than $n$ annihilates $T$ , and $\deg μ_{T} \geq n$ .

step 1.2L1L2

3.1

Combining steps 1.1 and 2.1, $\deg μ_{T} = n$ , and since $μ_{T}$ is monic and divides the monic $x^{n} - 1$ of the same degree, $μ_{T} = x^{n} - 1$ .

step 1.1step 2.1L2algebra

4.1

By [L3] the polynomial $χ_{T}$ is monic of degree $\dim_{F} K = n$ , and $μ_{T} ∣ χ_{T}$ by [L4]; two monic polynomials of the same degree, one dividing the other, are equal, so $μ_{T} = χ_{T}$ .

step 3.1L3L4given

5.1

By [L5] there is a cyclic vector $α \in K$ for $T$ , that is $Z (α; T) = K$ .

step 4.1L5

6.1

Let $d = \deg m_{T, α}$ . By [L7] the list $(α, T α, \dots, T^{d - 1} α)$ is an ordered basis of $Z (α; T) = K$ , which has dimension $n$ , so $d = n$ and $(α, σ α, \dots, σ^{n - 1} α)$ is an ordered $F$ -basis of $K$ .

step 5.1L6L7given

7.1

Since $Gal (K / F) = {i d, σ, \dots, σ^{n - 1}}$ , that list is exactly the family of conjugates of $α$ , so it is a normal basis.

step 1.2step 6.1given∎

Remarks

Why the minimal polynomial is forced to be $x^{n} - 1$ . The divisibility $μ_{T} ∣ x^{n} - 1$ is cheap; the content is the lower bound on its degree, and that is exactly Dedekind's independence of characters. Without it the minimal polynomial could be a proper divisor of $x^{n} - 1$ and no cyclic vector would be available.
The hypothesis is on the group, not on the base field. No finiteness or infiniteness of $F$ is used, so this proof also covers cyclic extensions of infinite fields, for which Every finite Galois extension of an infinite field has a normal basis gives a second and quite different argument.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every finite Galois extension has a normal basis

Statement

Every finite Galois extension $K / F$ has a normal basis (Normal bases of a finite Galois extension): there is $α \in K$ whose family of conjugates $(σ_{1} α, \dots, σ_{n} α)$ , indexed by $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ , is an ordered $F$ -basis of $K$ .

Facts & Assumptions

Given: A finite Galois extension $K / F$ of degree $n$ , so that $K$ is an $F$ -vector space of dimension $n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension, Equivalent characterizations of a finite Galois extension).

[L1]

Every finite Galois extension of an infinite field has a normal basis (Every finite Galois extension of an infinite field has a normal basis).

[L2]

Every finite Galois extension whose Galois group is cyclic has a normal basis (Every finite cyclic extension has a normal basis).

[L3]

An extension $E / F_{q}$ of finite fields of degree $m$ is Galois with $Gal (E / F_{q})$ cyclic of order $m$ , generated by $x \mapsto x^{q}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ ).

[L4]

A finite list $v$ of length $n$ is an ordered basis of $V$ if and only if every $x \in V$ has exactly one coordinate list $λ : n \to F$ with $x = \sum_{i < n} λ_{i} v_{i}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L5]

A finite field is a field whose underlying set is finite, and its order is that cardinality (Finite fields and their order).

Proof

technique · cases

1.1

In the case that $F$ is infinite, [L1] applies directly and $K / F$ has a normal basis.

assume-case infL1

1.2

In the case that $F$ is finite, fix an ordered $F$ -basis $v$ of $K$ of length $n$ ; by [L4] the map sending a coordinate list $λ : n \to F$ to $\sum_{i < n} λ_{i} v_{i}$ is a bijection onto $K$ , so $∣ K ∣ = ∣ F ∣^{n}$ is finite and $K$ is a finite field.

assume-case finL4L5given

2.1

In that same finite case, $K / ⁣ F$ is therefore an extension of finite fields of degree $n$ , so $Gal (K / F)$ is cyclic by [L3], and [L2] gives a normal basis.

step 1.2L2L3

3.1

The two cases are exhaustive, a field being finite or infinite and not both, so a normal basis exists in either case.

step 1.1step 2.1cases-exhaustive∎

Remarks

Two genuinely different proofs, not one proof with a case split. The infinite case runs on a determinant that is a nonzero polynomial (Every finite Galois extension of an infinite field has a normal basis); the finite case runs on a cyclic vector for the Frobenius acting linearly (Every finite cyclic extension has a normal basis). Neither argument covers the other case: the first fails because a polynomial can vanish on all of a finite field, the second because a Galois group need not be cyclic.
The finite case is not a hypothesis on the group. It is a hypothesis on the base field, which forces the group to be cyclic through [L3]. That is the whole reason the split is by the base field rather than by the group.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity

Definition

Let $K$ be a field and let $n \geq 1$ be an integer. An element $x \in K$ is an $n$ -th root of unity when $x^{n} = 1$ , that is when $x$ is a root of $t^{n} - 1 \in K [t]$ (Evaluation and roots of a polynomial in a commutative target ring). Write

$μ_{n} (K) : = {x \in K : x^{n} = 1} .$

This is a subgroup of $K^{\times}$ (Subgroup). Every $x \in μ_{n} (K)$ is invertible, with inverse $x^{n - 1}$ (Left inverse, right inverse, and invertible element of a monoid), so $μ_{n} (K) \subseteq K^{\times}$ ; it contains $1$ ; it is closed under multiplication, since $(x y)^{n} = x^{n} y^{n} = 1$ ; and it is closed under inverses, since $(x^{- 1})^{n} = (x^{n})^{- 1} = 1$ .

An element $ζ \in μ_{n} (K) \subseteq K^{\times}$ is a primitive $n$ -th root of unity when its order in the group $K^{\times}$ is exactly $n$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity):

$ord (ζ) = n .$

Equivalently, $ζ^{n} = 1$ and no exponent $k$ with $1 \leq k < n$ has $ζ^{k} = 1$ .

An $n$ -th root of unity need not be primitive. In $Q$ the element $- 1$ is a fourth root of unity of order two, not four; and $μ_{n} (K)$ may consist of $1$ alone, as $μ_{3} (Q)$ does. The two notions are separated deliberately, and the exact circumstances under which a primitive $n$ -th root of unity exists are the content of $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ and $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity.

Remarks

The concrete complex picture arrives later in the reading order. Over $C$ the $n$ -th roots of unity are the numbers $e^{2 π i k / n}$ , and that description is developed in The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ↗. Nothing on this page uses it: the definition above is purely algebraic and applies to every field, including those of positive characteristic, where the count of $n$ -th roots of unity can be smaller than $n$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$

Statement

Let $K$ be a field and $n \geq 1$ . Then $μ_{n} (K) = {x \in K : x^{n} = 1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity) is a finite cyclic subgroup of $K^{\times}$ whose order divides $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). It contains a primitive $n$ -th root of unity if and only if $∣ μ_{n} (K) ∣ = n$ , and in that case the primitive $n$ -th roots of unity in $K$ are exactly the generators of $μ_{n} (K)$ , of which there are $φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ , and the subgroup $μ_{n} (K)$ of $K^{\times}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L1]

Let $D$ be an integral domain. Every finite subgroup $G \leq D^{\times}$ of the unit group of $D$ is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic).

[L2]

A nonzero polynomial of degree $k$ over an integral domain has at most $k$ distinct roots in that domain (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L3]

For an element $x$ of finite order $m$ in a group: $x^{k} = e$ if and only if $m ∣ k$ ; and $∣ ⟨ x ⟩ ∣ = m = ord (x)$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L4]

In a cyclic group $⟨ g ⟩$ of finite order $m$ , the element $g^{a}$ generates the group if and only if $\gcd (a, m) = 1$ , and the group has exactly $φ (m)$ generators (A cyclic group of order $n$ has exactly $φ (n)$ generators).

Proof

technique · direct

1.1

$μ_{n} (K)$ is the set of roots in $K$ of the nonzero polynomial $t^{n} - 1$ , of degree $n$ ; a field is an integral domain, so $∣ μ_{n} (K) ∣ \leq n$ by [L2] and $μ_{n} (K)$ is finite.

L2given

2.1

Being a finite subgroup of $K^{\times}$ , $μ_{n} (K)$ is cyclic by [L1]; write $m : = ∣ μ_{n} (K) ∣$ and fix a generator $ζ_{0}$ , so that $ord (ζ_{0}) = m$ by [L3].

step 1.1L1L3

3.1

The order $m$ divides $n$ : $ζ_{0} \in μ_{n} (K)$ gives $ζ_{0}^{n} = 1$ , and [L3] turns this into $m ∣ n$ .

step 2.1L3

3.2

If $K$ contains a primitive $n$ -th root of unity $ζ$ , that is an element of order $n$ , then $ζ^{n} = 1$ puts $ζ$ in $μ_{n} (K)$ , and $⟨ ζ ⟩ \subseteq μ_{n} (K)$ has $n$ elements by [L3], so $n \leq m$ ; with step 1.1 this forces $m = n$ .

step 1.1step 2.1L3

3.3

Conversely, if $m = n$ then the generator $ζ_{0}$ of step 2.1 has order $n$ and so is a primitive $n$ -th root of unity in $K$ .

step 2.1L3

4.1

Suppose $m = n$ . An element $x \in K$ of order $n$ lies in $μ_{n} (K) = ⟨ ζ_{0} ⟩$ and satisfies $∣ ⟨ x ⟩ ∣ = n = ∣ μ_{n} (K) ∣$ by [L3], so $⟨ x ⟩ = μ_{n} (K)$ and $x$ is a generator; conversely a generator has order $n$ by [L3]. So the primitive $n$ -th roots of unity in $K$ are exactly the generators of $μ_{n} (K)$ , and [L4] counts them as $φ (n)$ .

step 2.1step 3.2step 3.3L3L4∎

Remarks

Order dividing $n$ , not equal to $n$ . The two failures are different in kind. Over $Q$ the polynomial $t^{3} - 1$ simply does not split, so $μ_{3} (Q) = {1}$ ; over a field of characteristic three it cannot split into distinct factors at all, since $t^{3} - 1 = (t - 1)^{3}$ there (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ). Only the first failure is repaired by passing to a splitting field.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity

Statement

Let $K$ be a field and $n \geq 1$ . Then

$t^{n} - 1 is separable over K ⟺ char K ∤ n$

(Repeated roots in extension fields and separable polynomials, The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise; divisibility of integers being that of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , under which $0 ∣ n$ holds only for $n = 0$ , so characteristic $0$ never divides $n \geq 1$ ).

When $char K ∤ n$ and $E$ is a splitting field of $t^{n} - 1$ over $K$ , the group $μ_{n} (E)$ is cyclic of order exactly $n$ , it contains exactly $φ (n)$ primitive $n$ -th roots of unity, and

$E = K (ζ)$

for every primitive $n$ -th root of unity $ζ \in E$ .

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ , the polynomial $f : = t^{n} - 1 \in K [t]$ , and the element $n \cdot 1_{K}$ obtained by adding $1_{K}$ to itself $n$ times.

[L1]

$f^{'} = \sum_{i \geq 1} i a_{i} x^{i - 1}$ for $f = \sum_{i} a_{i} x^{i}$ (The formal derivative of a polynomial); $(x^{k})^{'} = k x^{k - 1}$ for $k \geq 1$ and constants have derivative $0$ , and the derivative is additive (Linearity, power rule, Leibniz rule and the degree bound for the formal derivative). Hence $f^{'} = (n \cdot 1_{K}) t^{n - 1}$ .

[L2]

For a field $F$ and $0 \neq f \in F [x]$ : $f$ is separable over $F$ if and only if $\gcd (f, f^{'}) = 1$ in $F [x]$ (A nonzero polynomial over a field is separable exactly when its gcd with its derivative is $1$ ).

[L3]

For $f, g \in F [x]$ not both zero, $d = \gcd (f, g)$ is monic, divides both $f$ and $g$ , and every common divisor of $f$ and $g$ divides $d$ (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L4]

For a ring $R$ and $n \in Z$ , one has $n \cdot 1_{R} = 0_{R}$ if and only if $char (R)$ divides $n$ (The characteristic of a ring is the additive order of $1_{R}$ , with $0$ recording infinite order; $n \cdot 1_{R} = 0$ holds exactly when $char (R) ∣ n$ ; and in an integral domain every nonzero element has the same additive order as $1_{R}$ , The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise); the characteristic of a field is $0$ or a prime (The characteristic of a field is zero or a prime number).

[L5]

Every nonzero $f \in K [x]$ has a splitting field over $K$ (Every nonzero polynomial over a field has a splitting field); $f$ of degree $m$ splits over $E$ when $f = c \prod_{j = 1}^{m} (t - α_{j})$ with $c \in K^{\times}$ and $α_{j} \in E$ , repetitions allowed, and a splitting field is generated over $K$ by the roots (Polynomials that split and splitting fields of a polynomial or a family of polynomials), the notation $K (S)$ for the subfield generated by $K$ and a finite set $S$ being that of Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ .

[L6]

$μ_{n} (F)$ is a finite cyclic subgroup of $F^{\times}$ of order dividing $n$ ; it contains a primitive $n$ -th root of unity exactly when its order is $n$ , and there are then $φ (n)$ of them, namely the generators ( $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ , The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L7]

$f$ is separable over $K$ when it has no repeated root in any extension field of $K$ , where $a$ is a repeated root of $f$ in $E$ when $(t - a)^{2}$ divides the image of $f$ in $E [t]$ (Repeated roots in extension fields and separable polynomials).

Proof

technique · cases

1.1

In the case $n \cdot 1_{K} \neq 0$ , put $d : = \gcd (f, f^{'})$ . By [L1] $f^{'} = (n \cdot 1_{K}) t^{n - 1}$ with $n \cdot 1_{K}$ a unit of $K$ , so $d ∣ t^{n - 1}$ by [L3], hence $d ∣ t \cdot t^{n - 1} = t^{n}$ ; and $d ∣ f = t^{n} - 1$ , so $d$ divides $t^{n} - (t^{n} - 1) = 1$ and, being monic, $d = 1$ . By [L2] the polynomial $f$ is separable over $K$ .

assume-case posL1L2L3

1.2

In the case $n \cdot 1_{K} = 0$ , [L1] gives $f^{'} = 0$ , so every polynomial dividing $f$ is a common divisor of $f$ and $f^{'}$ ; by [L3] the monic $\gcd (f, 0)$ is divisible by every such divisor and divides $f$ , so it is $f$ itself, of degree $n \geq 1$ . Thus $\gcd (f, f^{'}) \neq 1$ and $f$ is not separable over $K$ by [L2].

assume-case zeroL1L2L3

2.1

The two cases are exhaustive and, by [L4], $n \cdot 1_{K} \neq 0$ says exactly $char K ∤ n$ ; so $f$ is separable over $K$ if and only if $char K ∤ n$ .

step 1.1step 1.2L4cases-exhaustive

3.1

Assume now $char K ∤ n$ and let $E$ be a splitting field of $f$ over $K$ , which exists by [L5]. Over $E$ one has $f = c \prod_{j = 1}^{n} (t - α_{j})$ with $α_{j} \in E$ , and comparing leading coefficients of the monic $f$ gives $c = 1$ .

step 2.1L5

4.1

The $α_{j}$ are pairwise distinct: if $α_{i} = α_{j}$ for $i \neq j$ then $(t - α_{i})^{2}$ divides $f$ in $E [t]$ , making $α_{i}$ a repeated root of $f$ in the extension $E$ of $K$ , which contradicts the separability supplied by step 2.1 through [L7].

step 2.1step 3.1L7

5.1

Hence $f$ has exactly $n$ distinct roots in $E$ , that is $∣ μ_{n} (E) ∣ = n$ ; by [L6] the group $μ_{n} (E)$ is cyclic of order $n$ and contains exactly $φ (n)$ primitive $n$ -th roots of unity.

step 3.1step 4.1L6

6.1

Fix such a $ζ$ . Every root of $f$ in $E$ lies in $μ_{n} (E) = ⟨ ζ ⟩$ , so is a power of $ζ$ ; since $E$ is generated over $K$ by those roots by [L5], $E = K (ζ)$ . At $n = 1$ the polynomial is $t - 1$ , $μ_{1} (E) = {1}$ , $φ (1) = 1$ , $ζ = 1$ and $E = K$ .

step 5.1L5L6∎

Remarks

The failing direction is inseparability, not a shortage of roots. When $char K = p$ divides $n$ , the derivative vanishes identically and no extension can separate the roots: writing $n = p^{k} m$ with $p ∤ m$ , one has $μ_{n} = μ_{m}$ in every field of characteristic $p$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ). Passing to a larger field does not help, which is why the hypothesis is carried on every later statement rather than removed by enlarging $K$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$

Statement

Let $K$ be a field of characteristic $p > 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise; $p$ is prime by The characteristic of a field is zero or a prime number) and let $k \geq 1$ . Then in $K [t]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution)

$t^{p^{k}} - 1 = (t - 1)^{p^{k}},$

and consequently $μ_{p^{k}} (K) = {1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

More generally, if $k \geq 0$ and $m \geq 1$ with $p ∤ m$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), and if $n : = p^{k} m$ , then

$μ_{n} (K) = μ_{m} (K) .$

Facts & Assumptions

Given: A field $K$ of characteristic $p > 0$ , so that $p \cdot 1_{K} = 0_{K}$ and hence $j \cdot c = (j \cdot 1_{K}) c = 0$ for every $c \in K [t]$ and every integer $j$ divisible by $p$ ; and an integer $k \geq 1$ .

[L1]

For a commutative ring $R$ , all $x, y \in R$ and $n \in N$ , $(x + y)^{n} = \sum_{j = 0}^{n} (\binom{n}{j}) x^{j} y^{n - j}$ , the natural-number coefficients acting by repeated addition (The binomial theorem over an arbitrary commutative ring).

[L2]

If $p$ is prime and $0 < j < p$ , then $p ∣ (\binom{p}{j})$ (A prime $p$ divides $(\binom{p}{k})$ for $0 < k < p$ ).

[L3]

$μ_{n} (K) = {x \in K : x^{n} = 1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

Proof

technique · direct

1.1

For every $u \in K [t]$ one has $(u - 1)^{p} = u^{p} - 1$ : by [L1] applied in the commutative ring $K [t]$ , $(u + (- 1))^{p} = \sum_{j = 0}^{p} (\binom{p}{j}) u^{j} (- 1)^{p - j}$ ; for $0 < j < p$ the coefficient $(\binom{p}{j})$ is a multiple of $p$ by [L2], so that term vanishes by the hypothesis on $K$ ; the surviving terms are $u^{p}$ and $(- 1)^{p}$ , and $(- 1)^{p} = - 1$ for odd $p$ while for $p = 2$ one has $1 = - 1$ in $K$ , so $(- 1)^{p} = - 1$ in either case.

L1L2given

2.1

Hence $(t - 1)^{p^{k}} = t^{p^{k}} - 1$ for every $k \geq 1$ , by induction on $k$ : at $k = 1$ this is step 1.1 with $u = t$ ; and if it holds at $k$ , then $(t - 1)^{p^{k + 1}} = ((t - 1)^{p^{k}})^{p} = (t^{p^{k}} - 1)^{p} = t^{p^{k + 1}} - 1$ , the last equality being step 1.1 with $u = t^{p^{k}}$ .

step 1.1algebra

3.1

Therefore $μ_{p^{k}} (K) = {1}$ : an $x \in K$ with $x^{p^{k}} = 1$ is a root of $t^{p^{k}} - 1$ , so $(x - 1)^{p^{k}} = 0$ by step 2.1, and a field has no nonzero element with a vanishing power, so $x = 1$ ; and $1^{p^{k}} = 1$ .

step 2.1L3

4.1

Let $k \geq 0$ and $m \geq 1$ with $p ∤ m$ , and put $n = p^{k} m$ . If $x^{n} = 1$ then $(x^{m})^{p^{k}} = x^{n} = 1$ , so $x^{m} \in μ_{p^{k}} (K)$ , which is ${1}$ by step 3.1 when $k \geq 1$ and is ${1}$ trivially when $k = 0$ ; either way $x^{m} = 1$ . Conversely $x^{m} = 1$ gives $x^{n} = (x^{m})^{p^{k}} = 1$ . Hence $μ_{n} (K) = μ_{m} (K)$ .

step 3.1L3algebra∎

Remarks

This is why every later hypothesis reads "the characteristic does not divide $n$ ". Nothing is lost by it: the $p$ -part of $n$ contributes no roots of unity at all in characteristic $p$ , so a statement about $μ_{n}$ there is already a statement about $μ_{m}$ for the prime-to- $p$ part $m$ . The hypothesis excludes a degenerate case rather than a genuine one.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$

Definition

Let $K$ be a field and $n \geq 1$ . A cyclotomic extension of $K$ of order $n$ is a splitting field $E$ of $t^{n} - 1$ over $K$ (Polynomials that split and splitting fields of a polynomial or a family of polynomials); one exists by Every nonzero polynomial over a field has a splitting field. It is written

$K (μ_{n}) : = E .$

The notation is accurate. The roots of $t^{n} - 1$ in $E$ are exactly the elements of $μ_{n} (E)$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity), and a splitting field is generated over $K$ by the roots, so

$E = K (μ_{n} (E))$

in the sense of Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ : $E$ is the smallest subfield of itself containing $K$ and the $n$ -th roots of unity it holds.

When the characteristic does not divide $n$ the extension has a single generator: by $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity the group $μ_{n} (E)$ is then cyclic of order $n$ and

$K (μ_{n}) = K (ζ)$

for any primitive $n$ -th root of unity $ζ \in E$ . Without that hypothesis the notation still names a splitting field, but $μ_{n} (E)$ can be much smaller than $n$ and no primitive $n$ -th root of unity need exist (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ).

Remarks

Which splitting field. Any two splitting fields of $t^{n} - 1$ over $K$ are $K$ -isomorphic (Any two splitting fields of a polynomial are isomorphic over the base field), and every statement made below about $K (μ_{n})$ is invariant under a $K$ -isomorphism, so the definite article is harmless; where a fixed ambient field matters, as in the compositum and intersection results, the statement says so and works inside one chosen extension of $K$ .

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise), and let $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then $E / K$ is a finite Galois extension, and there is an injective group homomorphism

$Gal (E / K) ⟶ (Z / n)^{\times}, σ ⟼ [a_{σ}]_{n},$

where $a_{σ}$ is any integer with $σ (ζ) = ζ^{a_{σ}}$ for a primitive $n$ -th root of unity $ζ \in E$ . The class $[a_{σ}]_{n}$ does not depend on which primitive $n$ -th root of unity is used, and $σ (x) = x^{a_{σ}}$ holds for every $x \in μ_{n} (E)$ .

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ with $char K ∤ n$ , the extension $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ), and a primitive $n$ -th root of unity $ζ \in E$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L1]

For $n \geq 1$ , $t^{n} - 1$ is separable over $K$ when $char K ∤ n$ ; in a splitting field $E$ the group $μ_{n} (E)$ is then cyclic of order $n$ , has $φ (n)$ primitive $n$ -th roots of unity, and $E = K (ζ)$ for every primitive $n$ -th root of unity $ζ \in E$ ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity).

[L2]

For a finite extension $L / F$ with $G = Aut (L / F)$ , the conditions " $L / F$ is Galois", " $L$ is the splitting field over $F$ of a separable polynomial", " $∣ G ∣ = [L : F]$ " and " $L^{G} = F$ " are equivalent (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and $Gal (K / F)$ , Relative field automorphisms and $Aut (K / F)$ ).

[L3]

$μ_{n} (E)$ is a finite cyclic subgroup of $E^{\times}$ of order dividing $n$ , and when its order is $n$ its generators are exactly the primitive $n$ -th roots of unity ( $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L4]

In a cyclic group $⟨ g ⟩$ of finite order $m$ , $g^{a}$ generates the group if and only if $\gcd (a, m) = 1$ (A cyclic group of order $n$ has exactly $φ (n)$ generators).

[L5]

For an element $x$ of finite order $m$ : $x^{k} = e$ if and only if $m ∣ k$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L6]

For $n \geq 1$ and $a \in Z$ , the class $[a]_{n} \in Z / n$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ) is a unit of $Z / n$ if and only if $\gcd (a, n) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ); the units form the group $(Z / n)^{\times}$ of order $φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L7]

If $a$ is algebraic over a field $K$ , then the simple extension $K (a) / K$ is finite, with degree equal to the degree of the minimal polynomial of $a$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

Proof

technique · direct

1.1

By [L1] the polynomial $t^{n} - 1$ is separable over $K$ , the group $μ_{n} (E)$ is cyclic of order $n$ generated by $ζ$ , and $E = K (ζ)$ . The element $ζ$ is algebraic because it is a root of $t^{n} - 1$ , so [L7] makes $E / K$ finite. Since $E$ is the splitting field of the separable polynomial $t^{n} - 1$ , [L2] now makes $E / K$ finite Galois.

L1L2L3L7given

2.1

Each $σ \in Gal (E / K)$ maps $μ_{n} (E)$ into itself, since $σ (x)^{n} = σ (x^{n}) = σ (1) = 1$ ; being injective on the finite set $μ_{n} (E)$ it restricts to a bijection, and it is multiplicative, so it restricts to a group automorphism of $μ_{n} (E)$ .

step 1.1given

3.1

Hence $σ (ζ)$ generates $μ_{n} (E) = ⟨ ζ ⟩$ , so $σ (ζ) = ζ^{a}$ for some integer $a$ , and [L4] gives $\gcd (a, n) = 1$ , so $[a]_{n} \in (Z / n)^{\times}$ by [L6]. The class is well defined: $ζ^{a} = ζ^{b}$ means $ζ^{a - b} = 1$ , which by [L5] and $ord (ζ) = n$ says $n ∣ a - b$ , that is $[a]_{n} = [b]_{n}$ . Write $[a_{σ}]_{n}$ for this class.

step 1.1step 2.1L4L5L6

4.1

For every $x \in μ_{n} (E)$ one has $x = ζ^{j}$ for some $j$ , so $σ (x) = σ (ζ)^{j} = ζ^{a_{σ} j} = x^{a_{σ}}$ .

step 1.1step 3.1

4.2

The map $σ \mapsto [a_{σ}]_{n}$ is a group homomorphism: $(σ τ) (ζ) = σ (ζ^{a_{τ}}) = σ (ζ)^{a_{τ}} = ζ^{a_{σ} a_{τ}}$ , so $[a_{σ τ}]_{n} = [a_{σ}]_{n} [a_{τ}]_{n}$ by the well-definedness of step 3.1.

step 3.1L6

4.3

It is injective: if $[a_{σ}]_{n} = [1]_{n}$ then $σ (ζ) = ζ$ by step 3.1, and since $E = K (ζ)$ and $σ$ fixes $K$ pointwise, $σ$ is the identity on $E$ .

step 1.1step 3.1

5.1

The class is independent of the chosen primitive $n$ -th root of unity: any other one is a generator $ζ^{'}$ of $μ_{n} (E)$ by [L3], so $ζ^{'} = ζ^{b}$ for some $b$ , and $σ (ζ^{'}) = σ (ζ)^{b} = ζ^{a_{σ} b} = (ζ^{b})^{a_{σ}} = (ζ^{'})^{a_{σ}}$ , which exhibits the same exponent class. With steps 4.1, 4.2 and 4.3 this proves the theorem.

step 1.1step 3.1step 4.1step 4.2step 4.3L3∎

Remarks

The embedding need not be onto. Surjectivity is exactly irreducibility of the $n$ -th cyclotomic polynomial over $K$ ( $Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto), and it fails over many base fields: over $F_{q}$ the image is the cyclic subgroup generated by the class of $q$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ ).

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Galois group of a cyclotomic extension is abelian

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise). Then $Gal (K (μ_{n}) / K)$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ) is abelian.

Facts & Assumptions

Given: A field $K$ and $n \geq 1$ with $char K ∤ n$ .

[L1]

$K (μ_{n}) / K$ is finite Galois and $σ \mapsto [a_{σ}]_{n}$ is an injective group homomorphism $Gal (K (μ_{n}) / K) \to (Z / n)^{\times}$ ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ).

[L2]

For every $n \geq 1$ , $(Z / n, \cdot, [1]_{n})$ is a commutative monoid (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold), and $(Z / n)^{\times}$ is the group of its invertible elements (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Proof

technique · direct

1.1

Multiplication on $Z / n$ is commutative by [L2], so the group $(Z / n)^{\times}$ , whose operation is that multiplication restricted to the units, is abelian.

2.1

By [L1] the group $Gal (K (μ_{n}) / K)$ is isomorphic to its image in $(Z / n)^{\times}$ , a subgroup of an abelian group; a subgroup of an abelian group is abelian, and a group isomorphic to an abelian group is abelian, so $Gal (K (μ_{n}) / K)$ is abelian.

step 1.1L1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$

Statement

Let $F_{q}$ be a finite field of order $q$ (Finite fields and their order) and let $n \geq 1$ with $\gcd (n, q) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ). Then the image of the embedding

$Gal (F_{q} (μ_{n}) / F_{q}) ⟶ (Z / n)^{\times}$

of $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ is the cyclic subgroup generated by $[q]_{n}$ , and

$[F_{q} (μ_{n}) : F_{q}] = ord ([q]_{n}),$

the multiplicative order of $[q]_{n}$ in $(Z / n)^{\times}$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q$ , of characteristic $p$ with $q$ a power of $p$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ), and an integer $n \geq 1$ with $\gcd (n, q) = 1$ ; the extension $E : = F_{q} (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[L1]

For a field $K$ and $n \geq 1$ with $char K ∤ n$ , the extension $K (μ_{n}) / K$ is finite Galois and $σ \mapsto [a_{σ}]_{n}$ , determined by $σ (ζ) = ζ^{a_{σ}}$ on a primitive $n$ -th root of unity, is an injective homomorphism into $(Z / n)^{\times}$ with $σ (x) = x^{a_{σ}}$ for every $x \in μ_{n}$ ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ).

[L2]

An extension $L / F_{q}$ of finite fields of degree $m$ is Galois with $Gal (L / F_{q}) = ⟨ σ_{q} ⟩$ cyclic of order $m$ , where $σ_{q} (x) = x^{q}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L3]

For $n \geq 1$ and $a \in Z$ , the class $[a]_{n}$ is a unit of $Z / n$ if and only if $\gcd (a, n) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L4]

For a finite Galois extension $L / F$ one has $∣ Gal (L / F) ∣ = [L : F]$ (Equivalent characterizations of a finite Galois extension, The degree $[K : F] = \dim_{F} K$ of a finite field extension).

Proof

technique · direct

1.1

The characteristic $p$ divides $q$ , and $\gcd (n, q) = 1$ , so $p ∤ n$ ; hence [L1] applies to $K = F_{q}$ and $E = F_{q} (μ_{n})$ is finite Galois over $F_{q}$ . Also $[q]_{n}$ is a unit of $Z / n$ by [L3].

L1L3given

2.1

$E$ is a finite field: it is a finite extension of the finite field $F_{q}$ by step 1.1, so it is a finite-dimensional $F_{q}$ -vector space over a finite field and therefore has finitely many elements. By [L2], $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ with $σ_{q} (x) = x^{q}$ .

step 1.1L2

3.1

The exponent attached to $σ_{q}$ by [L1] is $[q]_{n}$ , since $σ_{q} (ζ) = ζ^{q}$ for a primitive $n$ -th root of unity $ζ \in E$ . Because the embedding is a homomorphism and $Gal (E / F_{q})$ is generated by $σ_{q}$ , the image is the subgroup of $(Z / n)^{\times}$ generated by $[q]_{n}$ .

step 1.1step 2.1L1

4.1

The embedding is injective, so $∣ Gal (E / F_{q}) ∣$ equals the order of $⟨ [q]_{n} ⟩$ , which is $ord ([q]_{n})$ ; and $[E : F_{q}] = ∣ Gal (E / F_{q}) ∣$ by [L4]. Hence $[E : F_{q}] = ord ([q]_{n})$ .

step 3.1L1L4∎

Remarks

What the coprimality hypothesis does. It makes $[q]_{n}$ a unit, so that the subgroup it generates is defined, and it puts the extension inside the scope of [L1] by ruling out $char F_{q} ∣ n$ . Without it the group $μ_{n}$ collapses onto $μ_{m}$ for the prime-to- $p$ part $m$ of $n$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ), and the statement is about $m$ rather than $n$ .

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$

Definition

The cyclotomic polynomials $Φ_{n} \in Z [t]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution) are defined by recursion on $n \geq 1$ :

$Φ_{1} : = t - 1, Φ_{n} : = \frac{t^{n} - 1}{\prod_{\begin{array}{c} d ∣ n \\ 0 < d < n \end{array}} Φ_{d}} (n \geq 2),$

the divisors being the positive divisors of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). The product is formed in the commutative ring $Z [t]$ (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring): multiply the finitely many factors in any enumeration of the divisor set. Associativity and commutativity make the result independent of that enumeration.

What the fraction means. The denominator $P_{n} : = \prod_{d ∣ n, d < n} Φ_{d}$ is a product of monic polynomials in $Z [t]$ , hence itself monic (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree), so Division by a monic polynomial over a commutative ring supplies a unique pair $q, r \in Z [t]$ with $t^{n} - 1 = q P_{n} + r$ and $r = 0$ or $\deg r < \deg P_{n}$ . The definition sets $Φ_{n} : = q$ , and asserts $r = 0$ . That assertion, together with the consequences that each $Φ_{n}$ is monic of degree $φ (n)$ and that

$\prod_{d ∣ n} Φ_{d} = t^{n} - 1$

for every $n \geq 1$ , is discharged by The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ ↗, which is why that theorem is a numbered result and not a parenthesis: the division is carried out over $Z$ , not over a field, so exactness is a statement about integer coefficients and does not follow from the division algorithm.

Remarks

Why not define $Φ_{n}$ by its roots. The usual definition takes $Φ_{n}$ to be the monic polynomial whose roots are the primitive $n$ -th roots of unity in $C$ , or the minimal polynomial over $Q$ of a primitive $n$ -th root of unity. Neither is available at this point in the reading order, and neither would serve: the minimal-polynomial version makes $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ a tautology rather than a theorem, and over a base field where the reduction of $Φ_{n}$ is reducible it no longer controls the factorisation there (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ). The recursion above is a statement about $Z [t]$ alone, and the description of the roots is then a theorem (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity).
Reduction into another field. For a field $K$ the image of $Φ_{n}$ under the coefficientwise ring homomorphism $Z [t] \to K [t]$ induced by the canonical map $Z \to K$ and fixing $t$ is again written $Φ_{n}$ when no confusion arises; the identity $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ is preserved, since a ring homomorphism preserves finite products.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$

Statement

The recursion of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ is well posed: for every $n \geq 1$ the division defining $Φ_{n}$ is exact, $Φ_{n}$ is a monic element of $Z [t]$ ,

$\prod_{d ∣ n} Φ_{d} = t^{n} - 1,$

the product being over the positive divisors of $n$ , and

$\deg Φ_{n} = φ (n)$

(The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ). Moreover $(Φ_{n})_{n \geq 1}$ is the only family of monic polynomials in $Z [t]$ satisfying the displayed product identity for every $n \geq 1$ .

Facts & Assumptions

Given: The recursion of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ; the field $Q$ , which is an ordered field (The rationals form a totally ordered field), so that $m \cdot 1 > 0$ and in particular $m \cdot 1 \neq 0$ for every $m \geq 1$ , whence $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and $char Q$ divides no $n \geq 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

Let $R$ be a commutative ring and $g \in R [x]$ monic. For every $f \in R [x]$ there are unique $q, r \in R [x]$ with $f = q g + r$ and $r = 0$ or $\deg r < \deg g$ (Division by a monic polynomial over a commutative ring).

[L2]

$t^{n} - 1$ is separable over $K$ exactly when $char K ∤ n$ ; and then a splitting field $E$ has $μ_{n} (E)$ cyclic of order $n$ ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity, The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L3]

Every nonzero $f \in K [x]$ has a splitting field over $K$ (Every nonzero polynomial over a field has a splitting field); $f$ monic of degree $m$ splits over $E$ when $f = \prod_{j = 1}^{m} (t - α_{j})$ with $α_{j} \in E$ , repetitions allowed (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L4]

$f$ is separable over $K$ when it has no repeated root in any extension field, a repeated root of $f$ in $E$ being an $a$ with $(t - a)^{2}$ dividing the image of $f$ in $E [t]$ (Repeated roots in extension fields and separable polynomials).

[L5]

For an element $x$ of finite order $m$ in a group, $x^{k} = e$ if and only if $m ∣ k$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L6]

$\sum_{d ∣ n, d > 0} φ (d) = n$ for every positive integer $n$ (For every positive integer $n$ , $\sum_{d ∣ n, d > 0} φ (d) = n$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

[L7]

If $R$ is an integral domain then so is $R [x]$ (A polynomial ring over an integral domain is an integral domain), and for nonzero $f, g$ one has $\deg (f g) = \deg f + \deg g$ and $lc (f g) = lc (f) lc (g)$ (Over an integral domain, degrees add under multiplication of nonzero polynomials, Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

Proof

technique · induction

1.1

The assertion to be proved by strong induction on $n$ is: the division defining $Φ_{n}$ is exact, $Φ_{n} \in Z [t]$ is monic, $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , and $\deg Φ_{n} = φ (n)$ . At $n = 1$ the recursion sets $Φ_{1} = t - 1$ outright, the only positive divisor of $1$ is $1$ so the product is $Φ_{1} = t - 1$ , and $\deg Φ_{1} = 1 = φ (1)$ .

basegiven

1.2

Inductive hypothesis: fix $n \geq 2$ and assume the assertion for every $m$ with $1 \leq m < n$ .

1.3

Since $char Q = 0$ does not divide $n$ , [L2] and [L3] supply a splitting field $E$ of $t^{n} - 1$ over $Q$ in which $t^{n} - 1$ is separable and $μ_{n} (E)$ is cyclic of order $n$ . For a positive divisor $d$ of $n$ put $S_{d} : = {ζ \in μ_{n} (E) : ord (ζ) = d}$ and $Ψ_{d} : = \prod_{ζ \in S_{d}} (t - ζ) \in E [t]$ , a monic polynomial of degree $∣ S_{d} ∣$ .

L2L3given

2.1

For every positive divisor $m$ of $n$ one has $t^{m} - 1 = \prod_{ζ \in μ_{m} (E)} (t - ζ)$ with $∣ μ_{m} (E) ∣ = m$ : the polynomial $t^{m} - 1$ divides $t^{n} - 1$ in $Z [t]$ , since $t^{n} - 1 = (t^{m} - 1) (t^{n - m} + t^{n - 2 m} + \dots + 1)$ when $m ∣ n$ , so it splits over $E$ by [L3], and a repeated root of $t^{m} - 1$ in an extension would be a repeated root of $t^{n} - 1$ there, which [L4] excludes; hence its $m$ roots are distinct and they are by definition the elements of $μ_{m} (E)$ .

step 1.3L3L4algebra

3.1

For every positive divisor $m$ of $n$ , $μ_{m} (E)$ is the disjoint union of the $S_{d}$ over positive divisors $d$ of $m$ : an element $ζ \in μ_{n} (E)$ has finite order dividing $n$ , and $ζ^{m} = 1$ holds exactly when $ord (ζ) ∣ m$ by [L5]. Hence $\prod_{d ∣ m} Ψ_{d} = \prod_{ζ \in μ_{m} (E)} (t - ζ) = t^{m} - 1$ by step 2.1.

step 1.3step 2.1L5

4.1

For every positive divisor $d$ of $n$ with $d < n$ one has $Φ_{d} = Ψ_{d}$ , by induction on $d$ through the divisors of $n$ : at $d = 1$ both equal $t - 1$ , since $S_{1} = {1}$ ; and if $Φ_{e} = Ψ_{e}$ for every positive divisor $e$ of $d$ with $e < d$ , then step 1.2 and step 3.1 give $(\prod_{e ∣ d, e < d} Ψ_{e}) Φ_{d} = t^{d} - 1 = (\prod_{e ∣ d, e < d} Ψ_{e}) Ψ_{d}$ , and cancelling the nonzero left factor in the integral domain $E [t]$ ([L7]) yields $Φ_{d} = Ψ_{d}$ .

step 1.2step 3.1L7

5.1

Write $P : = \prod_{d ∣ n, d < n} Φ_{d}$ , monic in $Z [t]$ by step 1.2 and [L7]. By step 4.1 and step 3.1 applied with $m = n$ , in $E [t]$ one has $t^{n} - 1 = (\prod_{d ∣ n, d < n} Ψ_{d}) Ψ_{n} = P Ψ_{n}$ .

step 1.2step 3.1step 4.1L7

6.1

The division of $t^{n} - 1$ by $P$ in $Z [t]$ is exact and its quotient is $Ψ_{n}$ : by [L1] over $Z$ there are unique $q, r \in Z [t]$ with $t^{n} - 1 = q P + r$ and $r = 0$ or $\deg r < \deg P$ ; this is also a division by the monic $P$ in $E [t]$ , where step 5.1 exhibits the division with quotient $Ψ_{n}$ and remainder $0$ , so the uniqueness clause of [L1] over $E$ forces $q = Ψ_{n}$ and $r = 0$ . Hence $Φ_{n} = q = Ψ_{n}$ is monic in $Z [t]$ and $\prod_{d ∣ n} Φ_{d} = P Φ_{n} = t^{n} - 1$ .

step 5.1L1L7

7.1

Degrees: taking degrees in $t^{n} - 1 = P Φ_{n}$ with [L7] gives $n = \deg P + \deg Φ_{n}$ , and $\deg P = \sum_{d ∣ n, d < n} \deg Φ_{d} = \sum_{d ∣ n, d < n} φ (d)$ by step 1.2 and [L7]; so $\deg Φ_{n} = n - \sum_{d ∣ n, d < n} φ (d) = φ (n)$ by [L6].

step 1.2step 6.1L6L7

8.1

This is the assertion at $n$ , so the strong induction is complete and the assertion holds for every $n \geq 1$ . Uniqueness of the family follows by the same induction: if $(Φ_{m}^{'})_{m \geq 1}$ is monic in $Z [t]$ with $\prod_{d ∣ m} Φ_{d}^{'} = t^{m} - 1$ for all $m$ , then $Φ_{1}^{'} = t - 1 = Φ_{1}$ , and if $Φ_{m}^{'} = Φ_{m}$ for all $m < n$ then $P Φ_{n}^{'} = t^{n} - 1 = P Φ_{n}$ with $P \neq 0$ in the integral domain $Z [t]$ ([L7]), so $Φ_{n}^{'} = Φ_{n}$ .

step 6.1step 7.1L7discharge-induction∎

Remarks

Why a splitting field over $Q$ appears in a statement about $Z [t]$ . The exactness of the division is an identity between integer polynomials, but the only cheap reason for it is that the roots of $t^{n} - 1$ partition by order. The passage to $E$ produces that partition; the passage back is the uniqueness clause of monic division, which holds over $Z$ and over $E$ and pins the two computations to the same quotient.
The degree computation is where $φ$ enters. Nothing before step 7.1 mentions Euler's totient; it appears only through For every positive integer $n$ , $\sum_{d ∣ n, d > 0} φ (d) = n$ , and the same identity is what makes the count of primitive $n$ -th roots of unity match $\deg Φ_{n}$ in Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$Φ_{1} (0) = - 1$ and $Φ_{n} (0) = 1$ for $n \geq 2$

Statement

For the cyclotomic polynomials of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ,

$Φ_{1} (0) = - 1, Φ_{n} (0) = 1 for every n \geq 2.$

Facts & Assumptions

Given: The cyclotomic polynomials $Φ_{n} \in Z [t]$ and their defining identity; evaluation at $0$ is as in Evaluation and roots of a polynomial in a commutative target ring.

[L1]

For every $n \geq 1$ one has $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , the product being over the positive divisors of $n$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L2]

Evaluation at an element of a commutative ring is a unital ring homomorphism $Z [t] \to Z$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism), so it carries finite products to finite products.

Proof

technique · induction

1.1

$Φ_{1} = t - 1$ by The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , so $Φ_{1} (0) = - 1$ ; this is also the base of the induction below, taken at $n = 2$ , where [L1] and [L2] give $Φ_{1} (0) Φ_{2} (0) = 0^{2} - 1 = - 1$ , hence $- Φ_{2} (0) = - 1$ and $Φ_{2} (0) = 1$ .

baseL1L2given

1.2

Inductive hypothesis: fix $n \geq 3$ and assume $Φ_{m} (0) = 1$ for every $m$ with $2 \leq m < n$ .

2.1

Evaluating the identity of [L1] at $0$ and using [L2] gives $\prod_{d ∣ n} Φ_{d} (0) = - 1$ ; separating the factor $d = 1$ , which is $- 1$ by step 1.1, leaves $\prod_{d ∣ n, d > 1} Φ_{d} (0) = 1$ .

step 1.1L1L2

3.1

Every factor with $1 < d < n$ equals $1$ by step 1.2, so the product reduces to $Φ_{n} (0) = 1$ , which is the assertion at $n$ ; the induction is complete and $Φ_{n} (0) = 1$ for every $n \geq 2$ .

step 1.2step 2.1discharge-induction∎

Remarks

What this is for. The value $1$ at the origin makes $Φ_{n} (x) \equiv 1$ modulo $x$ for every integer $x$ , so no prime dividing $x$ divides $Φ_{n} (x)$ . That is the whole mechanism behind For every $n \geq 1$ there are infinitely many primes $p$ with $p \equiv 1 (m o d n)$ , and the exceptional value at $n = 1$ is why that theorem treats $n = 1$ separately.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), and let $E$ be a splitting field of $t^{n} - 1$ over $K$ , that is $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Write $Φ_{n}$ also for the image of the integer polynomial $Φ_{n}$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) in $K [t]$ . Then $Φ_{n}$ is separable over $K$ (Repeated roots in extension fields and separable polynomials), it splits over $E$ , and its roots in $E$ are exactly the $φ (n)$ primitive $n$ -th roots of unity in $E$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ with $char K ∤ n$ , a splitting field $E$ of $t^{n} - 1$ over $K$ , and the convention that for every finite subset $T \subseteq E$ the product $\prod_{ζ \in T} (t - ζ) \in E [t]$ means the finite product along any enumeration of $T$ ; because $E [t]$ is a commutative ring, the value is independent of the enumeration (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either, Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring). In particular, for each positive divisor $d$ of $n$ , let $S_{d} : = {ζ \in μ_{n} (E) : ord (ζ) = d}$ and $Ψ_{d} : = \prod_{ζ \in S_{d}} (t - ζ) \in E [t]$ .

[L1]

$t^{n} - 1$ is separable over $K$ when $char K ∤ n$ , and then $μ_{n} (E)$ is cyclic of order $n$ with exactly $φ (n)$ primitive $n$ -th roots of unity ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L2]

For every $m \geq 1$ one has $\prod_{d ∣ m} Φ_{d} = t^{m} - 1$ in $Z [t]$ , each $Φ_{d}$ monic of degree $φ (d)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ ); reduction into $K [t]$ preserves this identity.

[L3]

A monic $f$ of degree $m$ splits over $E$ when $f = \prod_{j = 1}^{m} (t - α_{j})$ with $α_{j} \in E$ , repetitions allowed, and a splitting field is generated over $K$ by the roots (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L4]

$f$ is separable over $K$ when no extension field of $K$ contains an $a$ with $(t - a)^{2}$ dividing the image of $f$ (Repeated roots in extension fields and separable polynomials).

[L5]

For an element $x$ of finite order $m$ , $x^{k} = e$ if and only if $m ∣ k$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L6]

$R [x]$ is an integral domain when $R$ is (A polynomial ring over an integral domain is an integral domain); and $f (a) = 0$ if and only if $x - a$ divides $f$ (Factor theorem over a commutative ring).

Proof

technique · direct

1.1

By [L1] the polynomial $t^{n} - 1$ is separable over $K$ and $μ_{n} (E)$ is cyclic of order $n$ ; so $t^{n} - 1$ has $n$ distinct roots in $E$ , namely the elements of $μ_{n} (E)$ , and $t^{n} - 1 = \prod_{ζ \in μ_{n} (E)} (t - ζ)$ by [L3].

L1L3

1.2

Each $ζ \in μ_{n} (E)$ has order dividing $n$ by [L5], and for a positive divisor $m$ of $n$ the condition $ζ^{m} = 1$ says exactly $ord (ζ) ∣ m$ ; so $μ_{m} (E)$ is the disjoint union of the $S_{d}$ over positive divisors $d$ of $m$ .

L5given

2.1

For every positive divisor $m$ of $n$ the polynomial $t^{m} - 1$ divides $t^{n} - 1$ in $Z [t]$ , since $t^{n} - 1 = (t^{m} - 1) (t^{n - m} + t^{n - 2 m} + \dots + 1)$ ; hence it splits over $E$ with distinct roots, which are the elements of $μ_{m} (E)$ , and $t^{m} - 1 = \prod_{ζ \in μ_{m} (E)} (t - ζ)$ with $∣ μ_{m} (E) ∣ = m$ .

step 1.1L3L4algebra

3.1

Consequently $\prod_{d ∣ m} Ψ_{d} = \prod_{ζ \in μ_{m} (E)} (t - ζ) = t^{m} - 1$ for every positive divisor $m$ of $n$ .

step 2.1step 1.2

4.1

For every positive divisor $d$ of $n$ the image of $Φ_{d}$ in $E [t]$ is $Ψ_{d}$ , by induction on $d$ through the divisors of $n$ : at $d = 1$ both are $t - 1$ , since $S_{1} = {1}$ ; and if the claim holds for every positive divisor $e$ of $d$ with $e < d$ , then [L2] and step 3.1 give $(\prod_{e ∣ d, e < d} Ψ_{e}) Φ_{d} = t^{d} - 1 = (\prod_{e ∣ d, e < d} Ψ_{e}) Ψ_{d}$ , and cancelling the nonzero left factor in the integral domain $E [t]$ ([L6]) gives $Φ_{d} = Ψ_{d}$ .

step 3.1L2L6

5.1

Taking $d = n$ : the image of $Φ_{n}$ in $E [t]$ is $\prod_{ζ \in S_{n}} (t - ζ)$ , so $Φ_{n}$ splits over $E$ and its roots there are exactly the elements of $S_{n}$ , which are the elements of order $n$ in $μ_{n} (E)$ , that is the primitive $n$ -th roots of unity in $E$ ; there are $φ (n)$ of them by [L1], in agreement with $\deg Φ_{n} = φ (n)$ from [L2].

step 4.1L1L2

6.1

$Φ_{n}$ is separable over $K$ . The product identity [L2] shows that the image of $Φ_{n}$ divides $t^{n} - 1$ in $K [t]$ . If an extension field $L / K$ contained an $a$ for which $(t - a)^{2}$ divided the image of $Φ_{n}$ , then the same square would divide the image of $t^{n} - 1$ in $L [t]$ , making $a$ a repeated root of $t^{n} - 1$ . This contradicts the separability of $t^{n} - 1$ supplied by [L1]. Thus [L4] applies.

L1L2L4algebra∎

Remarks

Both hypotheses are used, and neither can be dropped. If the characteristic divides $n$ then $t^{n} - 1$ is not separable and the roots of $Φ_{n}$ do not separate into orders at all; and the statement is about the roots in a splitting field, not in $K$ itself, since $Φ_{n}$ may have no root in $K$ at all, as $Φ_{3}$ over $Q$ shows.
Irreducibility is a separate question. The theorem says what the roots of $Φ_{n}$ are; it says nothing about whether $Φ_{n}$ factors over $K$ . Over $Q$ it does not ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ), over a finite field it usually does (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ), and in both cases the root description above is the same.

PropositionStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise), let $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ), and let $ζ \in E$ be a primitive $n$ -th root of unity (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity). The following are equivalent.

The image of $Φ_{n}$ in $K [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible (Irreducible and prime elements of an integral domain).
$[K (ζ) : K] = φ (n)$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).
The embedding $Gal (E / K) \to (Z / n)^{\times}$ of $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ is an isomorphism.

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ with $char K ∤ n$ , the extension $E = K (μ_{n})$ , a primitive $n$ -th root of unity $ζ \in E$ , and the minimal polynomial $m_{ζ} \in K [t]$ of $ζ$ over $K$ .

[L1]

The image of $Φ_{n}$ in $K [t]$ has as its roots in $E$ exactly the primitive $n$ -th roots of unity in $E$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity).

[L2]

$Φ_{n}$ is monic of degree $φ (n)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree); reduction into $K [t]$ preserves both.

[L3]

For $a$ algebraic over $K$ there is a unique monic irreducible $m_{a} \in K [t]$ with $f (a) = 0$ if and only if $m_{a} ∣ f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L4]

If $a$ is algebraic over $K$ with minimal polynomial of degree $m$ , then $[K (a) : K] = m$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L5]

For this $n \geq 1$ , $E / K$ is finite Galois and $σ \mapsto [a_{σ}]_{n}$ is an injective homomorphism $Gal (E / K) \to (Z / n)^{\times}$ ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ); moreover $E = K (ζ)$ for every primitive $n$ -th root $ζ$ ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity).

[L6]

$∣ (Z / n)^{\times} ∣ = φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and $∣ Gal (E / K) ∣ = [E : K]$ for a finite Galois extension (Equivalent characterizations of a finite Galois extension).

[L7]

In an integral domain, an irreducible element is a nonzero nonunit every one of whose factorisations has a unit factor (Irreducible and prime elements of an integral domain).

Proof

technique · direct

1.1

$ζ$ is a root of the image of $Φ_{n}$ in $K [t]$ by [L1], so $m_{ζ}$ divides that image by [L3]; both are monic, and $m_{ζ}$ has positive degree because $ζ \neq 0$ is not a root of a nonzero constant. Write the image of $Φ_{n}$ as $m_{ζ} g$ with $g \in K [t]$ monic.

L1L2L3

1.2

For the equivalence of clauses 2 and 3: by [L5] one has $E = K (ζ)$ and the embedding is injective into a group of order $φ (n)$ by [L6], so it is surjective if and only if $∣ Gal (E / K) ∣ = φ (n)$ ; and $∣ Gal (E / K) ∣ = [E : K] = [K (ζ) : K]$ by [L6]. An injective homomorphism onto its target is an isomorphism.

L5L6

2.1

For the implication from clause 1 to clause 2: if the image of $Φ_{n}$ is irreducible, then in the factorisation of step 1.1 one factor is a unit by [L7], and $m_{ζ}$ is not, so $g$ is a nonzero constant; both $m_{ζ} g$ and $m_{ζ}$ being monic forces $g = 1$ and $m_{ζ} = Φ_{n}$ . Hence $[K (ζ) : K] = \deg m_{ζ} = φ (n)$ by [L2] and [L4].

step 1.1L2L4L7

2.2

For the implication from clause 2 to clause 1: if $[K (ζ) : K] = φ (n)$ then $\deg m_{ζ} = φ (n) = \deg Φ_{n}$ by [L2] and [L4], so $g$ in step 1.1 is monic of degree $0$ , that is $g = 1$ and the image of $Φ_{n}$ equals $m_{ζ}$ , which is irreducible by [L3].

step 1.1L2L3L4

3.1

Steps 2.1 and 2.2 give the equivalence of clauses 1 and 2, and step 1.2 the equivalence of clauses 2 and 3; so all three are equivalent.

step 2.1step 2.2step 1.2∎

Remarks

This is a criterion, not a theorem about $Q$ . Over $Q$ all three clauses hold for every $n$ ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ), but over $F_{q}$ they hold only when the class of $q$ generates $(Z / n)^{\times}$ (The reduction of $Φ_{n}$ is irreducible over $F_{q}$ exactly when $[q]$ generates $(Z / n)^{\times}$ ).

PropositionStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-26Open item page →

$Φ_{p^{r}} (t) = \sum_{k < p} t^{k p^{r - 1}}$ , and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and $r \geq 1$ . Then

$Φ_{p^{r}} (t) = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}} = \frac{t^{p^{r}} - 1}{t^{p^{r - 1}} - 1},$

the polynomial $Φ_{p^{r}} (t + 1) \in Z [t]$ satisfies the Eisenstein criterion at $p$ (Eisenstein criterion over the integers), and consequently $Φ_{p^{r}}$ is irreducible in $Q [t]$ (Irreducible and prime elements of an integral domain).

The sum starts at $k = 0$ : its first term is the constant $1$ , and evaluation at $t = 1$ (Evaluation and roots of a polynomial in a commutative target ring) gives $Φ_{p^{r}} (1) = p$ .

Facts & Assumptions

Given: A prime $p$ and an integer $r \geq 1$ ; the cyclotomic polynomials of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ and the substitution homomorphisms of Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism, under which $t \mapsto t + 1$ is a ring automorphism of $Z [t]$ and of $Q [t]$ with inverse $t \mapsto t - 1$ .

[L1]

For every $n \geq 1$ , $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ with each $Φ_{d}$ monic in $Z [t]$ of degree $φ (d)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L2]

Let $f = a_{n} x^{n} + \dots + a_{0} \in Z [x]$ be primitive with $n \geq 1$ . If a prime $p$ satisfies $p ∤ a_{n}$ , $p ∣ a_{i}$ for every $i < n$ , and $p^{2} ∤ a_{0}$ , then $f$ is irreducible in $Q [x]$ (Eisenstein criterion over the integers).

[L3]

A nonzero integer polynomial is primitive exactly when no prime divides all of its coefficients (Content is the positive common divisor of the coefficients divisible by every common divisor, Content and primitive integer polynomials).

[L4]

For a fixed integer $k \geq 1$ , in a field $K$ of characteristic $p > 0$ one has $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ in $K [t]$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ); $Z / p$ is such a field (For every prime $p$ , the two operations on $Z / p$ make it a field, The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise).

[L5]

$φ (p^{k}) = p^{k} - p^{k - 1}$ for every prime $p$ and $k \geq 1$ (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ).

[L6]

Every integer greater than $1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime); and if a prime divides a finite product of integers it divides one of the factors (If a prime $p$ divides a finite product $\prod_{i < n} a_{i}$ of integers then $p ∣ a_{i}$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold).

[L7]

$R [x]$ is an integral domain when $R$ is (A polynomial ring over an integral domain is an integral domain).

Proof

technique · direct

1.1

The positive divisors of $p^{j}$ are exactly $p^{0}, \dots, p^{j}$ : a positive $d$ dividing $p^{j}$ with $d > 1$ has a prime divisor $q$ by [L6], and $q ∣ p^{j}$ forces $q ∣ p$ by [L6], hence $q = p$ since $p$ is prime and $q > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); writing $d = p d^{'}$ gives $d^{'} ∣ p^{j - 1}$ by cancellation, and repeating reduces $d$ to a power of $p$ not exceeding $p^{j}$ .

L6given

2.1

By [L1] at $n = p^{r}$ and at $n = p^{r - 1}$ , using step 1.1, $\prod_{j = 0}^{r} Φ_{p^{j}} = t^{p^{r}} - 1$ and $\prod_{j = 0}^{r - 1} Φ_{p^{j}} = t^{p^{r - 1}} - 1$ ; dividing, $(t^{p^{r - 1}} - 1) Φ_{p^{r}} = t^{p^{r}} - 1$ .

step 1.1L1

3.1

With $u : = t^{p^{r - 1}}$ the elementary identity $(u - 1) \sum_{k = 0}^{p - 1} u^{k} = u^{p} - 1$ gives $(t^{p^{r - 1}} - 1) \sum_{k = 0}^{p - 1} t^{k p^{r - 1}} = t^{p^{r}} - 1$ ; comparing with step 2.1 and cancelling the nonzero factor $t^{p^{r - 1}} - 1$ in the integral domain $Z [t]$ ([L7]) yields $Φ_{p^{r}} = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}}$ .

step 2.1L7algebra

3.2

Reducing step 2.1 modulo $p$ and applying [L4] twice in $(Z / p) [t]$ gives $\overline{Φ_{p^{r}}} (t - 1)^{p^{r - 1}} = (t - 1)^{p^{r}}$ , and cancelling in the integral domain $(Z / p) [t]$ ([L7]) gives $\overline{Φ_{p^{r}}} = (t - 1)^{p^{r} - p^{r - 1}}$ .

step 2.1L4L7

4.1

Evaluating step 3.1 at $t = 1$ gives $Φ_{p^{r}} (1) = p$ , since the sum has $p$ terms each equal to $1$ ; so the constant term of $Φ_{p^{r}} (t + 1)$ is $p$ , which is divisible by $p$ and not by $p^{2}$ .

step 3.1algebra

4.2

Substituting $t + 1$ , which commutes with reduction modulo $p$ because both are ring homomorphisms fixing the coefficients appropriately, gives $\overline{Φ_{p^{r}} (t + 1)} = t^{p^{r} - p^{r - 1}} = t^{φ (p^{r})}$ by [L5]. So every coefficient of $Φ_{p^{r}} (t + 1)$ other than the leading one is divisible by $p$ , while the leading coefficient is $1$ because $Φ_{p^{r}} (t + 1)$ is monic of degree $φ (p^{r})$ by [L1] and the substitution being degree preserving.

step 3.2L1L5given

5.1

$Φ_{p^{r}} (t + 1)$ is primitive by [L3], no prime dividing its leading coefficient $1$ , and its degree $φ (p^{r})$ is at least $1$ by [L5]; steps 4.1 and 4.2 supply the three Eisenstein conditions at $p$ , so [L2] makes it irreducible in $Q [t]$ .

step 4.1step 4.2L2L3L5

6.1

The substitution $t \mapsto t - 1$ is a ring automorphism of $Q [t]$ carrying $Φ_{p^{r}} (t + 1)$ to $Φ_{p^{r}}$ ; a ring automorphism preserves units and factorisations, so it carries irreducible elements to irreducible elements, and $Φ_{p^{r}}$ is irreducible in $Q [t]$ .

step 5.1given∎

Remarks

The term $k = 0$ is load bearing. Dropping it would change $Φ_{p^{r}} (1)$ from $p$ to $p - 1$ , and the Eisenstein constant-term condition would fail. The highest exponent would be unchanged, so the degree alone would not detect the wrong polynomial.
A self-contained route for prime powers. This gives irreducibility over $Q$ for $n$ a prime power without the general argument of $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ , and unlike that argument it exhibits an explicit polynomial to which a named criterion applies. The general theorem covers every $n$ and does not supersede this computation; the companion page works out the case $p = 7$ in $Φ_{7} (t + 1)$ is Eisenstein at seven ↗.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$

Statement

For every $n \geq 1$ the cyclotomic polynomial $Φ_{n}$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible in $Q [t]$ (Irreducible and prime elements of an integral domain).

Facts & Assumptions

Given: An integer $n \geq 1$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $m \cdot 1 > 0$ and in particular $m \cdot 1 \neq 0$ for $m \geq 1$ , whence $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no $n \geq 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); a splitting field $E$ of $t^{n} - 1$ over $Q$ (Every nonzero polynomial over a field has a splitting field); a primitive $n$ -th root of unity $ζ \in E$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity); and $f \in Q [t]$ the minimal polynomial of $ζ$ over $Q$ .

[L1]

$t^{n} - 1$ is separable over $Q$ and $μ_{n} (E)$ is cyclic of order $n$ with exactly $φ (n)$ primitive $n$ -th roots of unity, which are its generators ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

The roots of $Φ_{n}$ in $E$ are exactly the primitive $n$ -th roots of unity in $E$ ; $Φ_{n}$ is monic of degree $φ (n)$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity, The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

For $a$ algebraic over $K$ there is a unique monic irreducible $m_{a} \in K [t]$ with $h (a) = 0$ if and only if $m_{a} ∣ h$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L4]

If $ξ$ is a primitive $n$ -th root of unity in $E$ , $g \in Q [t]$ is its minimal polynomial and $p$ is a prime with $p ∤ n$ , then $g (ξ^{p}) = 0$ (If $p$ is a prime not dividing $n$ , a rational minimal polynomial of a primitive $n$ -th root of unity also kills its $p$ -th power).

[L5]

Every integer $a \geq 1$ is a finite product of primes, the empty product being $1$ (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ , Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L6]

In a cyclic group $⟨ g ⟩$ of finite order $m$ , $g^{a}$ generates the group if and only if $\gcd (a, m) = 1$ , that is when $a$ and $m$ are coprime (A cyclic group of order $n$ has exactly $φ (n)$ generators, Coprime integers: $\gcd (a, b) = 1$ ).

[L7]

A nonzero polynomial of degree $k$ over an integral domain has at most $k$ distinct roots in it (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

Proof

technique · direct

1.1

$ζ$ is a root of $Φ_{n}$ by [L2], so $f ∣ Φ_{n}$ in $Q [t]$ by [L3], and $f$ is monic and irreducible.

L2L3

1.2

Let $ξ \in E$ be any primitive $n$ -th root of unity. By [L1] it generates $μ_{n} (E) = ⟨ ζ ⟩$ , so $ξ = ζ^{a}$ for some integer $a$ , which may be taken with $a \geq 1$ after adding a multiple of $n$ ; and $\gcd (a, n) = 1$ by [L6]. Write $a = p_{1} p_{2} \dots p_{s}$ as a product of primes by [L5], with $s = 0$ when $a = 1$ . No $p_{i}$ divides $n$ , since $p_{i} ∣ a$ and $\gcd (a, n) = 1$ .

L1L5L6given

2.1

Put $ζ_{0} : = ζ$ and $ζ_{j} : = ζ_{j - 1}^{p_{j}}$ for $1 \leq j \leq s$ , so that $ζ_{s} = ζ^{a} = ξ$ . By induction on $j$ , each $ζ_{j}$ is a primitive $n$ -th root of unity and $f (ζ_{j}) = 0$ : at $j = 0$ this is the hypothesis on $ζ$ and step 1.1; and given it at $j - 1$ , the polynomial $f$ is monic irreducible and vanishes at the primitive $n$ -th root of unity $ζ_{j - 1}$ , so $f$ is the minimal polynomial of $ζ_{j - 1}$ by [L3], whence $f (ζ_{j}) = f (ζ_{j - 1}^{p_{j}}) = 0$ by [L4], while $ζ_{j}$ is primitive by [L6] because $\gcd (p_{j}, n) = 1$ .

step 1.1step 1.2L3L4L6

3.1

So $f$ vanishes at every primitive $n$ -th root of unity in $E$ , of which there are $φ (n)$ distinct ones by [L1]; hence $\deg f \geq φ (n)$ by [L7].

step 2.1L1L7

4.1

On the other hand $f ∣ Φ_{n}$ with $\deg Φ_{n} = φ (n)$ by [L2], so $\deg f \leq φ (n)$ ; therefore $\deg f = φ (n)$ , and $f$ and $Φ_{n}$ are monic with $f ∣ Φ_{n}$ , so $Φ_{n} = f$ is irreducible. At $n = 1$ this reads $Φ_{1} = t - 1$ , of degree $φ (1) = 1$ .

step 1.1step 3.1L2L3∎

Remarks

What the prime factorisation is doing. The lemma advances one prime at a time, and step 2.1 chains those advances along a factorisation of the exponent. The chain works only because each intermediate $ζ_{j}$ is again primitive and has the same minimal polynomial, which is what lets the lemma be reapplied rather than merely applied once.
Over other base fields the theorem is false. Irreducibility of $Φ_{n}$ depends on the base field, and over a finite field it usually fails (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ). The criterion that separates the cases is $Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$

Statement

Let $n \geq 1$ and let $ζ$ be a primitive $n$ -th root of unity (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity) in $Q (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then

$[Q (ζ) : Q] = φ (n)$

(The degree $[K : F] = \dim_{F} K$ of a finite field extension, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and the embedding

$Gal (Q (μ_{n}) / Q) ⟶ (Z / n)^{\times}$

of $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ is an isomorphism.

Facts & Assumptions

Given: An integer $n \geq 1$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no $n \geq 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

$Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ).

[L2]

For a field $K$ with $char K ∤ n$ and a primitive $n$ -th root of unity $ζ$ in a splitting field, irreducibility of the image of $Φ_{n}$ in $K [t]$ , the equality $[K (ζ) : K] = φ (n)$ , and surjectivity of the embedding $Gal (K (μ_{n}) / K) \to (Z / n)^{\times}$ are equivalent ( $Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto).

Proof

technique · direct

1.1

Since $char Q = 0$ does not divide $n$ , [L2] applies with $K = Q$ .

L2given

2.1

The image of $Φ_{n}$ in $Q [t]$ is $Φ_{n}$ itself, irreducible by [L1]; so the first clause of [L2] holds, and therefore so do the other two: $[Q (ζ) : Q] = φ (n)$ , and the embedding is onto, hence an isomorphism, being injective.

step 1.1L1L2∎

Remarks

The isomorphism is canonical. It sends $σ$ to the class of the exponent $a_{σ}$ with $σ (ζ) = ζ^{a_{σ}}$ , and $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ shows that class does not depend on which primitive $n$ -th root of unity is chosen. So every subgroup of $(Z / n)^{\times}$ names an intermediate field of $Q (μ_{n}) / Q$ without any choice being made.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$

Statement

Let $F_{q}$ be a finite field of order $q$ and let $n \geq 1$ with $\gcd (n, q) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ). Put $d : = ord ([q]_{n})$ , the multiplicative order of $[q]_{n}$ in $(Z / n)^{\times}$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ). Then the image of $Φ_{n}$ in $F_{q} [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is a product of pairwise distinct monic irreducible polynomials, each of degree $d$ , and there are $φ (n) / d$ of them.

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q$ , of characteristic $p$ with $q$ a power of $p$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ , The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise), and $n \geq 1$ with $\gcd (n, q) = 1$ ; hence $p ∤ n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). Write $E : = F_{q} (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ , Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ ) and ${\overset{ˉ}{Φ}}_{n}$ for the image of $Φ_{n}$ in $F_{q} [t]$ .

[L1]

Over a field $K$ with $char K ∤ n$ and a splitting field $E$ of $t^{n} - 1$ , the image of $Φ_{n}$ in $K [t]$ is separable, splits over $E$ , and its roots in $E$ are exactly the $φ (n)$ primitive $n$ -th roots of unity in $E$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity, The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L2]

$Φ_{n}$ is monic of degree $φ (n)$ , and reduction into $F_{q} [t]$ preserves both (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

$[F_{q} (μ_{n}) : F_{q}] = ord ([q]_{n})$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L4]

For $π \in F_{q} [t]$ monic irreducible of degree $e$ with a root $α$ in an extension field, $F_{q} (α)$ is a finite field of order $q^{e}$ and $[F_{q} (α) : F_{q}] = e$ (A monic irreducible of degree $d$ over $F_{q}$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$ ).

[L5]

$F [x]$ is a unique factorisation domain for every field $F$ (For every field $F$ , $F [x]$ is a unique factorisation domain); irreducible elements are as in Irreducible and prime elements of an integral domain.

[L6]

$f$ is separable over $K$ when no extension field contains an $a$ with $(t - a)^{2}$ dividing the image of $f$ (Repeated roots in extension fields and separable polynomials).

[L7]

Over an integral domain, $\deg (f g) = \deg f + \deg g$ for nonzero $f, g$ (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L8]

In any field $F$ , the group $μ_{n} (F)$ is cyclic of order dividing $n$ ; if it contains a primitive $n$ -th root of unity, then its order is $n$ , and in that case its generators are exactly the primitive $n$ -th roots of unity ( $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

Proof

technique · direct

1.1

Since $q$ is a power of $p$ and $\gcd (n, q) = 1$ , the prime $p$ does not divide $n$ ; so [L1] applies with $K = F_{q}$ and the splitting field $E = F_{q} (μ_{n})$ , and [L3] gives $[E : F_{q}] = d$ .

L1L3given

1.2

By [L2] the polynomial ${\overset{ˉ}{Φ}}_{n}$ is monic of degree $φ (n) \geq 1$ , so by [L5] it is a product of monic irreducible polynomials of $F_{q} [t]$ , say ${\overset{ˉ}{Φ}}_{n} = π_{1} \dots π_{r}$ with each $π_{i}$ monic irreducible.

L2L5

2.1

Each $π_{i}$ has a root in $E$ . By [L1] the polynomial ${\overset{ˉ}{Φ}}_{n}$ splits over $E$ into $φ (n)$ distinct linear factors, and $π_{i}$ divides it; since $E [t]$ is a unique factorisation domain by [L5], $π_{i}$ is, up to a unit, a product of some of those linear factors. Thus it has a root $α_{i} \in E$ , and $α_{i}$ is a primitive $n$ -th root of unity by [L1].

step 1.2L1L5

3.1

No two of the $π_{i}$ coincide. If $π_{i} = π_{j} = π$ for $i \neq j$ , then $π^{2}$ divides ${\overset{ˉ}{Φ}}_{n}$ . By step 2.1 the polynomial $π$ has a linear factor $t - α$ in $E [t]$ , so $(t - α)^{2}$ divides ${\overset{ˉ}{Φ}}_{n}$ there, contradicting the separability supplied by [L1] and [L6].

step 1.2step 2.1L1L5L6

3.2

Every $π_{i}$ has degree $d$ . Writing $e_{i} : = \deg π_{i}$ , [L4] gives $[F_{q} (α_{i}) : F_{q}] = e_{i}$ . Since $α_{i}$ is primitive, [L8] makes $μ_{n} (E)$ a cyclic group of order $n$ with generators exactly the primitive $n$ -th roots, so $α_{i}$ generates $μ_{n} (E)$ . Hence $μ_{n} (E) \subseteq F_{q} (α_{i})$ and therefore $E = F_{q} (μ_{n} (E)) \subseteq F_{q} (α_{i})$ ; and $α_{i} \in E$ gives $F_{q} (α_{i}) \subseteq E$ . Thus $F_{q} (α_{i}) = E$ and $e_{i} = [E : F_{q}] = d$ by step 1.1.

step 1.1step 2.1L4L8

4.1

Comparing degrees with [L7] and [L2], $φ (n) = \deg {\overset{ˉ}{Φ}}_{n} = \sum_{i = 1}^{r} e_{i} = r d$ , so $r = φ (n) / d$ ; with steps 3.1 and 3.2 this is the assertion.

step 1.2step 3.1step 3.2L2L7∎

Remarks

The degree of every factor is the same, and that is the content. A polynomial can factor into irreducibles of different degrees; here it cannot, because adjoining any primitive root produces the same field $F_{q} (μ_{n})$ . The primitive roots need not form a single Frobenius orbit: for $n = 7$ over $F_{2}$ they split into two orbits of size three, one for each irreducible cubic factor on the companion page.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The reduction of $Φ_{n}$ is irreducible over $F_{q}$ exactly when $[q]$ generates $(Z / n)^{\times}$

Statement

Let $F_{q}$ be a finite field of order $q$ and $n \geq 1$ with $\gcd (n, q) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ). The image of $Φ_{n}$ in $F_{q} [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible (Irreducible and prime elements of an integral domain) if and only if $[q]_{n}$ generates $(Z / n)^{\times}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ). In particular this can happen only when $(Z / n)^{\times}$ is cyclic.

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q$ and $n \geq 1$ with $\gcd (n, q) = 1$ ; write $d : = ord ([q]_{n})$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity) and ${\overset{ˉ}{Φ}}_{n}$ for the image of $Φ_{n}$ in $F_{q} [t]$ .

[L1]

${\overset{ˉ}{Φ}}_{n}$ is a product of pairwise distinct monic irreducible polynomials, each of degree $d$ , and there are $φ (n) / d$ of them (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ).

[L2]

[L3]

$∣ (Z / n)^{\times} ∣ = φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and $∣ ⟨ x ⟩ ∣ = ord (x)$ for an element of finite order (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · direct

1.1

By [L1] the number of monic irreducible factors of ${\overset{ˉ}{Φ}}_{n}$ , counted without repetition and with none repeated, is $r = φ (n) / d$ .

2.1

If $r = 1$ then ${\overset{ˉ}{Φ}}_{n}$ is itself one of those monic irreducible polynomials, hence irreducible; if $r \geq 2$ then ${\overset{ˉ}{Φ}}_{n}$ is a product of $r$ polynomials each of degree $d \geq 1$ , none of them a unit, so it is not irreducible. Hence ${\overset{ˉ}{Φ}}_{n}$ is irreducible exactly when $r = 1$ , that is exactly when $d = φ (n)$ .

step 1.1L1L2

3.1

By [L3] the subgroup $⟨ [q]_{n} ⟩$ has order $d$ and $(Z / n)^{\times}$ has order $φ (n)$ , so $d = φ (n)$ holds exactly when $⟨ [q]_{n} ⟩ = (Z / n)^{\times}$ , that is exactly when $[q]_{n}$ generates the unit group. With step 2.1 this proves the equivalence, and a group with a generator is cyclic.

step 2.1L3∎

Remarks

When the criterion cannot be met at all. If $(Z / n)^{\times}$ is not cyclic then no class generates it, so the reduction of $Φ_{n}$ is reducible over every finite field of order coprime to $n$ ; the smallest such $n$ is $8$ , where $(Z / 8)^{\times}$ has three elements of order two and no element of order four.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For $E / F$ finite Galois and $L / F$ finite inside a common field, $[E L : F] = [E : F] [L : F] / [E \cap L : F]$

Statement

Let $E / F$ be a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) and $L / F$ a finite extension (The degree $[K : F] = \dim_{F} K$ of a finite field extension), both subfields of a common field. Then the compositum $E L$ is finite over $F$ and

$[E L : F] = \frac{[E : F] [L : F]}{[E \cap L : F]} .$

Only one of the two extensions is required to be Galois.

Facts & Assumptions

Given: Subfields $E$ and $L$ of a common field, both containing $F$ , with $E / F$ finite Galois and $L / F$ finite; $E \cap L$ is a subfield containing $F$ and contained in $E$ , hence an intermediate field of $E / F$ .

[L1]

For $E / F$ finite Galois and $L / F$ an extension inside a common overfield, $E L / L$ is finite Galois and restriction gives $Gal (E L / L) ≅ Gal (E / E \cap L)$ (The Galois translation theorem).

[L2]

If $K / F$ is finite Galois and $F \subseteq E^{'} \subseteq K$ , then $K / E^{'}$ is finite Galois (A finite Galois extension is Galois over every intermediate field).

[L3]

For fields $F \subseteq K \subseteq M$ with $K / F$ and $M / K$ finite, $M / F$ is finite and $[M : F] = [M : K] [K : F]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

[L4]

For finite subextensions $E / F$ and $E^{'} / F$ of a common field, the compositum is finite and $[E E^{'} : F] \leq [E : F] [E^{'} : F]$ (For finite subextensions in a common field, $[E E^{'} : F] \leq [E : F] [E^{'} : F]$ ).

[L5]

For a finite Galois extension $M / K$ one has $∣ Gal (M / K) ∣ = [M : K]$ (Equivalent characterizations of a finite Galois extension).

Proof

technique · direct

1.1

$E \cap L$ is an intermediate field of $E / F$ , so $E / (E \cap L)$ is finite Galois by [L2].

L2given

1.2

By [L4] the compositum $E L$ is finite over $F$ .

L4given

1.3

Applying [L3] to $F \subseteq E \cap L \subseteq E$ gives $[E : F] = [E : E \cap L] [E \cap L : F]$ , so $[E : E \cap L] = [E : F] / [E \cap L : F]$ .

L3given

2.1

By [L1] the extension $E L / L$ is finite Galois with $Gal (E L / L) ≅ Gal (E / E \cap L)$ , so [L5] applied to both sides gives $[E L : L] = ∣ Gal (E L / L) ∣ = ∣ Gal (E / E \cap L) ∣ = [E : E \cap L]$ .

step 1.1step 1.2L1L5

3.1

Applying [L3] to $F \subseteq L \subseteq E L$ and substituting steps 2.1 and 1.3 gives $[E L : F] = [E L : L] [L : F] = [E : E \cap L] [L : F] = [E : F] [L : F] / [E \cap L : F]$ .

step 2.1step 1.3L3∎

Remarks

The Galois hypothesis is not decoration. Without it the formula fails: over $F = Q$ take $E = Q (\sqrt[3]{2})$ and $L = Q (ω \sqrt[3]{2})$ inside a splitting field of $t^{3} - 2$ , where $ω$ is a primitive cube root of unity. Both have degree three over $Q$ , since $t^{3} - 2$ is irreducible there. The compositum contains $ω = (ω \sqrt[3]{2}) / \sqrt[3]{2}$ , hence contains the splitting field $Q (\sqrt[3]{2}, ω)$ and equals it, so $[E L : Q] = 6$ . And $E \cap L = Q$ : its degree over $Q$ divides $3$ by the tower law, and it cannot be $3$ , since $E = E \cap L = L$ would put $ω$ in $E$ and force $[E : Q] \geq 6$ . The formula would predict $9$ . Neither $E$ nor $L$ is Galois over $Q$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$K (μ_{m}) K (μ_{n}) = K (μ_{lcm (m, n)})$

Statement

Let $K$ be a field and let $m, n \geq 1$ be integers such that $char K$ divides neither $m$ nor $n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). Put $ℓ : = lcm (m, n)$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ) and let $Ω$ be a splitting field of $t^{ℓ} - 1$ over $K$ (Every nonzero polynomial over a field has a splitting field). Then $char K ∤ ℓ$ ; the subfields $K (μ_{m} (Ω))$ and $K (μ_{n} (Ω))$ of $Ω$ are cyclotomic extensions of $K$ of orders $m$ and $n$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ); and their compositum inside $Ω$ is

$K (μ_{m}) K (μ_{n}) = K (μ_{ℓ}) = Ω .$

Facts & Assumptions

Given: A field $K$ , integers $m, n \geq 1$ with $char K$ dividing neither, $ℓ = lcm (m, n)$ , and a splitting field $Ω$ of $t^{ℓ} - 1$ over $K$ ; the characteristic of a field is $0$ or a prime (The characteristic of a field is zero or a prime number), and $0$ divides no positive integer.

[L1]

$lcm (a, b)$ is a common multiple of $a$ and $b$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ); every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) lcm (a, b) = ∣ a b ∣$ (Every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) \cdot lcm (a, b) = ∣ a b ∣$ ).

[L2]

If a prime $p$ divides $a b$ then $p ∣ a$ or $p ∣ b$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

$t^{k} - 1$ is separable over $K$ exactly when $char K ∤ k$ , and then a splitting field $E$ has $μ_{k} (E)$ cyclic of order $k$ and $E = K (ζ)$ for any primitive $k$ -th root of unity ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ , The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L4]

A polynomial is separable over $K$ when no extension field contains a repeated root (Repeated roots in extension fields and separable polynomials); a splitting field is generated over $K$ by the roots (Polynomials that split and splitting fields of a polynomial or a family of polynomials, Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ ).

[L5]

For a finite group $G$ and $H \leq G$ , $∣ H ∣$ divides $∣ G ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ); $⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups); and $∣ ⟨ x ⟩ ∣ = ord (x)$ for an element of finite order (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Proof

technique · direct

1.1

$char K ∤ ℓ$ . If $char K = 0$ this is immediate; if $char K = p$ is a prime dividing $ℓ$ , then $ℓ$ divides $m n$ because $\gcd (m, n) ℓ = m n$ by [L1] and $\gcd (m, n) \geq 1$ , so $p ∣ m n$ and [L2] gives $p ∣ m$ or $p ∣ n$ , contrary to hypothesis.

L1L2given

2.1

By [L3] and step 1.1 the polynomial $t^{ℓ} - 1$ is separable over $K$ , the group $μ_{ℓ} (Ω)$ is cyclic of order $ℓ$ , and $Ω = K (μ_{ℓ} (Ω))$ .

step 1.1L3L4

3.1

For every positive divisor $k$ of $ℓ$ one has $∣ μ_{k} (Ω) ∣ = k$ and $K (μ_{k} (Ω))$ is a splitting field of $t^{k} - 1$ over $K$ , hence a cyclotomic extension of $K$ of order $k$ : indeed $t^{k} - 1$ divides $t^{ℓ} - 1$ , since $t^{ℓ} - 1 = (t^{k} - 1) (t^{ℓ - k} + t^{ℓ - 2 k} + \dots + 1)$ , so it splits over $Ω$ , and a repeated root of it would be a repeated root of $t^{ℓ} - 1$ , excluded by step 2.1 through [L4]; so its $k$ roots are distinct and they are the elements of $μ_{k} (Ω)$ , which generate $K (μ_{k} (Ω))$ over $K$ .

step 2.1L3L4algebra

4.1

$m$ and $n$ are positive divisors of $ℓ$ by [L1], so step 3.1 applies to both: $K (μ_{m} (Ω))$ and $K (μ_{n} (Ω))$ are cyclotomic extensions of $K$ of orders $m$ and $n$ , and each contains a primitive root of unity of its order by [L3].

step 3.1L1L3

5.1

Both are contained in $Ω = K (μ_{ℓ} (Ω))$ , since $μ_{m} (Ω)$ and $μ_{n} (Ω)$ are subsets of $μ_{ℓ} (Ω)$ by $m ∣ ℓ$ and $n ∣ ℓ$ ; hence their compositum inside $Ω$ is contained in $K (μ_{ℓ})$ .

step 2.1step 4.1L1

5.2

For the reverse inclusion, fix a primitive $m$ -th root of unity $ζ_{m} \in μ_{m} (Ω)$ and a primitive $n$ -th root of unity $ζ_{n} \in μ_{n} (Ω)$ , and let $H : = ⟨ ζ_{m}, ζ_{n} ⟩ \leq μ_{ℓ} (Ω)$ . By [L5] the orders $m = ∣ ⟨ ζ_{m} ⟩ ∣$ and $n = ∣ ⟨ ζ_{n} ⟩ ∣$ both divide $∣ H ∣$ , so $∣ H ∣$ is a common multiple of $m$ and $n$ and therefore a multiple of $ℓ$ by [L1]; and $∣ H ∣$ divides $ℓ$ by [L5]. Hence $∣ H ∣ = ℓ$ and $H = μ_{ℓ} (Ω)$ .

step 2.1step 4.1L1L5

6.1

Both $ζ_{m}$ and $ζ_{n}$ lie in the compositum $K (μ_{m}) K (μ_{n})$ , which is a field, so $H \subseteq K (μ_{m}) K (μ_{n})$ and therefore $μ_{ℓ} (Ω) \subseteq K (μ_{m}) K (μ_{n})$ by step 5.2; hence $Ω = K (μ_{ℓ} (Ω)) \subseteq K (μ_{m}) K (μ_{n})$ . With step 5.1 this gives $K (μ_{m}) K (μ_{n}) = K (μ_{ℓ}) = Ω$ .

step 2.1step 5.1step 5.2∎

Remarks

The intersection is not the mirror image of this. The compositum identity holds over every base field of admissible characteristic, but the corresponding identity $K (μ_{m}) \cap K (μ_{n}) = K (μ_{\gcd (m, n)})$ does not: it is proved here only over $Q$ ( $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ ), and the companion page gives a finite-base-field failure in $F_{3} (μ_{5}) \cap F_{3} (μ_{7})$ is larger than $F_{3}$ although five and seven are coprime ↗.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$φ (m) φ (n) = φ (\gcd (m, n)) φ (lcm (m, n))$

Statement

For all integers $m, n \geq 1$ ,

$φ (m) φ (n) = φ (\gcd (m, n)) φ (lcm (m, n))$

(The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ , Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ , Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ). At $m = n = 1$ both sides are $1$ ; at $\gcd (m, n) = 1$ the identity is the multiplicativity $φ (m) φ (n) = φ (m n)$ .

Facts & Assumptions

Given: Integers $m, n \geq 1$ ; write $g : = \gcd (m, n)$ and $ℓ : = lcm (m, n)$ , both $\geq 1$ because $m$ and $n$ are nonzero. For a prime $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and an integer $e \geq 0$ put $w (p, e) : = p^{e} - p^{e - 1}$ when $e \geq 1$ and $w (p, 0) : = 1$ . For $k \geq 1$ write $D (k)$ for the set of primes dividing $k$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); it is finite, being contained in ${2, \dots, k}$ by If $d ∣ a$ and $a \neq 0$ then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ ; hence the set of divisors of a nonzero integer is bounded above by $∣ a ∣$ . Put $P : = D (m) \cup D (n)$ . Finite products over such sets are those of The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form.

[L1]

Let $k \geq 1$ and let $p_{0}, \dots, p_{r - 1}$ be an injective finite list consisting exactly of the prime divisors of $k$ ; put $e_{i} : = v_{p_{i}} (k)$ , so $e_{i} \geq 1$ . Then $φ (k) = \prod_{i < r} (p_{i}^{e_{i}} - p_{i}^{e_{i} - 1})$ (Euler's product formula $φ (n) = n \prod_{p ∣ n} (1 - 1 / p) = \prod_{p^{k} ∥ n} (p^{k} - p^{k - 1})$ for $n \geq 1$ , stated through a finite injective list of its prime divisors).

[L2]

For $a, b \geq 1$ and every prime $p$ : $v_{p} (\gcd (a, b)) = \min {v_{p} (a), v_{p} (b)}$ and $v_{p} (lcm (a, b)) = \max {v_{p} (a), v_{p} (b)}$ (For positive integers $a$ and $b$ and every prime $p$ : $v_{p} (\gcd (a, b)) = \min {v_{p} (a), v_{p} (b)}$ and $v_{p} (lcm (a, b)) = \max {v_{p} (a), v_{p} (b)}$ ; so the exponent-wise greatest common divisor is the $\gcd$ of the divisibility page and not a second notion).

[L3]

For a prime $p$ and a nonzero integer $a$ , $v_{p} (a)$ is the greatest $k \in N$ with $p^{k} ∣ a$ (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

Proof

technique · direct

1.1

For a prime $p$ and an integer $k \geq 1$ : $p ∣ k$ if and only if $v_{p} (k) \geq 1$ . If $v_{p} (k) \geq 1$ then $p$ divides $p^{v_{p} (k)}$ , which divides $k$ by [L3]; conversely $p ∣ k$ says $p^{1} ∣ k$ , so the greatest such exponent is at least $1$ .

1.2

For each $p \in P$ write $a : = v_{p} (m)$ and $b : = v_{p} (n)$ ; then $v_{p} (g) = \min {a, b}$ and $v_{p} (ℓ) = \max {a, b}$ by [L2], and the unordered pair ${\min {a, b}, \max {a, b}}$ is ${a, b}$ , so $w (p, a) w (p, b) = w (p, v_{p} (g)) w (p, v_{p} (ℓ))$ .

L2given

2.1

Consequently $D (g) = D (m) \cap D (n)$ and $D (ℓ) = D (m) \cup D (n)$ : by [L2] and step 1.1, $p \in D (g)$ says $\min {v_{p} (m), v_{p} (n)} \geq 1$ , that is $p \in D (m)$ and $p \in D (n)$ ; and $p \in D (ℓ)$ says $\max {v_{p} (m), v_{p} (n)} \geq 1$ , that is $p \in D (m)$ or $p \in D (n)$ .

step 1.1L2

3.1

The set $P$ is a finite set of primes containing $D (m)$ , $D (n)$ , $D (g)$ and $D (ℓ)$ by step 2.1. For every $k \in {m, n, g, ℓ}$ one has $φ (k) = \prod_{p \in P} w (p, v_{p} (k))$ : applying [L1] with the list $D (k)$ gives $φ (k) = \prod_{p \in D (k)} w (p, v_{p} (k))$ , and for $p \in P$ outside $D (k)$ step 1.1 gives $v_{p} (k) = 0$ , so the extra factors are $w (p, 0) = 1$ .

step 1.1step 2.1L1given

4.1

Multiplying the equalities of step 1.2 over the finite set $P$ and using step 3.1 four times gives $φ (m) φ (n) = \prod_{p \in P} w (p, v_{p} (m)) w (p, v_{p} (n)) = \prod_{p \in P} w (p, v_{p} (g)) w (p, v_{p} (ℓ)) = φ (g) φ (ℓ)$ .

step 3.1step 1.2∎

Remarks

Why the identity is not simply multiplicativity. For coprime $m$ and $n$ it reduces to $φ (m n) = φ (m) φ (n)$ , but the general case is what the intersection theorem needs: the degrees of $Q (μ_{m})$ and $Q (μ_{n})$ multiply to the degree of the compositum times the degree of the intersection, and it is the gcd–lcm form of the identity that turns that into $φ (\gcd (m, n))$ ( $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ ).

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

$Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$

Statement

Let $m, n \geq 1$ , put $d : = \gcd (m, n)$ and $ℓ : = lcm (m, n)$ (Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ , Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ), and let $Ω$ be a splitting field of $t^{ℓ} - 1$ over $Q$ (Every nonzero polynomial over a field has a splitting field), inside which the cyclotomic extensions $Q (μ_{k})$ for $k ∣ ℓ$ are taken (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then

$Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{d}) .$

Facts & Assumptions

Given: Integers $m, n \geq 1$ with $d = \gcd (m, n)$ and $ℓ = lcm (m, n)$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no positive integer (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); a splitting field $Ω$ of $t^{ℓ} - 1$ over $Q$ ; and, for each positive divisor $k$ of $ℓ$ , the subfield $Q (μ_{k}) : = Q (μ_{k} (Ω))$ of $Ω$ . Write $I : = Q (μ_{m}) \cap Q (μ_{n})$ .

[L1]

For a positive divisor $k$ of $ℓ$ , the subfield $Q (μ_{k} (Ω))$ generated by the $k$ -th roots of unity in $Ω$ is a cyclotomic extension of $Q$ of order $k$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ , The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity); because $char Q = 0$ divides no positive integer, $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity gives $μ_{k} (Ω)$ cyclic of order $k$ , with exactly $φ (k)$ primitive $k$ -th roots of unity.

[L2]

$[Q (μ_{k}) : Q] = φ (k)$ for every $k \geq 1$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L3]

$Q (μ_{k}) / Q$ is finite Galois ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ).

[L4]

For $E / F$ finite Galois and $L / F$ finite inside a common field, $[E L : F] = [E : F] [L : F] / [E \cap L : F]$ (For $E / F$ finite Galois and $L / F$ finite inside a common field, $[E L : F] = [E : F] [L : F] / [E \cap L : F]$ ).

[L5]

$Q (μ_{m}) Q (μ_{n}) = Q (μ_{ℓ})$ inside $Ω$ ( $K (μ_{m}) K (μ_{n}) = K (μ_{lcm (m, n)})$ ).

[L6]

$φ (m) φ (n) = φ (d) φ (ℓ)$ ( $φ (m) φ (n) = φ (\gcd (m, n)) φ (lcm (m, n))$ ).

[L7]

For fields $F \subseteq K \subseteq M$ with $K / F$ and $M / K$ finite, $[M : F] = [M : K] [K : F]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

Proof

technique · direct

1.1

$d$ divides $m$ and $n$ , and $m, n, d$ all divide $ℓ$ . For each positive divisor $k$ of $ℓ$ , write $ℓ = k q$ ; then $t^{ℓ} - 1 = (t^{k})^{q} - 1 = (t^{k} - 1) (t^{k (q - 1)} + \dots + t^{k} + 1)$ , so $t^{k} - 1$ splits over the splitting field $Ω$ of $t^{ℓ} - 1$ , and [L1] applies to $k$ . In particular all four cyclotomic extensions for $k = m, n, d, ℓ$ sit inside $Ω$ .

L1givenalgebra

2.1

For the inclusion $Q (μ_{d}) \subseteq I$ : since $d ∣ m$ , every $x$ with $x^{d} = 1$ satisfies $x^{m} = 1$ , so $μ_{d} (Ω) \subseteq μ_{m} (Ω)$ and hence $Q (μ_{d}) \subseteq Q (μ_{m})$ ; the same argument with $n$ gives $Q (μ_{d}) \subseteq Q (μ_{n})$ .

step 1.1givenalgebra

2.2

By [L3] the extension $Q (μ_{m}) / Q$ is finite Galois and $Q (μ_{n}) / Q$ is finite, both inside $Ω$ , so [L4] and [L5] give $φ (ℓ) = [Q (μ_{ℓ}) : Q] = [Q (μ_{m}) Q (μ_{n}) : Q] = φ (m) φ (n) / [I : Q]$ , using [L2] twice.

step 1.1L2L3L4L5

3.1

Hence $[I : Q] = φ (m) φ (n) / φ (ℓ) = φ (d)$ by [L6].

step 2.2L6

4.1

By step 2.1 the tower $Q \subseteq Q (μ_{d}) \subseteq I$ is defined, and [L7] with [L2] gives $φ (d) = [I : Q] = [I : Q (μ_{d})] φ (d)$ , so $[I : Q (μ_{d})] = 1$ and $I = Q (μ_{d})$ .

step 2.1step 3.1L2L7∎

Remarks

Where the base field is used. Only through [L2]: the equality $[Q (μ_{k}) : Q] = φ (k)$ for every $k$ , which is irreducibility of $Φ_{k}$ over $Q$ . Over a base field where some $Φ_{k}$ becomes reducible the degrees drop unevenly and the degree count in step 2.2 no longer forces the intersection down to $Q (μ_{d})$ .
The base field really matters. Over a general base field the same formula can fail; the companion page gives a finite-field witness in $F_{3} (μ_{5}) \cap F_{3} (μ_{7})$ is larger than $F_{3}$ although five and seven are coprime ↗.

CorollaryStatement: AI-generatedProof: AI-generatedprecheck passaudited 2026-08-26Open item page →

For an odd prime $p$ , $Q (ζ_{p})$ has exactly one intermediate field of degree two over $Q$

Statement

Let $p$ be an odd prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $ζ$ be a primitive $p$ -th root of unity (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity) in $Q (μ_{p})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then there is exactly one intermediate field $F$ with

$Q \subseteq F \subseteq Q (ζ), [F : Q] = 2$

(The degree $[K : F] = \dim_{F} K$ of a finite field extension).

Intermediate, not proper. At $p = 3$ the Galois group has order two, the unique subgroup of index two is the trivial one, and the field it names is $Q (ζ_{3})$ itself. Reading the statement as "proper subfield" would make it false at the smallest case in scope.

Facts & Assumptions

Given: An odd prime $p$ and a primitive $p$ -th root of unity $ζ$ in the cyclotomic extension $Q (μ_{p}) = Q (ζ)$ ; write $G : = Gal (Q (μ_{p}) / Q)$ .

[L1]

$Q (μ_{p}) / Q$ is finite Galois ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ) and the embedding $G \to (Z / p)^{\times}$ is an isomorphism ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L2]

Every finite subgroup of the unit group of an integral domain is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic); $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field, Field), hence an integral domain, and $(Z / p)^{\times}$ is its group of units (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L3]

$φ (p) = p - 1$ for a prime $p$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ), and $φ (p) = ∣ (Z / p)^{\times} ∣$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L4]

In a cyclic group of finite order $m$ there is exactly one subgroup of each order dividing $m$ , and every subgroup has that form (A finite cyclic group has exactly one subgroup of each order dividing its own).

[L5]

For $K / F$ finite Galois with $G = Gal (K / F)$ , the maps $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse bijections between subgroups and intermediate fields, and $[K^{H} : F] = [G : H]$ (The fundamental theorem of finite Galois theory).

[L6]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

By [L1] the group $G$ is isomorphic to $(Z / p)^{\times}$ , which is cyclic by [L2] and has order $p - 1$ by [L3]; so $G$ is cyclic of order $p - 1$ .

L1L2L3

2.1

Since $p$ is odd, $p - 1$ is even, so $2$ divides $p - 1$ and $(p - 1) / 2$ is a positive divisor of $p - 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). By [L4] there is exactly one subgroup $H \leq G$ with $∣ H ∣ = (p - 1) / 2$ .

step 1.1L4given

2.2

By [L5] and [L6], an intermediate field $F$ of $Q (μ_{p}) / Q$ has $[F : Q] = [G : H] = ∣ G ∣ / ∣ H ∣$ for its corresponding subgroup $H$ , so $[F : Q] = 2$ holds exactly when $∣ H ∣ = (p - 1) / 2$ .

step 1.1L5L6

3.1

The correspondence of [L5] is a bijection, so the intermediate fields of degree two over $Q$ are in bijection with the subgroups of order $(p - 1) / 2$ , of which there is exactly one by step 2.1. Hence there is exactly one such field.

step 2.1step 2.2L5∎

Remarks

Which field it is, is a different question. The argument counts intermediate fields; it produces no generator of the one it counts, and no claim is made here about identifying it. Naming that field concretely requires a computation this proof does not carry out.
Oddness is needed. At $p = 2$ the field $Q (μ_{2})$ is $Q$ itself, of degree $φ (2) = 1$ , and it has no intermediate field of degree two at all; the step that fails is step 2.1, where $p - 1 = 1$ is odd.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

For every $n \geq 1$ there are infinitely many primes $p$ with $p \equiv 1 (m o d n)$

Statement

For every integer $n \geq 1$ , the set

$T_{n} : = {p \in Z : p is prime and p \equiv 1 ⁣ ⁣ (m o d n)}$

(Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ) is not finite (Finite, countably infinite, countable, uncountable).

Facts & Assumptions

Given: An integer $n \geq 1$ and the cyclotomic polynomial $Φ_{n} \in Z [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ); evaluation at an integer is the ring homomorphism $Z [t] \to Z$ of Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism and Evaluation and roots of a polynomial in a commutative target ring.

[L1]

$Φ_{n}$ is monic in $Z [t]$ of degree $φ (n) \geq 1$ and $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

$Φ_{n} (0) = 1$ for every $n \geq 2$ ( $Φ_{1} (0) = - 1$ and $Φ_{n} (0) = 1$ for $n \geq 2$ ).

[L3]

Over a field $K$ with $char K ∤ n$ and a splitting field $E$ of $t^{n} - 1$ over $K$ , the roots of the image of $Φ_{n}$ in $E$ are exactly the primitive $n$ -th roots of unity in $E$ , that is the elements of order $n$ in $E^{\times}$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity, The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L4]

$Z / p$ is a field of characteristic $p$ for every prime $p$ (For every prime $p$ , the two operations on $Z / p$ make it a field, The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise, The characteristic of a field is zero or a prime number, The congruence class $[a]_{n}$ and the quotient set $Z / n$ ), and every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

[L5]

$∣ (Z / p)^{\times} ∣ = φ (p) = p - 1$ for a prime $p$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ); and for a finite group $G$ with $H \leq G$ , $∣ H ∣$ divides $∣ G ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ), while $∣ ⟨ x ⟩ ∣ = ord (x)$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L6]

Every integer greater than $1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L7]

A nonzero polynomial of degree $k$ over an integral domain has at most $k$ distinct roots in it (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L8]

The set of primes is not finite (Euclid's theorem: for every $n \in N$ and every list $p : n \to Z$ of primes there is a prime not among $p_{0}, \dots, p_{n - 1}$ ; consequently the set of primes is not finite).

Proof

technique · cases

1.1

In the case $n = 1$ , every integer is congruent to $1$ modulo $1$ , since $1$ divides every integer, so $T_{1}$ is the set of all primes, which is not finite by [L8].

assume-case oneL8given

1.2

In the case $n \geq 2$ , let $S \subseteq T_{n}$ be any finite set and put $M : = n \prod_{p \in S} p$ , an integer with $M \geq 2$ ; the goal is to produce a prime in $T_{n}$ outside $S$ .

assume-case biggiven

2.1

There is an integer $k \geq 1$ with $∣ Φ_{n} (k M) ∣ > 1$ : the three polynomials $Φ_{n}$ , $Φ_{n} - 1$ and $Φ_{n} + 1$ are nonzero of degree $φ (n) \geq 1$ by [L1], so by [L7] over the integral domain $Z$ at most $3 φ (n)$ integers $x$ satisfy $Φ_{n} (x) \in {- 1, 0, 1}$ ; the integers $M, 2 M, 3 M, \dots$ are pairwise distinct, so some $k M$ with $k \geq 1$ avoids that finite set.

step 1.2L1L7

2.2

Since $Φ_{n} (0) = 1$ by [L2], there is $h \in Z [t]$ with $Φ_{n} = 1 + t h$ ; evaluating at $k M$ gives $N : = Φ_{n} (k M) = 1 + k M h (k M)$ .

step 1.2L1L2given

3.1

By step 2.1 the integer $∣ N ∣$ exceeds $1$ , so it has a prime divisor $p$ by [L6], and $p ∣ N$ . That prime does not divide $k M$ : otherwise $p$ would divide $k M h (k M)$ and hence $N - k M h (k M) = 1$ , which is impossible for a prime. In particular $p ∤ M$ , so $p ∤ n$ and $p \notin S$ , both $n$ and every member of $S$ dividing $M$ .

step 2.1step 2.2L6given

4.1

Reduce modulo $p$ . Since $p ∣ Φ_{n} (k M)$ , the class $α : = [k M]_{p}$ is a root of the image of $Φ_{n}$ in $(Z / p) [t]$ , and $α \neq 0$ because $p ∤ k M$ by step 3.1. Let $E$ be a splitting field of $t^{n} - 1$ over $Z / p$ , which exists by [L4]; as $char (Z / p) = p$ does not divide $n$ , [L3] applies and $α$ , lying in $Z / p \subseteq E$ , has order exactly $n$ in $E^{\times}$ .

step 3.1L3L4

5.1

The order of $α$ in the subgroup $(Z / p)^{\times}$ of $E^{\times}$ is the same $n$ , so $n$ divides $∣ (Z / p)^{\times} ∣ = p - 1$ by [L5]; that is $p \equiv 1 (m o d n)$ , so $p \in T_{n}$ and $p \notin S$ .

step 3.1step 4.1L5given

6.1

In the case $n \geq 2$ , then, no finite subset of $T_{n}$ exhausts it, so $T_{n}$ is not finite; with step 1.1 the two cases are exhaustive and cover every $n \geq 1$ .

step 1.1step 5.1cases-exhaustive∎

Remarks

What replaces the archimedean estimate. The usual proof chooses $x$ large enough that $∣ Φ_{n} (x) ∣ > 1$ by a growth estimate. Step 2.1 replaces that by a root count, which needs no order structure on $Z$ beyond the distinctness of the multiples of $M$ , and gives exactly the same conclusion.
The case $n = 1$ is not a degenerate instance. For $n = 1$ one has $Φ_{1} (0) = - 1$ , not $1$ ( $Φ_{1} (0) = - 1$ and $Φ_{n} (0) = 1$ for $n \geq 2$ ), so step 2.2 is unavailable; the congruence is vacuous there and the statement is Euclid's theorem.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-26Open item page →

Every finite abelian group is a quotient of $(Z / n)^{k}$ for some $n$ and $k$

Statement

For every finite abelian group $G$ there are positive integers $n$ and $k$ and a surjective group homomorphism

$(Z / n)^{k} ⟶ G,$

where $(Z / n)^{k}$ denotes the set of $k$ -tuples of classes in $Z / n$ , with componentwise addition, for the additive group $Z / n$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold). Equivalently, $G ≅ (Z / n)^{k} / H$ for a subgroup $H$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

Facts & Assumptions

Given: A finite abelian group $G$ .

[L1]

For every finite abelian group $G$ there is a unique list $1 < n_{1} ∣ n_{2} ∣ \dots ∣ n_{r}$ with $G ≅ C_{n_{1}} \times \dots \times C_{n_{r}}$ ; the trivial group corresponds to the empty list (Fundamental theorem of finite abelian groups: invariant-factor form, Invariant-factor data for a finite abelian group).

[L2]

A cyclic group of finite order $m$ is isomorphic to $(Z / m, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[L3]

For every natural number $m$ , one has $[a]_{m} = [b]_{m}$ exactly when $m ∣ a - b$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), addition of classes is represented by addition of representatives, and $(Z / m, +)$ is an abelian group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L4]

For a homomorphism $f : A \to B$ , the rule $a \ker f \mapsto f (a)$ is an isomorphism $A / \ker f \to im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ , The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · cases

1.1

In the case that the invariant-factor list of [L1] is empty, $G$ is trivial; take $n = 1$ and $k = 1$ , so that $(Z / 1)^{1}$ is a one-element group and the unique map to $G$ is a surjective homomorphism.

assume-case trivialL1

1.2

In the case $r \geq 1$ , put $n : = n_{r}$ and $k : = r$ . Each $n_{i}$ divides $n$ , the list being a divisibility chain by [L1].

assume-case nontrivialL1

2.1

For each $i$ the rule $[a]_{n} \mapsto [a]_{n_{i}}$ is a well-defined surjective group homomorphism $Z / n \to Z / n_{i}$ : if $[a]_{n} = [b]_{n}$ then $n ∣ a - b$ , hence $n_{i} ∣ a - b$ because $n_{i} ∣ n$ , so $[a]_{n_{i}} = [b]_{n_{i}}$ by [L3]; it respects addition by [L3]; and every class $[a]_{n_{i}}$ is the image of $[a]_{n}$ .

step 1.2L3

3.1

Let $P : = (Z / n)^{r}$ and $Q : = (Z / n_{1}) \times \dots \times (Z / n_{r})$ , both with componentwise addition. By [L3] each coordinate $(Z / n, +)$ and $(Z / n_{i}, +)$ is an abelian group, so $P$ and $Q$ are abelian groups under these coordinatewise operations. Define $π ⁣ : P \to Q$ by $π ([a_{1}]_{n}, \dots, [a_{r}]_{n}) : = ([a_{1}]_{n_{1}}, \dots, [a_{r}]_{n_{r}})$ . Step 2.1 gives each coordinate map $Z / n \to Z / n_{i}$ as a well-defined surjective homomorphism, so $π$ is a well-defined surjective group homomorphism. Fact [L2] identifies each coordinate group $(Z / n_{i}, +)$ with $C_{n_{i}}$ , and [L1] identifies $C_{n_{1}} \times \dots \times C_{n_{r}}$ with $G$ up to isomorphism. Composing with that isomorphism gives a surjective homomorphism $(Z / n)^{k} \to G$ .

step 1.2step 2.1L1L2L3

4.1

The two cases are exhaustive, the invariant-factor list being empty or not, so such $n$ and $k$ exist for every finite abelian $G$ ; and [L4] turns any such surjection into an isomorphism $G ≅ (Z / n)^{k} / H$ with $H$ its kernel.

step 1.1step 3.1L4cases-exhaustive∎

Remarks

Why the invariant-factor form is convenient. The invariant factors form a divisibility chain, so the single modulus $n = n_{r}$ works immediately. A primary decomposition also proves the statement: take $n$ to be the least common multiple of the finitely many prime-power orders and reduce $Z / n$ onto each cyclic factor. The invariant-factor form simply avoids that extra choice of modulus.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every finite abelian group is the Galois group of some finite Galois extension of $Q$

Statement

For every finite abelian group $G$ there is a finite Galois extension $L / Q$ (Finite Galois extensions and $Gal (K / F)$ ) with

$Gal (L / Q) ≅ G,$

and $L$ may be taken inside a cyclotomic field $Q (μ_{N})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

Facts & Assumptions

Given: A finite abelian group $G$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no positive integer (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

There are positive integers $n, k$ and a surjective group homomorphism $(Z / n)^{k} \to G$ (Every finite abelian group is a quotient of $(Z / n)^{k}$ for some $n$ and $k$ , The external direct product $G \times H$ with componentwise multiplication).

[L2]

For every $n \geq 1$ the set of primes $p$ with $p \equiv 1 (m o d n)$ is not finite (For every $n \geq 1$ there are infinitely many primes $p$ with $p \equiv 1 (m o d n)$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Finite, countably infinite, countable, uncountable, Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

For a finite pairwise-coprime list $n_{0}, \dots, n_{r - 1}$ of positive integers with $N = \prod_{i} n_{i}$ , the map $[x]_{N} \mapsto ([x]_{n_{i}})_{i}$ is a bijection $Z / N \to \prod_{i} Z / n_{i}$ preserving addition, multiplication, $[0]$ and $[1]$ (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication, The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, Coprime integers: $\gcd (a, b) = 1$ , Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L4]

$(Z / p)^{\times}$ is a finite subgroup of the unit group of the field $Z / p$ , hence cyclic (Every finite subgroup of the unit group of an integral domain is cyclic, For every prime $p$ , the two operations on $Z / p$ make it a field, Field), of order $φ (p) = p - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L5]

A cyclic group of finite order $m$ is isomorphic to $(Z / m, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[L6]

$Gal (Q (μ_{N}) / Q) ≅ (Z / N)^{\times}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ), and this group is abelian (The Galois group of a cyclotomic extension is abelian).

[L7]

For $K / Q$ finite Galois with group $G^{'}$ and $H \leq G^{'}$ , the field $K^{H}$ is an intermediate field (The fundamental theorem of finite Galois theory); it is Galois over $Q$ exactly when $H$ is normal (Normal subgroup: invariance under conjugation), and then $Gal (K^{H} / Q) ≅ G^{'} / H$ (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L8]

For a homomorphism $f : A \to B$ the rule $a \ker f \mapsto f (a)$ is an isomorphism $A / \ker f \to im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

Proof

technique · direct

1.1

Fix $n, k \geq 1$ and a surjection $π ⁣ : (Z / n)^{k} \to G$ by [L1].

2.1

Choose pairwise distinct primes $p_{1}, \dots, p_{k}$ with $p_{i} \equiv 1 (m o d n)$ : the set of such primes is not finite by [L2], so at each of the $k$ steps one may pick a prime outside the finitely many already chosen. Put $N : = p_{1} \dots p_{k}$ .

step 1.1L2

3.1

Distinct primes are coprime: a positive common divisor of $p_{i}$ and $p_{j}$ is $1$ or $p_{i}$ , and is $1$ or $p_{j}$ , so if it is not $1$ then $p_{i} = p_{j}$ . Hence $p_{1}, \dots, p_{k}$ is a pairwise-coprime list.

step 2.1givenalgebra

3.2

For each $i$ there is a surjective homomorphism $(Z / p_{i})^{\times} \to Z / n$ : by [L4] the group $(Z / p_{i})^{\times}$ is cyclic of order $m_{i} = p_{i} - 1$ , which $n$ divides by step 2.1; [L5] identifies it with $(Z / m_{i}, +)$ , and $[a]_{m_{i}} \mapsto [a]_{n}$ is well defined because $n ∣ m_{i}$ , is a homomorphism, and is onto.

step 2.1L4L5

4.1

By [L3] the map $[x]_{N} \mapsto ([x]_{p_{i}})_{i}$ is a bijection $Z / N \to \prod_{i} Z / p_{i}$ preserving multiplication and $[1]$ , so it carries units to units bijectively and restricts to a group isomorphism $(Z / N)^{\times} \to \prod_{i} (Z / p_{i})^{\times}$ .

step 3.1L3

5.1

Taking the product of the maps of step 3.2 and composing with step 4.1 and with $π$ gives a surjective group homomorphism $Ψ_{0} ⁣ : (Z / N)^{\times} \to G$ ; composing with the isomorphism of [L6] gives a surjective homomorphism $Ψ ⁣ : Gal (Q (μ_{N}) / Q) \to G$ .

step 1.1step 4.1step 3.2L6

6.1

Put $G^{'} : = Gal (Q (μ_{N}) / Q)$ and $H : = \ker Ψ$ . The group $G^{'}$ is abelian by [L6], so every subgroup is normal, $g H g^{- 1} = H$ holding for all $g$ ; hence $L : = Q (μ_{N})^{H}$ is an intermediate field, $L / Q$ is finite Galois, and $Gal (L / Q) ≅ G^{'} / H$ by [L7].

step 5.1L6L7

7.1

By [L8] applied to $Ψ$ , which is surjective, $G^{'} / H ≅ im Ψ = G$ ; hence $Gal (L / Q) ≅ G$ , with $L$ inside $Q (μ_{N})$ .

step 5.1step 6.1L8∎

Remarks

What is produced is a subfield, not a cyclotomic field. The construction realises $G$ as the Galois group of an intermediate field of $Q (μ_{N}) / Q$ , and it must: the Galois group of $Q (μ_{N})$ itself is $(Z / N)^{\times}$ , whose order $φ (N)$ is even for $N \geq 3$ , so most finite abelian groups are not of that form. The companion page spells out that failure in FALSE: every finite abelian group is $Gal (Q (μ_{n}) / Q)$ for some $n$ ↗.
The distinctness of the primes is needed twice. It makes the list pairwise coprime so that the Chinese remainder theorem applies, and it makes the product $N$ have exactly the intended unit group. Repeating a prime would collapse two factors into one.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Every intermediate field of $Q (μ_{n}) / Q$ is Galois over $Q$ with abelian Galois group

Statement

Let $n \geq 1$ and let $F$ be an intermediate field of $Q (μ_{n}) / Q$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then $F / Q$ is a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) and $Gal (F / Q)$ is abelian.

Facts & Assumptions

Given: An integer $n \geq 1$ and an intermediate field $Q \subseteq F \subseteq Q (μ_{n})$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no positive integer (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). Write $G : = Gal (Q (μ_{n}) / Q)$ .

[L1]

$Q (μ_{n}) / Q$ is finite Galois ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ) with $G ≅ (Z / n)^{\times}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ), and $G$ is abelian (The Galois group of a cyclotomic extension is abelian).

[L2]

For $K / F_{0}$ finite Galois with group $G$ , the maps $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse bijections between subgroups of $G$ and intermediate fields (The fundamental theorem of finite Galois theory).

[L3]

With $K / F_{0}$ finite Galois, $G = Gal (K / F_{0})$ , $H \leq G$ and $E = K^{H}$ : the extension $E / F_{0}$ is Galois exactly when $H$ is normal in $G$ (Normal subgroup: invariance under conjugation), and then restriction gives $Gal (E / F_{0}) ≅ G / H$ (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1

By [L1] the extension $Q (μ_{n}) / Q$ is finite Galois and $G$ is abelian.

2.1

By [L2] there is a subgroup $H \leq G$ with $F = Q (μ_{n})^{H}$ .

step 1.1L2

3.1

Since $G$ is abelian, $g H g^{- 1} = H$ for every $g \in G$ , so $H$ is normal in $G$ ; hence $F / Q$ is Galois and $Gal (F / Q) ≅ G / H$ by [L3].

step 1.1step 2.1L3

4.1

A quotient of an abelian group is abelian, since the images of two commuting elements commute and every element of $G / H$ is such an image; so $Gal (F / Q)$ is abelian.

step 1.1step 3.1L3∎

Remarks

This is the proved half of the Kronecker–Weber picture on this page. Every subfield of a rational cyclotomic field is abelian over $Q$ ; the converse is recorded separately as Recorded, not proved: every finite abelian extension of $Q$ lies in a cyclotomic field ‡.

RemarkRemark: Literature-sourcedProof: Not supplied‡ sources checked 2026-08-26‡ not proved hereOpen item page →

‡ Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Recorded, not proved: every finite abelian extension of $Q$ lies in a cyclotomic field

Statement

Kronecker–Weber theorem. Let $L / Q$ be a finite Galois extension whose Galois group is abelian. Then there is an integer $n \geq 1$ with

$L \subseteq Q (μ_{n})$

(The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

The statement fails over larger base fields. Over $K = Q (i)$ the extension obtained by adjoining a fourth root of $1 + i$ is abelian over $K$ and is contained in no cyclotomic extension of $K$ .

Remarks

What this library does prove. The converse half is Every intermediate field of $Q (μ_{n}) / Q$ is Galois over $Q$ with abelian Galois group: every intermediate field of $Q (μ_{n}) / Q$ is Galois over $Q$ with abelian Galois group. Together with $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , which computes $Gal (Q (μ_{n}) / Q)$ exactly, that half says the subfields of cyclotomic fields are abelian; Kronecker–Weber says there are no others.

What would prove it, and which track that belongs to. The standard argument reduces the global statement to a local one at each prime that ramifies in $L$ , and then analyses the higher ramification groups of the $p$ -adic rationals to show that the local extension is contained in a local cyclotomic extension. Every ingredient of that reduction — valuations, local fields, the ring of integers of a number field, decomposition and inertia groups — belongs to algebraic number theory, and none of it is developed anywhere in this library. An alternative route through class field theory needs strictly more. Neither source consulted here proves the theorem: Conrad calls it deep and states it without proof, and Milne states it only in a footnote to an exercise.

Why it is recorded rather than omitted. The arithmetic consequences of the theorem are stated elsewhere in terms of it, so a page that builds the cyclotomic machinery and then says nothing about its sharpest classical application would leave that seam pointing at nothing. Recording it with the fuchsia not-proved-here marking is the honest form: the reader is told, at the point of contact, that the proof belongs to a later algebraic-number-theory track.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

$Gal (F_{8} / F_{2})$ is cyclic of order three with no proper intermediate field

Example

Let $K : = F_{2} [t] / (t^{3} + t + 1)$ and let $α$ be the class of $t$ . Then $K$ is a field of order $8$ , the squaring map $σ (x) = x^{2}$ generates

$Gal (K / F_{2}) = {i d, σ, σ^{2}},$

cyclic of order three, its two orbits on $K ∖ F_{2}$ are

${α, α^{2}, α^{2} + α} and {α + 1, α^{2} + 1, α^{2} + α + 1},$

and $K / F_{2}$ has no intermediate field other than $F_{2}$ and $K$ .

Facts & Assumptions

Given: The polynomial $π : = t^{3} + t + 1 \in F_{2} [t]$ , the ring $K = F_{2} [t] / (π)$ and the class $α$ of $t$ , so that $α^{3} = α + 1$ because $α^{3} + α + 1 = 0$ and $- 1 = 1$ in characteristic two.

[L1]

[L2]

For a field $F$ and nonconstant $p \in F [x]$ , $p$ is irreducible if and only if $F [x] / (p)$ is a field (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

[L3]

If $a$ is algebraic over $F$ with minimal polynomial $m_{a}$ of degree $n$ , then $F (a)$ has power basis $1, a, \dots, a^{n - 1}$ and $[F (a) : F] = n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension); a monic irreducible vanishing at $a$ is that minimal polynomial (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L4]

An extension $E / F_{q}$ of finite fields of degree $n$ is Galois with $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ cyclic of order $n$ , where $σ_{q} (x) = x^{q}$ , and $∣ E ∣ = q^{n}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields, For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ , Finite fields and their order).

[L5]

The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ for the positive divisors $d$ of $n$ , one for each divisor (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

Verification

technique · direct

1.1

$π$ has no root in $F_{2}$ : $π (0) = 1$ and $π (1) = 1 + 1 + 1 = 1$ . By [L1] it is irreducible, so $K$ is a field by [L2].

L1L2given

2.1

$π$ is monic irreducible with $π (α) = 0$ , so it is the minimal polynomial of $α$ over $F_{2}$ and $[K : F_{2}] = 3$ with power basis $1, α, α^{2}$ by [L3]; hence $∣ K ∣ = 2^{3} = 8$ by [L4].

step 1.1L3L4

3.1

By [L4] the extension $K / F_{2}$ is Galois with Galois group generated by $σ (x) = x^{2}$ and of order three.

step 2.1L4

4.1

The orbit of $α$ : $α^{2}$ is $α^{2}$ ; $α^{4} = α \cdot α^{3} = α (α + 1) = α^{2} + α$ ; and $(α^{2} + α)^{2} = α^{4} + α^{2} = (α^{2} + α) + α^{2} = α$ , using that squaring is additive in characteristic two. So ${α, α^{2}, α^{2} + α}$ is one orbit of size three.

step 2.1step 3.1given

5.1

The orbit of $α + 1$ : $(α + 1)^{2} = α^{2} + 1$ , $(α^{2} + 1)^{2} = α^{4} + 1 = α^{2} + α + 1$ , and $(α^{2} + α + 1)^{2} = α^{4} + α^{2} + 1 = α + 1$ . So ${α + 1, α^{2} + 1, α^{2} + α + 1}$ is the other orbit of size three, and together with ${0}$ and ${1}$ these account for all eight elements.

step 2.1step 3.1step 4.1given

6.1

The positive divisors of three are $1$ and $3$ , so by [L5] the intermediate fields are exactly two: $F_{2}$ and $K$ itself. There is no field strictly between them.

step 2.1step 3.1L5∎

Remarks

Why the two nontrivial orbits have the same size. Each is an orbit of a group of prime order acting without fixed points outside $F_{2}$ : a fixed point of $σ$ is an element with $x^{2} = x$ , and those are exactly the two elements of $F_{2}$ (The elements of a finite extension fixed by the $q$ -power map are exactly the base field).

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The intermediate fields of $F_{2^{12}} / F_{2}$ match the divisors of twelve

Example

Let $E$ be a field with $F_{2}$ as a subfield and $[E : F_{2}] = 12$ , so that $∣ E ∣ = 2^{12}$ . Its intermediate fields over $F_{2}$ are exactly

$F_{2}, F_{2^{2}}, F_{2^{3}}, F_{2^{4}}, F_{2^{6}}, F_{2^{12}} = E,$

one for each of the six positive divisors $1, 2, 3, 4, 6, 12$ of twelve, with $F_{2^{d}} \subseteq F_{2^{e}}$ exactly when $d$ divides $e$ . Neither of $F_{2^{4}}$ and $F_{2^{6}}$ contains the other, and

$F_{2^{4}} \cap F_{2^{6}} = F_{2^{2}}, F_{2^{4}} F_{2^{6}} = F_{2^{12}} .$

Facts & Assumptions

Given: A field $E$ with subfield $F_{2}$ and $[E : F_{2}] = 12$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension); divisibility as in Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , with $\gcd$ and $lcm$ as in Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ and Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ .

[L1]

$E / F_{2}$ is Galois with cyclic Galois group generated by $x \mapsto x^{2}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ ), and $∣ E ∣ = 2^{12}$ (For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ , Finite fields and their order).

[L2]

The intermediate fields of $F_{q^{n}} / F_{q}$ are exactly the $F_{q^{d}} = {x : x^{q^{d}} = x}$ for the positive divisors $d$ of $n$ , one for each divisor, with $[F_{q^{d}} : F_{q}] = d$ and $F_{q^{d}} \subseteq F_{q^{e}}$ if and only if $d ∣ e$ (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).

[L3]

A field of order $p^{m}$ has, for each positive divisor $e$ of $m$ , exactly one subfield of order $p^{e}$ , namely ${a : a^{p^{e}} = a}$ , and these are all of its subfields (The subfields of $F_{p^{n}}$ are the unique fields $F_{p^{d}}$ for positive divisors $d$ of $n$ ).

[L4]

Every common divisor of two integers divides their greatest common divisor, and their least common multiple divides every common multiple (Every common divisor of $a$ and $b$ divides $\gcd (a, b)$ ; consequently $d = \gcd (a, b)$ exactly when $d \geq 0$ , $d ∣ a$ , $d ∣ b$ , and every common divisor of $a$ and $b$ divides $d$ — a characterisation that holds at $(a, b) = (0, 0)$ as well, Every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) \cdot lcm (a, b) = ∣ a b ∣$ ).

Verification

technique · direct

1.1

The positive divisors of twelve are $1, 2, 3, 4, 6, 12$ , six in all, since a positive divisor $d$ of $12$ satisfies $d \leq 12$ and direct inspection of $1, \dots, 12$ leaves exactly these.

givenalgebra

2.1

By [L1] and [L2] the intermediate fields of $E / F_{2}$ are the $F_{2^{d}}$ for those six $d$ , one for each, with $F_{2^{d}} \subseteq F_{2^{e}}$ exactly when $d ∣ e$ .

step 1.1L1L2

3.1

Neither $4 ∣ 6$ nor $6 ∣ 4$ , so by step 2.1 neither of $F_{2^{4}}$ and $F_{2^{6}}$ contains the other.

step 2.1given

4.1

Their intersection is an intermediate field of $E / F_{2}$ , being a subfield of $E$ containing $F_{2}$ , so it is $F_{2^{c}}$ for a unique divisor $c$ of $12$ by step 2.1; from $F_{2^{c}} \subseteq F_{2^{4}}$ and $F_{2^{c}} \subseteq F_{2^{6}}$ one gets $c ∣ 4$ and $c ∣ 6$ , so [L4] gives $c ∣ \gcd (4, 6) = 2$ ; and $2 ∣ 4$ and $2 ∣ 6$ put $F_{2^{2}}$ inside both, so $2 ∣ c$ . Hence $c = 2$ and the intersection is $F_{2^{2}}$ .

step 2.1step 3.1L4

5.1

Their compositum is likewise an intermediate field $F_{2^{c^{'}}}$ , and it contains both, so $4 ∣ c^{'}$ and $6 ∣ c^{'}$ by step 2.1. Thus [L4] gives $lcm (4, 6) = 12 ∣ c^{'}$ ; since $c^{'} ∣ 12$ , $c^{'} = 12$ and the compositum is $E = F_{2^{12}}$ .

step 2.1step 4.1L4

6.1

The same six fields are what [L3] produces for $E$ , whose order is $2^{12}$ : its subfields are the ${a : a^{2^{e}} = a}$ for the divisors $e$ of $12$ , and these are the sets named in [L2]. So the Galois indexing and the elementary one agree here.

step 2.1step 5.1L2L3∎

Remarks

Where the two descriptions coincide. Because the base field is the prime field, the divisors of $[E : F_{2}]$ and the divisors of $\log_{2} ∣ E ∣$ are the same list; over a larger base field the two indexings differ by the factor $[F_{q} : F_{p}]$ , and The Galois description of the subfields of a finite field and the elementary divisibility criterion agree records the translation.

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The divisor-sum identity at $q = 2$ , $n = 3$ finds exactly two monic irreducible cubics

Example

Over $F_{2}$ the divisor-sum identity ( $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$ ) at $n = 3$ reads

$N_{2} (1) + 3 N_{2} (3) = 2^{3} = 8,$

and $N_{2} (1) = 2$ , so $N_{2} (3) = 2$ . The two monic irreducible cubics in $F_{2} [t]$ are

$t^{3} + t + 1 and t^{3} + t^{2} + 1.$

Facts & Assumptions

Given: The field $F_{2}$ with two elements and the counts $N_{2} (d)$ of monic irreducible polynomials of degree $d$ in $F_{2} [t]$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L1]

$\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for every $n \geq 1$ , the sum over the positive divisors of $n$ ( $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L2]

[L3]

For a commutative ring $R$ , $a \in R$ and $f \in R [x]$ : $f (a) = 0$ if and only if $x - a$ divides $f$ (Factor theorem over a commutative ring, Evaluation and roots of a polynomial in a commutative target ring).

Verification

technique · direct

1.1

The monic polynomials of degree one in $F_{2} [t]$ are $t$ and $t + 1$ , and each is irreducible, having degree one; so $N_{2} (1) = 2$ .

givenalgebra

1.2

A monic cubic over $F_{2}$ is $f = t^{3} + a t^{2} + b t + c$ with $a, b, c \in F_{2}$ , so there are eight of them. Such an $f$ has no root in $F_{2}$ exactly when $f (0) = c \neq 0$ and $f (1) = 1 + a + b + c \neq 0$ , that is exactly when $c = 1$ and $a + b = 1$ .

givenalgebra

2.1

The positive divisors of $3$ are $1$ and $3$ , so [L1] at $q = 2$ and $n = 3$ reads $N_{2} (1) + 3 N_{2} (3) = 8$ ; with step 1.1 this gives $3 N_{2} (3) = 6$ and $N_{2} (3) = 2$ .

step 1.1L1algebra

3.1

The pairs $(a, b)$ with $a + b = 1$ in $F_{2}$ are $(0, 1)$ and $(1, 0)$ , so exactly two monic cubics have no root in $F_{2}$ , namely $t^{3} + t + 1$ and $t^{3} + t^{2} + 1$ ; by [L2] these two are irreducible and by [L2] and [L3] the other six are not, each having a root and hence a linear factor. This agrees with the count $N_{2} (3) = 2$ of step 2.1.

step 2.1step 1.2L2L3algebra∎

Remarks

The identity is a recursion, not a formula. It determines $N_{2} (3)$ only because $N_{2} (1)$ is already known; at $n = 4$ it would read $N_{2} (1) + 2 N_{2} (2) + 4 N_{2} (4) = 16$ and would need $N_{2} (2)$ first.

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The four roots of $t^{4} + t + 1$ over $F_{2}$ are the Frobenius powers of any one of them

Example

Let $K : = F_{2} [t] / (t^{4} + t + 1)$ and let $α$ be the class of $t$ , so $α^{4} = α + 1$ . Then $K$ is a field of order $16$ , and the four conjugates of $α$ over $F_{2}$ ,

$α, α^{2}, α^{4} = α + 1, α^{8} = α^{2} + 1,$

are pairwise distinct, are exactly the roots of $t^{4} + t + 1$ in $K$ , and satisfy $α^{16} = α$ , so the Frobenius orbit closes at length four.

Facts & Assumptions

Given: The polynomial $π : = t^{4} + t + 1 \in F_{2} [t]$ , the ring $K = F_{2} [t] / (π)$ , and the class $α$ of $t$ , so $α^{4} = α + 1$ since $- 1 = 1$ in characteristic two; squaring is additive there.

[L1]

[L2]

$f (a) = 0$ if and only if $x - a$ divides $f$ (Factor theorem over a commutative ring); and over an integral domain $\deg (f g) = \deg f + \deg g$ for nonzero $f, g$ (Over an integral domain, degrees add under multiplication of nonzero polynomials, Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

[L4]

A monic irreducible vanishing at $a$ is the minimal polynomial of $a$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), and then $F (a)$ has power basis $1, a, \dots, a^{n - 1}$ with $[F (a) : F] = n$ its degree (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L5]

An extension of finite fields of degree $n$ over $F_{q}$ is Galois with cyclic Galois group generated by $σ_{q} (x) = x^{q}$ , and has $q^{n}$ elements (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields, For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L6]

A monic irreducible $π$ of degree $d$ over $F_{q}$ with a root $α$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$ and $π = \prod_{i < d} (t - α^{q^{i}})$ (A monic irreducible of degree $d$ over $F_{q}$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$ ).

Verification

technique · direct

1.1

$π$ has no root in $F_{2}$ , since $π (0) = 1$ and $π (1) = 1 + 1 + 1 = 1$ ; so by [L2] it has no factor of degree one.

L2given

1.2

The only monic irreducible quadratic in $F_{2} [t]$ is $t^{2} + t + 1$ : the four monic quadratics are $t^{2}$ , $t^{2} + 1$ , $t^{2} + t$ and $t^{2} + t + 1$ , and the first three have the root $0$ , $1$ and $0$ respectively, so [L1] leaves only the last.

L1L2given

2.1

$π$ is irreducible. A factorisation of $π$ into two nonconstant factors has degrees summing to four by [L2], so it is either $1 + 3$ , excluded by step 1.1, or $2 + 2$ ; and every monic quadratic factor would have to be irreducible, hence equal to $t^{2} + t + 1$ by step 1.2, giving $π = (t^{2} + t + 1)^{2} = t^{4} + t^{2} + 1$ , which is not $π$ .

step 1.1step 1.2L2given

3.1

By [L3] the ring $K$ is a field; $π$ is monic irreducible with $π (α) = 0$ , so $[K : F_{2}] = 4$ with power basis $1, α, α^{2}, α^{3}$ by [L4], and $∣ K ∣ = 2^{4} = 16$ by [L5].

step 2.1L3L4L5

4.1

Compute the conjugates in that basis: $α^{4} = α + 1$ by hypothesis, and $α^{8} = (α^{4})^{2} = (α + 1)^{2} = α^{2} + 1$ . So the four elements $α, α^{2}, α^{4}, α^{8}$ have coordinate lists $(0, 1, 0, 0)$ , $(0, 0, 1, 0)$ , $(1, 1, 0, 0)$ and $(1, 0, 1, 0)$ , which are pairwise different, so the four elements are pairwise distinct.

step 3.1given

5.1

$α^{16} = (α^{8})^{2} = (α^{2} + 1)^{2} = α^{4} + 1 = (α + 1) + 1 = α$ , so the orbit closes after four steps.

step 4.1given

6.1

By [L6] applied to $π$ , of degree four, the elements $α, α^{2}, α^{4}, α^{8}$ are exactly the roots of $π$ and $π = (t - α) (t - α^{2}) (t - α^{4}) (t - α^{8})$ in $K [t]$ ; steps 4.1 and 5.1 verify the distinctness and the closing of the orbit directly.

step 2.1step 4.1step 5.1L6∎

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A normal basis of $F_{8}$ over $F_{2}$

Example

Let $K : = F_{2} [t] / (t^{3} + t + 1)$ , a field of order $8$ , let $α$ be the class of $t$ , and let $σ (x) = x^{2}$ generate $Gal (K / F_{2})$ . Put

$β : = α + 1.$

Then the conjugate list

$(β, β^{2}, β^{4}) = (α + 1, α^{2} + 1, α^{2} + α + 1)$

is a normal basis of $K$ over $F_{2}$ (Normal bases of a finite Galois extension).

Not every element works. The generator $α$ itself does not: its conjugate list is $(α, α^{2}, α^{2} + α)$ , whose three members sum to $0$ , so they are linearly dependent over $F_{2}$ and are not a basis.

Facts & Assumptions

Given: The field $K = F_{2} [t] / (t^{3} + t + 1)$ with $α$ the class of $t$ , so $α^{3} = α + 1$ and $α^{4} = α^{2} + α$ ; squaring is additive in characteristic two.

[L1]

$t^{3} + t + 1$ has no root in $F_{2}$ , so it is irreducible (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field) and $K$ is a field (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible); it is the minimal polynomial of $α$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), so $[K : F_{2}] = 3$ with power basis $1, α, α^{2}$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L2]

$K / F_{2}$ is Galois with $Gal (K / F_{2}) = {i d, σ, σ^{2}}$ cyclic of order three, $σ (x) = x^{2}$ , and $∣ K ∣ = 8$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields, For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L3]

A list $v$ of length $n$ is an ordered basis of $V$ if and only if every $x \in V$ has exactly one coordinate list $λ$ with $x = \sum_{i < n} λ_{i} v_{i}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L4]

For a linear map $T : V \to W$ with $V$ finite-dimensional, $\dim V = \dim \ker T + \dim im T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L5]

Every finite Galois extension with cyclic Galois group has a normal basis (Every finite cyclic extension has a normal basis).

Verification

technique · direct

1.1

By [L1] and [L2] the space $K$ is a three-dimensional $F_{2}$ -vector space with basis $1, α, α^{2}$ , and $Gal (K / F_{2}) = {i d, σ, σ^{2}}$ acts by $x \mapsto x, x^{2}, x^{4}$ .

L1L2

2.1

The conjugate list of $α$ is $(α, α^{2}, α^{4})$ with $α^{4} = α \cdot α^{3} = α (α + 1) = α^{2} + α$ ; hence $α + α^{2} + α^{4} = α + α^{2} + α^{2} + α = 0$ . A vanishing combination with all coefficients $1$ is nontrivial, so this list is linearly dependent over $F_{2}$ and is not a basis.

step 1.1L3given

3.1

The conjugates of $β = α + 1$ are $β$ , $β^{2} = (α + 1)^{2} = α^{2} + 1$ and $β^{4} = (β^{2})^{2} = (α^{2} + 1)^{2} = α^{4} + 1 = α^{2} + α + 1$ .

step 1.1step 2.1given

4.1

The seven nonzero $F_{2}$ -combinations of $β, β^{2}, β^{4}$ are nonzero: the three single terms are $α + 1$ , $α^{2} + 1$ and $α^{2} + α + 1$ ; the three pairwise sums are $β + β^{2} = α^{2} + α$ , $β + β^{4} = α^{2}$ and $β^{2} + β^{4} = α$ ; and the total sum is $β + β^{2} + β^{4} = 1$ . None of these seven is $0$ , as each has a nonzero coordinate list in the basis $1, α, α^{2}$ .

step 1.1step 3.1given

5.1

So the $F_{2}$ -linear map $T ⁣ : F_{2}^{3} \to K$ sending $(c_{1}, c_{2}, c_{3})$ to $c_{1} β + c_{2} β^{2} + c_{3} β^{4}$ has trivial kernel by step 4.1; both spaces have dimension three by step 1.1, so [L4] makes $T$ surjective as well, hence bijective, and [L3] makes $(β, β^{2}, β^{4})$ an ordered basis of $K$ over $F_{2}$ .

step 1.1step 4.1L3L4

6.1

That list is the family of conjugates of $β$ under $Gal (K / F_{2})$ by step 3.1, so it is a normal basis, as [L5] guarantees exists for this cyclic extension.

step 3.1step 5.1L2L5∎

Remarks

A conjugate family of the right size can still fail. The list $(α, α^{2}, α^{4})$ has three distinct members and is a single Galois orbit, yet it is not a basis; what fails is independence, not the orbit condition. The normal basis theorem asserts that some element works, never that every element does (FALSE: every basis of a finite field over a subfield is a normal basis).

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${1 + i, 1 - i}$ is a normal basis of $C / R$ while ${1, i}$ is not

Example

The extension $C / R$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ) is finite Galois of degree two with $Gal (C / R) = {i d, z \mapsto \overline{z}}$ (Real and imaginary parts, complex conjugation, and modulus). For it:

the conjugate list $(1 + i, 1 - i)$ is a normal basis (Normal bases of a finite Galois extension);
$(1, i)$ is an $R$ -basis of $C$ that is not a conjugate list of any element;
$(i, - i)$ is the conjugate list of $i$ but is not a basis.

The last two show that the two conditions in the definition of a normal basis are independent of each other.

Facts & Assumptions

Given: The complex field with $i^{2} = - 1$ and conjugation $\overline{a + b i} = a - b i$ for $a, b \in R$ (Real and imaginary parts, complex conjugation, and modulus); in $R$ one has $2 \neq 0$ .

[L1]

$C = R (i)$ is a simple algebraic extension with power basis $1, i$ and $[C : R] = 2$ ( $C / R$ has power basis $1, i$ and degree $2$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L2]

Every field automorphism of $C$ fixing $R$ pointwise is either the identity or complex conjugation, and these two are distinct (The only real-field automorphisms of $C$ are the identity and complex conjugation, Relative field automorphisms and $Aut (K / F)$ ).

[L3]

Complex conjugation is a real-field automorphism (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L4]

A finite extension $K / F$ with $G = Aut (K / F)$ is Galois exactly when $∣ G ∣ = [K : F]$ (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and $Gal (K / F)$ ).

[L5]

A list $v$ of length $n$ is an ordered basis of $V$ if and only if every $x \in V$ has exactly one coordinate list with respect to it (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis); and $\dim V = \dim \ker T + \dim im T$ for a linear $T$ on a finite-dimensional $V$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L6]

Every finite Galois extension of an infinite field has a normal basis (Every finite Galois extension of an infinite field has a normal basis).

Verification

technique · direct

1.1

By [L2] and [L3] the group $Aut (C / R)$ has exactly the two elements $i d$ and conjugation, so its order is $2 = [C : R]$ by [L1]; hence $C / R$ is finite Galois with that Galois group by [L4].

L1L2L3L4

2.1

The conjugate list of $1 + i$ is $(1 + i, 1 - i)$ , whose members have coordinate lists $(1, 1)$ and $(1, - 1)$ in the ordered basis $(1, i)$ of [L1]. For $a, b \in R$ , $a (1 + i) + b (1 - i) = (a + b) + (a - b) i$ vanishes exactly when $a + b = 0$ and $a - b = 0$ , hence when $2 a = 0$ , that is $a = 0$ and then $b = 0$ .

step 1.1L1given

2.2

$(1, i)$ is a basis by [L1], but no $z \in C$ has conjugate list $(z, \overline{z})$ with underlying set ${1, i}$ : such a $z$ would lie in ${1, i}$ , and the set for $z = 1$ is ${1}$ while for $z = i$ it is ${i, - i}$ , neither of which is ${1, i}$ .

step 1.1L1given

3.1

So the linear map $R^{2} \to C$ sending $(a, b)$ to $a (1 + i) + b (1 - i)$ has trivial kernel; both spaces have dimension two by [L1], so [L5] makes it bijective and $(1 + i, 1 - i)$ an ordered $R$ -basis of $C$ . Being the conjugate list of $1 + i$ , it is a normal basis, in agreement with [L6].

step 1.1step 2.1L1L5L6

4.1

$(i, - i)$ is the conjugate list of $i$ , since $\overline{i} = - i$ , and its two members are distinct; but $1 \cdot i + 1 \cdot (- i) = 0$ is a vanishing combination with nonzero coefficients, so the list is not independent and by [L5] is not a basis. With steps 3.1 and 2.2 this establishes all three claims.

step 1.1step 3.1step 2.2L5given∎

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FALSE: every basis of a finite field over a subfield is a normal basis

Statement

False claim. For every extension $E / F_{q}$ of finite fields, every $F_{q}$ -basis of $E$ is a normal basis (Normal bases of a finite Galois extension).

Facts & Assumptions

Given: The ring $L : = F_{2} [t] / (t^{2} + t + 1)$ with $α$ the class of $t$ , so that $α^{2} = α + 1$ because $α^{2} + α + 1 = 0$ and $- 1 = 1$ in characteristic two.

[L1]

A polynomial of degree $2$ or $3$ over a field is irreducible if and only if it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field); and $F [x] / (p)$ is a field exactly when $p$ is irreducible (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

[L2]

A monic irreducible vanishing at $a$ is the minimal polynomial of $a$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), and then $F (a)$ has power basis $1, a, \dots, a^{n - 1}$ with $[F (a) : F] = n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L3]

An extension of finite fields of degree $n$ over $F_{q}$ is Galois with cyclic Galois group generated by $σ_{q} (x) = x^{q}$ , of order $n$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields, For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L4]

A normal basis of $K / F$ is an ordered $F$ -basis of the form $(σ_{1} γ, \dots, σ_{n} γ)$ for a single $γ \in K$ , indexed by $Gal (K / F)$ (Normal bases of a finite Galois extension).

[L5]

Every finite Galois extension has a normal basis (Every finite Galois extension has a normal basis).

Refutation

technique · direct

1.1

$t^{2} + t + 1$ has no root in $F_{2}$ , its values at $0$ and $1$ both being $1$ , so it is irreducible and $L$ is a field by [L1]; it is the minimal polynomial of $α$ , so $[L : F_{2}] = 2$ with ordered basis $(1, α)$ by [L2], and $L$ has four elements $0, 1, α, α + 1$ .

L1L2given

2.1

By [L3] the extension $L / F_{2}$ is Galois with $Gal (L / F_{2}) = {i d, σ}$ , where $σ (x) = x^{2}$ .

step 1.1L3

3.1

The conjugate lists of the four elements are $(0, 0)$ , $(1, 1)$ , $(α, α + 1)$ and $(α + 1, α)$ , using $α^{2} = α + 1$ and $(α + 1)^{2} = α^{2} + 1 = α$ . Their underlying sets are ${0}$ , ${1}$ and ${α, α + 1}$ .

step 1.1step 2.1given

4.1

The list $(1, α)$ is an $F_{2}$ -basis of $L$ by step 1.1, but its underlying set ${1, α}$ is none of the three sets in step 3.1, so it is not the conjugate list of any element and hence is not a normal basis by [L4]. The false claim therefore fails already for $L / F_{2}$ .

step 1.1step 3.1L4

5.1

What is true is the existential statement: some element of $L$ generates a normal basis, and $α$ does, since $(α, α + 1)$ is a list of two distinct elements whose only vanishing $F_{2}$ -combinations are trivial, as $α \neq 0$ , $α + 1 \neq 0$ and $α + (α + 1) = 1 \neq 0$ . That is the content of [L5], which asserts existence and never universality.

step 3.1step 4.1L4L5∎

Remarks

Where the false claim comes from. The normal basis theorem is an existence statement, and its proofs single out an element by a nonvanishing condition — a determinant in the infinite case, a cyclic vector in the finite case. Both conditions genuinely exclude some elements, as A normal basis of $F_{8}$ over $F_{2}$ shows over $F_{8}$ .

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$Φ_{1}$ through $Φ_{12}$ computed from the divisor recursion

Example

Running the recursion of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ gives

$\begin{aligned} Φ_{1} & = t - 1, & Φ_{2} & = t + 1, & Φ_{3} & = t^{2} + t + 1, \\ Φ_{4} & = t^{2} + 1, & Φ_{5} & = t^{4} + t^{3} + t^{2} + t + 1, & Φ_{6} & = t^{2} - t + 1, \\ Φ_{7} & = t^{6} + t^{5} + t^{4} + t^{3} + t^{2} + t + 1, & Φ_{8} & = t^{4} + 1, & Φ_{9} & = t^{6} + t^{3} + 1, \\ Φ_{10} & = t^{4} - t^{3} + t^{2} - t + 1, & Φ_{11} & = \sum_{k = 0}^{10} t^{k}, & Φ_{12} & = t^{4} - t^{2} + 1, \end{aligned}$

each monic in $Z [t]$ , with degrees

$1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4$

matching $φ (1), \dots, φ (12)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Facts & Assumptions

Given: The recursion $Φ_{1} = t - 1$ and $Φ_{n} = (t^{n} - 1) / \prod_{d ∣ n, d < n} Φ_{d}$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); and the elementary identity $(t^{a} - 1) (t^{a (b - 1)} + \dots + t^{a} + 1) = t^{a b} - 1$ for $a, b \geq 1$ .

[L1]

For every $n \geq 1$ , $Φ_{n}$ is monic in $Z [t]$ with $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ and $\deg Φ_{n} = φ (n)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L2]

For a prime $p$ and $r \geq 1$ , $Φ_{p^{r}} = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}}$ ( $Φ_{p^{r}} (t) = \sum_{k < p} t^{k p^{r - 1}}$ , and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$ ).

[L3]

$Z [t]$ is an integral domain (A polynomial ring over an integral domain is an integral domain), so a nonzero factor may be cancelled; and division by a monic polynomial has a unique quotient and remainder (Division by a monic polynomial over a commutative ring).

[L4]

$φ (1) = 1$ and $φ (p) = p - 1$ for a prime $p$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ); $φ (p^{k}) = p^{k} - p^{k - 1}$ (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ); and for $n \geq 1$ with prime divisors $p_{0}, \dots, p_{r - 1}$ and $k_{i} = v_{p_{i}} (n)$ , $φ (n) = \prod_{i < r} (p_{i}^{k_{i}} - p_{i}^{k_{i} - 1})$ (Euler's product formula $φ (n) = n \prod_{p ∣ n} (1 - 1 / p) = \prod_{p^{k} ∥ n} (p^{k} - p^{k - 1})$ for $n \geq 1$ , stated through a finite injective list of its prime divisors).

Verification

technique · direct

1.1

$Φ_{1} = t - 1$ is the base clause of the recursion, of degree $1 = φ (1)$ by [L4].

L4given

1.2

The prime powers among $2, \dots, 12$ are $2, 3, 4, 5, 7, 8, 9, 11$ , and [L2] gives their cyclotomic polynomials directly: $Φ_{2} = 1 + t$ , $Φ_{3} = 1 + t + t^{2}$ , $Φ_{4} = 1 + t^{2}$ , $Φ_{5} = 1 + t + t^{2} + t^{3} + t^{4}$ , $Φ_{7} = \sum_{k = 0}^{6} t^{k}$ , $Φ_{8} = 1 + t^{4}$ , $Φ_{9} = 1 + t^{3} + t^{6}$ and $Φ_{11} = \sum_{k = 0}^{10} t^{k}$ .

2.1

For $n = 6$ : the positive divisors of $6$ are $1, 2, 3, 6$ and those of $3$ are $1, 3$ , so [L1] gives $Φ_{1} Φ_{2} Φ_{3} Φ_{6} = t^{6} - 1$ and $Φ_{1} Φ_{3} = t^{3} - 1$ ; dividing and using the given identity with $a = 3$ , $b = 2$ yields $Φ_{2} Φ_{6} = (t^{6} - 1) / (t^{3} - 1) = t^{3} + 1$ . Since $(t + 1) (t^{2} - t + 1) = t^{3} + 1$ and $Φ_{2} = t + 1$ is nonzero, cancelling in $Z [t]$ by [L3] gives $Φ_{6} = t^{2} - t + 1$ .

step 1.2L1L3given

2.2

For $n = 10$ : the divisors of $10$ are $1, 2, 5, 10$ and those of $5$ are $1, 5$ , so $Φ_{2} Φ_{10} = (t^{10} - 1) / (t^{5} - 1) = t^{5} + 1$ by [L1] and the given identity with $a = 5$ , $b = 2$ ; and $(t + 1) (t^{4} - t^{3} + t^{2} - t + 1) = t^{5} + 1$ , so cancelling gives $Φ_{10} = t^{4} - t^{3} + t^{2} - t + 1$ .

step 1.2L1L3given

3.1

For $n = 12$ : the divisors of $12$ are $1, 2, 3, 4, 6, 12$ and those of $6$ are $1, 2, 3, 6$ , so $Φ_{4} Φ_{12} = (t^{12} - 1) / (t^{6} - 1) = t^{6} + 1$ by [L1] and the given identity with $a = 6$ , $b = 2$ ; and $(t^{2} + 1) (t^{4} - t^{2} + 1) = t^{6} + 1$ with $Φ_{4} = t^{2} + 1$ , so cancelling gives $Φ_{12} = t^{4} - t^{2} + 1$ .

step 1.2step 2.1L1L3given

4.1

The degrees read off the displayed polynomials are $1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4$ . By [L4] these are $φ (1) = 1$ , $φ (2) = 1$ , $φ (3) = 2$ , $φ (4) = 2^{2} - 2 = 2$ , $φ (5) = 4$ , $φ (6) = (2 - 1) (3 - 1) = 2$ , $φ (7) = 6$ , $φ (8) = 2^{3} - 2^{2} = 4$ , $φ (9) = 3^{2} - 3 = 6$ , $φ (10) = (2 - 1) (5 - 1) = 4$ , $φ (11) = 10$ and $φ (12) = (2^{2} - 2) (3 - 1) = 4$ , so every degree matches [L1].

step 1.1step 1.2step 2.1step 2.2step 3.1L1L4∎

Remarks

Every division in the recursion is exact and stays over $Z$ . That is not visible from the table and is not a coincidence of small $n$ : it is The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , and it is what makes the recursion a definition rather than a computation that might fail.

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$Φ_{7} (t + 1)$ is Eisenstein at seven

Example

The translated seventh cyclotomic polynomial is

$Φ_{7} (t + 1) = t^{6} + 7 t^{5} + 21 t^{4} + 35 t^{3} + 35 t^{2} + 21 t + 7.$

Its leading coefficient is $1$ , every other coefficient is divisible by $7$ , and its constant term $7$ is not divisible by $7^{2}$ . So it satisfies Eisenstein's criterion at the prime $7$ , and therefore $Φ_{7}$ is irreducible over $Q$ .

Facts & Assumptions

Given: The prime-power cyclotomic formula and the polynomial $Φ_{7}$ .

[L1]

For a prime $p$ and $r \geq 1$ , $Φ_{p^{r}} (t) = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}},$ and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$ ( $Φ_{p^{r}} (t) = \sum_{k < p} t^{k p^{r - 1}}$ , and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$ ).

[L2]

Eisenstein's criterion: if a prime $p$ divides every non-leading coefficient of a polynomial in $Z [t]$ , does not divide the leading coefficient, and $p^{2}$ does not divide the constant term, then the polynomial is irreducible over $Q$ (Eisenstein criterion over the integers).

Verification

technique · direct

1.1

Applying [L1] at $p = 7$ and $r = 1$ gives $Φ_{7} (t) = 1 + t + t^{2} + t^{3} + t^{4} + t^{5} + t^{6}$ .

2.1

Therefore $Φ_{7} (t + 1) = \frac{(t + 1)^{7} - 1}{t} = t^{6} + 7 t^{5} + 21 t^{4} + 35 t^{3} + 35 t^{2} + 21 t + 7,$ by the binomial theorem.

step 1.1algebra

3.1

In the polynomial of step 2.1 the leading coefficient is $1$ , the remaining coefficients $7, 21, 35, 35, 21, 7$ are all divisible by $7$ , and the constant term $7$ is not divisible by $49$ ; so [L2] applies at the prime $7$ .

step 2.1L2algebra

4.1

Hence $Φ_{7} (t + 1)$ is irreducible over $Q$ , and this is exactly the degree-one prime-power case of [L1].

step 3.1L1∎

Remarks

Why this example matters later. The explicit coefficients are what the counterexample page uses when it says the Eisenstein route already proves irreducibility for prime-power cyclotomic polynomials before the general Dedekind argument is built.

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$Φ_{5}$ has four roots in $F_{11}$

Example

Over $F_{11}$ the fifth cyclotomic polynomial

$Φ_{5} (t) = t^{4} + t^{3} + t^{2} + t + 1$

splits into four distinct linear factors:

$Φ_{5} (t) = (t - 3) (t - 4) (t - 5) (t - 9) .$

The four roots $3, 4, 5, 9$ are exactly the primitive fifth roots of unity in $F_{11}$ .

Facts & Assumptions

Given: The field $F_{11}$ and the polynomial $Φ_{5} (t) = t^{4} + t^{3} + t^{2} + t + 1$ .

[L1]

If $\gcd (n, q) = 1$ , the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ).

[L2]

For $\gcd (n, q) = 1$ , the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ , so the extension degree is the order of $[q]$ modulo $n$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ ).

Verification

technique · direct

1.1

In $(Z / 5)^{\times}$ one has $[11] = [1]$ , so the order of $[11]$ modulo $5$ is $1$ . Hence [L1] and [L2] say every irreducible factor of $Φ_{5}$ over $F_{11}$ is linear, and the factors are distinct.

L1L2algebra

1.2

The powers of $3$ in $F_{11}^{\times}$ are $3^{2} = 9$ , $3^{3} = 5$ , $3^{4} = 4$ and $3^{5} = 1$ , so the four nontrivial fifth roots of unity in $F_{11}$ are $3, 9, 5, 4$ .

givenalgebra

2.1

Each of $3, 4, 5, 9$ is therefore a root of $t^{5} - 1$ and is not $1$ , so each is a root of $Φ_{5} (t) = (t^{5} - 1) / (t - 1)$ . Since $Φ_{5}$ is monic of degree $4$ , it follows that $Φ_{5} (t) = (t - 3) (t - 4) (t - 5) (t - 9) .$

step 1.2algebra

3.1

The roots $3, 4, 5, 9$ all have multiplicative order $5$ by step 1.2, so they are exactly the primitive fifth roots of unity in $F_{11}$ .

step 1.2algebra∎

Remarks

This is the order-one case of the finite-field factorisation theorem. When $[q]$ has order one modulo $n$ , every irreducible factor has degree one and the whole cyclotomic polynomial splits over the base field.

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$Φ_{7}$ factors over $F_{2}$ into the two monic irreducible cubics

Example

Over $F_{2}$ one has

$Φ_{7} (t) = t^{6} + t^{5} + t^{4} + t^{3} + t^{2} + t + 1 = (t^{3} + t + 1) (t^{3} + t^{2} + 1) .$

These are the two monic irreducible cubic factors, and over a splitting field the two factors correspond to the Frobenius orbits

${ζ, ζ^{2}, ζ^{4}} and {ζ^{3}, ζ^{6}, ζ^{5}}$

of a primitive seventh root of unity $ζ$ .

Facts & Assumptions

Given: The field $F_{2}$ and a primitive seventh root of unity $ζ$ in a splitting field of $Φ_{7}$ .

[L1]

[L2]

For $\gcd (n, q) = 1$ , the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ ).

[L3]

The roots of a monic irreducible polynomial of degree $d$ over $F_{q}$ are one Frobenius orbit $α, α^{q}, \dots, α^{q^{d - 1}}$ (A monic irreducible of degree $d$ over $F_{q}$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$ ).

Verification

technique · direct

1.1

In $(Z / 7)^{\times}$ the class $[2]$ has order $3$ , since $2^{3} = 8 \equiv 1 (m o d 7)$ and $2 \equiv̸ 1$ , $2^{2} = 4 \equiv̸ 1 (m o d 7)$ . So [L1] and [L2] say every irreducible factor of $Φ_{7}$ over $F_{2}$ is a distinct cubic.

L1L2algebra

2.1

The product of the two monic cubics is $(t^{3} + t + 1) (t^{3} + t^{2} + 1) = t^{6} + t^{5} + t^{4} + t^{3} + t^{2} + t + 1 = Φ_{7} (t)$ in $F_{2} [t]$ . Since both factors are monic of degree $3$ , step 1.1 makes them the two irreducible factors of $Φ_{7}$ .

step 1.1algebra

3.1

Frobenius acts by $ζ \mapsto ζ^{2}$ , so the orbit of $ζ$ is ${ζ, ζ^{2}, ζ^{4}}$ because $ζ^{8} = ζ$ , and the orbit of $ζ^{3}$ is ${ζ^{3}, ζ^{6}, ζ^{5}}$ because $(ζ^{3})^{2} = ζ^{6}$ , $(ζ^{6})^{2} = ζ^{12} = ζ^{5}$ and $(ζ^{5})^{2} = ζ^{10} = ζ^{3}$ . These are the two size-three Frobenius orbits of primitive seventh roots, and [L3] identifies them as the respective root sets of the two irreducible cubic factors from step 2.1.

step 1.1step 2.1L3algebra∎

Remarks

This is the degree-three case of the theorem, not a coincidence of cubics. The orbit size and the factor degree are both the order of $[2]$ modulo $7$ .

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$Gal (Q (ζ_{12}) / Q) ≅ (Z / 12)^{\times}$ and its three quadratic subfields

Example

Let $ζ : = ζ_{12}$ . Then

$Gal (Q (ζ) / Q) ≅ (Z / 12)^{\times} = {[1], [5], [7], [11]},$

every nonidentity element has order two, and the three order-two subgroups have fixed fields

$Q (i), Q (\sqrt{3}), Q (\sqrt{- 3}) .$

So $Q (ζ_{12})$ has exactly three quadratic intermediate fields.

Facts & Assumptions

Given: A primitive twelfth root of unity $ζ = ζ_{12}$ and the automorphisms $σ_{a} (ζ) = ζ^{a}$ for $[a] \in (Z / 12)^{\times}$ .

[L1]

$[Q (ζ_{12}) : Q] = φ (12) = 4$ and $Gal (Q (ζ_{12}) / Q) ≅ (Z / 12)^{\times}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L2]

For a finite Galois extension $E / F$ , subgroups of $Gal (E / F)$ correspond bijectively to intermediate fields, and the fixed field of a subgroup $H$ has degree $[E^{H} : F] = [Gal (E / F) : H]$ (The fundamental theorem of finite Galois theory).

Verification

technique · direct

1.1

The units modulo $12$ are $[1], [5], [7], [11]$ , since these are exactly the residue classes in ${1, \dots, 11}$ coprime to $12$ . Their squares are $[25] = [1]$ , $[49] = [1]$ and $[121] = [1]$ , so every nonidentity element has order two.

L1algebra

2.1

Therefore $(Z / 12)^{\times}$ is the Klein four-group, with three order-two subgroups: $H_{i} = {[1], [5]}, H_{- 3} = {[1], [7]}, H_{3} = {[1], [11]} .$ By [L1] and [L2], each fixed field has degree $2$ over $Q$ .

step 1.1L1L2

3.1

The subgroup $H_{i}$ fixes $i = ζ^{3}$ , because $σ_{5} (ζ^{3}) = ζ^{15} = ζ^{3}$ . Since $i \notin Q$ and the fixed field has degree $2$ by step 2.1, that fixed field is $Q (i)$ .

step 2.1algebra

3.2

The subgroup $H_{- 3}$ fixes $2 ζ^{2} - 1$ , because $σ_{7} (ζ^{2}) = ζ^{14} = ζ^{2}$ ; and $(2 ζ^{2} - 1)^{2} = 4 ζ^{4} - 4 ζ^{2} + 1 = - 3,$ since $ζ^{2}$ is a root of $t^{2} - t + 1$ . So the fixed field contains $\sqrt{- 3}$ , and again step 2.1 makes it exactly $Q (\sqrt{- 3})$ .

step 2.1algebra

3.3

The subgroup $H_{3}$ fixes $ζ + ζ^{- 1}$ , because $σ_{11}$ is complex conjugation. Moreover $(ζ + ζ^{- 1})^{2} = ζ^{2} + 2 + ζ^{- 2} = 3,$ since $ζ^{2} + ζ^{- 2} = 1$ . So the fixed field contains $\sqrt{3}$ , and step 2.1 makes it exactly $Q (\sqrt{3})$ .

step 2.1algebra

4.1

The three order-two subgroups of step 2.1 therefore yield the three quadratic intermediate fields $Q (i)$ , $Q (\sqrt{- 3})$ and $Q (\sqrt{3})$ , and there are no others because [L2] gives a bijection between subgroups and intermediate fields.

step 2.1step 3.1step 3.2step 3.3L2∎

Remarks

The same phenomenon already occurs at order eight. The field $Q (ζ_{8})$ also has Klein four Galois group and three quadratic subfields. The order-twelve calculation is useful because its three fields are the familiar $Q (i)$ , $Q (\sqrt{3})$ and $Q (\sqrt{- 3})$ .

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In characteristic three, $t^{3} - 1 = (t - 1)^{3}$ and $μ_{6}$ coincides with $μ_{2}$

Example

Let $K$ be a field of characteristic $3$ . Then

$t^{3} - 1 = (t - 1)^{3}, μ_{3} (K) = {1}, μ_{6} (K) = μ_{2} (K) = {1, - 1},$

so $t^{6} - 1$ has only the two distinct roots $1$ and $- 1$ rather than six.

Facts & Assumptions

Given: A field $K$ of characteristic $3$ .

[L1]

For a fixed integer $k \geq 1$ , in characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ).

[L2]

$μ_{n} (K) = {x \in K : x^{n} = 1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

Verification

technique · direct

1.1

Applying [L1] at $p = 3$ and $k = 1$ gives $t^{3} - 1 = (t - 1)^{3}$ and $μ_{3} (K) = {1}$ .

1.2

Since $(1)^{2} = 1$ and $(- 1)^{2} = 1$ , one has ${1, - 1} \subseteq μ_{2} (K)$ . Conversely, if $x \in μ_{2} (K)$ then $x^{2} = 1$ , so $x^{2} - 1 = (x - 1) (x + 1) = 0$ and therefore $x = 1$ or $x = - 1$ in the field $K$ ; hence $μ_{2} (K) = {1, - 1}$ .

L2algebra

2.1

If $x \in μ_{6} (K)$ then $(x^{2})^{3} = x^{6} = 1$ , so $x^{2} \in μ_{3} (K)$ by [L2]; step 1.1 gives $x^{2} = 1$ , hence $x \in μ_{2} (K)$ by [L2]. Thus $μ_{6} (K) \subseteq μ_{2} (K)$ .

step 1.1step 1.2L2

2.2

Conversely, if $x \in μ_{2} (K)$ then $x^{6} = (x^{2})^{3} = 1$ , so $x \in μ_{6} (K)$ . Therefore $μ_{6} (K) = μ_{2} (K) = {1, - 1}$ .

step 1.2L2algebra

3.1

Using step 1.1, $t^{6} - 1 = (t^{3} - 1) (t^{3} + 1) = (t - 1)^{3} (t + 1)^{3},$ so its only distinct roots are $1$ and $- 1$ , exactly the two elements of step 2.2.

step 1.1step 2.2algebra∎

Remarks

This is why the characteristic hypothesis is load-bearing. The statement " $∣ μ_{n} ∣ = n$ " fails here for two different reasons at once: the polynomial $t^{3} - 1$ is inseparable, and the extra cube roots never appear even after passing to a splitting field because the splitting field is already the base field.

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$F_{3} (μ_{5}) \cap F_{3} (μ_{7})$ is larger than $F_{3}$ although five and seven are coprime

Statement refuted

That the rational intersection theorem $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ holds over every base field: that for every field $K$ and all positive integers $m, n$ with the characteristic of $K$ dividing neither,

$K (μ_{m}) \cap K (μ_{n}) = K (μ_{\gcd (m, n)}) .$

The witness below takes $K = F_{3}$ , $m = 5$ and $n = 7$ , and realizes both splitting fields inside one fixed field $Ω$ of order $3^{12}$ . Since $\gcd (5, 7) = 1$ , the right-hand side is $K (μ_{1}) = F_{3}$ , while the intersection on the left is the common subfield $F_{9} \subseteq Ω$ .

Facts & Assumptions

Given: The base field $K = F_{3}$ and a field $Ω$ of order $3^{12}$ , which exists by For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists. The two cyclotomic splitting fields will be identified with their base-field-isomorphic copies inside $Ω$ .

[L1]

For $\gcd (n, q) = 1$ , the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ , so the degree of $F_{q} (μ_{n}) / F_{q}$ is the order of $[q]$ modulo $n$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ ).

[L2]

The intermediate fields of $F_{q^{N}} / F_{q}$ are exactly the $F_{q^{d}}$ for the positive divisors $d$ of $N$ , one for each divisor, with $F_{q^{d}} \subseteq F_{q^{e}}$ exactly when $d ∣ e$ (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).

[L3]

Over $Q$ one has $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ ( $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ ).

[L4]

$K (μ_{r})$ is the splitting field of $t^{r} - 1$ over $K$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[L5]

Finite fields of the same order are isomorphic by an isomorphism fixing their common prime field (Finite fields of the same order are isomorphic).

Counterexample

technique · direct

1.1

In $(Z / 5)^{\times}$ , the class $[3]$ has order $4$ , since $3^{4} = 81 \equiv 1 (m o d 5)$ and no smaller positive power of $3$ is congruent to $1$ modulo $5$ . So [L1] gives $[F_{3} (μ_{5}) : F_{3}] = 4$ , hence this splitting field has order $3^{4}$ and [L5] lets us identify it over $F_{3}$ with the unique subfield $F_{3^{4}}$ of $Ω$ .

L1L2L4L5algebra

1.2

In $(Z / 7)^{\times}$ , the powers of $[3]$ are $[3], [2], [6], [4], [5], [1]$ , so $[3]$ has order $6$ . Thus [L1] gives $[F_{3} (μ_{7}) : F_{3}] = 6$ , hence this splitting field has order $3^{6}$ and [L5] lets us identify it over $F_{3}$ with the unique subfield $F_{3^{6}}$ of $Ω$ .

L1L2L4L5algebra

2.1

Under the fixed identifications of steps 1.1 and 1.2, both $F_{3^{4}}$ and $F_{3^{6}}$ are subfields of $Ω = F_{3^{12}}$ by [L2], since $4 ∣ 12$ and $6 ∣ 12$ . Their intersection is then an intermediate field of $Ω / F_{3}$ , so by [L2] it is $F_{3^{d}}$ for some divisor $d$ of $12$ . Because the intersection lies in both fields, [L2] gives $d ∣ 4$ and $d ∣ 6$ , hence $d ∣ 2$ ; and since $F_{3^{2}}$ lies in both fields, [L2] gives $2 ∣ d$ . Therefore $d = 2$ and $F_{3} (μ_{5}) \cap F_{3} (μ_{7}) = F_{3^{2}} = F_{9} .$

step 1.1step 1.2L2

3.1

Since $\gcd (5, 7) = 1$ , the right-hand side of the refuted identity is $K (μ_{1})$ , which is just $K = F_{3}$ because $t - 1$ already splits over $K$ . So the claimed equality would read $F_{9} = F_{3}$ , which is false.

step 2.1L4algebra

4.1

The refuted statement therefore fails over the base field $F_{3}$ , even though the rational theorem [L3] is true.

step 3.1L3∎

Remarks

Why the rational hypothesis matters. Over finite fields the intersection is controlled by the gcd of the extension degrees, not by the gcd of the orders of the roots of unity.

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

FALSE: every cyclotomic polynomial has all coefficients in ${- 1, 0, 1}$

Statement

False claim. Every coefficient of every cyclotomic polynomial $Φ_{n} \in Z [t]$ lies in ${- 1, 0, 1}$ .

The failure first appears at $n = 105$ : the coefficient of $t^{7}$ in $Φ_{105} (t)$ is $- 2$ .

Facts & Assumptions

Given: The cyclotomic recursion $\prod_{d ∣ n} Φ_{d} (t) = t^{n} - 1$ for $n \geq 1$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ).

[L1]

For every $n \geq 1$ , $Φ_{n}$ is a monic polynomial in $Z [t]$ and the displayed divisor recursion holds (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ ).

Refutation

technique · direct

1.1

Running the divisor recursion for $n = 105 = 3 \cdot 5 \cdot 7$ and truncating modulo $t^{8}$ gives $Φ_{105} (t) \equiv \frac{(1 - t^{3}) (1 - t^{5}) (1 - t^{7})}{1 - t} \equiv 1 + t + t^{2} - t^{5} - t^{6} - 2 t^{7} (m o d t^{8}) .$ The first congruence is the defining recursion with every factor of degree at least $15$ dropped modulo $t^{8}$ , and the second comes from expanding $(1 - t)^{- 1} = 1 + t + t^{2} + t^{3} + t^{4} + t^{5} + t^{6} + t^{7} (m o d t^{8})$ .

givenL1algebra

2.1

Step 1.1 shows that the coefficient of $t^{7}$ in $Φ_{105} (t)$ is $- 2$ , and $- 2 \notin {- 1, 0, 1}$ . So the false claim fails.

step 1.1algebra∎

Remarks

Why this is a real pattern and not a silly claim. For many small values of $n$ the coefficients do lie in ${- 1, 0, 1}$ , so the first counterexample is not visually obvious from the recursion alone.

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FALSE: $Φ_{n}$ is irreducible over every field

Statement

False claim. For every field $K$ and every $n \geq 1$ with the characteristic of $K$ not dividing $n$ , the image of $Φ_{n}$ in $K [t]$ is irreducible.

What is true is different in two directions: over $Q$ every $Φ_{n}$ is irreducible, but over a finite field the factor degree is governed by the order of the Frobenius class modulo $n$ .

Facts & Assumptions

Given: The rational irreducibility theorem and the finite-field factorisation theorem.

[L1]

For every $n \geq 1$ , the cyclotomic polynomial $Φ_{n}$ is irreducible in $Q [t]$ ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ).

[L2]

Refutation

technique · direct

1.1

In $(Z / 5)^{\times}$ one has $[11] = [1]$ , so the order of $[11]$ modulo $5$ is $1$ . Therefore [L2] says that over $F_{11}$ every irreducible factor of $Φ_{5}$ has degree $1$ .

L2algebra

2.1

Hence the reduction of $Φ_{5}$ in $F_{11} [t]$ is a product of distinct linear factors, so it is reducible there. This contradicts the false claim.

step 1.1algebra

3.1

The contradiction does not touch [L1]: irreducibility over $Q$ is a theorem, but it does not persist over arbitrary base fields.

step 2.1L1∎

Remarks

The finite-field theorem is the correct replacement. The question over $F_{q}$ is not "irreducible or not?" in the abstract, but "what is the order of $[q]$ modulo $n$ ?"

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FALSE: $μ_{n} (K)$ has $n$ elements in every field $K$

Statement

False claim. For every field $K$ and every $n \geq 1$ , the group $μ_{n} (K)$ of $n$ -th roots of unity in $K$ has exactly $n$ elements.

The witness below shows two different failures: over $Q$ there is no primitive cube root of unity in the field at all, while in characteristic $3$ the equation $x^{3} = 1$ is inseparable and has only one root.

Facts & Assumptions

Given: The groups $μ_{n} (K)$ of roots of unity.

[L1]

$μ_{n} (K)$ is cyclic of order dividing $n$ , and it has a primitive $n$ -th root of unity exactly when its order is $n$ ( $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L2]

$[Q (ζ_{3}) : Q] = φ (3) = 2$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L3]

In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ).

Refutation

technique · direct

1.1

If $μ_{3} (Q)$ had three elements, then [L1] would give a primitive cube root of unity $ζ_{3}$ in $Q$ . But then $Q (ζ_{3}) = Q$ , contradicting [L2], which says this extension has degree $2$ . So $μ_{3} (Q)$ does not have three elements.

L1L2

1.2

If $K$ has characteristic $3$ , then [L3] gives $μ_{3} (K) = {1}$ , so again $μ_{3} (K)$ does not have three elements.

2.1

The false claim fails already at $n = 3$ , both over $Q$ and over every field of characteristic $3$ .

step 1.1step 1.2∎

Remarks

The two failures have different causes. Over $Q$ the polynomial $t^{3} - 1$ is separable but its nontrivial roots lie in a quadratic extension; in characteristic $3$ the polynomial itself collapses to $(t - 1)^{3}$ .

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FALSE: every finite abelian group is $Gal (Q (μ_{n}) / Q)$ for some $n$

Statement

False claim. For every finite abelian group $G$ there is an $n \geq 1$ with

$Gal (Q (μ_{n}) / Q) ≅ G .$

The obstruction is visible already at the level of cardinality: the cyclic group $C_{3}$ occurs as a Galois group over $Q$ , but never as the Galois group of a cyclotomic field.

Facts & Assumptions

Given: Cyclotomic Galois groups and the theorem realising finite abelian groups over $Q$ .

[L1]

$Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , so its order is $φ (n)$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

Every finite abelian group is the Galois group of some finite Galois extension of $Q$ (Every finite abelian group is the Galois group of some finite Galois extension of $Q$ ).

[L3]

In a finite group, the order of every subgroup divides the order of the group (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Refutation

technique · direct

1.1

For $n = 1$ and $n = 2$ , the group $(Z / n)^{\times}$ is trivial, so $φ (1) = φ (2) = 1$ .

L1algebra

1.2

If $n \geq 3$ , then the unit class $[- 1]$ in $(Z / n)^{\times}$ has order $2$ : one has $[- 1]^{2} = [1]$ , and $[- 1] \neq [1]$ because otherwise $n$ would divide $2$ , contrary to $n \geq 3$ . Therefore [L3] makes the group order $φ (n) = ∣ (Z / n)^{\times} ∣$ even.

L1L3algebra

2.1

Steps 1.1 and 1.2 show that $φ (n)$ is never $3$ . So no cyclotomic field $Q (μ_{n})$ has Galois group of order $3$ , and in particular none has Galois group isomorphic to $C_{3}$ .

step 1.1step 1.2L1

3.1

By [L2], however, some finite Galois extension of $Q$ does have Galois group $C_{3}$ . Hence the false claim fails.

step 2.1L2∎

Remarks

What the true theorem says instead. The proved result is that every finite abelian group is the Galois group of a subfield of a cyclotomic field, not of the cyclotomic field itself.

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A degree-three Galois extension of $Q$ inside $Q (ζ_{7})$

Example

Let $ζ : = ζ_{7}$ . Then the fixed field of the unique order-two subgroup of $Gal (Q (ζ) / Q)$ is

$Q (ζ + ζ^{- 1}),$

a degree-three Galois extension of $Q$ with cyclic Galois group, and the element $ζ + ζ^{- 1}$ has minimal polynomial

$t^{3} + t^{2} - 2 t - 1.$

Facts & Assumptions

Given: A primitive seventh root of unity $ζ = ζ_{7}$ .

[L1]

$Gal (Q (ζ_{7}) / Q) ≅ (Z / 7)^{\times}$ and has order $φ (7) = 6$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L2]

A finite cyclic group has exactly one subgroup of each order dividing its own (A finite cyclic group has exactly one subgroup of each order dividing its own).

[L3]

For a finite Galois extension, subgroups correspond to intermediate fields, and the fixed field of a subgroup $H$ has degree equal to the subgroup index (The fundamental theorem of finite Galois theory).

[L4]

Every finite subgroup of the unit group of an integral domain is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic).

[L5]

Under the finite Galois correspondence, a normal subgroup $H$ has a Galois fixed field and restriction gives $Gal (F / Q) ≅ G / H$ (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence).

[L6]

Every finite group of prime order is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it).

Verification

technique · direct

1.1

By [L1] the Galois group of $Q (ζ) / Q$ is isomorphic to the finite subgroup $(Z / 7)^{\times}$ of the unit group of the field $Z / 7$ , so [L4] makes it cyclic; its order is $6$ . Thus [L2] gives a unique subgroup $H$ of order $2$ , and [L3] makes its fixed field $F$ have degree $[F : Q] = 6 / 2 = 3$ .

L1L2L3L4

2.1

The subgroup $H$ is generated by the class $[- 1]$ , so it acts by complex conjugation. Therefore $ζ + ζ^{- 1}$ is fixed by $H$ and lies in $F$ .

step 1.1algebra

3.1

Put $x : = ζ + ζ^{- 1}$ . Then $ζ^{2} + ζ^{- 2} = x^{2} - 2, ζ^{3} + ζ^{- 3} = x^{3} - 3 x .$ Since $1 + ζ + ζ^{2} + ζ^{3} + ζ^{4} + ζ^{5} + ζ^{6} = 0$ , dividing by $ζ^{3}$ gives $1 + (ζ + ζ^{- 1}) + (ζ^{2} + ζ^{- 2}) + (ζ^{3} + ζ^{- 3}) = 0.$ Substituting the expressions above yields $x^{3} + x^{2} - 2 x - 1 = 0.$

step 2.1algebra

4.1

The element $x$ is not rational: if it were, then $ζ$ would satisfy the quadratic polynomial $t^{2} - x t + 1 \in Q [t]$ , which would force $[Q (ζ) : Q] \leq 2$ , contradicting [L1]. Since $x \in F$ , the prime degree $[F : Q] = 3$ from step 1.1 leaves only the subfields $Q$ and $F$ , so $Q (x) = F$ . Therefore the minimal polynomial of $x$ has degree $3$ , and the cubic from step 3.1 is that minimal polynomial.

step 1.1step 3.1L1

5.1

The ambient Galois group is cyclic and hence abelian, so $H$ is normal. By [L5], the fixed field $F / Q$ is Galois and $Gal (F / Q)$ is isomorphic to the quotient by $H$ , which has order $3$ . The group is cyclic by [L6], so $F / Q$ is a cyclic cubic Galois extension.

step 1.1L5L6algebra∎

Remarks

This is the smallest nontrivial case of the subfield theorem. The subgroup lattice of $(Z / 7)^{\times}$ has one index-two subgroup, and the fixed field is already visible through the real element $ζ + ζ^{- 1}$ .

Finite Fields and Cyclotomic Extensions

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

A finite cyclic group has exactly one subgroup of each order dividing its own

Statement

Facts & Assumptions

Proof

Remarks

A cyclic group of order nn has exactly φ(n)φ(n) generators

Statement

Facts & Assumptions

Proof

Remarks

The relative Frobenius x↦xqx↦xq of an extension of finite fields

Definition

Remarks

The elements of a finite extension fixed by the qq-power map are exactly the base field

Statement

Facts & Assumptions

Proof

Remarks

For a degree-nn extension of a field of order qq, the qq-power map has order exactly nn

Statement

Facts & Assumptions

Proof

A finite extension of a finite field of order qq is Galois with cyclic Galois group generated by x↦xqx↦xq

Statement

Facts & Assumptions

Proof

Remarks

The intermediate fields of Fqn/FqFqn​/Fq​ are the FqdFqd​, one for each positive divisor dd of nn

Statement

Facts & Assumptions

Proof

Remarks

The Galois description of the subfields of a finite field and the elementary divisibility criterion agree

Statement

A monic irreducible of degree dd over FqFq​ has the dd distinct roots α,αq,…,αqd−1α,αq,…,αqd−1

Statement

Facts & Assumptions

Proof

Remarks

∑d∣nd Nq(d)=qn∑d∣n​dNq​(d)=qn for the counts Nq(d)Nq​(d) of monic irreducibles of degree dd over FqFq​

Statement

Facts & Assumptions

Proof

Remarks

Normal bases of a finite Galois extension

Definition

A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial

Statement

Facts & Assumptions

Proof

Remarks

For a finite Galois extension, (αj)(αj​) is a base-field basis exactly when the matrix (σiαj)(σi​αj​) is invertible

Statement

Facts & Assumptions

Proof

Remarks

Over an infinite base field, no nonzero polynomial vanishes at the conjugate tuple of every element

Statement

Facts & Assumptions

Proof

Remarks

Every finite Galois extension of an infinite field has a normal basis

Statement

Facts & Assumptions

Proof

Remarks

Every finite cyclic extension has a normal basis

Statement

Facts & Assumptions

Proof

Remarks

Every finite Galois extension has a normal basis

Statement

Facts & Assumptions

Proof

A cyclic group of order $n$ has exactly $φ (n)$ generators

The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields

The elements of a finite extension fixed by the $q$ -power map are exactly the base field

For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$

A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$

The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$

A monic irreducible of degree $d$ over $F_{q}$ has the $d$ distinct roots $α, α^{q}, \dots, α^{q^{d - 1}}$

$\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$

For a finite Galois extension, $(α_{j})$ is a base-field basis exactly when the matrix $(σ_{i} α_{j})$ is invertible

The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity

$μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$

$t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity

In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$

The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$

$K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$

For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$

The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$

The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$

$Φ_{1} (0) = - 1$ and $Φ_{n} (0) = 1$ for $n \geq 2$

Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity

$Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto

$Φ_{p^{r}} (t) = \sum_{k < p} t^{k p^{r - 1}}$ , and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$

If $p$ is a prime not dividing $n$ , a rational minimal polynomial of a primitive $n$ -th root of unity also kills its $p$ -th power

$Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$