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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

Statement

Let $n \geq 1$ , let $E$ be a splitting field of $t^{n} - 1$ over $Q$ , 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), let $f \in Q [t]$ be the minimal polynomial of $ζ$ over $Q$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), and let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) with $p ∤ n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). Then

$f (ζ^{p}) = 0.$

Facts & Assumptions

Given: The data of the statement; $Q$ 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 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), which divides no $n \geq 1$ , so $t^{n} - 1$ is separable over $Q$ and $μ_{n} (E)$ is cyclic of order $n$ generated by $ζ$ ( $t^{n} - 1$ 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, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ); $Φ_{n}$ denotes the cyclotomic polynomial (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) and also its images in $Q [t]$ and in $(Z / p) [t]$ , where reduction is the ring homomorphism of Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism and $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L1]

The roots of $Φ_{n}$ in $E$ are 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); $Φ_{n}$ is monic in $Z [t]$ of degree $φ (n)$ and divides $t^{n} - 1$ there (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 $a$ algebraic over a field $K$ 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).

[L3]

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

[L4]

For nonzero $u, v \in Z [x]$ , $cont (u v) = cont (u) cont (v)$ (The product of primitive integer polynomials is primitive, and contents multiply); content is the nonnegative gcd of the coefficients and primitive means content $1$ (Content and primitive integer polynomials); a nonzero integer polynomial is primitive exactly when no prime divides all its coefficients (Content is the positive common divisor of the coefficients divisible by every common divisor).

[L5]

In a field $L$ of characteristic $p > 0$ the map $x \mapsto x^{p}$ is a field endomorphism (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields); and every $c$ in a field with $p$ elements satisfies $c^{p} = c$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L6]

$t^{n} - 1$ is separable over $Z / p$ when $p ∤ n$ , that is it has no repeated root in any extension field ( $t^{n} - 1$ 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, Repeated roots in extension fields and separable polynomials); every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

[L7]

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

[L8]

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).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $f (ζ^{p}) \neq 0$ .

assume-contra

1.2

$ζ^{p}$ is again a primitive $n$ -th root of unity: $\gcd (p, n) = 1$ because $p$ is prime and $p ∤ n$ , so [L8] makes $ζ^{p}$ a generator of $μ_{n} (E)$ , of order $n$ . Hence $Φ_{n} (ζ^{p}) = 0$ by [L1].

L1L8given

1.3

Since $Φ_{n} (ζ) = 0$ by [L1], [L2] gives $f ∣ Φ_{n}$ in $Q [t]$ ; write $Φ_{n} = f g$ with $g \in Q [t]$ , monic because $f$ and $Φ_{n}$ are.

L1L2

2.1

Both $f$ and $g$ lie in $Z [t]$ . Let $b \geq 1$ be least with $b f \in Z [t]$ and $d \geq 1$ least with $d g \in Z [t]$ ; these exist because clearing denominators gives some such integer. If a prime $q$ divided every coefficient of $b f$ then $q$ would divide its leading coefficient $b$ , and $(b / q) f$ would lie in $Z [t]$ , contradicting minimality; so $b f$ is primitive by [L4], and likewise $d g$ . Then [L4] gives $cont (b d Φ_{n}) = cont ((b f) (d g)) = 1$ , while $cont (b d Φ_{n}) = b d cont (Φ_{n}) = b d$ because $Φ_{n}$ is monic in $Z [t]$ and hence primitive by [L4]; so $b d = 1$ and $b = d = 1$ .

step 1.3L1L4

2.2

From step 1.2 and step 1.3, $0 = Φ_{n} (ζ^{p}) = f (ζ^{p}) g (ζ^{p})$ in the field $E$ , and $f (ζ^{p}) \neq 0$ by step 1.1, so $g (ζ^{p}) = 0$ .

step 1.1step 1.2step 1.3

3.1

Hence $ζ$ is a root of the polynomial $g (t^{p})$ , so $f ∣ g (t^{p})$ in $Q [t]$ by [L2]; writing $g (t^{p}) = f h$ with $h \in Q [t]$ , the division of the monic $g (t^{p}) \in Z [t]$ by the monic $f \in Z [t]$ has quotient and remainder in $Z [t]$ by [L3], and by the uniqueness clause of [L3] read in $Q [t]$ that quotient is $h$ and the remainder is $0$ ; so $h \in Z [t]$ .

step 2.1step 2.2L2L3

4.1

Reduce modulo $p$ and let $L$ be a splitting field of $\overset{ˉ}{f}$ over $Z / p$ , which exists by [L6]; $\overset{ˉ}{f}$ is monic of the same degree as $f$ , which is at least $1$ , so it has a root $a \in L$ .

step 2.1step 3.1L6given

5.1

Evaluating the reduction of step 3.1 at $a$ gives $\overset{ˉ}{g} (a^{p}) = \overset{ˉ}{f} (a) \overset{ˉ}{h} (a) = 0$ . Writing $\overset{ˉ}{g} = \sum_{i} c_{i} t^{i}$ with $c_{i} \in Z / p$ and using [L5] twice, $\overset{ˉ}{g} (a)^{p} = \sum_{i} c_{i}^{p} a^{i p} = \sum_{i} c_{i} (a^{p})^{i} = \overset{ˉ}{g} (a^{p}) = 0$ , so $\overset{ˉ}{g} (a) = 0$ because $L$ is a field.

step 3.1step 4.1L5

6.1

Thus $t - a$ divides both $\overset{ˉ}{f}$ and $\overset{ˉ}{g}$ in $L [t]$ by [L7], so $(t - a)^{2}$ divides $\overset{ˉ}{f} \overset{ˉ}{g} = \overline{Φ_{n}}$ , which divides $t^{n} - 1$ in $(Z / p) [t]$ by [L1]. Then $a$ is a repeated root of $t^{n} - 1$ in the extension $L$ of $Z / p$ , contradicting [L6], since $p ∤ n$ . The assumption of step 1.1 is therefore untenable and $f (ζ^{p}) = 0$ .

step 1.1step 4.1step 5.1L1L6L7discharge-contradiction∎

Remarks

Where $p ∤ n$ is used. Twice, and both uses are essential: in step 1.2, to know that $ζ^{p}$ is still primitive, and in step 6.1, to know that $t^{n} - 1$ is separable modulo $p$ . If $p$ divided $n$ the reduction $\overline{Φ_{n}}$ could genuinely have a repeated factor and the argument would produce no contradiction.
Why the passage to $Z [t]$ is not cosmetic. Reduction modulo $p$ is defined on integer polynomials, so the factorisation $Φ_{n} = f g$ has to be known to happen over $Z$ before step 4.1 can start. That is exactly what step 2.1 supplies, and it is where Gauss's content lemma enters.

Depends on

Used by

Φₙ is irreducible in ℚ[t] for every n≥1 Theorem

Dependency tree · two levels

107 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.10 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Theorem 2.5 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Theorem 9.8 (standard reference, not scraped)

If pp is a prime not dividing nn, a rational minimal polynomial of a primitive nn-th root of unity also kills its pp-th power