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

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$t^{n} - 1$ 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

1.1assume-case posL1L2L3

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 $K$ .

1.2assume-case zeroL1L2L3

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].

2.1step 1.1step 1.2L4cases-exhaustive

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 $char K ∤ n$ .

3.1step 2.1L5

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 $c = 1$ .

4.1step 2.1step 3.1L7

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].

5.1step 3.1step 4.1L6

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.

6.1step 5.1L5L6∎

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 $E = K$ .

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 $K$ .

Depends on

Used by

Dependency tree · two levels

66 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

P. L. Clark, Field Theory (course notes/monograph), Proposition 9.4 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 1 (standard reference, not scraped)

tn−1tn−1 is separable over KK exactly when the characteristic does not divide nn, and then a splitting field carries nn distinct nn-th roots of unity