PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1L1L2given

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.

2.1step 1.1algebra

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 $u = t^{p^{k}}$ .

3.1step 2.1L3

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 $1^{p^{k}} = 1$ .

4.1step 3.1L3algebra∎

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 $μ_{n} (K) = μ_{m} (K)$ .

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.

Depends on

Used by

In characteristic three, t³-1=(t-1)³ and μ₆ coincides with μ₂ Example
The Hopf algebra of a split torus and its root-of-unity subgroups Example
FALSE: μₙ(K) has n elements in every field K False statement
Φ_pʳ(t)=∑_k<pt^kpʳ⁻¹, and Φ_pʳ(t+1) is Eisenstein at p Proposition

Dependency tree · two levels

33 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

K. Conrad, Cyclotomic Extensions (expository blurb), Section 1 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Chapter 9, Section 1 (standard reference, not scraped)

In characteristic pp the only pkpk-th root of unity is 11, and tpk−1=(t−1)pktpk−1=(t−1)pk