LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

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.

If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$

Statement

Let $F$ have characteristic $p > 0$ , let $a \in F$ not be a $p$ th power in $F$ , and let $n \geq 1$ . Then $x^{p^{n}} - a$ is irreducible in $F [x]$ .

Facts & Assumptions

Given: A field $F$ of characteristic $p > 0$ , an element $a \notin F^{p}$ , and a natural number $n \geq 1$ .

[L1]

Frobenius is injective and $(u - v)^{p^{r}} = u^{p^{r}} - v^{p^{r}}$ in characteristic $p$ (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields).

[L2]

A nonzero polynomial is separable exactly when it is coprime to its derivative (A nonzero polynomial over a field is separable exactly when its gcd with its derivative is $1$ ).

[L3]

Every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

[L4]

Every irreducible polynomial in characteristic $p$ is uniquely a separable irreducible polynomial in a power $x^{p^{e}}$ (In characteristic $p$ , every irreducible polynomial is uniquely $g (x^{p^{e}})$ with $g$ irreducible and separable).

Proof

technique · direct

1.1L1L3

In a splitting field supplied by [L3], choose a root $b$ of $x^{p^{n}} - a$ ; [L1] gives $x^{p^{n}} - a = (x - b)^{p^{n}}$ , so $b$ is its only distinct root.

2.1step 1.1L2L4

Let $q$ be the minimal polynomial of $b$ over $F$ . By [L4], write $q (x) = g (x^{p^{r}})$ with $g$ irreducible and separable. Every root of $q$ is also a root of $x^{p^{n}} - a$ , hence equals $b$ by step 1.1; separability of $g$ and [L2] therefore force $g$ to be linear. Thus $q (x) = x^{p^{r}} - c$ for some $c = b^{p^{r}} \in F$ and some $0 \leq r \leq n$ .

3.1step 2.1algebra

If $r < n$ , then $a = b^{p^{n}} = c^{p^{n - r}}$ is a $p$ th power in $F$ , contrary to the hypothesis; hence $r = n$ and $q = x^{p^{n}} - a$ .

4.1step 3.1∎

Therefore $x^{p^{n}} - a$ is the minimal polynomial of $b$ and is irreducible. The hypothesis excludes $a = 0$ because $0 = 0^{p}$ , and the same argument includes $n = 1$ .

If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

If aa is not a ppth power in a characteristic-pp field, then xpn−axpn−a is irreducible for every n≥1n≥1

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$