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I have seen some quite related qustions $x^p-x+a$ irreducible for nonzero $a\in K$ a field of characteristic $p$ prime

$x^p -x-c$ is irreducible over a field of characteristic $p$ if it has no root in the field

But may I please ask how to prove this statement? Assume that $f(x)$ is not irreducible, then we need to show that splits. But it seems that the proof is base on the fact that $f(x)$ has a root. So how can I see that fact that it has a root? Or otherwise, may I ask for proving it?

Y.X.
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  • See here. Some of the answers cover this version as well. The next to highest degree term of any factor has coefficient of the form $k\theta +m$ with integers $k, m$. There $k$ is the degree of the said factor so $0<k<p$. – Jyrki Lahtonen Apr 27 '17 at 04:27

1 Answers1

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If you meant $a \in \mathbb{F}_p^*$ then this answers the question.

Let $g(x) =x^p-x = \prod_{n \in \mathbb{F}_p} (x-n)$. For any $m \in \mathbb{F}_p$ : $g(m) = 0$. Take $a \in \mathbb{F}_p^*$. Then $f(x) = g(x)-a$ has no roots in $\mathbb{F}_p$.

Let $\beta$ be one of its root in $\mathbb{F}_{p^k}$ the splitting field of $f$. Since $g(\beta+n) = g(\beta)$, the other roots of $f$ are $\beta+n,n \in \mathbb{F}_p$.

Thus, $\sigma \in Gal(\mathbb{F}_{p^k}/\mathbb{F}_{p})$ sends $\beta$ to $\beta+n$. Since $f$ doesn't split completely in $\mathbb{F}_p$ we can choose $\sigma$ such that $\sigma(\beta) \ne \beta$ so that $n\ne 0$. Therefore $n$ generates $(\mathbb{F}_p,+)$ and the iterates of $\sigma$ act transitively on the roots of $f$, which means $f$ is irreducible over $\mathbb{F}_p$.

All this to say $\mathbb{F}_p(\beta)=\mathbb{F}_p[x]/(f(x))= \mathbb{F}_{p^{\,p}}$ which is the splitting field of $f$.

Since $p$ is prime, there is no subfield between $\mathbb{F}_p$ and $\mathbb{F}_{p^{\,p}}$, and hence there are two possible cases if $K$ is an extension of $\mathbb{F}_p$ :

  • Or $\mathbb{F}_{p^p} \subseteq K$ and $f$ splits completely over $K$,

  • Or $\mathbb{F}_{p^p} \not\subseteq K$ and $f$ is irreducible over $K$.

If you meant $a \in K^*$ is algebraic over $\mathbb{F}_p$ then it depends on $f$ having a root in $\mathbb{F}_p(a)=\mathbb{F}_{p^N}$ or not. If it has no root, repeat the same argument with $\sigma \in Gal(\mathbb{F}_{p^{N}}(\beta)/\mathbb{F}_{p^N})$. You get again that the $p$ roots of $f$ are $\beta+n$ and the iterates of $\sigma$ act transively on those, therefore $f$ is irreducible over $\mathbb{F}_{p^N}$ and splits completely in $\mathbb{F}_{p^{N}}(\beta) = \mathbb{F}_{p^{N}}|x]/(f(x))=\mathbb{F}_{p^{pN}}$. Again $[\mathbb{F}_{p^{pN}}:\mathbb{F}_{p^{N}}]= p$ is prime so there are no subfield between the two, and there are two possible cases if $K$ is an extension of $\mathbb{F}_p(a)=\mathbb{F}_{p^{N}} $ :

  • Or $\mathbb{F}_{p^{pN}} \subseteq K$ and $f$ splits completely over $K$,

  • Or $\mathbb{F}_{p^{pN}} \not\subseteq K$ and $f$ is irreducible over $K$.

reuns
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