Let $f(x)$ be irreducible in $F[x]$, $F$ of characteristic $p>0$. Show that $f(x)$ can be written as $g(x^{p^e})$ where $g(x)$ is irreducible and separable. Use this to show that every root of $f(x)$ has the same multiplicity $p^e$ (in a splitting field).

I have been able to prove that $f$ can be expressed in terms of $g(x^{p^e})$, where $g$ is irreducible and separable. Also what is the relation between the splitting field of $g(x)$ over $F$ and the splitting field of $f(x)$ over $F$?

  • $\begingroup$ to get your case, put g=f and e=0 $\endgroup$ – tony Apr 9 at 16:26
  • $\begingroup$ Note that since $p$ is prime, $p\mid {p^e\choose n}$ for all $0 < n < p^e$. Then consider $(x - \omega)^{p^e} \mod p$. $\endgroup$ – Paul Sinclair Apr 10 at 0:24

Hint: Prove that for all $h\in F[x]$ you have $h(x^{p^e})=\operatorname{Frob}^eh(x)^{p^e}$.

Details: For every prime number $p$ and every positive integer $e$ the binomial coefficient $\tbinom{p^e}{k}$ is divisible by $p$ for all integers $k$ with $1<k<p^e$. It follows that for all $c\in\Bbb{F}_{p^e}$ and all $r\in\Bbb{F}_{p^e}[x]$ you have $$(cx^n+r(x))^{p^e}=c^{p^e}(x^n)^{p^e}+r(x)^{p^e}=\operatorname{Frob}^e(c)(x^{p^e})^n+r(x)^{p^e}.$$ By induction on the degree of $h$ it follows that $h(x)^{p^e}=\operatorname{Frob}^eh(x^{p^e})$ for all $h\in\Bbb{F}_{p^e}[x]$. In particular, this shows that every root of $h(x^{p^e})$ has multiplicity $p^e$, or a multiple thereof, and if $g\in F[x]$ is separable then every root of $g(x^{p^e})$ has the same multiplicity $p^e$.

  • $\begingroup$ How do you show that h(x^(p^e))=h(x)^(p^e)? I'm trying but I'm not getting it. $\endgroup$ – tony Apr 11 at 8:55
  • $\begingroup$ @tony I have added a few details. I see that I assumed that $F\subset\Bbb{F}_{p^e}$ earlier, which is of course not true in general; I've edited to cover your more general setting. $\endgroup$ – Servaes Apr 11 at 9:09
  • $\begingroup$ I have come across the Frob concept. Is there no other way to proceed? $\endgroup$ – tony Apr 11 at 14:27
  • $\begingroup$ @tony All you need to know for this question is that the map $c\ \mapsto\ c^{p^e}$ is an injective endomorphism of $F$, which is not hard to show. $\endgroup$ – Servaes Apr 12 at 15:26

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