If $f \in F[x]$ irreducible, ${\rm char}(F) = p$, then $f(x) = g(x^{p^e})$ and every root of $f$ has multiplicity $p^e$ in some splitting field

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 $$F$$?

• to get your case, put g=f and e=0 – tony Apr 9 at 16:26
• Note that since $p$ is prime, $p\mid {p^e\choose n}$ for all $0 < n < p^e$. Then consider $(x - \omega)^{p^e} \mod p$. – 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 $$1. It follows that for all $$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$$.
• @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. – Servaes Apr 11 at 9:09
• @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. – Servaes Apr 12 at 15:26