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In Field and Galois theory of Patrick Morandi there's a proof about this proposition:

Let $f(x)\in F(x)$ be an irreducible polynomial. If $char(F)=p>0$, then $f(x)=g(x^{p^m})$ for some $m\geq 0$ and some $g(x)\in F(x)$ that is irreducible and separable over $F$.

My question is about the proof used:

Suppose that $char(F) = p$, and let $f(x)\in F[x]$. Let $m$ be maximal such that $f(x)\in F[x^{p^m}]$. Such an $m$ exists, since $f\in F[x^{p^0}]$ and $f$ lies in $F[x^{p^r}]$ for only finitely many $r$ because any nonconstant polynomial in $F[x^{p^r}]$ has degree at least $p^r$. Say $f(x)=g(x^{p^m})$. Then $g(x)\notin F[x^{p}]$ by maximality of $m$. Moreover, $g(x)$ is irreducible over $F$, since if $g(x) = h(x) \cdot k(x)$, then $f (x) = h(x^{p^m}) \cdot k(x^{p^m} )$ is reducible over $F$. Because of the statement, $g$ is separable over $F$.

This are my doubts:

  • The affirmation he gives about the existence of $m$ isn't saying that $\deg(f)=p^k$ for certain integer $k$?
  • If the first question is a no, why exists that $m$?
  • Why because of the statement the polynomial $g$ is separable over $F$?

The proof can be read in the page $44$, but is the same as the posted here.

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    For the first two bullets: $f(x)$ has a finite degree (as do all polynomials), and the argument is simply relying on that piece of information: $\deg f(x)=\deg g(x^{p^r})=p^r\deg g(x)\ge p^r$, because $g$ cannot be a constant, and this can hold for only finitely many $r$. The last bullet comes from the fact that if $g$ is NOT separable, then $g(x)\in F[x^p]$. – 2017-01-16
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    But the fact of $f(x)\in F[x^{p^m}]$ doesn't mean that $\deg(f)$ is a power of $p$? – 2017-01-16
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    No it doesn't. Why would it? It isn't needed. We simply use $p^r\le \deg f(x)$ to show that $f\in F[x^{p^r}]$ can hold only for finitely many $r$. – 2017-01-16
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    It makes me think that $\deg(f)$ is a multiple or a power of $p$ if $f(x)$ is a polynomial that can be seen as a polynomial $g(x^{p^m})$. I don't know if it's true, but it makes me think that happens. – 2017-01-16
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    Yes, that is true. If $f(x)=g(x^{p^m})$ then certainly $\deg f(x)=p^m\deg g(x)$ is a multiple of $p^m$. – 2017-01-16
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    So the only polynomials of $\mathbb{F}_p[x]$ that are irreducible are the ones with degree multiple of $p$? – 2017-01-16
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    Most emphatically NO! It could well be that the maximal $r$ is $m=0$, when $f(x)=g(x)$. This happens every time $\deg f(x)$ is not a multiple of $p$, and may also happen when $p\mid \deg f(x)$. We have $f(x)\in F[x^p]$ if and only if ALL THE TERMS of $f(x)$ have degrees that are multiples of $p$. – 2017-01-16
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    For example, when $p=2$, and $F=\Bbb{F}_2(T)$, then both $f_1(x)=x^2+T$ and $f_2(x)=x^2+x+T$ are irreducible. Here $f_1(x)\in F[x^2]$, and we can write $f_1(x)=g_1(x^2)$ with $g_1(x)=x+T$. OTOH $f_2(x)\notin F[x^2]$, and in that case we would have $g_2(x)=f_2(x)$. – 2017-01-16
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    Thanks for the clarifications, I appreciate the time you took to answer them. Now I understand all. – 2017-01-16

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