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Let $F$ be a finite field. Let $E/F$ be a field extension of degree $m\in\mathbb{N}$. Show that $E/F$ is Galois.

I would like to varify my answer because I am absolutely sure about it;

$F$ is finite so $p=\text{char}(F)>0$. Thus $\mathbb{F}_p\subseteq F$, and this extension is finite because $F$ is finite. Suppose $n=[F:\mathbb{F}_p]$. Hence $F\cong\mathbb{F}_{p^n}$.

$$[E:\mathbb{F}_p]=[E:F]\cdot[F:\mathbb{F}_p]=m\cdot n$$ Thus, $E\cong\mathbb{F}_{p^{mn}}$ and $E$ is the splitting field of $x^{p^{nm}}-x$ over $\mathbb{F}_p$. So $E/\mathbb{F}_p$ is Galois. ( C.f. Hungerford's Algebra, Chapter V, Theorem 3.11. ; note that $x^{p^{nm}}-x$ is separable over $\mathbb{F}_p$ since the finite field $\mathbb{F}_p$ is perfect .)

Hence, $E/F$ is Galois.

Thanks in advance.

Plantation
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J. Doe
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    Yes, it looks correct. – Mark Jul 29 '19 at 19:05
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    Well, just a short note: you didn't mention that $x^{p^{nm}}-x$ is separable. I hope you remember that it's important here. – Mark Jul 29 '19 at 19:12
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    @Mark Not any more important than it would be over $\Bbb Q$. Finite fields are also perfect. – Arthur Jul 29 '19 at 19:17
  • @Arthur Yes, I know. You need to write one of the following: the polynomial is separable or that you know that finite fields are perfect. – Mark Jul 29 '19 at 19:22
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    @Arthur Though at this point, when working through such an exercise, this may not be known yet. So it seems worth while making explicit. – Servaes Jul 29 '19 at 19:22
  • Indeed, writing down the polynomial you did is probably the best possible way of showing that every algebraic extension of a finite field is separable. – Lubin Jul 30 '19 at 00:31

1 Answers1

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Extension is Galois iff it is Normal+Separable. One can use following Lemmas: Let $k \subset \mathbb{F} \subset \mathbb{E}$ be a finite extension.

  1. $\mathbb{E}/k$ is separable iff $\mathbb{F}/k$ and $\mathbb{E}/\mathbb{F}$ are.
  2. If $\mathbb{E}/k$ is normal, so is $\mathbb{E}/\mathbb{F}$ (Here we don't need finiteness of extensions)

Thanks to those two lemmas, we can limit ourselves to the case $\mathbb{F}_q/\mathbb{F}_p$. Here we can use the fact that $\mathbb{F}_q$ is a splitting field of the polynomial $x^q-x$ over the field $\mathbb{F}$.

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