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The question is this:

Prove that the Galois group of the algebraic closure of a finite field is Abelian and the order of every element in the Galois group is infinite.

I have a trouble with the first part. How can I prove the Galois group is Abelian?

(Sorry, I'm very beginner in Galois theory.)

user26857
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mathvc_
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    Every finite extension of finite fields is Abelian. – Angina Seng Jan 06 '18 at 19:27
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    No matter what the extension $k\supset\mathbb F_p$, the Galois group of this extension is generated by the frobenius, sending $z\in k$ to $z^p$. All hangs together for the frobenius to generate your infinite Galois group. – Lubin Jan 06 '18 at 19:28
  • @Lubin can you please make your comment into an answer? I can't really follow your statements. I'm really noob. – mathvc_ Jan 06 '18 at 19:36
  • I guess what Lord Shark is saying is that if you find two automorphisms that won't commute, then their restrictions to some finite field would not commute either (because the algebraic closure is a union of such things). But the Galois groups of extensions of finite fields are known to be cyclic, in particular abelian, so this is absurd. – Jyrki Lahtonen Jan 06 '18 at 19:50
  • Your second question was handled here. – Jyrki Lahtonen Jan 06 '18 at 21:50
  • @Jyrki: the point is that the Galois group of an infinite extension can be defined as a cofiltered limit of Galois groups of its finite subextensions, and the cofiltered limit of abelian groups is abelian. I don't think your argument works because two elements of the Galois group need not jointly fix a finite subextension. – Qiaochu Yuan Jan 07 '18 at 00:04
  • @QiaochuYuan In a more general case, sure. Here, with the base field $k=\Bbb{F}_q$ finite, all its finite extensions are Galois, so they are fixed by all the automorphisms. Or, more simply, if the commutator $\delta:=[\sigma,\tau]\neq id$ for some automorpisms $\sigma,\tau$, then $\delta(\alpha)\neq \alpha$ for some element $\alpha$. Then $K:=k(\alpha)$ is a finite field, and consequently $\sigma$ and $\tau$ A) preserve $K$, and B) their restrictions to $K$ commute as $Gal(K/k)$ is abelian. – Jyrki Lahtonen Jan 07 '18 at 16:13

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Although I think the comments do nicely to answer your question, since you’ve asked for an expansion of my comment, here goes:

Let $k\supset \Bbb F_q$ be an extension of degree $n$, so that $\vert k\vert=q^n$, and every element of $k$ is a root of $f(X)=X^{q^n}-X$. This happens because the (cyclic) group $k^*$ has order $q^n-1$. Thus $k$ is all the roots of $f$, and only these.

Now $\varphi_q$, defined by $\varphi_q(z)=z^q$ is an $\Bbb F_q$-automorphism of $k$, in other words, an element of $\text{Gal}^k_{\Bbb F_q}$. You see, I think easily, that the first power of $\varphi_q$ to be identity on $k$ is the $n$-th power. So the powers of $\varphi_q$ fill out $\text{Gal}^k_{\Bbb F_q}$, in other words, this Galois group is cyclic, generated by $\varphi_q$.

Well, this is quite wonderful. The same formula, $\varphi_q(z)=z^q$, works no matter the size of $k$, i.e. no matter the value of $n=[k:\Bbb F_q]$. This formula even works on the algebraic closure $\Omega$ of $\Bbb F_q$, which is just the union of all the finite extensions of $\Bbb F_q$. I won’t go into detail here, but in some sense (fairly easily made precise, but I don’t want to go into details here), $\text{Gal}^\Omega_{\Bbb F_q}$ is topologically generated by $\varphi_q$, it is “pro-cyclic”. And definitely abelian.

( Maybe I might say that the significance of the very apposite comments of Lord Shark the Unknown and Jyrki Lahtonen is this theorem that presumably is proved in every basic course on Galois Theory: the compositum of two (and even infinitely many) abelian extensions is abelian. )

Lubin
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