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If $N$ is the algebraic closure of a finite field $F$, prove that $\operatorname{Gal}(N/F)$ is an abelian group and that any element of the Galois group has infinite order.

If $N$ were a finite extension of $F$, then it would be easy to see that $\operatorname{Gal}(N/F)$ is a cyclic group. But I don't know if I can proceed using this idea in the above case.

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  • $\begingroup$ See this question for lack of elements of finite order. Abelian? How will a commutator act on any finite extension $E$ of $F$? Recall that $N$ is the union of such extensions $E$. $\endgroup$ – Jyrki Lahtonen Jun 16 '15 at 11:11
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Let F be your finite field, say of order q (where q is necessarily a power of a prime p). It is known from the elementary theory of finite fields that any extension E/F of degree n is Galois, with cyclic Galois group isomorphic to (Z/nZ, +), generated by the so called Frobenius automorphism, sending x to x^q. "The" algebraic closure of F being the inductive limit of the finite extensions of F, its Galois group over F is the projective limit of the Z/nZ, usually denoted Z^, the "profinite completion" of Z. By construction (check on projective limits, or just argue that Z is dense in Z^ with respect to the profinite topology), Z^ is abelian and has no non null element of finite order.

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