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I have some questions on the field extension $\overline{\mathbb{F}_p}|\mathbb{F}_p$ for some prime number $p$: Are the any other intermediate fields besides $\mathbb{F}_{p^n}$ for $n \in \mathbb{N}$ or $\overline{\mathbb{F}_p}$? What is the Galois group of the extension $\overline{\mathbb{F}_p}|\mathbb{F}_p$? Do we have that $\overline{\mathbb{F}_p}= \cup_{n=1}^{\infty} \mathbb{F}_{p^n}$?

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if $\alpha$ is algebraic over $F_p$ then $\alpha\in F_{p^n}$ for some $n$, so the algebraic closure is the union (direct limit) of the $F_{p^n}$. there are intermediate fields such as $\cup F_{p^{2^n}}$ – yoyo Aug 13 '11 at 20:37
the galois group is the inverse limit of the galois groups of the finite extensions – yoyo Aug 13 '11 at 20:45
up vote 8 down vote accepted

Yes, the algebraic closure of $\mathbb{F}_p$ is the union (or more carefully, the direct limit) of the finite fields: first, this makes sense because given any $n$ and $m$, both $\mathbb{F}_{p^n}$ and $\mathbb{F}_{p^m}$ are subfields of $\mathbb{F}_{p^{\mathrm{lcm}(n,m)}}$, so you will indeed get a field. And given any polynomial $f(x)$ with coefficients in $\mathbb{F}_p$, $f(x)$ has at least one root in a finite extension of $\mathbb{F}_p$, hence in the union. And, furthermore, we clearly need at least all the finite fields, since they are the splitting fields of $x^{p^m}-x$ over $\mathbb{F}_p$. So the union is an algebraic closure of $\mathbb{F}_p$.

The Galois group is the inverse limit of the Galois groups of the finite subextensions (this is true in any infinite Galois extension), which are cyclic; you end up with the inverse limit of the cyclic groups, and that is called $\widehat{\mathbb{Z}}$, the profinite completion of $\mathbb{Z}$.

Yes, there are other intermediate fields. For example, take the union of $\mathbb{F}_{p^{2^k}}$ for $k=1,2,\ldots$. These fields form a chain (since $\mathbb{F}_{p^m}\subseteq\mathbb{F}_{p^n}$ if and only if $m|n$), so their union is a field. It is clearly not finite. And it does not contain any element of degree not a power of $2$ over $\mathbb{F}_p$ (in particular, no root of an irreducible cubic), so it cannot be the whole algebraic closure.

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That saves me writing my answer. The only comment I have is that if you are unfamiliar with inverse limits, you can note two points: for each positive integer $n$, the field $\overline{\mathbb{F}_{p}}$ has a unique subfield of order $n$, which has a cyclic Galois group of order $n$, and any automorphism of $\overline{\mathbb{F}_{p}}$ must stabilise each of these finite fields (and its total effect is determined by what it does to each of them). – Geoff Robinson Aug 13 '11 at 20:51
Dear @Geoff: Instead of "subfield of order $n$", don't you mean "subfield of order $p^n$"? Or "sub-extension of degree $n$? – Pierre-Yves Gaillard Aug 13 '11 at 21:32
@phil: An element of $\widehat{\mathbb{Z}}$ is given by a family $(a_n)_{n > 1}$, such that, for all $n$, $a_n$ is in $\mathbb Z/n\mathbb Z$, and $a_n\equiv a_d$ mod $d$ whenever $d$ divides $n$. This is a topological ring, but we only view it here as a topological additive group. (The topology is induced by that of the product of the $\mathbb Z/n\mathbb Z$. - in particular $\widehat{\mathbb{Z}}$ is compact.) – Pierre-Yves Gaillard Aug 13 '11 at 21:45
@Pierre: yes, thanks: of course I meant subfield of order $p^n$, not subfield of order $n$- another stupid typo, which seems harder to fix in comments without a complete rewrite. – Geoff Robinson Aug 13 '11 at 22:01
@Arturo Magidin: Thank you for your convincing and complete answer to my question. – phil Nov 28 '11 at 13:37

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