What is the algebraic closure of $\mathbb F_q$?

What is the algebraic closure of $\mathbb F_q$ with $q$ being some power of a prime $p$ ?

I wrote, ''the algebraic closure'' because, they're the same up to isomorphism right ?

It cannot be finite, otherwise it is not algebraically closed, so how does it look like ?

• What sort of answer would help you? I don't think there is any "nice" description of this, other than the fact that it will be the same for any power of $p$. Commented Jul 28, 2016 at 12:15
• Perhaps this has the information you're looking for Commented Jul 28, 2016 at 12:16
• @TobiasKildetoft I'm studying finite fields math.umn.edu/~garrett/m/algebra/notes/09.pdf#page=2 (Proof on the top of page 2) it is written that Frobenius stabilizes all fields between $F_1$ and $E$, but $E$ should be infinite, i find this a bit strange Commented Jul 28, 2016 at 12:18
• What is strange about $E$ being infinite? Commented Jul 28, 2016 at 12:24

Given finite fields $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$ with $\gcd(m, n) = 1$ then the compositum is the finite field $\mathbb{F}_{p^{mn}}$. This allows us to define the algebraic closure of $\mathbb{F}_{p}$ as the union $$\overline{\mathbb{F}_{p}}=\bigcup_{k\ge 1} \mathbb{F}_{p^k}.$$ For prime powers $q=p^n$ the algebraic closure $\overline{\mathbb{F}_{q}}$ can be constructed by building and gluing $\ell$-adic towers $$\mathbb{F}_{q}\subset \mathbb{F}_{q^{\ell}}\subset \mathbb{F}_{q^{\ell^2}}\subset \cdots$$ see here for an algorithm and an impressive picture on page $5$.
• If you have $$\overline{\mathbb{F}_{p}}=\bigcup_{k\ge 1} \mathbb{F}_{p^k}$$, why do you still bother about $\overline{\mathbb{F}_{q}}$? It is just $\overline{\mathbb{F}_{p}}$, since $$\mathbb{F}_{q}/\mathbb{F}_{p}$$ is algebraic...
• Because these $\ell$-adic towers are interesting in itself, e.g., for computing roots in $\mathbb{F}_q$ quickly. Commented Jul 28, 2016 at 12:54
• Instead of using a single prime $\ell$ at a time, an alternative is to use the tower of extensions $\Bbb{F}_{q^{m!}}\subset\Bbb{F}_{q^{(m+1)!}}$. The factorials cover all the prime powers and their gcds in due time. +1 of course Commented Jul 29, 2016 at 5:30