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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 ?

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  • $\begingroup$ 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 $p$. $\endgroup$ Commented Jul 28, 2016 at 12:15
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    $\begingroup$ Perhaps this has the information you're looking for $\endgroup$ Commented Jul 28, 2016 at 12:16
  • $\begingroup$ @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 $\endgroup$
    – user257
    Commented Jul 28, 2016 at 12:18
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    $\begingroup$ What is strange about $E$ being infinite? $\endgroup$ Commented Jul 28, 2016 at 12:24

1 Answer 1

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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 $5$.

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  • $\begingroup$ 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... $\endgroup$
    – MooS
    Commented Jul 28, 2016 at 12:45
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    $\begingroup$ Because these $\ell$-adic towers are interesting in itself, e.g., for computing roots in $\mathbb{F}_q$ quickly. $\endgroup$ Commented Jul 28, 2016 at 12:54
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    $\begingroup$ 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 $\endgroup$ Commented Jul 29, 2016 at 5:30

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