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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 ?

user26857
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user257
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  • 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 $p$. – Tobias Kildetoft Jul 28 '16 at 12:15
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    Perhaps this has the information you're looking for – Ben Grossmann Jul 28 '16 at 12:16
  • @TobiasKildetoft I'm studying finite fields http://www.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 – user257 Jul 28 '16 at 12:18
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    What is strange about $E$ being infinite? – Tobias Kildetoft Jul 28 '16 at 12:24
  • @TobiasKildetoft: $\mathbb Q$ is countable and so is $ \overline {\mathbb Q}$, $\mathbb R$ has cardinality $\mathfrak c$ and so does $\mathbb {\overline R}=\mathbb C$, but $\mathbb F_q$ is finite and $E=\overline{\mathbb F_q}$ is not – J. W. Tanner Jun 20 '24 at 02:22

1 Answers1

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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 $5$.

user26857
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Dietrich Burde
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  • 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... – MooS Jul 28 '16 at 12:45
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    Because these $\ell$-adic towers are interesting in itself, e.g., for computing roots in $\mathbb{F}_q$ quickly. – Dietrich Burde Jul 28 '16 at 12:54
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    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 – Jyrki Lahtonen Jul 29 '16 at 05:30
  • Perhaps you should say what $\ell$ is – J. W. Tanner Jun 20 '24 at 01:47