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EDIT I've now posted this question on mathoverflow. It probably makes sense to post answers over there, unless someone prefers posting here.

Let $\mathrm{ACF}_p$ denote the category of algebraically closed fields of characteristic $p$, with all homomorphisms as morphisms. The question is: when is there an equivalence of categories between $\mathrm{ACF}_p$ and $\mathrm{ACF}_l$ (with the expected answer being: only when $p=l$)?

Here are some easy observations:

  1. First, any equivalence of categories must do the obvious thing on objects, preserving transcendence degree because $K$ has a smaller transcendence degree than $L$ if and only if there is a morphism $K \to L$ but not $L \to K$.

  2. We can distinguish the case $p=0$ from the case $p \neq 0$ because $\mathrm{Gal}(\mathbb{Q}) \not \cong \mathrm{Gal}(\mathbb{F}_p)$. But $\mathrm{Gal}(\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ for any prime $p$. So we can't distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ for different primes $p \neq l$ in such a simple-minded way. So for the rest of this post, let $p,l$ be distinct primes.

  3. The next guess is that maybe we can distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ by seeing that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$. To this end, note that there is a tower $\mathbb{F}_p \subset \overline{\mathbb{F}_p} \subset \overline{\mathbb{F}_p}(t) \subset \overline{\mathbb{F}_p(t)}$. The automorphism groups of these intermediate extensions are respectively $\hat{\mathbb{Z}}$, $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$, and a free profinite group (the last one is according to wikipedia).

From (3), there is at least a subquotient $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$ which looks different for different primes. But I don't see how to turn this observation into a proof that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$ specifically, or that $\mathrm{ACF}_p \not \simeq \mathrm{ACF}_l$ more generally.

Note that if we change "algebraically closed fields of characteristic $p$" to "fields of characteristic $p$", then its easy to distinguish these categories because their subcategories of finite fields look very different.

Also, there is a natural topological enrichment of $\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence. I'd be interested to hear of a way to distinguish these topologically-enriched categories.

tcamps
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  • Note that if we change "algebraically closed fields of characteristic $p$" to "fields of characteristic $p$", then its easy to distinguish these categories because their subcategories of finite fields look very different. – tcamps Mar 29 '16 at 17:51
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    I don't agree. The subcategories of finite fields are all equivalent to the poset $(\Bbb{N} , \mbox{divides})$. – Crostul Mar 29 '16 at 17:58
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    I'm taking all field homomorphisms as morphisms. So the subcategory of finite fields includes the automorphism group of each finite field. But this... actually is not different! So I think you're right to disagree after all! – tcamps Mar 29 '16 at 18:00
  • No, not the same morphisms (even though it's true that the categories of finite fields for all $p$ are equivalent). – Captain Lama Mar 29 '16 at 18:00
  • @Crostul The automorphism group of the four element field is not isomorphic to the automorphism group of the nine element field. But just looking at the four element field suffices to prove your assertion is false. – egreg Mar 29 '16 at 18:09
  • I see. My bad, I always forget morphisms. – Crostul Mar 29 '16 at 18:11
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    @egreg Both these automorphism groups are cyclic of order 2, generated by a Frobenius element, right? – tcamps Mar 29 '16 at 18:11
  • @tcamps Yeah, right; but the main thing is that there are automorphisms – egreg Mar 29 '16 at 18:53
  • Your first observation is not as trivial as you make it sound: it uses the fact that the totally ordered class of all cardinal numbers has no automorphisms besides the identity. – Eric Wofsey Apr 05 '16 at 03:29
  • @EricWofsey yeah, I guess I did gloss over that. – tcamps Apr 08 '16 at 00:13

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