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 $p=l$)?

I'd also be interested in the answer with the words "algebraically closed" deleted -- I'm not sure if this is easier or harder. Also, there is a natural topological enrichment of $\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence (which yields the usual topology on the absolute Galois group). I'd be interested to hear of a way to distinguish these topologically-enriched categories, which in principle might be easier.

Here are some easy observations (with the "algebraically closed" condition):

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 $L \to K$ (and using the fact that the category of ordinals has no nonidentity automorphisms, as pointed out by Eric Wofsey).

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.

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 $\mathrm{PGL}_2(\overline{\mathbb{F}_p}) \subset \mathrm{Aut}(\overline{\mathbb{F}_p(t)})$ 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.

I initially posted this on math.SE, thinking that there must be some easy way to resolve it lying just beyond the reaches of my algebraic competency, but after the question has languished there for a week I think maybe I should try it out here.

finitefields of characteristic $p$ doesn't depend on the prime $p$ (up to equivalence). $\endgroup$ – HeinrichD Sep 12 '16 at 13:18