# Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?

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):

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

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

• I'm not sure what you mean by your part 1. I think you can say something along these lines, but the statement that there is a map $K\to L$ if $L$ has higher transcendence degree is in general false. (There are no maps from an extension of $\mathbb{F}_p$ to the function field of $\mathbb{P}^1_{\mathbb{F}_p}.$) Commented Apr 8, 2016 at 0:51
• @Dmitry Vaintrob (1) is only in the algebraically closed setting. I agree it's not clear what happens to objects for categories of (possibly non-algebraically-closed) fields. Commented Apr 8, 2016 at 0:53
• Here is an almost trivial remark: The category of finite fields of characteristic $p$ doesn't depend on the prime $p$ (up to equivalence). Commented Sep 12, 2016 at 13:18
• Commented Feb 23, 2020 at 14:37

If you do not impose an algebraically closed condition, no two are equivalent. This basically follows from your (3). Namely, observe that

• An extension is finite if and only if it has finitely many subextensions (otherwise, it contains $\mathbb{F}_p(x)$, which then contains $\mathbb{F}_p(x^n)$ for all $n$).
• $\bar{\mathbb{F}}_p$ is the unique extension that can be written as a direct limit involving all algebraic extensions.
• The smallest field containing but not isomorphic to $\bar{\mathbb{F}}_p$ is the rational function field $\bar{\mathbb{F}}_p(x)$ (in the sense that any map $\bar{\mathbb{F}}_p\to K$ contains a transcendental element, hence factors through $\bar{\mathbb{F}}_p(x)$, and a map $\bar{\mathbb{F}}_p\to \bar{\mathbb{F}}_p(x)$ cannot factor through another field that is not isomorphic to one of these).
• Now as you've observed, the automorphism group of $\bar{\mathbb{F}}_p(x)$ over $\bar{\mathbb{F}}_p$ is $PGL_2(\bar{\mathbb{F}}_p)$. These groups are nonisomorphic for different primes. To see this, e.g. note that $PGL_2(\bar{\mathbb{F}}_p)$ contains a maximal abelian subgroup isomorphic to $\mathbb{F}_p$ (of unipotent upper-triangular matrices), but not $\mathbb{F}_\ell$ for $\ell \ne p$ (as all other maximal commutative subgroups are conjugate to diagonal matrices, which have lots of different prime components).

The algebraically closed case seems to be more difficult from this point of view.

• OK, great this would seem to be a proof for the case of all fields. But I don't quite follow. You attempt to characterize $\overline{\mathbb{F}_p}$ by the fact that it's a direct limit of objects with finite automorphism group -- but why should the finite fields be the only fields (of finite characteristic) with finite automorphism groups? Also, are you claiming that $\overline{\mathbb{F}_p}(t)$ can be characterized as a minimal extension of $\overline{\mathbb{F}_p}$? What about the intermediate field $\overline{\mathbb{F}_p}(t^2)$? Commented Apr 8, 2016 at 1:08
• Okay -- on the second point, assuming that $\overline{\mathbb{F}_p}$ can be characterized then I buy that we win, because $\overline{\mathbb{F}_p}$ has the property that any extension has an intermediate field with relative automorphism group $PGL_2(\overline{\mathbb{F}_p})$. But I'm still not convinced that we can characterize $\overline{\mathbb{F}_p}$. Commented Apr 8, 2016 at 1:11
• Yeah, by "minimal" I mean in the to say that we impose the relation $K<L$ if there exists a map $K\to L$. (In which case $k(x) \cong k(x^2)$.) But you're absolutely right about there being more extensions with finite automorphism group. Instead, I think you can characterize finite extensions by the property that they have finitely many sub-extensions. Commented Apr 8, 2016 at 1:18
• Answer edited to fix this. Commented Apr 8, 2016 at 1:31
• I'm more interested in the algebraically closed case, but I initially thought that the case of all fields would be easier, but then I got stuck, so it's good to see a solution. A few points: I think in your first two bullets, it should say "finite" instead of "algebraic", right (and so in the first bullet, I suppose there should be a separate case for $\overline{\mathbb{F}_p}$)? On the last part I'm having second thoughts (probably because of my weak field theory) -- is it really the case that all intermediate fields between $k$ and $k(t)$ are isomorphic to $k(t)$? Commented Apr 8, 2016 at 2:36

The answer is "only when $p=l$" if you work with the category of algebraically closed fields enriched over the category of topological spaces. That means I want to put a topology on the set of homomorphisms from one field to another. I will do this with the compact-open topology for the discrete topology for each fields - i.e. a basis for the open sets consists of the sets $B_{x,y} =\{ f \mid f(x)=y\}$. This is the standard way of topologing Galois groups, for instance.

We will construct $PGL_2(\overline{\mathbb F}_p)$ from this category in several steps. (This is sufficient, because of what you noted about maximal $p$-subgroups.)

First, as you noted, we can find in this category the unique algebraically closed field of transcendence degree $0$, namely $\overline{\mathbb F_p}$ and of transcendence degree $1$, namely $\overline{\mathbb F_p(t)}$. Take any map from the first to the second. Consider the group of automorphisms of the second that form a commutative triangle with this map. This is the topological group $\operatorname{Aut}(\overline{\mathbb F_p(t)}/\overline{\mathbb F}_p)$.

Next we classify the compact open subsets of this group. Any open subgroup contains some nonempty open set of the form: $\{ f \mid f(x_1)=y_1,\dots, f(x_n)=y_n\}$. By composing with element of that set, it contains $\{f \mid f(x_1)=x_1,\dots, f(x_n)=x_n\}$. This is an open subgroup, so it is finite-index. It follows that this subgroup contains the Galois group of the curve with function field $\mathbb F_p(x_1,\dots,x_n)$ as a finite index subgroup, so it contains the Galois group of some cover of that curve as a finite index normal subgroup, so it is the Galois group of a quotient of that cover by a finite subgroup of its automorphism group. So every compact open subgroup is the Galois group of some curve.

Let $H$ be a compact open subgroup fixing some field $F$. The normalizer $N(H)$ of $H$ acts naturally on $F$, and the kernel of that action is $H$, so $N(H)/H$ is a subgroup of the automorphisms of $F$. In fact it is all the automorphisms of $F$, as these clearly normalize $H$. So we may construct the set of all automorphism groups of curves over $\overline{\mathbb F}_p$.

Among these, the only infinite nonsolvable groups are the automorphism groups of curves of genus $0$, which are all isomorphic to $PGL_2(\overline{\mathbb F}_p)$. So we may construct $PGL_2(\overline{\mathbb F}_p)$.

• Great, thanks! Dmitry Vaintrob had some thoughts along similar lines in the chat that evolved from our discussion in the comments to his answer. I'm grateful for all the details you've included because my field theory is pretty weak. When I asked the question I had supposed that the topological case wouldn't be that different from the pure categorical case, but now it's not so clear how to remove the use of the topology. By the way, where do you use compactness of your open subgroup? Commented Apr 8, 2016 at 15:12
• @TimCampion Open subgroups of compact topological groups are finite index, which I use to make the Galois group finite index in the subgroup and thus ensure the subgroup is itself a Galois group. Commented Apr 8, 2016 at 15:23
• @WillSawin I am sure that this is trivial, but how can you conclude, exactly? I mean, once you construct $PGL_2(\overline{\mathbf F}_p)$, how do you get a contradiction assuming there is an equivalence of categories between the two? Commented Apr 8, 2016 at 18:14
• @FilippoAlbertoEdoardo I understand Will to be saying that the argument is concluded exactly as in Dmitry Vaintrob's answer. Commented Apr 8, 2016 at 19:33
• @TimCampion Ah, yes, I see. Very elegant. Commented Apr 8, 2016 at 22:55

The pure category theory reading of this question is: is there some universal property of objects and morphisms that distinguishes amongst categories of fields?

Of course the nitty-gritty comes down to Galois theory and finite group theory, but I will leave these to algebraists and just give the categorical property.

Algebraists may be aware of the 1980s work by Yves Diers on multi-colimits. I think he was primarily interested in categories of commutative rings or maybe integral domains rather than fields.

In such categories a diagram may fail to have a colimit in the standard sense but instead have a set of them, with the universal property that, for each cocone, exactly one member of the set has the usual property.

For this problem with categories of fields, we have to generalise a little further to what François Lamarche and I called poly-colimits. Now the set of colimit candidates becomes a groupoid.

The simplest way to understand the definition of multi- or poly-colimit is probably to look at slice categories. In the case of categories of fields, these are just lattices of subfields. The problem with this is that the set of colimit candidates is not very clear and it is probably quite difficult to see the groupoid structure.

A better way, at least for a categorist, is to see this situation as a factorisation system, which is how it is presented in my papers.

I think the way you calculate the poly-coproduct of two fields is to form their tensor product qua modules over the common prime field and express that as a direct sum of fields. (I learned this idea from Charles Matthews.)

I think if you start with two "disjoint" normal separable extensions, this construction will give a single field. However, this is not the coproduct, but the poly-coproduct, indexed by a group (one-object groupoid) that must pretty closely related to some Galois group.

I am sorry for the handwaving, but I hope that there is sufficient novelty in what I have said to pique the appetite of some categorically minded algebraist. It was well over 20 years ago when I explored this topic, many of my papers were never finished and I lost interest.)

• Could you give some reference for the definition of poly-colimit? Commented Apr 13, 2016 at 1:25
• @DenisNardin Quick answer: see my papers or web search "polycolimit". Long answer: later or email me. Commented Apr 13, 2016 at 11:26