Can one recover an algebraically closed field $k$ from the dots and arrows of its category of finitely generated $k$-algebras?

You are gracious enough to host me for a few days while I attend a conference. After I leave, you're surprised to see a gift on the kitchen table. It's a box with a category inside! The objects aren't labeled so it's a little hard to tell what's going on with it, but you can see, for example, that there's a terminal object. There's also a note:

Dear X,

I've always been very fond of this category, and I thought you might like it too. It's a category of finitely generated algebras over some algebraically closed field - unfortunately I've forgotten which one! But I'm sure you'll figure it out.

Regards,

Cory

Can you determine what the field in question is?

More generally, given just the dots and arrows of a category (no taking sections!) and the information that for some unknown ring $$R$$ the category is the category of

• Classical varieties over $$R$$
• Affine schemes over $$R$$
• Schemes over $$R$$
• etc.

in what cases is there a unique ring $$R$$ (up to isomorphism) which produces my category (up to equivalence)?

(I posted this on stack exchange, but it seems like a reasonable overflow question so I am crossposting it: https://math.stackexchange.com/questions/3002833/can-one-recover-an-algebraically-closed-field-k-from-its-category-of-finitely)

• I think you need to leave a question at interpersonal.stackexchange.com about your choice of post-conference gifts. :-) – LSpice Nov 19 '18 at 21:06
• I would like a brief, non-flourished and precise way to formulate the question. – YCor Nov 19 '18 at 22:21
• @YCor, this phrasing seems a little complicated. I think the intended family of questions is reasonably clear: are there any two distinct rings with equivalent categories of, say, finitely generated algebras over that ring, or schemes over spec of that ring, etc? – Sarah Griffith Nov 19 '18 at 22:56
• I agree with @Ycor --- the question as stated is entirely unclear. – Steven Landsburg Nov 20 '18 at 12:14
• @CoryGriffith That is not what you've done. You very emphatically said "only know the dots and arrows", which does not by any means sound the same as knowing the category up to equivalence. – Kevin Carlson Nov 20 '18 at 23:38

I think the following works for any commutative ring: Consider the category of abelian group objects in the overcategory above the initial element. This is equivalent to the category of finitely generated $$R$$-Modules (see for example https://ncatlab.org/nlab/show/module). By the general theory of Morita equivalence, this determines $$R$$.

• Specifically, you can recover $R$ as the endomorphism ring of the identity endomorphism on the category of finitely generated $R$-modules: see qchu.wordpress.com/2012/02/06/… for details. If $R$ is noncommutative this recovers $Z(R)$. – Qiaochu Yuan Nov 19 '18 at 22:58

Here is a crude attempt. I am working with the opposite category: affine schemes of finite type over $$k$$. We make the following definitions.

Let $$T$$ be the terminal object. Let us call a map $$X\to Y$$ injective/surjective/bijective/constant if it becomes so after taking $${\rm Hom}(T, -)$$. Note that bijective does not imply isomorphism.

Say that $$X$$ is irreducible if whenever we have a map $$Y\sqcup Z\to X$$ which is surjective then one of $$Y\to X$$, $$Z\to X$$ is surjective.

Say $$X$$ is a fat point if $$X\to T$$ is bijective.

Say $$X$$ is an irreducible curve if whenever we have an injective $$Y\to X$$ with $$Y$$ irreducible, then either $$Y$$ is a fat point or there exist finitely many maps $$T_i\to X$$ ($$i=1, \ldots, r$$) with $$T_i\simeq T$$ such that $$Y\sqcup T_1 \sqcup \ldots \sqcup T_r \to X$$ is bijective.

Finally, say $$A$$ is an affine line if it is an irreducible curve and has the property that every irreducible curve $$Y$$ admits a map $$Y\to A$$ which is not constant.

Now, let $$A$$ be an affine line. Fix two different points $$0\colon T\to A$$ and $$1\colon T\to A$$. There exists a unique structure of a ring object on $$A$$ with $$0$$ as zero and $$1$$ as one.

Then $$k = {\rm Hom}(T, A)$$ as rings.

Added later. One can extend this to perfect fields:

Say that $$S$$ is a separable field extension if it is not the disjoint union of two non-initial objects, and if $$S\times S$$ is isomorphic to a finite disjoint union of copies of $$S$$.

We change the first definition for $$k$$ algebraically above to read:

Let us call a map $$X\to Y$$ injective/constant if it becomes so after taking $${\rm Hom}(S, -)$$ for every separable field extension. Call it surjective if for every $$S\to Y$$ from a separable field extension there exists a separable field extension $$S'$$ and a commutative square $$S'\to S, \quad S'\to X \quad X\to Y, \quad S\to Y$$ (forgot how to draw diagrams). Call it bijective if it is both injective and surjective.

Now the rest of the definitions is unchanged, except that in the definition of an irreducible curve we allow the $$T_i$$ to be separable field extensions.