Complements of subobjects and algebraic geometry

Question

This is an idle question, I don't really have any particular application in mind. In any category $C$, for any object $c \in C$ we can consider its subobjects, namely monomorphisms $d \hookrightarrow c$. If the objects of $C$ are "sufficiently spacelike" we might hope that subobjects have complements, in the following sense. Say that two subobjects are disjoint if their intersection (pullback) is trivial (the initial object).

Definition: The complement of a subobject $d \hookrightarrow c$, if it exists, is the terminal subobject $\neg d \hookrightarrow c$ which is disjoint from $d$.

This recovers the usual notion of the complement of a subset of a set, for example. The definition makes sense more generally in any preorder and if applied to a Heyting algebra, such as the open subsets of a topological space, recovers the Heyting complement.

Here is the example which motivates this question. Take $C$ to be the category of functors $\text{CRing} \to \text{Set}$ and consider first the usual forgetful functor $U$ which sends a commutative ring $R$ to its underlying set. This functor has a subfunctor $\{ 0 \}$ sending a commutative ring $R$ to the set consisting of its additive identity $ \{ 0 \}$. Now we can ask: what is the complement of this subfunctor, if any?

There is a "naive complement" given by $R \mapsto R \setminus \{ 0 \}$, but this is not a functor, because elements that are nonzero in some commutative ring may become zero after applying a homomorphism. So we have to "functorialize" this construction: we need to know what elements $r$ of a commutative ring $R$ are not only nonzero but remain nonzero after the application of any nonzero homomorphism. (Something goes wrong here with the zero ring; I am just going to ignore this for now.) This is equivalent to requiring that $r$ is not in any proper ideal, which means that the ideal $(r)$ generated by $r$ must be the unit ideal; hence this condition is equivalent to requiring that $r$ be invertible. We conclude that the complement of the zero subfunctor is the group $R^{\times}$ of units.

In terms of functors of points, this lets us describe abstractly the sense in which the punctured affine line $\mathbb{A}^1 \setminus \{ 0 \} = \text{Spec } \mathbb{Z}[x, x^{-1}]$ is the complement of the origin in $\mathbb{A}^1$!

Now we can take complements again, and ask: what is the complement of $R^{\times}$ in $R$? (Something again goes wrong here with the zero ring, which I am again going to ignore.) There is a "naive complement" $R \mapsto R \setminus R^{\times}$ sending a ring to its non-units, it is not a functor because a non-unit may become a unit after applying a homomorphism, so we need to "functorialize" and understand what elements $r \in R$ are not only non-units but remain non-units after the application of any nonzero homomorphism. If $r$ is not contained in some prime ideal $P$ then it becomes a unit in the fraction field $\text{Frac}(R/P)$; therefore $r$ must be contained in every prime ideal, and hence in their intersection, which is the nilradical. So $r$ must be nilpotent. Conversely nilpotents clearly satisfy the desired property, and so the complement of $R^{\times}$ is $\text{Nil}(R)$.

$\text{Nil}(R)$ is not representable but it is close: it is "pro-representable" by the rings $\mathbb{Z}[x]/x^n$ which organize into a cofiltered limit $\mathbb{Z}[[x]]$ (equipped with the $x$-adic topology). Geometrically this says that the complement of the punctured affine line in $\mathbb{A}^1$ is not the origin again but its formal neighborhood, which is quite nice and seems to have something to do with the Beauville-Laszlo theorem.

If only I knew what was going on with the zero ring! The problem is that all of these functors assign the zero ring its unique element $\{ 0 \}$ (since it is both invertible and nilpotent), so none of them are technically disjoint!

So, now my actual

Question: Has anyone seen this notion of the complement of a subobject applied to functors of points in algebraic geometry in particular? Also, what's going on with the zero ring above and how do we fix it?

I have not thought about whether we should really be working in the category of Zariski sheaves or whatever else; if someone can explain how that might fix the zero ring issue I'd be grateful.

Edit: Okay, I have a guess. Everything would be fixed if we worked in Zariski sheaves and the initial sheaf was not the initial presheaf, but was instead the sheaf that assigns the empty set to every nonzero ring but assigns $1$ to the zero ring. That alters the definition of disjointness in exactly the right way to allow all of the above functors to be disjoint. Is that right? Then we'd just need to check that $\text{Nil}(R)$ is a Zariski sheaf, which I think is true?

Answer 1

I don't know if you're still interested in the question and this might not be a very satisfactory answer, but these notes for a course taught by Sam Raskin explicitly use your notion of complement in the context of functorial algebraic geometry. He doesn't give your general definition but applies the construction directly to functors $\mathbf{CRing}\to\mathbf{Set}$.

Concerning the zero ring, since he is defining Zariski topology, he can't use Zariski sheaves from the start, so his solution is a bit more brutal: he simply stipulates that $\mathbf{CRing}$ is the category of non-zero commutative rings, then he defines an affine scheme as either a representable presheaf on $\mathbf{CRing}^{\mathrm{op}}$ or the empty presheaf. (The latter will turn out to be a sheaf because now $\mathbf{CRing}^{\mathrm{op}}$ does not have an initial object). He goes on to define Zariski-closed subpresheaves of affine schemes (as duals to ring epimorphisms $R\to R/I$), then Zariski-closed subpresheaves of arbitrary presheaves (as those morphisms which pulled back along any map from an affine give a closed morphism), and finally Zariski-open subpresheaves of arbitrary presheaves as complements of closed subpresehaves, using exactly the definition you give in your question.

As an aside, one surprising thing to me when I first saw these notes is that he doesn't define sheaves in the "orthodox" way, that is, via a Grothendieck topology on $\mathbf{CRing}^{\mathrm{op}}$, but gives instead a direct definition using covers by open subpresheaves. The definition seems harder to work with than the usual one, at least to a non-expert like me, but it's probably just an impression (the proofs are often too sketchy to judge, and I never went very far in the notes anyway).

Answer 2

In the earliest version of my thesis I actually tried reconstructing the Zariski topology based on this observation. It works, but to be honest I have my doubts about whether it can be generalised.

Anyway, you can see how to "fix" the definition of complement in definition 2.9 of these notes. The key point is that $\textbf{CRing}^\textrm{op}$ already has a strict initial object and we want our notion of disjointness to respect that, so we say that two sub(pre)sheaves are disjoint if their intersection is inhabited only over (strict) initial objects of $\textbf{CRing}^\textrm{op}$. This basically amounts to restricting to sheaves for the Grothendieck topology where the empty family covers $\operatorname{Spec} {\{ 0 \}}$. It is equivalent to just omit $\{ 0 \}$ from the category of rings – if you accept the law of excluded middle.

Yves Diers has done a lot of work trying to generalise algebraic geometry by looking at "codisjunctors", which are basically representable (pseudo)complements.