# Is the category of schemes wellpowered? regularly wellpowered?

Wellpowered means that for every scheme $X$, the subobject lattice of monormophisms $Y \to X$ is essentially small; regularly wellpowered means that for every scheme $X$, the regular subobject lattice of regular monomorphisms $Y \to X$ (being a regular mono means that $Y \to X$ is the equalizer of some pair of maps) is essentially small. Wellpoweredness implies regular wellpoweredness, but not conversely.

Let me sheepishly admit that I ask this question knowing next to nothing about algebraic geometry. My motivation comes from this MO discussion where it was clarified that the category of schemes is concretizable (i.e. admits a faithful functor to $\mathbf{Set}$) if and only if it is regularly wellpowered.

Here's what I know: This MO question quotes SGA giving a characterization of monomorphisms locally of finite type, but not arbitrary monomorphisms. This MO question indicates that every regular mono is a locally closed immersion. So I would be very happy if someone could tell me whether a scheme can have a large number of locally closed immersions into it.

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I've accepted Laurent's answer which shows regular wellpoweredness but leaves general wellpoweredness open. This is because I was more interested in the regular case anyway. If anyone has further observations concerning general monomorphisms of schemes, I'd be interested to hear about it: I suspect there is a wide range of behavior possible which might be regarded as "pathological" for the purposes of actually doing algebraic geometry. But schemes might nevertheless be wellpowered: after all, monomorphisms of affine schemes are complicated, but $\mathbf{Aff}$ is still wellpowered. –  Tim Campion Mar 18 at 21:09
Do you have a reference for the wellpoweredness of $\mathbf{Aff}$? –  Laurent Moret-Bailly Mar 19 at 8:45
Well, it's equivalent to say that the category of commutative rings is co-wellpowered: each object has a set of equivalence classes of epimorphisms out. In fact, every locally presentable category (including the category of rings) is co-wellpowered, although this is nontrivial. It is proved e.g. in Chapter 1.D of Adámek and Rosický, Locally Presentable and Accessible Categories. –  Tim Campion Mar 19 at 14:30
For rings in particular, Andrej Bauer gives a more direct argument here. Ring epis are subtle (as opposed to regular ring epis, which are just quotients), and were apparently extensively studied by the Séminaire Samuel -- see this volume devoted to them. In particular, the first article after the introduction, by Mazet, apparently provides the basis for Andrej's argument: a ring epi, though not necessarily surjective, can't have a domain with larger cardinality than its codomain. –  Tim Campion Mar 19 at 14:35
But this implies the same property for the category of schemes; in fact this is stated here but I could not find a proof there, so I am giving one in a second answer. –  Laurent Moret-Bailly Mar 19 at 19:59

Now I am told that the category of affine schemes is wellpowered (equivalently, the category of commutative rings is cowellpowered). Let me deduce from this that the category of schemes is wellpowered. So let $X$ be a scheme. For each monomorphism $f:Y\to X$ you can find an affine open covering $(V_i)_{i\in I}$ of $Y$, and a family $(U_i)_{i\in I}$ of affine open subschemes of $X$ such that $f(V_i)\subset U_i$ for all $i$. Clearly you can bound the cardinality of $I$ by that of the underlying space of $X$. Each induced map $V_i\to U_i$ is a monomorphism, so by the result on affine schemes the set of possible data $(I,(V_i\to X))$ is essentially small. Hence it suffices to prove that the family $(V_i\to X)$ determines $Y$. In fact:

Claim. $Y=\sup_{i\in I} V_i$ (as a subobject of $X$).

Proof. Clearly $V_i\leq Y$ for each $i$. Conversely, if $Z\to X$ is a subobject containing each $V_i$, the (unique) $X$-morphisms $V_i\to Z$ must agree on each intersection $V_i\cap V_j$ ($i,j\in I$) since $V_i\cap V_j\to X$ is a monomorphism. But since $(V_i)$ is an open covering of $Y$, this gives rise to a (unique) $X$-morphism $Y\to Z$. QED.

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Very nice! As a technical point: how do you get just one $V_i$ for each $U_i$? I see that I could take an open affine cover of each $f^{-1}(U_i)$ and the argument would still work, but I would end up with more $V$'s than $U$'s. –  Tim Campion Mar 20 at 0:51
You can take $I=\vert Y\vert$ (the underlying space of $Y$, which is smaller than $\vert X\vert$). For each $y\in I$, choose $V_y$ and $U_y$ such that $y\in V_y$. (We don't need the $U$'s to cover $X$). –  Laurent Moret-Bailly Mar 20 at 7:35
Oh -- of course: we choose $V_y$ as a neighborhood of $f(y)$, and then choose $U_y$ as an affine neighborhood of $y$ in $f^{-1}(U_y)$. I was thinking that maybe it came down to a principle like "the image of an affine scheme is always contained in an affine open set," but I guess this may be false in general. Thanks for taking the time to address elementary issues, especially since everything has turned out to be elementary here modulo the affine case! –  Tim Campion Mar 20 at 13:10

If I understand the question correctly, the category of schemes is regularly wellpowered.

A locally closed immersion factors as $Y\xrightarrow{i}U\xrightarrow{j}X$ where $j$ (resp. $i$) is an open (resp. closed) immersion. Now open immersions correspond bijectively to open subspaces of the underlying space of $X$ (this is clearly small), while closed immersions into $U$ are indexed by quasicoherent ideals in $\mathcal{O}_U$. These also form a small set, in fact a subset of $\prod_V 2^{\Gamma(V,\mathcal{O}_U)}$ where the product ranges over open subsets $V$ of $U$.

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Beautiful. Thanks so much! Before I accept, I'd just like to confirm that there are no hidden technical conditions required for the argument to go through. For example, these Vakil notes say (as Exercise 2D, p. 15) that a locally closed immersion factors as an open immersion followed by a closed immersion, but only under the additional hypothesis that that the morphism be quasicompact or else that the codomain be reduced. So apparently these technical conditions can be lifted? –  Tim Campion Mar 18 at 16:37
The quasicompact (or reduced) assumption is needed to embed $Y$ as an open subscheme of a closed subscheme of $X$. To factor the oher way, there is no condition (and this follows easily from the definition of an immersion). –  Laurent Moret-Bailly Mar 18 at 17:30
Great, thanks again. For completeness, here again is the link I gave in the question to Martin Brandenberg's argument that regular monos are locally closed immersions: it's because a regular mono is a pullback of a diagonal map; the latter is a l.c.i. and l.c.i.'s are stable under pullback. –  Tim Campion Mar 18 at 19:08
We could also prove this in an even more elementary way by imitating Laurent's other answer to reduce regular wellpoweredness of $\mathbf{Sch}$ to $\mathbf{Aff}$. Unlike general wellpoweredness of $\mathbf{Aff}$, regular wellpoweredness of $\mathbf{Aff}$ is easy to see: it's just the fact that a regular epi of rings is a quotient map, and there are only set-many quotients. The analog of this statement in Laurent's argument is the bound on the number of quasicoherent ideals. –  Tim Campion Mar 20 at 13:25
Unfortunately, a morphism of affine schemes which is a regular mono of schemes (i.e. an immersion) is not necessarily a regular mono in $\mathbf{Aff}$ (i.e. a closed immersion). –  Laurent Moret-Bailly Mar 20 at 14:29