[Wellpowered](http://ncatlab.org/nlab/show/well-powered+category) 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](http://ncatlab.org/nlab/show/regular+monomorphism) $Y \to X$ (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](https://mathoverflow.net/a/159995/2362) where it was clarified that the category of schemes is [concretizable](http://ncatlab.org/nlab/show/concrete+category) (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](https://mathoverflow.net/questions/56591/what-are-the-monomorphisms-in-the-category-of-schemes) quotes SGA giving a characterization of monomorphisms *locally of finite type*, but not arbitrary monomorphisms. [This MO question](https://mathoverflow.net/questions/66279/regular-monomorphisms-of-schemes?) 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.



Is the category of schemes wellpowered? regularly wellpowered?
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. 


