# Is the functor of points of a scheme cofinally small?

Background: In functorial algebraic geometry one would like to consider the category of all functors $\mathsf{CRing} \to \mathsf{Set}$ and define/characterize the category of schemes as a full subcategory. However, there are serious set-theoretic difficulties: this is not a category, it is too large. One possible solution is suggested in Demazure-Gabriel's book. They use three nested universes. I wonder if we may stay in one universe. In general, one constructs the cocompletion $\widehat{\mathcal{C}}$ of a category $\mathcal{C}$ (not assumed to be small) as the category of functors $F : \mathcal{C}^{\mathrm{op}} \to \mathsf{Set}$ which are cofinally small, i.e. the category of elements $\int F$ is cofinally small. Then $\widehat{\mathcal{C}}$ is a category. We may apply this to $\mathcal{C}^{\mathrm{op}}=\mathsf{CRing}$ and develop functorial algebraic geometry inside $\widehat{\mathcal{C}}$. But the question is if the usual category of (geometric) schemes embeds into it. This leads to the following questions:

Question. Let $X$ be a scheme. Is there a set of commutative rings $\{A_i\}$ with the property that for every commutative ring $A$, every morphism $\mathrm{Spec}(A) \to X$ factors through some $\mathrm{Spec}(A_i)$?

The answer is yes if $X$ is quasi-compact quasi-separated. In fact, in that case $X(-) : \mathsf{CRing} \to \mathsf{Set}$ commutes with filtered colimits so that we may take a skeleton of all finitely generated commutative rings (which is a set). If $X$ is arbitrary, I guess that $X(-)$ commutes with $\lambda$-directed colimits for some cardinal number $\lambda$, which would answer the question. But I'm not sure about that.

Question. In case the answer to the first question is "Yes": Can we choose that set in such a way that for every $\mathrm{Spec}(A) \to X$ the category of morphisms $\mathrm{Spec}(A) \to \mathrm{Spec}(A_i)$ over $X$ is connected?

• I haven't thought about your question but there is something else you can do to stay in only one universe : when $C$ is a category endowed with a Grothendieck topology (here $C = Ring^{op}$ with the Zariski topology) you might want to consider the category of "small sheaves", that is the category of sheaves which are small colimit of representable sheaves (when the topology is trivial it gives back $\hat{C}$). If your topology satisfies a smallness condition (basically that the class of covering sieve of any object is generated by a set) then the category of small sheaves is locally small. – Simon Henry Jan 19 '15 at 9:56
• I might be very confused w.r.t. the first question, but what about the set $\{B\}$ if $\operatorname{Spec}(B)\rightarrow X$ is an atlas? – Fernando Muro Jan 19 '15 at 10:13
• @Fernando: Well, this doesn't work. Not every morphism $\mathrm{Spec}(A) \to X$ factors through $\mathrm{Spec}(B)$ when $\mathrm{Spec}(B) \to X$ is an atlas. – Martin Brandenburg Jan 19 '15 at 10:20
• @Simon Henry: This is equivalent to the condition of being cofinally small. – Martin Brandenburg Jan 19 '15 at 10:21
• Being a small colimit of representable in the category of presheaves is equivalent to being cofinally small. What I mean is "being a small colimit of representable in the categories of sheaves" that is why we need a condition on the topology to obtain a locally small category. – Simon Henry Jan 19 '15 at 10:32

Maybe I'm missing something, but it seems to me like the first question is just a simple definition chase. A morphism $\mathrm{Spec}(A)\to X$ is determined by finitely many affine open subsets $\mathrm{Spec}(B_n) \subseteq X$ and some ring maps $B_n\to A_{f_n}$ to localizations of $A$ satisfying certain compatibility properties. There are not so many elements of $A$ that are needed to witness all of this. Specifically, we need to witness the $f_n$, numerators of everything in the images of the maps $B_n\to A_{f_n}$, some elements annihilated by powers of the $f_n$ that witness that the maps $B_n\to A_{f_n}$ are actually homomorphisms and are appropriately compatible, and elements of $A$ that witness that some $f_n$ are in the ideal generated by others (and that $1$ is in the ideal generated by all of them). In total, it is clear that some subring $A_0\subseteq A$ of cardinality at most $\aleph_0+\sum |B_n|$ contains all the needed witnesses, and our morphism will factor through $\mathrm{Spec}(A_0)$.
For the second question, consider two factorizations $\mathrm{Spec}(A)\to\mathrm{Spec}(A_i)\to X$ ($i=0,1$) where $|A_i|<\kappa_0$ for some cardinal $\kappa_0$ depending only on $X$. Let $Y=\mathrm{Spec}(A_0)\times_X \mathrm{Spec}(A_1)$; this may not be affine, but it will still satisfy some cardinality bound depending on $X$. Applying the previous paragraph with $Y$ in place of $X$, we can factor the map $\mathrm{Spec}(A)\to Y$ through $\mathrm{Spec}(A_2)$, and we can bound $A_2$ by some cardinal $\kappa_1$ depending only on $X$. Now repeat this argument allowing $A_0$ and $A_1$ to have cardinality up to $\kappa_1$, and get some new bound $\kappa_2$ on the cardinality of $A_2$. Repeating this inductively, we get a sequence of cardinals $\kappa_n$; let $\kappa=\sup \kappa_n$. Then whenever $|A_0|<\kappa$ and $|A_1|<\kappa$, we can find $A_2$ with $|A_2|<\kappa$ that connects them. (Actually, I'm pretty sure that if you choose $\kappa_0$ in the obvious way, $\kappa_1$ will just be the same as $\kappa_0$, so none of this iteration is actually necessary.)
• Why are there finitely many affine open subsets $\mathrm{Spec}(B_{n}) \subset X$? Or do they depend on $A$? – jmc Jan 19 '15 at 12:04
• They depend on the particular morphism $\mathrm{Spec}(A)\to X$, but there are only a set of different affine open subsets of $X$ in total, so there are only a set of possible values of the $B_n$. – Eric Wofsey Jan 19 '15 at 12:07
• Thank you for adding the second part. I still have to digest all this. In the first part, I think we have to be a little bit more careful as for the compatibility properties. Not every subring of $A$ such that the $B_n$ factors through localizations will induce a morphism. – Martin Brandenburg Jan 19 '15 at 13:53