Is the functor of points of a scheme cofinally small?

Question

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?

Answer 1

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.)