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Background

Let $\mathcal{F} : \textbf{CRing} \to \textbf{Set}$ be a functor and denote by $\textbf{P}_\mathcal{F}$ the category of points of $\mathcal{F}$ whose objects are pairs $(R , \rho)$ where $R$ is a ring and $\rho : h^R \to \mathcal{F}$ is a morphism (here $h^R := \operatorname{Hom}_{\textbf{CRing}}(R , -) \in [\textbf{CRing} , \textbf{Set}]$ is the Yoneda embedding $h^-$ applied to $R$), and whose morphisms $(R , \rho) \to (S , \sigma)$ are ring homomorphisms $\varphi: R \to S$ such that $\rho \circ h^\varphi = \sigma$. Define the diagram $D_\mathcal{F} : \textbf{P}_\mathcal{F}^\text{op} \to \textbf{LRS}$ to be the composition of the forgetful functor $\textbf{P}_\mathcal{F}^\text{op} \to \textbf{CRing}^\text{op}$ and the spec functor $\operatorname{Spec} : \textbf{CRing}^{\text{op}} \to \textbf{LRS}$. Explicitly it sends $(R , \rho) \mapsto \operatorname{Spec}(R)$.

In Demazure & Gabriel's Introduction to Algebraic Geometry and Algebraic Groups, the geometric realization of the functor $\mathcal{F}$ is defined by $| \mathcal{F} | := \operatorname{colim} D_\mathcal{F}$. In this book it is proved that the geometric realization functor $| - | : [\textbf{CRing} , \textbf{Set}] \to \textbf{LRS}$ is left adjoint to the functor $\textbf{LRS} \to [\textbf{CRing} , \textbf{Set}]$ which sends $X \mapsto \operatorname{Hom}_{\textbf{LRS}}(\operatorname{Spec}(-) , X)$.

In particular if $\mathcal{F}$ is representable by some scheme $Y$, i.e. $\mathcal{F} \cong \operatorname{Hom}_{\textbf{LRS}}(\operatorname{Spec}(-) , Y)$, then I think that $|\mathcal{F}| \cong Y$.

Question

My question is how does one go about computing the geometric realization of a functor which is not representable by schemes? For example what is the geometric realization of the functor $R \mapsto R^{\oplus \mathbb{N}}$? (This is basically the only non-representable functor I know.)

Small colimits in $\textbf{Set}$ are easy to construct (take the disjoint union and mod out by some equivalence relation), but the colimit in question here is certainly not small. Of course Demazure & Gabriel use Grothendieck universes throughout their book, so then $\textbf{P}_\mathcal{F}$ is a small category, however I'm not sure how to find a suitable Grothendieck universe to explicitly calculate this colimit for my example.

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    $\begingroup$ You need to pay more attention to the definitions being used. Demazure and Gabriel consider functors $\mathcal{M} \to \mathcal{E}$, where $\mathcal{M}$ is a category of commutative rings that is small relative to $\mathcal{E}$, the category of sets. In particular, every functor $\mathcal{M} \to \mathcal{E}$ is a colimit of a small diagram of representables, as usual. $\endgroup$ – Zhen Lin Mar 23 at 22:28
  • $\begingroup$ Cool, thanks! I was hoping to get away with "ignoring set theoretic issues" (as people often say) since I don't really understand them, but it seems I need to bite the bullet! We can restrict the functor $R \mapsto R^{\oplus \mathbb{N}}$ to any such $\mathcal{M}$, though, and then compute the geometric realization of this functor by taking a small colimit, right? Does this depend on our choice of $\mathcal{M}$? (Perhaps this is answered in your paper "Universes for category theory"). And if not, what locally ringed space do we get? $\endgroup$ – Nate Gallup Mar 23 at 22:43
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    $\begingroup$ I do not know about your specific functor. It can happen that the colimit depends on the choice of $\mathcal{M}$ – there is a somewhat infamous example involving the fpqc sheafification of a certain presheaf. What I discuss in my paper is more about the choice of $\mathcal{E}$ – thankfully enlarging $\mathcal{E}$ is usually harmless. By the way, using universes is a way of ignoring set-theoretic issues: not using universes is much more delicate! $\endgroup$ – Zhen Lin Mar 23 at 23:28
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If your goal is to read Demazure and Gabriel's book, then (as I explained in the comments) your question is based on false premises and the solution is to read the definitions carefully. But let me address your question as written, since it will illuminate why Demazure and Gabriel use the definitions they do.

First, as you observe, the category of elements of an arbitrary functor $F : \textbf{CRing} \to \textbf{Set}$ is not always small. Actually, it is almost never small, because $\textbf{CRing}$ itself is not small: as soon as $F (A)$ is non-empty for all rings $A$, then the category of elements of $F$ will be at least as big as $\textbf{CRing}$. This is not actually fatal for the problem at hand (though it does introduce many complications). It sometimes happen that the functor $F$ you are interested in is a colimit of a small diagram of representable functors, i.e. there is a small diagram $A : \mathcal{I}^\textrm{op} \to \textbf{CRing}$ such that $F (B) \cong \varinjlim_\mathcal{I} \textbf{CRing} (A, B)$ naturally in $B$. In that case, you can compute the colimit $\left| F \right|$ you seek as $\varinjlim_\mathcal{I} \operatorname{Spec} A$.

The biggest complication is that $\left| F \right|$ is not well defined for arbitrary $F$, so you only get a partially defined functor $\left| - \right|$. If you try to restrict to a full subcategory of functors $\textbf{CRing} \to \textbf{Set}$ on which $\left| - \right|$ is well defined everywhere, you then have the complication that the putative right adjoint may not have image contained in that subcategory. I am not aware of any good way to resolve this dilemma; I think you have no choice but to settle for a partially defined adjoint.

Now for some good news: there is a clean necessary and sufficient condition for a functor $F : \textbf{CRing} \to \textbf{Set}$ to be a colimit of a small diagram of representable functors.

Definition. Let $\kappa$ be an infinite regular cardinal. A $\kappa$-accessible functor is a functor that preserves $\kappa$-filtered colimits.

Proposition. Let $F : \textbf{CRing} \to \textbf{Set}$ be a functor. The following are equivalent:

  • $F$ is $\kappa$-accessible.
  • $F$ is the left Kan extension of a functor $\textbf{CRing}_\kappa \to \textbf{Set}$ along the inclusion $\textbf{CRing}_\kappa \hookrightarrow \textbf{CRing}$, where $\textbf{CRing}_\kappa$ is the full subcategory of $\kappa$-presentable rings (i.e. rings presentable by $< \kappa$ generators and $< \kappa$ relations).
  • There is a small diagram $A : \mathcal{I}^\textrm{op} \to \textbf{CRing}$ such that $F \cong \varinjlim_\mathcal{I} \textbf{CRing} (A, -)$ and, for each $i$ in $\mathcal{I}$, $A (i)$ is a $\kappa$-presentable ring.

The functor $R \mapsto R^{\oplus \mathbb{N}}$ you mention is easily seen to preserve filtered colimits (i.e. be an $\aleph_0$-accessible functor). It is just as easy to see that it is the colimit of a small (indeed, countable!) diagram of representable functors, namely, $$\textbf{CRing} (\mathbb{Z}, -) \longrightarrow \textbf{CRing} (\mathbb{Z} [x_1], -) \longrightarrow \textbf{CRing} (\mathbb{Z} [x_1, x_2], -) \longrightarrow \cdots$$ where the maps are the ones induced by the homomorphisms $\mathbb{Z} [x_1, \ldots, x_n, x_{n+1}] \to \mathbb{Z} [x_1, \ldots, x_n]$ that send $x_i$ to $x_i$ for $1 \le i \le n$ and $x_{n+1}$ to $0$. Thus, the geometric realisation of $R \mapsto R^{\oplus \mathbb{N}}$ is the colimit $\varinjlim_n \mathbb{A}^n$.

I suppose I owe you an example of a functor $\textbf{CRing} \to \textbf{Set}$ that is not accessible. Choose an ordinal-indexed sequence of fields, $K_\alpha$, such that $K_\alpha$ is strictly smaller in cardinality than $K_\beta$ whenever $\alpha < \beta$. Let $F (R) = \coprod_{\alpha} \textbf{CRing} (K_\alpha, R)$ for non-zero rings $R$ and let $F (\{ 0 \}) = 1$. Since any ring homomorphism $K_\alpha \to R$ is injective when $R$ is non-zero, $\textbf{CRing} (K_\alpha, R)$ is empty for sufficiently large $\alpha$, so $F (R)$ is indeed a set. On the other hand, it is clear that $F$ cannot be the left Kan extension of any functor $\textbf{CRing}_\kappa \to \textbf{Set}$: if it were, it would be impossible to distinguish between this $F$ and the one where we cut off the disjoint union at some ordinal $\beta$ such that $K_\beta$ is not $\kappa$-presentable.

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  • $\begingroup$ Thank you so much for this excellent answer! It indeed is extremely helpful for understanding the motivations of D&G, which I will now read much more closely! $\endgroup$ – Nate Gallup Mar 24 at 14:23

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