# Can the Category of Schemes be Concretized?

If not, are there any interesting subcategories that can be concertized? If I am not mistaken, the category of reduced finite type varieties over the complex numbers would be an example, where the forgetful functor to sets would be given by looking at the underlying map of points.

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Can you define "concretized"? It's not a common term in category theory. Sometimes people call a category concrete if there exists (or if it comes equipped with) a faithful functor to the category of sets. But usage varies. –  Tom Leinster Oct 23 '09 at 2:45
Yup, that's exactly what I mean. –  Dinakar Muthiah Oct 23 '09 at 2:50
@Tom: I think "concretized" is a great word. The "concrete category" should mean "category equipped with a faithful functor to SET", rather than the existence of such a thing. –  Theo Johnson-Freyd Oct 23 '09 at 4:55

The category of schemes is not small-concrete.

Let $S$ be a generating set. Let $U$ be the set of all rings $A \neq 0$ such that $\mathrm{Spec}(A)$ is an open subscheme of a scheme in $S$. Let $X$ be a set whose cardinality is larger than any element of $U$, for example, $2^{\bigsqcup_{A \in U} A}$. Let $K$ be the field $\mathbb{Q}(t_x)_{x \in X}$, where $t_x$ are a collection of algebraically independent generators indexed by $X$. So $|K|$ is larger than $|A|$ for any $A \in U$. Since ring maps from a field to a nontrivial ring are always injective, $\mathrm{Hom}(\mathrm{Spec}(A),\mathrm{Spec}(K))=\emptyset$ for every $A \in U$, and therefore $\mathrm{Hom}(s,\mathrm{Spec}(K))=\emptyset$ for every $s \in S$.

There is only one map from the empty set to itself. But $\mathrm{Spec}(K)$ has nontrivial isomorphisms, coming from permuting the generators. So

$\mathrm{Hom}(\mathrm{Spec}(K),\mathrm{Spec}(K)) \longrightarrow \mathrm{Hom}_{\mathrm{Set}^{S^\mathrm{op}}}( (\mathrm{Spec}(K))(-), (\mathrm{Spec}(K))(-))$

is not injective.

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Should the second U in the second sentence be an S ? –  Peter Arndt Oct 26 '09 at 12:24
Fixed, thank you very much! –  David Speyer Oct 26 '09 at 13:34

I'd like to suggest that this isn't quite the right question. At least, it seems to me that modifying the question (in a direction that Theo was hinting) would be more interesting.

The problem with the question as asked is that, for a given category $C$, the mere existence of a faithful functor $C \to \mathbf{Set}$ tells you very little indeed. Perhaps you have some reason for wanting to know that I can't see. But a condition that seems to have more bite is 'small-concreteness', defined as follows.

Let C be a category. A set-valued functor $U: C \to \mathbf{Set}$ is small if it can be expressed as a small colimit of representables. Call a category $C$ small-concrete if there exists a small, faithful functor $C \to \mathbf{Set}$. In the special case that $C$ is small, all set-valued functors on $C$ are small and small-concrete = concrete.

It's not too hard to show that a category is small-concrete if and only if it admits a generating set. (A generating set in a category $C$ is a [small] set $S$ of objects such that, for any distinct maps $f, g: a \to b$ in $C$, there exist $s \in S$ and $q: s \to a$ such that $fq \neq gq$.) The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see Categories for the Working Mathematician.

You can exploit this as follows. Suppose you want to show that the category of affine schemes is not small-concrete (which would imply that the category of all schemes isn't either). Assuming for a contradiction that it is small-concrete, the category $\mathbf{Ring}$ of commutative rings has a cogenerating set. Since $\mathbf{Ring}$ is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from $\mathbf{Ring}$ to a locally small category has a left adjoint. I guess it's possible to cook up (or look up) an example of a limit-preserving functor out of $\mathbf{Ring}$ that doesn't have a left adjoint. That would produce the desired contradiction.

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What is an example of a continuous functor $\mathsf{Ring} \to \mathsf{Set}$ which doesn't have a left adjoint? –  Martin Brandenburg Sep 6 '13 at 9:16
Just a quick idea for a candidate. Start with a continuous functor $\Phi:\mathfrak{Group}\rightarrow\mathfrak{Set}$ which has no left adjoint (for example, the product of $Hom_{\mathfrak{Group}}(\Gamma_\alpha,\cdot)$ where $\Gamma_\alpha$ is a simple group of cardinality $\aleph_\alpha$) and compose with the group of units functor $U:\mathfrak{Ring}\rightarrow\mathfrak{Group}$ which has left adjoint given by the group ring functor $Z:\mathfrak{Group}\rightarrow\mathfrak{Ring}$. –  Adam Epstein Sep 6 '13 at 10:49
@AdamEpstein: Unfortunately, this idea doesn't quite work since $U$ factors through the continuous inclusion $i: \textbf{Ab} \to \textbf{Grp}$, and continuous functors of the form $\textbf{Ab} \to \textbf{Set}$ (including for example $\Phi \circ i$) are representable, hence are right adjoints. I'm hoping the basic idea can be modified (and tried and deleted something, which those with 10k rep can see). –  Todd Trimble Sep 10 '13 at 2:33
(Undeleted now; I hope I've fixed my earlier try.) –  Todd Trimble Sep 10 '13 at 3:15
Indeed. I flubbed the obvious point of how $\Phi$ is all about nonabelian groups. –  Adam Epstein Sep 10 '13 at 10:41

This "answer" is meant to supplement Tom Leinster's answer, and is really in response to Martin Brandenburg's comment below Tom's answer, where he asks for an example of a continuous (i.e., limit-preserving) functor $\textbf{CRing} \to \textbf{Set}$ that is not a right adjoint. Adam Epstein's idea suggested the following possibility.

Choose, for each infinite cardinal $\alpha$, a field $F_\alpha$ of that cardinality (say, for definiteness, the characteristic zero algebraically closed field of transcendence degree $\alpha$ over $\mathbb{Q}$), and put $A_\alpha = \mathbb{Z} \times F_\alpha$. Each non-trivial quotient ring (corresponding to a regular epi) of $A_\alpha$ either contains a copy of $F_\alpha$, or is a quotient ring of $\mathbb{Z}$ (possibly $\mathbb{Z}$ itself).

This has the following consequence: for any (commutative) ring $R$ of cardinality less than $\alpha$, there is exactly one map $f: A_\alpha \to R$. For if we have a (regular epi)-mono factorization $A_\alpha \to Q \to R$ where $Q \to R$ is monic, then the possibility where $Q$ contains a copy of $F_\alpha$ is ruled out, hence the factorization must take the form

$$A_\alpha \stackrel{\text{epi}}{\to} \mathbb{Z}/(n) \stackrel{\text{mono}}{\to} R$$

where $(n)$ is uniquely determined as the annihilator of the identity in $R$.

Now form the functor

$$G = \prod_{\alpha \in \text{Card}} \hom(A_\alpha, -): \textbf{CRing} \to \textbf{Set}$$

As soon as $\alpha \gt \text{Card}(R)$, we have that $\hom(A_\alpha, R)$ is a one-element set. Thus for each $R$, $G(R)$ is a set even though $G$ itself is a class-sized product. Being a product of continuous functors, $G$ is continuous. But $G$ cannot be representable (just by simple cardinality considerations; e.g., $G(A_\alpha)$ has size greater than $\alpha$, for any $\alpha$, since algebraically closed fields have lots of automorphisms).

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Very nice example. –  Adam Epstein Sep 10 '13 at 10:47