# 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.

• 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
• @TheoJohnson-Freyd interesting. I think that "concrete category" should not mean anything, and what you say should be conveyed by "concretizable category". Though I am not really sure what other people think. – user138661 May 16 at 12:58

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.

• Should the second U in the second sentence be an S ? – Peter Arndt Oct 26 '09 at 12:24

Apparently, there is an abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof.

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.

• Incidentally, the same proof shows that the category of ringed spaces is concretisable. – Zhen Lin Mar 18 '14 at 19:49
• Was it really that simple all along? Very nice! – Tim Campion Mar 18 '14 at 20:17

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.

• 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).

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

Conclusion (Added 3/18/14): Even though $\mathbf{Sch}$ is not small-concretizable, it is concretizable more generally. However the construction of a faithful functor to $\mathbf{Set}$ is convoluted simple if you use Zhen Lin's approach!

The answers to this question jumped to using a modified notion of concreteness. But there are classic results about concretizability in the original sense of this question -- admitting a faithful functor to $\mathbf{Set}$. The most famous, of course is Freyd's result that the homotopy category of spaces is not concrete.

But Freyd wrote another paper on concreteness. In it he shows, (Thm 4.1.iii), that a finitely-complete category is concretizable iff it is regularly-well-powered. That is, iff every object has a small set of isomorphism classes of regular subobjects, where a regular subobject means an equalizer of two morphisms.

I know nothing about schemes, but I'll go out on a limb and assume that category is finitely complete and regularly-well-powered. Therefore it is concretizable.

EDIT (3/18/14): After discussing this issue with Zhen below, I decided it was worth its own MO question. It turns out that $\mathbf{Sch}$ is regularly wellpowered. A simple argument given by Martin Brandenberg in this question statement shows that regular monos are locally closed immersions, and Laurent Moret-Bailly showed, by a factorization into a closed and an open immersion, that the there can only be a small set of the latter into a given scheme.

(The following is superseded by Zhen's answer:) So there exists a faithful functor $\mathbf{Sch} \to \mathbf{Set}$. The obvious question is: what does it look like? Is it perhaps a certain \emph{large} colimit of representables? Unfortunately, Freyd's construction is very artificial -- it involves embedding $\mathbf{Sch}$ into an abelian category and then using a brute-force transfinite induction. Anyway, I suspect it might give some insight into the category of schemes to try to construct a more natural faithful functor to $\mathbf{Set}$.

• Finitely complete – yes. But monomorphisms are quite subtle and I would not dare to speculate on wellpoweredness... – Zhen Lin Mar 11 '14 at 8:21
• Well, in the category of rings, epimorphisms are subtle, but regular epis are surjective, so rather less pathological. In particular, affine schemes are regularly well-powered, though of course we already knew from David Speyer's answer that affine schemes are concretizable... Are there additional sources of subtlety that affect even the regular monos of schemes? – Tim Campion Mar 11 '14 at 11:14
• A scheme is much more than just a ring. It's not even obvious to me whether the underlying continuous map of a regular monomorphism has to be injective, let alone be a topological embedding; and on the ring side, it's not obvious that the morphism of structure sheaves has to be an epimorphism, let alone a regular one. – Zhen Lin Mar 11 '14 at 11:41

Here is a concretization of Ring^{op}: For any ring R, let 2^R be the set of subsets of R. Given a map f: R --> S , we define 2^f : 2^S --> 2^R by I --> f^{-1}(I). I claim that this is faithful. Proof: let f and g be two different maps R --> S. So there is some r in R with f(r) \neq g(r). But then 2^f( {f(r)} ) contains r and 2^g( {f(r)} ) does not contain r, so 2^f \neq 2^g.

I want to say that I can extend the functor 2^* to schemes by taking a directed limit over all open affines, but I am nervous about the details.

• It is a fact that a category C is concretizable iff C^op is concretizable. Your argument is a proof of this fact. However, I don't think this will still hold for Tom's more restrictive definition of small-concrete – Dinakar Muthiah Oct 23 '09 at 17:59

Ok, ignoring my deleted answer where I misread the question, I don't have a full solution (working on some details), but to extend the functor you describe much beyond actual, literal varieties over C, you'll need to handle the case of Spec(k[x]/x^2), which has the property that a map from it to a variety is a tangent vector, so you'll need to find some set S such that hom(S,V), as sets, has a map for every tangent vector to V, for any V. I don't see how to satisfy it, but I've not yet got a proof that you can't, and it doesn't say anything about functors not being the forgetful functor on C-varieties, though it gives a possible way to show that it can't extend much.