Rigidity of the category of schemes - MathOverflow

Rigidity of the category of schemes

2011-02-28T11:23:50Z

Call a category $C$ rigid if every equivalence $C \to C$ is isomorphic to the identity. I don't know if this is standard terminology. Many of the usual algebraic categories are rigid, for example sets, commutative monoids, groups, abelian groups, commutative rings, but also the category of topological spaces. The category of monoids (or rings) is not rigid because $M \mapsto M^{\mathrm{op}}$ is an equivalence which is not isomorphic to the identity. See [M] for a survey and the general strategy for proving rigidity. The case of commutative rings was discussed recently on MO here. The philosophy is that a category is rigid if every object can be defined in a categorical way, which is a quite interesting property.

Question: Is the category of schemes rigid?

Here is what I've done so far: The initial scheme is $\emptyset$ and the terminal scheme is $\text{Spec}(\mathbb{Z})$. Spectra of fields are characterized by the property that they are non-initial and and every morphism from a non-initial object to them is an epimorphism, see Kevin's answer here. The underlying set $|X|$ of a scheme is the set of equivalence classes of morphisms $Y \to X$, where $Y$ is the spectrum of a field. So this recovers $|X|$ from $X$ in a categorical manner. If $x \in |X|$, then $\text{Spec}(\kappa(x))$ is the terminal spectrum of a field which maps to $X$ and has (set) image $x$.

However, I'm not able to recover the topology from $X$. I don't know how to characterize open or closed immersions. They are exactly the étale resp. proper monomorphisms, see this MO question, but it seems to be hard to characterize étale and proper categorically. After all, if are able to characterize affine schemes, then we will be done, since the category of affine schemes is rigid and every scheme is the canonical colimit of the affine schemes mapping into it.

In order to characterize affine schemes, it is enough to characterize the ring object $\mathbb{A}^1_\mathbb{Z}$ in the category of schemes, since we can then define the ring of global sections of a scheme categorically and then say that affine schemes $Y$ are characterized by the property that for all schemes $X$ the map $Hom(X,Y) \to Hom(\mathcal{O}(Y),\mathcal{O}(X))$ is bijective.

Other approaches: 1. First show that the category of fields is rigid. I've already shown that the notions of prime field, $\mathbb{F}_p$, $\mathbb{Q}$, finite, characteristic, normal, separable, algebraic, galois, transcendent, transcendence degree are categorical, but this is not enough to distinguish, for example, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. If $F$ is a self-equivalence of the category of fields, then $F$ maps $K(X)$ to $F(K)(X)$, so taking automorphisms there is a natural isomorphism $\text{PGL}(2,K) \cong \text{PGL}(2,F(K))$, but I wonder if this already implies that $K \cong F(K)$ naturally. 2. Characterize local schemes as a special full reflective subcategory containing the spectra of fields. 3. Try to categorify cohomology theory and use Serre's criterion for affineness.

EDIT (May '11): I've restarted this project in the last days. If $k$ is a field with only trivial endomorphisms, then I can show that every self-equivalence of $\text{Sch}/k$ preserves $\text{Spec}(k[\epsilon]/\epsilon^2)$, but also $\text{Spec}(k[[t]])$. But I still have no idea how to approach $\text{Spec}(k[t])$ categorically. Even basic notions such as "closed point" or "quasicompact" remain unclear.

EDIT (Feb '12): Let's work with $\mathrm{Sch}/k$ for some algebraically closed field $k$. Then $F$ maps $\mathbb{A}^1_k$ to a ring object in $\mathrm{Sch}/k$. If we already knew that it is of finite type over $k$ and irreducible, then a Theorem by Greenberg (Cor. 4.4 in Algebraic Rings, Trans. AMS, Vol. 111, No. 3, pp. 472 - 481) will imply that the underlying scheme is just $\mathbb{A}^n_k$ for some $n$. Now using my question about factorization we should be able to conclude $n=1$. Of course, many details are missing here; for example it is not clear at all why $F$ should preserve schemes of finite type.

Any ideas concerning the categorical characterization of other properties / objects are appreciated. Feel free to add every piece as a single answer even if it does not answer the whole question.

[M] E. Makai jun, Automorphisms and Full Embeddings of Categories in Algebra and Topology, online

Answer by Harry Gindi for Rigidity of the category of schemes

2011-02-28T14:19:54Z

This is just a comment that was getting out of hand in terms of length:

Toen's notes on stacks characterize etaleness (resp. properness) categorically, but the definition is not pretty and involves some very annoying questions of representability (of maps) and dealing with similarly annoying properties of atlases. Lurie has also provided a topos-theoretic description of etaleness:

Let $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$ be $\mathcal{G}$-structured toposes by a geometry $\mathcal{G}$ (if we take this $\mathcal{G}=\mathcal{G}_{Zar}$ to be the opposite category of commutative rings of finite presentation over $\mathbf{Z}$ with admissible morphisms being maps induced by localization of a single element and the topology on the admissible subcategory given by collections of admissible morphisms $A\to A_i$ (each determined by a single element $a_i\in A$) such that their associated elements generate the unit ideal of $A$, then a $\mathcal{G}$-structured topos, a pair consisting of a topos $X$ and a lex functor $\mathcal{O}:\mathcal{G}\to X$ sending covering sieves composed of admissible morphisms in $\mathcal{G}$ to jointly effective epimorphic families in $X$, is a locally ringed topos).

Then we say that a left-geometric morphism $f^*:(X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$ (left-geometric meaning that we are using the opposite convention for direction of morphisms) of G-structured toposes (this means that there is a natural transformation $\alpha:f^*\mathcal{O}_X\to \mathcal{O}_Y$ such that the square naturality diagram $\alpha(U)\to \alpha(A)$ in $Y$ induced by an admissible morphism $U\to A$ in $\mathcal{G}$ is a cartesian square) is etale if the following two properties hold:

The left-geometric morphism $f^*$ is a left-local homeomorphism of toposes, or that its right adjoint is a local homeomorphism of toposes (some people call this, confusingly, an etale geometric morphism, or even more confusingly, simply an etale morphism (Lurie does this, so be careful)).
The distinguished map $\alpha:f^*\mathcal{O}_X\to \mathcal{O}_Y$ is an equivalence of $\mathcal{G}$-structures on $Y$.

For the Zariski geometry $\mathcal{G}_{Zar}$ described above, and $f^*:(X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$ the induced left-geometric map between gros Zariski toposes induced by a map of schemes $f:y\to x$ to be etale, it is necessarily a Zariski-open immersion.

There is also an etale geometry (now you can see why the choice of the term "etale" for the general concept is unfortunate!) for which the etale morphisms (of $\mathcal{G}_{et}$-structured toposes) correspond exactly to etale morphisms of schemes (when restricted to schemes).

This geometry, $\mathcal{G}_{et}$ is defined to have underlying category the same as $\mathcal{G}_{Zar}$ , but the admissible morphisms are now the morphisms corresponding to the etale ring maps, and the topology on the admissible subcategory is given by the appropriate restriction of the etale topology (in the opposite category, these are finite collections of etale ring maps that are jointly faithfully flat).

This is actually pretty useful for the following reason: It allows us to define etale morphisms without requiring the very cumbersome condition of representability of a map, and more importantly, it turns out that this condition is "really local" in the sense that we can talk about it without lugging around an atlas everywhere we go.

For proper morphisms, I think that we can tell a similar story, but I don't know exactly how to do it, and I don't have time right now to figure it out.

Answer by John Iskra for Rigidity of the category of schemes

2011-02-28T14:50:30Z

Boy, I like this line of inquiry. I've been thinking about similar issues myself for quite a while now, but I did not realize that so much work had already been done. I'm not sure what you're looking for exactly, but I believe that categorical definitions for open and proper might be found in either Z. Luo's inspirational page www.geometry.net/cg or in Dier's book Caegories of Commutative Algebras upon which Luo's work is based. Both give answers to your question in the case of affine schemes. I'm not sure if the answers translate to the full category of schemes, which I think is what you're looking for.

Answer by S. Carnahan for Rigidity of the category of schemes

2011-03-01T09:21:40Z

This is a rather small fraction of an answer, but I think I have a way to distinguish $\operatorname{Spec} \mathbb{Q}(\sqrt{2})$ from $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$ in a way that is invariant under autoequivalences.

First, since autoequivalences preserve fiber products and coproducts, connectedness of schemes can be defined canonically by checking maps to a coproduct of final objects (i.e., $\operatorname{Spec} \mathbb{Z} \coprod \operatorname{Spec} \mathbb{Z}$). For each prime $p$, the spectrum of the local ring $\mathbb{Z}_{(p)}$ is the universal connected scheme that receives a map from the spectrum of $\mathbb{F}_p$ and the spectrum of $\mathbb{Q}$.

Now, let $X$ be the image of $\operatorname{Spec} \mathbb{Q}(\sqrt{2})$ under a given autoequivalence. We know that $X$ is the spectrum of some quadratic extension of the rationals. To distinguish $X$ from $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$, it suffices to show that the prime 3 is inert. To do this, we check that there is a connected scheme $Y$ (the image of the spectrum of the local ring over 3) that admits maps from $X$ and $\operatorname{Spec} \mathbb{F}_9$ and a map to $\operatorname{Spec} \mathbb{Z}_{(3)}$ , such that it does not admit a map from $\operatorname{Spec} \mathbb{Q}$ , and the map from $\operatorname{Spec} \mathbb{F}_9$ does not factor through $\operatorname{Spec} \mathbb{F}_3$ . If $X$ were isomorphic to $\operatorname{Spec} \mathbb{Q}(\sqrt{3})$, such $Y$ would not exist.

I think you can use similar methods to distinguish any nonisomorphic number fields.

Answer by Laurent Moret-Bailly for Rigidity of the category of schemes

2012-02-18T15:22:20Z

A scheme $X$ is reduced if and only if the natural map $$ \coprod_{x\in X}\mathrm{Spec} \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed, let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local, irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}\,(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.