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

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    $\begingroup$ This is a better use of the word "rigid" than is normally used. The old word "rigid category" refers to a monoidal category with well-behaved duals [en.wikipedia.org/wiki/Rigid_category]. It's not a very good word. In general, particularly bad words in category theory come from Australia, and some of the good ones are Russian, but in this case the name is from two Germans (Dold and Puppe). Another reasonable meaning for "rigid" is "no nontrivial infinitesimal deformations", whereas you mean "no nontrivial automorphisms". $\endgroup$ – Theo Johnson-Freyd Feb 28 '11 at 15:46
  • $\begingroup$ Oh, +1 by the way. $\endgroup$ – Theo Johnson-Freyd Feb 28 '11 at 15:46
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    $\begingroup$ Interestingly, the category of categories is almost rigid, in that it only has one autoequivalence (up to iso, I suppose) which isn't isomorphic to the identity, namely the functor $C \mapsto C^{op}$. $\endgroup$ – David Roberts Mar 1 '11 at 3:54
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    $\begingroup$ Theo, I've heard people say "rigid monoidal category", but not "rigid category" to mean the same thing. (If a Wikipedian wrote that, then I think that's a mistake!) $\endgroup$ – Todd Trimble May 11 '11 at 23:13
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    $\begingroup$ Martin, re your third paragraph, the category of semigroups isn't rigid, is it? It has an automorphism sending each semigroup to its opposite. But this automorphism is not isomorphic to the identity. Ditto (not necessarily commutative) rings. $\endgroup$ – Tom Leinster Feb 17 '12 at 16:59

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\operatorname{Spec}  \kappa( x )\to X$$ 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 $\operatorname{Spec}(K)$ (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 $\operatorname{Spec} (R/t^n)\to\operatorname{Spec}\ (n\geq1)$ are distinct monomorphisms with image $a$.)

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

  • $\begingroup$ Martin, you are right about specialization, but somethig can be saved. I'll edit my answer. $\endgroup$ – Laurent Moret-Bailly Feb 19 '12 at 13:32
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    $\begingroup$ Thanks for providing the details. I've deleted the irrelevant comments above. a) In the proof, "only one monomorphism" / "infinitely many monomorphisms" are meant to be "up to isomorphism". b) Let's say that $x$ is a strong specialization of $y$ if this is witnessed by a two-point scheme as above. Then it is easy to verify that every strong specialization is a specialization and that every immediate specialization is an immediate strong specialization. Now you claim that also the converse holds, but I have difficulties with the proof: $\endgroup$ – Martin Brandenburg Feb 20 '12 at 7:36
  • $\begingroup$ Let $x$ be an intermediate strong specialization of $y$. Choose some affine neighborhood $\mathrm{Spec}(A)$ of $x$, then $y,x$ correspond to prime ideals $\mathfrak{p} \subseteq \mathfrak{q} \subseteq A$. There is no prime ideal $\mathfrak{r}$ such that $\mathfrak{p} \subseteq \mathfrak{r} \subseteq \mathfrak{q}$ is an intermediate chain, because otherwise two two maps $A \to (A/\mathfrak{p})_\mathfrak{q}$, $(A/\mathfrak{r})_\mathfrak{q}$ contradict the asumption that $\mathfrak{p} \subseteq \mathfrak{q}$ is immediate strong. But what about longer chains? Perhaps the followig adjustment works: $\endgroup$ – Martin Brandenburg Feb 20 '12 at 7:41
  • $\begingroup$ [Lemma: $x$ is an intermediate specialization of $y$ if and only if $x$ is a strong specialization of $y$ and there is no proper finite chain of strong specializations $x,x_1,...,x_n,y$.] But I can only prove this lemma for noetherian schemes. $\endgroup$ – Martin Brandenburg Feb 20 '12 at 7:46
  • $\begingroup$ What about the following: [$X$ is irreducible of dimension $\leq d$ iff there is some point $y$ such that every point $x$ can be reached by $\leq d$ many strong specializations, but not by $>d$ many proper strong specializations.] It is easy to prove $\Rightarrow$, but for $\Leftarrow$ it is unclear whether $X$ has finite dimension at all. The basic problem is that it is unclear if we can compose a specialization into finitely many intermediate (and thus strong) specializations. $\endgroup$ – Martin Brandenburg Feb 20 '12 at 12:26

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:

  1. 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)).
  2. 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.

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    $\begingroup$ I cannot tell what is the relation between this and Martin's question! $\endgroup$ – Mariano Suárez-Álvarez Feb 28 '11 at 14:25
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    $\begingroup$ @Harry: This is offtopic and just again showing off your knowledge about topos theory etc. ... please read the question more carefully. Your answer is not connected with my question. $\endgroup$ – Martin Brandenburg Feb 28 '11 at 15:09
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    $\begingroup$ Not sure why this is getting down votes... $\endgroup$ – Steven Gubkin Feb 28 '11 at 16:15
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    $\begingroup$ Some people seem to have skipped the first sentence: "This is a just a comment". As such it is very interesting (even though it could have quickly recalled what Toen means by "a geometry"). It doesn't bother me that it doesn't solve the problem (because we don't know how to construct the Zariski topoi without defining open immersions in ring theoretic terms first. The down votes seem unecesssary. $\endgroup$ – AFK Feb 28 '11 at 19:17
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    $\begingroup$ As far as I can tell, the problem is that Martin wants a characterization of open immersion and étale morphism that is invariant under autoequivalences of the category of schemes, and you have provided a characterization of open immersion that depends on the Zariski topos (and the corresponding geometry), which is not obviously invariant under such autoequivalences. You claim that we know how to construct everything ring-theoretically, but it is not obvious that affine schemes are taken to affine schemes under all autoequivalences of schemes (and in fact, if they were, he would be done). $\endgroup$ – S. Carnahan Mar 1 '11 at 8:12

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.

  • $\begingroup$ I can't prove your universal property of $\text{Spec } \mathbb{Z}_{(p)}$, but the following works: It is the unique connected scheme together with a bijektive morphism from $\text{Spec } \mathbb{F}_p \coprod \text{Spec } \mathbb{Q}$. $\endgroup$ – Martin Brandenburg Mar 16 '11 at 23:35
  • $\begingroup$ I think you can also use the Neukirch–Uchida theorem, which states that number fields with isomorphic Galois groups are isomorphic. I think we can recover the Galois group with its profinite group structure from the category of schemes as well. $\endgroup$ – R. van Dobben de Bruyn Dec 18 '17 at 4:16
  • $\begingroup$ Maybe you can use this to determine all fields by viewing which finitely generated fields sit inside it. The idea is to use Pop's extension of the Neukirch–Uchida theorem to all finitely generated infinite fields of arbitrary characteristic (up to purely inseparable morphisms, which are not detected by the Galois group), cf. Pop's article in the 1997 Obergurgl proceedings on Resolution of Singularities. I haven't worked this out carefully, and I anticipate some problems (e.g. purely separable extensions). $\endgroup$ – R. van Dobben de Bruyn Dec 18 '17 at 4:42

It appears to me that my recent answer here (I posted as a guest, hence the different account) also applies to this question: As Laurent pointed out, we can distinguish nilpotent thickenings. For any scheme $S$, consider the opposite of the category of abelian cogroup objects in $\text{Sch}_{S/}$ which are nilpotent thickenings of $S$ (This last condition is actually automatic, but is a bit of a hassle to prove if we don't assume the counit to be separated). In particular, every object is affine over $S$ and the usual argument works to show that the category is equivalent to the category of quasicoherent modules on $S$.

Now we'd like to apply the Gabriel-Rosenberg theorem to reconstruct $S$. However, the consensus seems to be that the theorem has only been proven for quasi-separated schemes. Perhaps it is at least true that the module category of a non-affine scheme does not have a compact projective generator? In any case, the following is a more elementary way to proceed:

We can distinguish the structure sheaf of $\text{Spec }\mathbb Z$. For instance, it is the unique compact object of $\text{Mod }\mathbb Z$ with endomorphism ring $\mathbb Z$. We can also define quasicoherent pullback of modules by cartesian pullback of schemes, so we can distingush the structure sheaf of any scheme. Taking endomorphism rings reconstructs the functor $\Gamma(-,\mathcal O)$, which, as Martin points out in the question, is enough to distinguish affine schemes and prove rigidity of the category of schemes.

  • $\begingroup$ Dear user131755, I am finishing up an article proving more generally that $\operatorname{\underline{Isom}}(\operatorname{\underline{Sch}}_S,\operatorname{\underline{Sch}}_{S'}) = \operatorname{\underline{Isom}}(S',S)$, which is a non-Noetherian version of a result by Mochizuki (in particular answering the question of the OP). Your argument looks correct, and the observation that thickenings are automatically affine would give a substantial shortcut to my existing argument. Could you please get in touch with me so that we can discuss how to proceed? $\endgroup$ – R. van Dobben de Bruyn Nov 27 '18 at 18:58
  • $\begingroup$ @R.vanDobbendeBruyn I sent you an email $\endgroup$ – user131755 Nov 30 '18 at 15:03
  • $\begingroup$ Let $X$ be a scheme defined over a field of characteristic zero (I don't know if this is necessary). If $QCoh(X)$ has a compact projective generator, then $X$ is affine. For if $P \twoheadrightarrow \mathcal O_X$ is an epimorphism from a compact projective, then evaluation and coevaluation form maps $\mathcal O_X{}^\to_\leftarrow P \otimes P^\vee$ whose composite is $n \cdot id_{\mathcal O_X}$ where $n$ is the rank of $P$. Since we're in characteristic 0, this is invertible, so $\mathcal O_X$ is a retract of the projective $P$ and hence also projective. So $X$ is affine by Serre's criterion. $\endgroup$ – Tim Campion Dec 27 '18 at 6:05
  • $\begingroup$ @TimCampion Serre's criterion also assumes quasicompactness of the scheme, though. $\endgroup$ – user131755 Jan 7 at 12:38

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.

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    $\begingroup$ I also found Luo's site recently. But I have not found anything which answery my question. It is very interesting that the category of schemes can be constructed categorically from the category of affine schemes (geometry.net/cg/scheme.html), but I don't need that here. Where are open immersions defined categorically? $\endgroup$ – Martin Brandenburg Feb 28 '11 at 15:03
  • $\begingroup$ I don't know that he does so explicitly. Read Categorical Geometry. Toward the end (in 5.3, I think) he provides sufficient conditions for an arrow f to induce an open map Spec(f). The conditions are entirely categorical. It should be possible, given Luo's characterization of schemes to give similar sufficient conditions for open immersions in the category of schemes over affine schemes. My advice is to read cg, if you haven't already. $\endgroup$ – John Iskra Feb 28 '11 at 16:06
  • $\begingroup$ According to Diers (I switch directions here so we are working on the geometric side): An object $X$ is pre-neat if for all local objects $L$, any two arrows $f$ and $g$ from $L$ to $X$ are either equal or disjoint. Take the definition of local object to be that given in Luo's categorical geometry in \S 3.4, disjoint arrows are those whose pullback is the initial object. An etale arrow is then one which is finitely presentable, neat and coflat. I'll work on the extension of this to schemes. $\endgroup$ – John Iskra Feb 28 '11 at 22:06
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    $\begingroup$ Then isn't that a problem you need to overcome anyway, regardless of whether we have an acceptable definition of etale? Maybe I don't understand your strategy. $\endgroup$ – John Iskra Feb 28 '11 at 23:43
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    $\begingroup$ I've started to study Luo's categorical geometry, and also Dier's book. I'm pretty convinced that these approaches won't answer my question: The categories in study are always similar to the category of rings (Dier) resp. affine(!) schemes (Luo); you can define notions of commutative algebra in pure categorical terms, but not the ones of algebraic geometry within the category of schemes. Dier just defines a scheme as a space together with a sheaf of "generalized" rings, and within the category of schemes no pure categorical definitions are made $\endgroup$ – Martin Brandenburg May 12 '11 at 10:04

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