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I've found many remarks that the category of schemes is not cocomplete. The category of locally ringed spaces is cocomplete, and in some special cases this turns out to be the colimit of schemes, but in other cases not (which is, of course, no evidence that the colimit does not exist). However, I want to understand in detail a counterexample where the colimit does not exist, but I hardly found one. In FGA explained I've found the reference, that Example 3.4.1 in Hartshorne, Appendix B is a smooth proper scheme over $\mathbb{C}$ with a free $\mathbb{Z}/2$-action, but the quotient does not exist (without proof). To be honest, this is too complicated to me. Are there easy examples? You won't help me just giving the example, because there are lots of them, but the hard part is to prove that the colimit really does not exist.

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I'm going to attempt to clarify your question, because I misunderstood it twice. Your question is not about locally-ringed spaces. Your question is not about the famous free Z/2Z-action on the smooth proper 3-fold. Your question is simply to give, with full proof, an example of a diagram in the category of schemes, with no colimit in the category of schemes. Right? –  Kevin Buzzard Dec 29 '09 at 11:44
    
yes, exactly :-). besides, I'm interested in special cases where the colimit exists (see the comments in emertons answer). –  Martin Brandenburg Dec 29 '09 at 12:18
    
If you take the naive approach and just google "no categorical quotient" then you seem to get lots of examples. I just looked through a few and perhaps the one you'll like most is the one in "push-out of schemes" by Li (p538, example 2). I should stress that I did not check anything here though. –  Kevin Buzzard Dec 29 '09 at 13:38
    
there is a gap, li does not prove that X/W = X -> Y' is an isomorphism, which may be very hard: how do get a map into the quotient? I don't know what he means with "(locally, point by point," and can't find "Remark 4.2". –  Martin Brandenburg Dec 29 '09 at 15:49
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As another clarification, if $R \rightrightarrows U$ is an etale equivalence relation in schemes s.t. the alg space $X = U/R$ is not a scheme, for it to be an "example" one must prove there's no "initial map" from $X$ to schemes (i.e., map $\pi:X \rightarrow S$, or equivalently map $U \rightarrow S$ inducing same composites back to $R$, to a scheme $S$ which is initial among all such maps). But just as for "quotients" of schemes, one needs properties beyond "categorical" for it to be useful. So this is an exercise in pathology. Not that there's anything wrong with that... –  BCnrd Apr 24 '10 at 20:25
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5 Answers

up vote 24 down vote accepted

Edit: BCnrd gave a proof in the comments that this example works, so I've edited in that proof.

A possible proven example

I suspect There is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.

I don't have a proof, but I thought other people might have ideas if I posted this here.

If a coequalizer $P$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $P$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.

On the other hand, let $U$ be an affine open around the image of the generic point in $P$. $U$ has dense open preimages $V$ and $V'$ in both affine lines. Let $W=V\cap V'$ inside the affine line, so we have two maps from $W$ to the affine $U$ which coincide at the generic point of $W$, and hence are equal (as $U$ is affine). In particular, the two maps from affine line to categorical pushout $P$ coincide at each "common pair" of closed points of the two copies of $W$, contradicting the previous paragraph.


Edit: The questions below are no longer relevant, but I'd like to leave them there for some reason.

Here are some questions that might be helpful to answer:

If the coequalizer above does exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?

(see the related question Can a coequalizer of schemes fail to be surjective?)

Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of separated schemes equal to $\mathbb A^1$? (probably)

 

What are some ways to determine that a functor $Sch\to Set$ is not corepresented by a scheme?

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I have one answer to the last question. If F is a contravariant functor from Schemes to Sets such that F(Spec(ℤ)) is more than one point, then F cannot be corepresentable. –  Anton Geraschenko May 8 '10 at 22:03
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An easier example is to divide the "double-headed snake" $\mathbb A^1\cup \mathbb A^1$ by $\mathbb Z_2$. That is not a scheme either, but it is an algebraic space. In other words, replace your $Spec\ k(t)$ by $Spec\ k[t,1/t]$ . That feels more ordinary. Koll'ar called this algebraic space this a "bug-eyed cover" of 1. He has a paper with "bug-eyed" in the title on these. –  VA. May 8 '10 at 23:52
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@VA: can you explain your quotient suggestion more fully, so I can see how it is different from the line with a doubled point? Maybe I am not seeing what $\mathbf{Z}/2\mathbf{Z}$-action you have in mind. I did consider trying to do something more ordinary like this, but got stuck on not removing enough closed points. I agree it is much better if the example will be an algebraic space. –  BCnrd May 9 '10 at 0:08
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The proof above shows more generally: Let $X$ be an integral scheme with non-closed generic point $\eta$, such that the closed points are dense in $X$ (for example a variety over a field). Then the coequalizer of $\eta \rightrightarrows X \sqcup X$ does not exist. I think a slogan for the proof is that identifying the two generic points implies that some neighborhoods should be identified as well, but closed points are not identified because you can use test schemes $X$ and $X$ with a doubled point. –  Martin Brandenburg May 9 '10 at 0:19
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The prototypical bug-eyed cover I know of is the quotient of $\mathbb A^1$ by the ℤ/2 action given by x↦-x, except you "remove the action at 0" from the relation. The pushout in the category of algebraic spaces is "the right one", but there is also a pushout in the category of schemes. If there were an example which is an algebraic space, it would also answer David Brown's question mathoverflow.net/questions/4587/…. –  Anton Geraschenko May 9 '10 at 0:31
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This is not an answer to the question, but is intended to show that Martins's concern about the possible distinction between the colimit in the category of schemes vs. in the category of locally ringed spaces is a valid one. Indeed, if I haven't blundered below, then it seems that in some circumstances at least the direct limit (colimit) of schemes exists, but does not coincide with the direct limit in the category of locally ringed spaces.

For example, suppose that $X_n$ is the Spec of $k[x]/(x^n),$ for some field $k$ (and the transition maps are the obvious ones). Then the direct limit of the $X_n$ in the category of locally ringed spaces is a formal scheme which is not a scheme, whose underlying topological space is a point, and whose structure sheaf (which is in this context simply a ring, namely the stalk at the unique point) is $k[[x]]$.

On the other hand, suppose given compatible maps from the $X_n$ to a scheme $S$. These must all map the common point underlying the $X_n$ to some point $s \in S$, which lies in some affine open Spec $A$. Thus the maps from the $X_n$ factor through Spec $A$, and correspond to compatible maps $A \rightarrow k[x]/(x^n),$ i.e. to a map $A \rightarrow k[[x]].$ This in turn gives a map Spec $k[[x]] \rightarrow$ Spec $A\subset X,$ and so we see that the natural compatible maps from the $X_n$ to Spec $k[[x]]$ identify Spec $k[[x]]$ with the direct limit of the $X_n$ in the category of schemes.

EDIT: As is noted in a comment of David Brown's attached to his answer, this example generalizes, e.g. if $I$ is an ideal in a ring $A$, then the direct limit in the category of schemes of Spec $A/I^n$ coincides with Spec $\hat A$, where $\hat A$ is the $I$-adic completion of $A$.

FURTHER EDIT: I am no longer certain about the claim of the previous paragraph. If $A/I$ (and hence each $A/I^n$) is local then for any scheme $S$ the maps Spec $A/I^n \to S$ factor through an affine open subscheme, so one reduces to computations in the category of rings, and hence finds that indeed the direct limit of the Spec $A/I^n$ equals Spec $\hat{A}$. More generally, I'm currently unsure ... .

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in general, the colimit of local affine schemes and local transition maps exists and is a local affine scheme. this is because you can describe morphisms on local schemes via points and local homomorphisms on the stalks. in particular, the colimit of the $Spec A/I^n$ is $Spec \hat{A}$, when $I$ is a maximal ideal. I try to prove the general case: if $f_n : Spec A/I^n \to X$ are compatible morphisms, we want to glue them to $Spec \hat{A} \to X$. if $X$ is affine, this is trivial. –  Martin Brandenburg Dec 29 '09 at 10:41
    
in the general case, let $U \subseteq X$ be an open affine. since the transition maps $Spec A/I^n \to Spec A/i^{n+1}$ are homeomorphisms, the images of the $f_n$ are all equal. consider the preimage of $U$ in $Spec A/I$. let $f \in A$ such $D(\overline{f})$ is a basic open subset of this preimage. then all the $f_n$ restrict to compatible $Spec A_f / (I_f)^n \to U$. the affine case yields $Spec \hat{A_f} \to U$. now we want to glue these morphisms to $Spec \hat{A} \to X$. so it would be nice that $Spec \hat{A_f}$ is an open cover of $Spec \hat{A}$, but this seems to be unlikely ... –  Martin Brandenburg Dec 29 '09 at 10:42
    
I still have not understood the claim in the EDIT if $I$ is not maximal. –  Martin Brandenburg May 8 '10 at 19:42
    
Dear Martin, I will try to reconstruct the proof, and then post it. –  Emerton May 9 '10 at 4:20
    
Dear Martin, My reconstruction effort has failed, at least for the moment, and so I have added a further edit, scaling back the claim of the previous one. –  Emerton Dec 10 '10 at 4:17
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Sounds to me like you don't want to hear the proof, but you want to hear "the point". The point is that a scheme must by definition be covered by affine schemes, and sometimes when doing exercises in Hartshorne the proofs go like this: first do the question for affine schemes, where the question becomes ring theory, and then do it for all schemes by glueing together. So here is how you might want to try and quotient out a scheme by a free action of $Z/2Z$: first let's say the scheme is affine, so $Spec(A)$, with $A$ having an action of $Z/2Z$, and try and figure out if the quotient exists, and then move onto the general case.

In the affine case we have a ring $A$ with an action of $Z/2Z$, and if $B$ is the invariants, then you can convince yourself that $Spec(B)$ is the quotient.

Now let' s do the general case. Say $X$ is a scheme with an action of $Z/2Z$. Choose a point $x$ in $X$. Now let $Spec(A)$ be an affine containing $X$. Now $Z/2Z$ acts on $Spec(A)\ldots$oh wait, no, no it doesn't, because the action will maybe move $Spec(A)$ to another affine $Spec(B)$. So let's try intersecting $Spec(A)$ and $Spec(B)$, The intersection will often be affine so let's consider this and call it $Spec(C)$ and$\ldots$oh no, wait, that doesn't work either, because $x$ might not be in $Spec(C)$. Hmm.

The well-known 3-fold you mention in your question behaves in this way. You can't cover it by affines each one of which is preserved by the action, so you get stuck.

Does that help?

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that the naive approach does not work, does not prove anything. in fact, there are categorical constructions which are in some sense unexpected. in this case, I want to understand a prove of a counterexample. –  Martin Brandenburg Dec 28 '09 at 19:03
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again: we have a full subcategory C of D and a diagram F in C, whose colimit in D does not lie in C. this is no reason that the colimit of F in C does not exist. of course, this situation is likely to be an example, but one has to prove it. –  Martin Brandenburg Dec 28 '09 at 20:14
    
"that the naive approach does not work, does not prove anything." Of course it doesn't. But it does prove that I misunderstood what you were asking. Let me try again then: just go and read Mumford's "Geometric Invariant Theory" and see his careful explanation of why the quotient isn't a scheme. Will this do? If not, what are you asking? I'm trying my best ;-) –  Kevin Buzzard Dec 28 '09 at 23:29
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I think he's asking for an example of a diagram of schemes where there is no categorical colimit, and not for an example where there is no categorical colimit that looks "correct" geometrically. E.g. if E is an elliptic curve over the complex numbers with a point x of infinite order, then translation by x induces an action of Z on E, and I believe the categorical colimit exists but it is Spec(C), because there are not many translation-invariant functions. So the idea is to construct an example where there are not too few functions out that the colimit exists in some trivial way. –  Tyler Lawson Dec 29 '09 at 4:17
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Direct limits of schemes fail to exist. A good example is the following: let X be a scheme and Z a closed subscheme defined by an ideal I. Then for any n we get the nth infinitesmal neighborhood Z^(n) defined by the ideal I^(n+1) and a diagram $Z \to Z^2 \to \cdots \to Z^n \to \cdots$ and in general the direct limit of this diagram does not exist in the category of schemes. (It does exist however in the category of formal schemes). Knutson's Algebraic Spaces, Chapter 5, section 1 (near the end) explains this well (for algebraic spaces, but for this point it is fine to think of everything as a scheme).

A simple example is P^1 over a dvr (R,t), with Z the closed subscheme of P^1 defined by t.

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1) seems that something in your answer is not visible. anyway, is there are a easy prove that the colimit does not exist in this example? 2) as I already said, this is no prove that the pushout does not exist. –  Martin Brandenburg Dec 28 '09 at 19:06
    
Ah. You are right, too hasty on my part. I edited my answer. –  David Zureick-Brown Dec 28 '09 at 19:17
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thank you. how can we prove that the colimit in your example (infinitesemal neighborhood in P^1) does not exist? so we have to conclude a contradiction, if X = colim(Z^n) exists. I have no idea how this can be done. I know nothing about algebraic spaces. sorry that I repeat my question again and again, but as I said the problem is not really in "finding" examples, but checking them. –  Martin Brandenburg Dec 28 '09 at 20:19
    
for example, we could try that A^1 -> A^2 -> ... (the canonical closed immersions) has no colimit in the categors of schemes. assume X is a colimit. the global sections are morphisms to A^1, thus we get that the global sections of X are k[[x_1,x_2,...]]. I don't know how to go on. –  Martin Brandenburg Dec 29 '09 at 0:54
    
For affine schemes the colimit does exist, it is just the completion. The point behind the P^1 example is that you can't glue the `completed A^1's' together to get a completed P^1. I'll see if I can think of a succinct way to see this. –  David Zureick-Brown Dec 29 '09 at 8:02
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Direct limits include in particular quotients.

I think that if $Z \subset X$ is a closed subscheme, then $X/Z$ usually doesn't exist.

Let's take for $Z$ a divisor on $X$, such that there exist a rational function $f: X \to \mathbb P^1$ with poles only on $Z$. Then according to the universal property of quotients such a function descends to a function $f \bar: X/Z \to \mathbb P^1$, with the pole consisting of one point (I'm little uncertain, why it's a point, but perhaps it must be a point). But that's impossible unless $dim(X)=1$.

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Let X = Spec k[x,y] and Z = Spec k[x] (the divisor y=0). Then the quotient X/Z exists in affine schemes: it is the pullback of k[x,y] -> k[x] <- k, which is the non-Noetherian subring k[y,yx,yx^2,...]. I'm not clear whether Spec of this ring is still the quotient in the category of all schemes, but I think that more is necessary to make your argument work. –  Tyler Lawson Dec 29 '09 at 13:06
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"that's impossible unless dim(X)=1." I think you mean that's impossible unless dim(X/Z)=1. It may be that something with the right universal property does exist in the category of schemes, but that Z is not the only thing that gets crushed. Why should the dimension of X/Z agree with the dimension of X? –  Anton Geraschenko May 8 '10 at 17:05
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