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Let $Z \to X$ be a closed immersion of schemes. Is it true that for every morphism $Z \to Y$, the pushout $X \cup_Z Y$ in the category of schemes exists? If yes, a) does it turn out to be simply sthe pushout in the category of locally ringed spaces, b) is the natural morphism $Y \to X \cup_Z Y$ a closed immersion?

In his paper "Gluing Schemes and a Scheme Without Closed Points", Karl Schwede studies such questions. In particular, he gives an affirmative answer if everything is affine (Theorem 3.4), but also for arbitrary schemes if we also assume that $Z \to Y$ is a closed immersion (Corollary 3.9). My intuition says that it should be also true if we drop this condition, but on the other hand the paper shows with some examples that our intuition may be wrong in the context of pushouts. I'm aware that colimits of schemes are not well-behaved in general, but in my research it would be useful to construct pushouts also when just one inclusion is a closed immersion.

As with this MO question, I think it is not sufficient just to say that some pushout does not exist just because you don't see one. I'm interested in rigorous proofs.

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@the usual downvoter: Please leave a comment how I can improve my question. –  Martin Brandenburg May 8 '11 at 13:00
Apart from my answer below, and Karl Schwede's paper which I did not know, here are two other references for the case where both maps are closed immersions: S. Anantharaman, "Schémas en groupes..." Mem. SMF 33 (1973), avalilable at Numdam: numdam.org/numdam-bin/feuilleter?id=MSMF_1973__33_ and my paper "Construction de revêtements...", J. Alg. 240 (2001), 505-534, Proposition 3.2 (giving some complements to Anantharaman's result). –  Laurent Moret-Bailly May 9 '11 at 9:41

4 Answers 4

up vote 4 down vote accepted

In the positive direction, see D. Ferrand, "Conducteur, descente et pincement", Bull. SMF 131 (4), 2003, 553-585, especially Th. 5.4: the answer to all questions is yes if $Z\to Y$ is finite and every finite set in $X$ (resp. $Y$) is contained in an affine open subset.

More generally, assuming $Z\to Y$ affine, Theorem 7.1 gives a necessary and sufficient condition for the following to hold: (i) the pushout $X\cup_Z Y$ (as locally ringed spaces) is a scheme; (ii) $Y\to X\cup_Z Y$ is a closed immersion; (iii) $X\to X\cup_Z Y$ is affine.

The condition is this: for each $y\in Y$, the inverse image of $\mathrm{Spec(\mathcal{O}_{Y,y})}$ in $Z$ has a basis of affine open neighborhoods in $X$.

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Thanks! Although no counterexample is given, this subtle study indicates that pushouts along closed subschemes don't exist in general, or at least, that we cannot do it with the LRS pushout. Very interesting conditions! –  Martin Brandenburg May 9 '11 at 8:48
The statement 5.4 doesn't seem quite right although I'm probably being dumb (or maybe I'm misreading the French). In particular, see Ravi's comment to mathoverflow.net/questions/5143/… There Ravi chooses two fibers and glues them to each other but in a funny way (certainly a finite map). On the other hand, everything in Ravi's example is quasi-projective, so the condition about every finite set of points contained in an open affine is obvious. –  Karl Schwede May 9 '11 at 12:57
The obstruction to Ravi's example I always thought was that there is no way to make open affines line up. So if $X \to X \cup_Z Y$ is affine and $Y \to X \cup_Z Y$ is a closed immersion, then the inverse image of an open affine along both of these maps is still open affine. But such a thing does not exist I thought because the open sets can't possibly line up. Perhaps I'm misremembering. –  Karl Schwede May 9 '11 at 13:02
Ravi's example actually is a scheme (this is example XIII.3.1 in Raynaud's thesis, LNM 119). –  Laurent Moret-Bailly May 9 '11 at 14:18
Huh, so any gluing involving quasi-projective schemes will always satisfy this condition. Thanks! –  Karl Schwede May 9 '11 at 14:39

It seems to me that you are asking for something far too general. This fails badly already when Z is a curve lying on an algebraic surface X, and Y is a point. In this case you are asking whether it is possible to blow down the curve, which is only true if Z is rational and has self-intersection -1. (See Sándor Kovács's comment.) Also, it seems that there are plenty of cases where a pushout does exist in the category of schemes but which is not "geometrically correct", e.g. if you let $X = \mathbf P^1 \times \mathbf P^1$, $Z = \mathrm{pt} \times \mathbf P^1$ and $Y$ a point, then the pushout exists in schemes and is equal to $\mathbf P^1$. (This is a very special case of the "rigidity lemma" whose only application I know is to show that an abelian variety is commutative.)

Edit: As Martin points out, this only deals with showing that the colimit in schemes is not the colimit in locally ringed spaces, not whether a colimit in schemes exists.

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Thank you. Could you please indicate why the LRS pushout is not a scheme in these examples? –  Martin Brandenburg May 8 '11 at 14:49
@Dan: while the general philosophy of your answer is correct, one of the statements you make is not: It is not true that only $(-1)$-curves can be blown down. It it true that they are the only one that can be blown down to a smooth point. In general what you get is that the self-intersection of the curve is negative, but neither the genus nor the actual absolute value of the self-intersection is limited in general. For an example see my next comment. (Running out of characters).... –  Sándor Kovács May 8 '11 at 17:55
... and the example Take an arbitrary curve $C$ embedded in $\mathbb P^n$ and let $d$ denote its degree. Take the cone over this curve and blow up the vertex. The exceptional curve is isomorphic to the original and has self-intersection $−d$. This shows that for all $(g,d)$ such that there exists a projective curve of genus g and degree $d$ in some projective space, there also exists a curve of genus $g$ and self-intersection $−d$ that can be contracted. Of course, this does not mean that only such pairs are possible. –  Sándor Kovács May 8 '11 at 18:00
@Sándor: Thanks for the correction. @Martin: If you take the $\mathbf P^1 \times \mathbf P^1$ example, then it is possible to contract one of the horizontal or vertical fibers to a point in the category LRS, but the rigidity lemma shows that this is not possible in Sch. So the LRS pushout can not be a scheme. –  Dan Petersen May 8 '11 at 20:56

See also Theorem 6.1 in Artin's Algebraization of Formal Moduli: II. Existence of Modifications, Ann. of Math. (2) 91 1970 88–135.

Theorem 6.1: Let $Y'$ be a closed algebraic subspace of an algebraic space $X'$, and let $f_0: Y' \to Y$ be a finite morphism. There is a unique maximal modification $f: X' \to X$, $Y \subset X$ whose restriction to $Y'$ is $f_0$. It is the amalgamated sum $X = X' \coprod_{Y'} Y$ in the category of algebraic spaces.

I had always thought that gluings in the generality mentioned above in Laurent Moret-Bailly's answer above only existed in the category of algebraic spaces. I suspect I'm missing something obvious but I don't know what it is. I'd deeply appreciate it if someone could set me straight on this.

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The key assumption in Ferrand's result (which is not necessary if you work with algebraic spaces) is the existence of affine neighborhoods for finite sets. To construct quotients which are schemes you typically need this kind of condition. –  Laurent Moret-Bailly May 9 '11 at 14:24
I see. Thanks. –  Karl Schwede May 9 '11 at 14:28

Not an answer, but there is a beautiful instance of such a gluing used by Mazur in his famous Eisenstein ideal paper (page 50)

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@SGP: Well there the pushout with $Z=Spec(\mathbb{F}_p)$ is just used without any definition / existence issue. –  Martin Brandenburg May 8 '11 at 19:58

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