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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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    $\begingroup$ @the usual downvoter: Please leave a comment how I can improve my question. $\endgroup$
    – Martin Brandenburg
    Commented May 8, 2011 at 13:00
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    $\begingroup$ 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). $\endgroup$
    – Laurent Moret-Bailly
    Commented May 9, 2011 at 9:41

4 Answers 4

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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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  • $\begingroup$ 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! $\endgroup$
    – Martin Brandenburg
    Commented May 9, 2011 at 8:48
  • $\begingroup$ 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. $\endgroup$
    – Karl Schwede
    Commented May 9, 2011 at 12:57
  • $\begingroup$ 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. $\endgroup$
    – Karl Schwede
    Commented May 9, 2011 at 13:02
  • $\begingroup$ Ravi's example actually is a scheme (this is example XIII.3.1 in Raynaud's thesis, LNM 119). $\endgroup$
    – Laurent Moret-Bailly
    Commented May 9, 2011 at 14:18
  • $\begingroup$ Huh, so any gluing involving quasi-projective schemes will always satisfy this condition. Thanks! $\endgroup$
    – Karl Schwede
    Commented May 9, 2011 at 14:39
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To complement Laurent Moret-Bailly and Karl Schwede's answers (10 years ago!), the pushout of a closed immersion $Z\to X$ along an affine morphism $Z\to Y$ always exists in the category of algebraic spaces. This is a special case of Theorem 1.8 of https://arxiv.org/abs/2205.08623.

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  • $\begingroup$ Thanks for adding this, David. $\endgroup$
    – Martin Brandenburg
    Commented May 20, 2022 at 20:31
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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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  • $\begingroup$ 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. $\endgroup$
    – Laurent Moret-Bailly
    Commented May 9, 2011 at 14:24
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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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  • $\begingroup$ @SGP: Well there the pushout with $Z=Spec(\mathbb{F}_p)$ is just used without any definition / existence issue. $\endgroup$
    – Martin Brandenburg
    Commented May 8, 2011 at 19:58

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