Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

When does it make sense to glue schemes together along subschemes?

In particular: is there a way to glue two schemes together along a closed point (say we're working over a field)? Can you glue two closed points of the same scheme together? Is it easier to glue in the category of algebraic spaces?

share|improve this question
add comment

4 Answers 4

Given schemes $X,Y$ and $Z$ such that $Z$ is a closed subscheme of both $X$ and $Y$ the pushout exists in the category of schemes. So in particular one can glue schemes along a closed point. A reference for this (carried out via the category of locally ringed spaces) is given in this paper of Schwede (Corollary 3.9).

In general though the pushout in the category of locally ringed spaces need not be a scheme even if one pushes out along a subscheme - see for instance Example 3.3 in Schwede's paper.

share|improve this answer
    
Wow.. interesting work there by Schwede! :) –  Jose Capco Nov 12 '09 at 7:29
1  
Schwede's Example 3.3 doesn't really show that the coproduct doesn't exist in the scheme category; just that the coproduct in the ringed space category isn't a scheme. –  Andrew Critch Nov 12 '09 at 7:40
    
Fixed this. Do you know if the pushout in the category of sheaves is representable in this example? I can think of plenty of non-representable pushouts but none I am sure about along a subscheme. –  Greg Stevenson Nov 12 '09 at 7:45
3  
I'm sure I'm missing something, but: how does your first sentence imply that "in particular one can ... glue a scheme to itself along a pair of closed points"? Example I have in mind that gives me pause without being a counterexample to your statement: if you try to glue an elliptic curves times the $\mathbb{P}^1$ (over $\mathbb{C}$) to itself, by gluing one fiber over $\mathbb{P}^1$ to another, not by the identity but by a nontorsion element, you just get an algebraic space that is not a scheme. –  Ravi Vakil Apr 8 '10 at 5:19
1  
Ravi, you are absolutely right. I assume the author meant just choose two closed points, and glue them (but not glue other things), which is true say for quasi-projective schemes I think. You should always be able to do this assuming both points can be represented as points in the same affine patch (if not more generally). Hm, if you have two points which can't live on the same affine patch due to failure of separatedness, I'm not sure what happens if you try to glue them. –  Karl Schwede Apr 16 '10 at 16:33
show 3 more comments

is there a way to glue two schemes together along a closed point (say we're working over a field)? Is it easier to glue in the category of algebraic spaces?

For this particular pushout, the geometric intuition is quite simple: given two algebraic varieties, one of which lives in $\mathbb A^m$, another in $\mathbb A^n$, combine them in two complementary hyperplanes in $\mathbb A^{m+n}$. Algebraically, this easily generalizes to an affine scheme $\mathrm{Spec}\\, R\_1\times R\_2/{\mathrm{relationship}}$ and then you glue everything together.

As correctly said above, general pushouts of schemes may not be schemes themselves.

share|improve this answer
add comment

Consider a commutative local ring R, say a valuation domain, with maximal ideal M. Consider the fiber product $R \times_M R$ (I wrote M instead of R/M), coming from the pullback in commutative rings $R\rightarrow R/M$. Then the corresponding prime spectra of this fibered product (in rings) is actually a form of gluing of the same same (affine) scheme Spec R along the closed point M. So this is the case where this happens.

So I think you can do such things for affine Schemes. For affine schemes, you can at least reverse the topology (they are sometimes called inverse spectrum) and you can form a sheaf over this topology similar to the canonical structure sheaf, but the closed points becomes the generic points in this topology. I cannot recall correctly, but I think the stalks of this sheaves become integral domains (so it is some form of dual to the affine schemes, local becomes integral and so on)

share|improve this answer
add comment

In the affine case, let $X=Spec A$, $Y=Spec B$, and $Z=Spec R$. If you have morphisms $f:Z\rightarrow X$ coming from $\phi:A\rightarrow R$ and $g:Z\rightarrow Y$ coming from $\psi: B\rightarrow R$ (because $Aff$ is anti-equivalent to
$CRing$), then the pushout $X \coprod_{Z} Y$, gluing $X$ and $Y$ along $Z$ is given by $Spec D$, where

$D=A\times_{ R} B:=$ { $(a,b) \in A\times B | \phi (a)= \psi (b) $ } .

share|improve this answer
    
The coproduct in CRing is given by the tensor product over $\mathbb{Z}$, and the pushout over $R$ is given by the tensor product over $R$. I suspect you meant the pullback of A and B over R, written $A\times_R B$. Also, the "coproduct" that you're referring to is called the pushout, the gluing, or the "fibered coproduct", although this last one is nonstandard. The coproduct of affine schemes is specifically the disjoint union. –  Harry Gindi Apr 17 '10 at 5:52
    
I think I meant something like the "fibered coproduct of schemes". That is, the opposite notion to the actual fibered product in CRings (and the latter corresponds, if I'm not mistaken, to my definition of $D$). –  Qfwfq Apr 17 '10 at 6:36
    
I fixed it for you. –  Harry Gindi Apr 17 '10 at 7:16
    
@Qfwfg: Your pushout is really only the pushout in the category of affine schemes. For non-affine schemes, in general, your pushout does noes have the desired universal property. One has to require, for example, that $\phi$ or $\psi$ is surjective (see the paper by Karl Schwede). –  Martin Brandenburg Nov 12 '11 at 11:22
    
@Martin: yes, my answer was only meant to be a (very) partial answer. –  Qfwfq Nov 12 '11 at 13:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.