Pushouts in the Category of Schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:51:49Z http://mathoverflow.net/feeds/question/5143 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes Pushouts in the Category of Schemes Dinakar Muthiah 2009-11-12T03:01:57Z 2010-04-17T07:16:13Z <p>When does it make sense to glue schemes together along subschemes?</p> <p>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?</p> http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes/5160#5160 Answer by Greg Stevenson for Pushouts in the Category of Schemes Greg Stevenson 2009-11-12T05:54:08Z 2010-04-17T00:41:02Z <p>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 <a href="http://www-personal.umich.edu/~kschwede/SchemeWithoutPoints.pdf" rel="nofollow"> this</a> paper of Schwede (Corollary 3.9).</p> <p>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.</p> http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes/5165#5165 Answer by Jose Capco for Pushouts in the Category of Schemes Jose Capco 2009-11-12T07:24:26Z 2009-11-12T09:32:05Z <p>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. </p> <p>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)</p> http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes/5714#5714 Answer by Ilya Nikokoshev for Pushouts in the Category of Schemes Ilya Nikokoshev 2009-11-16T16:38:49Z 2009-11-16T16:38:49Z <blockquote> <p>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?</p> </blockquote> <p>For this particular pushout, the <strong>geometric intuition</strong> 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.</p> <p>As correctly said above, general pushouts of schemes may not be schemes themselves.</p> http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes/21645#21645 Answer by Qfwfq for Pushouts in the Category of Schemes Qfwfq 2010-04-17T05:02:23Z 2010-04-17T07:16:13Z <p>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<br> $CRing$), then the pushout $X \coprod_{Z} Y$, gluing $X$ and $Y$ along $Z$ is given by $Spec D$, where </p> <p>$D=A\times_{ R} B:=$ { $(a,b) \in A\times B | \phi (a)= \psi (b)$ } .</p>