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Let $R_1,R_2$, and $S$ be commutative rings with maps $R_1,R_2 \to S$ and form the fiber product $R = R_1 \times_S R_2$. A well-known theorem of Milnor says that under certain assumptions the category of finitely generated projective $R$-modules is equivalent to a category of "patching data": an object in this category consists of finitely generated projective modules $P_1$ and $P_2$ over $R_1$ and $R_2$ respectively, together with an $S$-isomorphism $P_1 \otimes_{R_1} S \cong P_2 \otimes_{R_2} S$. This may be applied, for instance, to derive a Meyer-Vietoris sequence for Picard groups of rings.

My question: does this generalize to pushouts of schemes? For instance, if I am computing the Picard group of a projective nodal cubic, which can be obtained from $\mathbb{P}^1$ by identifying two points, can I apply a Meyer-Vietoris sequence similar to the one that works in the affine case?

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Can I ask about the assumptions on Milnor's theorem? (if pushouts are along nice (flat?) morphisms then it should be true that the category of modules is the fibre product you'd expect, I've never checked this and I would love to be proved wrong if false) – Yosemite Sam Jun 26 '12 at 21:21
I think it's enough to assume $R_1 \to S$ is surjective and $R \to R_1$ is injective. From what I understand, there are various hypotheses that work: if the natural functor from finitely generated projective $R$-modules to patching data is an equivalence, the fiber square is called a Milnor square. – Justin Campbell Jun 26 '12 at 22:50
Justin Campbell: $R_i\rightarrow S$ surjective is all you need. – Steven Landsburg Jun 27 '12 at 0:24
This is used all the time in deformation theory. For example, you can find a treatment in my paper with Mattia Talpo in the arxiv. – Angelo Jun 27 '12 at 6:49
up vote 4 down vote accepted

First of all, you have to show that these pushouts exist at all. More precisely, if $A \hookrightarrow X$ is a closed immersion and $A \to Y$ is an arbitrary morphism, then the pushout $X \cup_A Y$ exists in the category of schemes. Moreover, the underlying ringed space is the pushout of the underlying ringed spaces (thus topologically it's a pushout in the usual sense, and algebraically it is given by a fiber product of structure sheaves); remark that this is not trivial at all. One can prove that the canonical map $Y \to X \cup_A Y$ is a closed immersion. You can find all that in Karl Dahlke's paper "Gluing schemes and a scheme without closed points".

Milnor's Theorem states that the functor $\mathrm{Vect}(X \cup_A Y) \to \mathrm{Vect}(X) \times_{\mathrm{Vect}(A)} \mathrm{Vect}(Y)$ is an equivalence of categories, where the right hand side is the $2$-fiber product and $\mathrm{Vect}(-)$ denotes the category of algebraic vector bundles. At least when everything is affine, but since everything in sight is local, we immediately also get it for arbitrary schemes.

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For existence and properties of the pushout, see also D. Ferrand, "Conducteur, descente et pincement", Bull. Soc. math. France (2003, but the results were obtained in 1970). – Laurent Moret-Bailly Jun 27 '12 at 12:25
@Martin Brandenburg: I think you mean Karl Schwede's paper, do you? – Ronald Bernard Nov 15 '13 at 12:07

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