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The following question is motivated by a desire to find a rough analog in algebraic geometry of the usual notion of gluing of smooth bordisms.

Suppose k is an algebraically closed field of characteristic 0, A, B, and C are smooth varieties over k, and f: A→B and g: A→C are closed regular embeddings of varieties. Although the category of schemes (or varieties over k) is not cocomplete, Corollary 3.9 in Karl Schwede's paper Gluing schemes and a scheme without closed points says that the pushout of f and g in the category of schemes exists. Is this pushout a (not necessarily irreducible) variety over k, i.e., a reduced separated scheme of finite type over k?

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  • $\begingroup$ If you glue together varieties $B$ and $C$ along $A$, you don't necessarily get something irreducible right? $\endgroup$
    – J.C. Ottem
    Commented Dec 11, 2012 at 16:07
  • $\begingroup$ @J. C. Ottem: No, as the example of A=Spec k, B=C=k[t] demonstrates. $\endgroup$
    – Dmitri Pavlov
    Commented Dec 11, 2012 at 16:21
  • $\begingroup$ Ok, thanks for clarifying your question. $\endgroup$
    – J.C. Ottem
    Commented Dec 11, 2012 at 17:59
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    $\begingroup$ I believe this is true but it's been a long time since I thought it. I think separatedness is ok once you know Noetherian, by the valuative criterion. Reducedness is local and then obvious (remember the pullback of the rings is a subring of the product of the two rings you are gluing). So the only question is whether it is finite type. I don't know where to find the finite type condition written down though (it's also local). But see 5.3.2 in Ferrand's paper Conduteur, Descente et Pincement. In particular, the $B \coprod C$ is finite over the gluing, so this should get you close. $\endgroup$
    – Karl Schwede
    Commented Dec 13, 2012 at 14:44
  • $\begingroup$ This interesting question boilds down to: If $A,B,C$ are finitely generated $k$-algebras and $B \twoheadrightarrow A$, $C \twoheadrightarrow A$ are surjective homomorphisms, is then the fiber product $B \times_C A$ also finitely generated? $\endgroup$
    – Martin Brandenburg
    Commented Dec 18, 2012 at 8:42

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