Timeline for Is the pushout of smooth varieties along a smooth closed subvariety again a variety?
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| May 23, 2019 at 3:59 | comment | added | sdey | @DmitriPavlov: Since you are assuming all your schemes $A,B,C$ are smooth, so in particular they are regular, so regular embedding is same as usual embedding : see stacks.math.columbia.edu/tag/0E9J. I guess, in the formulation of Martin Brandenburg's comment, the only thing he is missing is that $A,B,C$ are regular and integral domains. | |
| Dec 18, 2012 at 15:12 | comment | added | Dmitri Pavlov | @Martin: Sure, but your statement is quite a bit stronger (e.g., no conditions on smoothness and regularity). In your formula for the fiber product you should exchange A and C. | |
| Dec 18, 2012 at 8:42 | comment | added | Martin Brandenburg | 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? | |
| Dec 13, 2012 at 14:44 | comment | added | Karl Schwede | 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. | |
| Dec 11, 2012 at 17:59 | comment | added | J.C. Ottem | Ok, thanks for clarifying your question. | |
| Dec 11, 2012 at 16:25 | history | edited | Dmitri Pavlov | CC BY-SA 3.0 |
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| Dec 11, 2012 at 16:24 | history | rollback | Dmitri Pavlov |
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| Dec 11, 2012 at 16:22 | history | edited | Dmitri Pavlov | CC BY-SA 3.0 |
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| Dec 11, 2012 at 16:21 | comment | added | Dmitri Pavlov | @J. C. Ottem: No, as the example of A=Spec k, B=C=k[t] demonstrates. | |
| Dec 11, 2012 at 16:07 | comment | added | J.C. Ottem | If you glue together varieties $B$ and $C$ along $A$, you don't necessarily get something irreducible right? | |
| Dec 11, 2012 at 15:31 | history | asked | Dmitri Pavlov | CC BY-SA 3.0 |