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Nov 8, 2015 at 17:28 answer added Mahdi Majidi-Zolbanin timeline score: 4
Nov 7, 2015 at 15:55 comment added Andrea No, any proof of that fact would be welcome, though I would be particularly interested in seeing one that does not use localizations.
Nov 7, 2015 at 15:29 comment added Mahdi Majidi-Zolbanin So you know one way of showing $Z(s)=Z(I_s)$ by passing to the localizations of $A(V)$ and $M$ but you want to see another argument? Because you wrote: "possibly without passing through the localizations of $A(V)$ and $M$".
Nov 7, 2015 at 7:40 history edited Andrea
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Nov 6, 2015 at 17:59 comment added Andrea I agree that $Z=Z(s)$ is Zariski closed, but I want to show that it is precisely the zero set of the ideal $I_s$.
Nov 6, 2015 at 17:08 comment added Daniel Barter Have you thought about what happens geometrically? Suppose that you have an algebraic variety $X$ and a vector bundle $ \pi : E \to X$. If $s$ is a section that $Z = \{ s(x) = 0 \} $ defines a subset of $X$. If $E$ is trivial over $U$, then $Z \cap U$ is cut out by ${\rm rank} \, E$ equations. It follows that the whole set $Z$ is closed in the Zariski topology.
Nov 6, 2015 at 14:39 review First posts
Nov 6, 2015 at 14:48
Nov 6, 2015 at 14:39 history asked Andrea CC BY-SA 3.0