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I want to understand better the notion of zero scheme of a section of a vector bundle. For simplicity I will consider the case of affine varieties.

Let $\mathbb{K}$ be an algebraically closed field, $V\subset\mathbb{K}^n$ be an affine algebraic variety (maybe reducible) and $A(V)=\mathbb{K}[x_1,...,x_n]/I(V)$ be its coordinate ring.

By Serre's theorem, taking global sections gives a bijective correspondence between isomorphism classes of algebraic vector bundles over $V$ and projective $A(V)$-modules of finite type.

So consider a projective $A(V)$-module $M$ of finite type, and let $s\in M$. Then we have:

  • the zero set of $s$, $$Z(s):=\{z\in V\ |\ s(z)=0\},$$ where $s(z)$ is the image of $s$ in the fiber $M(\mathfrak{m}_z)=M\otimes_{A(V)}k(\mathfrak{m}_z)$, $\mathfrak{m}_z\subset A(V)$ being the maximal ideal corresponding to $z\in V$ and $k(\mathfrak{m}_z)=A(V)/\mathfrak{m}_z$ being its residue field;
  • the ideal associated to $s$, $I_s:=\text{im}\, \iota_s$, where $\iota_s:M^\vee\rightarrow A(V)$ is the $A(V)$-linear map given by evaluation on $s$; this defines a (maybe nonreduced) closed subscheme of $V$.

I expect that $Z(s)$ coincides with the zero set of $I_s$, i.e. with $$Z(I_s):=\{z\in V\ |\ a(z)=0\ \forall a\in I_s\}.$$ How to see this (possibly without passing through the localizations of $A(V)$ and $M$)? Does everything make sense even when $M$ is not projective? Does the dual $M^\vee$ have some geometric interpretation in this context?

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  • $\begingroup$ 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. $\endgroup$ Nov 6, 2015 at 17:08
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    $\begingroup$ I agree that $Z=Z(s)$ is Zariski closed, but I want to show that it is precisely the zero set of the ideal $I_s$. $\endgroup$
    – Andrea
    Nov 6, 2015 at 17:59
  • $\begingroup$ 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 $M$". $\endgroup$ Nov 7, 2015 at 15:29
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    $\begingroup$ No, any proof of that fact would be welcome, though I would be particularly interested in seeing one that does not use localizations. $\endgroup$
    – Andrea
    Nov 7, 2015 at 15:55

1 Answer 1

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Consider the following conditions:

  1. $z\in Z(s)$
  2. $s(z)=0$
  3. $s\in\mathfrak{m}_zM$
  4. the image of the map $A(V)\stackrel{\varphi_s}{\longrightarrow}M$ defined by $1\mapsto s$, is in $\mathfrak{m}_zM$
  5. $\forall f\in M^{\vee}$ we have $(f\circ\varphi_s)(1)\in\mathfrak{m}_z$
  6. $\forall f\in M^{\vee}$ we have $ f(s)\in \mathfrak{m}_z$
  7. $I_s\subseteq\mathfrak{m}_z$.

Then it is clear that $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Rightarrow(5)\Leftrightarrow(6)\Leftrightarrow (7)$. Assuming $M$ is projective we will show $(6)\Rightarrow (3)$, which will achieve what you want to prove. Assume condition $(6)$ holds. If $s\not\in\mathfrak{m}_zM$, then $\overline{s}$, the image of $s$ in $M/\mathfrak{m}_zM$ is not zero. Since $M/\mathfrak{m}_zM$ is a finite-dimensional vector space over $A(V)/\mathfrak{m}_z$, one can define a linear map $\sigma\colon M/\mathfrak{m}_zM\rightarrow A(V)/\mathfrak{m}_z$ such that $\sigma(\overline{s})\neq0$. This $\sigma$ can also be considered as a map of $A(V)$-modules. Composing $\sigma$ with the map $M\rightarrow M/\mathfrak{m}_zM$ we get an $A(V)$-linear map $\overline{f}\colon M\rightarrow A(V)/\mathfrak{m}_z$ such that $\overline{f}(s)\neq0$. Since $M$ is projective, $\overline{f}$ can be lifted to an $A(V)$-linear map $f\colon M\rightarrow A(V)$ and we have $f(s)\not\in\mathfrak{m}_z$. This contradict with condition $(6)$, which we had assumed holding. enter image description here

P.S. If $M$ is not projective you will get $Z(s)\subseteq Z(I_s)$, but I don't know how one would get $Z(I_s)\subseteq Z(s)$ without using that $M$ is projective.

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  • $\begingroup$ Very clear, thanks a lot. Interesting that $(6)\Rightarrow (3)$ does not work for general $M$, also considered what it means geometrically. It would be nice to understand whether there are counterexamples. $\endgroup$
    – Andrea
    Nov 9, 2015 at 9:17

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