Zero scheme of global sections of vector bundles on affine varieties

Question

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?

Answer 1

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.

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.