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?