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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.