# Zero scheme of global sections of vector bundles on affine varieties

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?

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

## 2 Answers

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.

• 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. – Andrea Nov 9 '15 at 9:17

Here is "another proof": Let $$E:=A\{e_1,..,e_n\}$$ and $$E^*:= A\{x_1,..,x_n\}$$with $$x_i:=e_i^*$$ the dual basis. Here $$A$$ is a commutative unital ring. Let $$e:=\sum_i a_ie_i \in E$$ and let $$\phi_e: A \rightarrow E$$ be defined by $$\phi_e(a):=ae$$. Dualizing we get a map

Z1. $$\phi^*_e: E^* \rightarrow A$$

with $$\phi^*_e(x_i)=a_i$$. Hence $$Im(\phi^*_e)=\{a_1,..,a_n\}:=I(e) \subseteq A$$.

You want to define the "zero scheme" of the global section $$e$$ and to prove that this scheme agrees set theoretically with the "set of zeros" of $$e$$ mentioned above.

Definition. The zero scheme of $$e$$ is by definition $$Z(e):=V(I(e))$$ where $$I(e)$$ is the ideal defined above.

Let $$Z_e:=\{x \in Spec(A):$$ such that $$ev^E_x(e)=0 \in E(x)\}$$.

Here $$E(x):=E\otimes_A \kappa(x)$$ is the fiber of $$E$$ at $$x$$ and $$\kappa(x)$$ is the residue field of the point $$x$$. There is a canonical "evaulation map"

$$ev^E_x:E \rightarrow E(x)$$

defined by $$ev_x^E(u):=u\otimes 1 \in E\otimes_A \kappa(x)$$.

Note: If $$\mathfrak{p}_x\subseteq A$$ is the prime ideal corresponding to $$x$$ and $$a\in A$$, it follows $$ev^A_x(a)=0$$ iff $$a\in \mathfrak{p}_x$$.

Lemma 1. $$x\in Z_e$$ iff $$\mathfrak{p}_x\in V(I(e))$$.

proof: Let $$ev^E_x(e)=0$$ it follows $$a_i\in \mathfrak{p}_x$$ for all $$i$$ hence $$\mathfrak{p}_x\in V(I(e))$$. The converse is similar.

Lemma 1 holds for any finite rank projective $$A$$-module $$E$$.

Hence the "zero scheme" $$Z(e):=V(I(e))$$ agrees with your intuition: It's topological space equals the "zero set" $$Z_e$$.

Question: "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 $$M^*$$ have some geometric interpretation in this context?"

Answer: Given any $$A$$-module $$E$$ and any element $$e \in E$$ you get a canonical map

$$\phi_e: A \rightarrow E$$ and a dual map $$\phi^*_e:E^* \rightarrow A$$ and you may define $$I(e):=Im(\phi^*_e) \subseteq A$$. It follows $$I(e)$$ is an ideal. A natural candidate for the "zero scheme of $$e$$" is the closed subscheme $$Z(e):=V(I(e))$$ defined above. Hence the zero scheme of $$e$$ may be defined in complete generality.