What is the zero subscheme of a section

Question

Let $X$ be a scheme, $\mathcal F$ a locally free sheaf of rank $r$ and $s \in \Gamma(X, \mathcal F)$ a global section of $\mathcal F$.

Question: What is the zero subscheme of $s$?

I can't believe that pouring through Hartshorne hasn't turned up a definition of this. It should be some subscheme of $X$. The only thing I can think of is the set of points $x \in X$ where $s$ goes to $0$ in the stalk $\mathcal F_x$, i.e., the complement of the support of $s$. But that would make the zero subscheme of $s$ open and that doesn't make sense because in what I'm reading there is a hypothesis that $s$ is a regular section, and that this has something to do with the codimension of the zero scheme in $X$ (which would always be $0$ if the zero scheme were open). Which leads me to question $2$:

Question 2: What is a regular section? Is it a section whose zero subscheme is regular? Cause that would be great if it were true.

Answer 1

You are right, the zero scheme of a section is not the complement of the support. In other words, the condition is not "$s_x=0$ in $\mathcal F_x$", but rather "$s(x)=0$". To answer your question, I have to make sense of the latter expression.

Locally around $x$, a section $s\in \Gamma(X,\mathcal F)$ is represented by an $r$-tuple of regular functions (holomorphic, if you work in the category of complex manifolds) $$f_1,\dots,f_r:U\to \mathbb A^1,$$ for some open neighborhood $U\subset X$ of $x$. (After all, to say that $\mathcal F$ is a locally free sheaf of rank $r$ boils down to saying that locally around every point there is a trivializing open set, namely some $U\subset X$ as above such that $\mathcal F|_U\cong \mathscr O_X^r|_U$; hence $s$ corresponds to a certain $r$-tuple of regular functions under this trivialization.)

For such functions $f_i$, it makes sense to ask whether or not $f_i(x)=0$. If the latter condition is satisfied for $i=1,\dots,r$, then we say that $s(x)=0$ (and this does not depend on the open neighborhood $U$. The locus of such $x$'s is closed.

Finally, a section $s$ is called regular if the codimension of its zero scheme $Z(s)\subset X$ inside $X$ is the expected one, namely if $$\textrm{codim}(Z(s),X)=r.$$ This is equivalent, algebraically, to $(f_1,\dots,f_r)$ being a regular sequence in the ring $\mathscr O_X(U)$.

Answer 2

Let $V$ be the total space of $\mathcal{F}$, i.e. the global spectrum of the quasicoherent sheaf of algebras $\text{Sym}(\mathcal{F}^{\vee})$. There is a natural projection $V \to X$. Then a global section of $\mathcal{F}$ can be thought of as a morphism $s : X \to V$ such that the composition $X \to V \to X$ is the identity on $X$ (literally a section of the projection). In particular, the morphism $s : X \to V$ is a closed embedding. Let $Z \subset V$ be the image of the zero section of $\mathcal{F}$: then the zero subscheme of $s \in \Gamma(X,\mathcal{F})$ is the scheme-theoretic preimage of $Z$ by the morphism $s : X \to V$.

Edit: Write $Z(s) = s^{-1}(Z)$ for the zero subscheme of $s$. I claim that Brenin's definition gives the underlying (closed) set of $Z(s)$, which determines the maximal reduced subscheme of $Z(s)$ but not the subscheme $Z(s)$ itself. We are trying to prove that two subsets of $X$ are equal, which is obviously a local question, so we may replace $X$ by an open subset where $\mathcal{F}$ is trivial and $X = \text{Spec } A$ is affine. Let $M = \Gamma(X,\mathcal{F})$ be the $A$-module corresponding to $\mathcal{F}$ and choose an $A$-basis $m_1,\cdots,m_r \in M$. Then our given section $s \in M$ can be written $s = f_1m_1 + \cdots f_rm_r$ for some $f_1,\cdots,f_r \in A$. I'll leave it to you to check that the ideal $I$ of $Z(s)$ is generated by the $f_i$ (this is a matter of unraveling definitions). But Brenin's vanishing set is defined as the subset of $X$ where the $f_i$ vanish, i.e. the closed subset corresponding to $I$.