Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The zero set of $s$ is the set of points $x$ such that $s$ vanishes in $F_x\otimes k(x)$. If you write locally $s$ as a linear combination $a_1e_1+\cdots +a_re_r$ in a basis $e_1, \dots, e_r$, then the zero set is the common zero locus of the regular functions $a_1, \dots, a_r$. For Question 2, regular section usually means section of the sheaf, by opposition to rational sections which are sections on some open subset. – Cantlog Feb 26 '14 at 22:28
The definition of "regular" depends on context. Can you give us some? – Justin Campbell Feb 26 '14 at 22:43
@JustinCampbell: The statement in what I'm reading is "Assume the section $s$ is regular. Then the codimension of the zero subscheme of $s$ is $r$". I am unclear if the "in particular" statement is the definition of regular, or implied by the definition of regular. – Jim Feb 26 '14 at 23:07
I see, so it's just what Brenin says in his or her answer. – Justin Campbell Feb 26 '14 at 23:13
up vote 7 down vote accepted

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 $\mathscr O_X(U)$.

share|cite|improve this answer
This definition doesn't take into account the scheme structure on the vanishing locus, which need not be reduced. – Justin Campbell Feb 26 '14 at 22:42
Nice explanations Brenin ! However I am a little skeptical about your definition of regular sections. Is there a place where this terminology is used ? – Cantlog Feb 26 '14 at 22:50
I am sure I read it more than once but I cannot find out where, now. For sure, Liu's book contains the concept of a regular immersion, which generalizes our situation: one could say $s$ is regular if the immersion $Z(s)\to X$ is "a regular immersion of codimension $r$". Sometimes one also calls such sections transverse sections. – Brenin Feb 26 '14 at 23:09
Just to show that I'm not being a (complete) pedant: let $X$ be the affine line over a field $k$ and $\mathcal{F} = \mathcal{O}_X = k[t]$ the structure sheaf. Then the sections $t$ and $t^2$ have the same zero set, namely the point $0$, but their zero subschemes are different! The zero subscheme of $t^2$, which is defined by the ideal $(t^2)$, is nonreduced. Since the OP is looking at Hartshorne, this seems like a point worth emphasizing. – Justin Campbell Feb 26 '14 at 23:42
@JustinCampbell: it is fine to be pedant. I agree $Z(s)$ might be nonreduced. Hartshorne also treats general schemes, and so does the question, but of course if one looks at varieties, the reduced scheme structure has to be put on $Z(s)$. I do not know whether under some assumptions the zero locus of a section is reduced. – Brenin Feb 27 '14 at 0:03

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 $I$.

share|cite|improve this answer
Do you know of a source that has more info on this? Such as the equivalence of your definition with Brenin's description of the points? – Jim Feb 27 '14 at 7:06
Actually, I don't really know of a good place, but it's probably in EGA somewhere (I'm sort of kidding). Brenin's definition actually does not agree with mine: he defined the zero set of a section as opposed to the zero subscheme. What is true is that the underlying set of the zero subscheme of a section is the zero set in Brenin's sense. I'll add this to my answer when I get a chance. – Justin Campbell Feb 27 '14 at 22:52
See my second comment on Brenin's answer for a simple example where the zero subscheme is nonreduced, so this distinction actually arises. – Justin Campbell Feb 27 '14 at 22:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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