Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 $D=\sum_{i=0}^n\left.n_iD_i\right.$ be a Weil divisor on a nonsingular, projective variety $X$. Then, it corresponds to a Cartier divisor $D$, which corresponds to a line bundle $\mathcal{L}:=\mathcal{L}(D)$ and there exists a section $s\in\mathcal{L}(X)$ such that $D$ is the divisor of zeros of $s$. Now, I am wondering: What is the vanishing set of $s$? It feels like it should be

$\displaystyle Z(s)=\bigcup_{\substack{0\le i\le n\\n_i \ge 0}} D_i$

Is this true and if yes, can you provide a proof? It feels like this should be well-known, but I cannot find any good references. If you could provide any, that would also be nice.

share|cite|improve this question
What distinction do you make between "the divisor of zeros of the section $s$" of which you write that it is $D$ and "the vanishing set of $s$" of which you conjecture that it is $Z(s)$, the positive part of $D$? – Georges Elencwajg Jun 28 '11 at 14:14
This comment made me realize my mistake and my confusion xD. There is really nothing more to say. I will accept your answer and remove my comment, it would just be confusing to anyone else. – Jesko Hüttenhain Jun 28 '11 at 15:12
All right. However your deleted comment made me realize that my answer wasn't optimal and I have just edited it. So thanks for your constructive criticism! – Georges Elencwajg Jun 28 '11 at 15:48
up vote 4 down vote accepted

Consider the Cartier divisor $D=\Sigma p_kD_k-\Sigma n_lE_l$ ( with $p_k, n_l>0$ ) and a covering $(U_i)$ of $X$ such that on each $U_i$ the Cartier divisor $D\cap U_i$ is given by a rational function $f_i\in \mathcal {Rat}(U_i)$. The line bundle $\mathcal L=\mathcal O(D)$ may be defined by the Cech cocycle $g_{ij}=f_i/f_j \in \mathcal O^\ast(U_i\cap U_j)$.
A better, more concrete, description of $\mathcal L =\mathcal O(D)$ is that as a sheaf it is the sub-$\mathcal O_X$-Module $\mathcal L\subset \mathcal {Rat}_X$ which restricted to $U_i$ equals $\mathcal L |U_i=\frac{1}{f_i} \mathcal O_{U_i}$.

The line bundle $\mathcal L$ then has a canonical rational section $s=1 \in \Gamma(X, \mathcal L)\subset \mathcal Rat(X)$. We have trivializations of $\mathcal L$ attached to our data, namely $\mathcal L|U_i \stackrel {\simeq}{\longrightarrow} \mathcal O_{U_i}:g\mapsto gf_i$, mapping $s|u_i=1\in \Gamma(U_i, \mathcal L)$ to $f_i\in \Gamma (U_i,\mathcal {Rat}(U_i))$ And now it is completely tautological that the divisor of $s$ is $D$, since that divisor is given by the $f_i$'s on $U_i$.

1) As you see, the fact that $X$ is projective is actually irrelevant.
2) The construction works on any scheme, however singular it may be, as long as you consider Cartier divisors.
3) If your scheme is separated, noetherian, integral and locally factorial, you can indeed associate to each Weil divisor a unique Cartier divisor and then apply the preceding considerations. But this is rather a different issue: your question is really about cartier divisors.

share|cite|improve this answer
I have just edited my answer, taking into account your (now deleted) comment which made me realize that I should have expressed myself more clearly. – Georges Elencwajg Jun 28 '11 at 15:47
I feel that you need to take $D$ to be an effective divisor for this global section to exist. If such a section $s$ existed then $s|U_i=1\rightarrow f_i\in\Gamma(U_i,Rat(U_i))$ but this map should be a trivialization of $\mathcal L$ so the image should lie in $\Gamma(U_i,O_{U_i}).$ This implies that $D$ is effective. – Lalit Jain Jun 29 '11 at 15:13

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.