2
$\begingroup$

This is from Liu's book. Let $X$ be an irreducible quasiprojective variety over an infinite field. Let $D_1 \ldots , D_n$ where $n=\dim X$ be Cartier divisors on X. Show that there exists $D_i' \equiv D_i$ (linear equivalence) such that $$\text{dim} \bigcap_{1\leq i \leq r} \operatorname{Supp} D_i' = n-r$$ for every $r \leq n$.

i am able to do this if I assume that none of the $D_i$ has empty support and that all are effective. So my question is in two parts:

  1. How can I produce a divisor with nonempty support? What if all sections of the structure sheaf is invertible?

  2. In the case where all divisors are effective, the support has dimension $n-1$ ( essentially by Krull) - how do we extend this to the general case?

$\endgroup$
  • $\begingroup$ Dear Heidar, Do you know about ample and very ample divisors? (I'm not familiar with Liu's book, so I don't know what tools he is giving you to use.) Regards, $\endgroup$ – Matt E Jun 16 '13 at 14:08
  • $\begingroup$ Dear Matt: I do know about ample and very ample divisors - But I fail to see how I could apply it here. Any further hint? Thank you!!! $\endgroup$ – Heidar Svan Jun 16 '13 at 14:14
2
$\begingroup$

If $D$ is a divisor, let $H$ be a very ample divisor for which $D + H$ is also very ample. (For quasi-projective varities, such $H$ always exists.)

Then write $D = (D + H) - H$. Now we can vary $D+H$ and $H$ as much as we like, so that they meet properly.

(Remember the very ample just means hyperplane sections with respect to some projective embedding, and we can always move a hyperplane (in this case the hyperplane whose intersectino with $X$ equals $D+H$) so that it meets any given subvariety (in this case $H$) properly.)

Now continue by induction on $n$.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Just a brief check: The way we are ensuring that D has non-empty support is through varying the hyperplane divisor? Another ignorant question: Why does this imply non-empty support? I think I follow the other things fine. $\endgroup$ – Heidar Svan Jun 16 '13 at 15:06
  • $\begingroup$ @HeidarSvan: Dear Heidar, I'm not sure I understand your question. $D+H$ is a very ample divisor, the intersection of $X$ with a hyperplane under some projective embedding. Certainly we can choose the hyperplane so that this intersection is non-empty. But I think your concern about non-empty support is a bit misplaced (e.g. in the case that $X$ is actually projective, the intersection with a hyperplane will never be non-empty). The problem is that a priori a divisor may not move at all; that is why we add a very ample divisor to it in the first place. Regards, $\endgroup$ – Matt E Jun 16 '13 at 15:34
  • $\begingroup$ Let me try to be more precise: Let us consider the case where $n=1$ . Then my worry is that the support of D would be empty. We need to find a linearly equivalent divisor with non-empty support. In the $n=1$ case, how do we find such a linearly equivalent divisor? $\endgroup$ – Heidar Svan Jun 16 '13 at 15:58
  • $\begingroup$ @Heidear: Dear Heidar, The divisor $D$ has empty support precisely if it vanishes. To find a linearly equivalent divisor with non-empty support, just consider an projective embedding $X \hookrightarrow \mathbb P^N$, let $D_1$ and $D_2$ be two distinct hyperplane sections of $X$, and consider $D_1 - D_2$. (This is just my above argument with $H$ being the very ample divisor corresponding to the given projective embedding, with $D$ taken to be zero.) As I wrote, this is not the real point of the problem. The more substantial part is to show that you can move divisors enough so that ... $\endgroup$ – Matt E Jun 17 '13 at 0:20
  • $\begingroup$ ... the intersections are proper (so that the dimension is $n - r$ and not something bigger). I would spend more time thinking about that, rather than focusing on the somewhat tangential point of dealing with case of the zero divisor. Regards, $\endgroup$ – Matt E Jun 17 '13 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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