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?

  • $\begingroup$ 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

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 $n$.

  • $\begingroup$ 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

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