Can effective divisors generate all Cartier divisors?

Question

Let $X$ be a Noetherian scheme. My question is that: for any Cartier divisor $D$, can we write it as $D_1-D_2$ where $D_1$,$D_2$ are effective? What about further assume $X$ is integral?

I can see $D$ is locally generated by fractions of sections of $O_X$. But it seems not easy to depart the denominators and numerators explicitly. Could you provide some help? Thanks.

Related, but unanswered: https://math.stackexchange.com/questions/3992498/example-of-a-cartier-divisor-that-cannot-be-written-as-the-difference-of-effecti — KReiser, Jan 22 '22 at 08:25
@Sasha Do you mean $X$ has an ample invertible sheaf? And could you give a proof or share a reference? Thanks! — Richard, Jan 22 '22 at 12:52
@Richard: This can be found in Hartshorne, but you can also see a simple argument below. — Sasha, Jan 22 '22 at 18:33
@Richard: this may also help https://stacks.math.columbia.edu/tag/0AYM — Ben C, Jan 22 '22 at 20:56

score 1 · Answer 1 · answered Jan 22 '22 at 18:32

1

Assume $X$ has an ample (Cartier) divisor class $H$ (equivalently, an ample invertible sheaf). Then for each Cartier divisor $D$ there exists an integer $n \gg 0$ such that the linear systems $|nH|$ and $|D + nH|$ contain effective (Cartier) divisors, say $D_1$ and $D_2$. Then $$ D_2 - D_1 = (D + nH) - nH = D. $$

answered Jan 22 '22 at 18:32

Sasha

20,727

I can't understand how to deduce these two linear systems contain effective Cartier divisors. Could you provide a reference? Thanks! – Richard Jan 23 '22 at 14:39
Read the part about ample divisors in the Hartshorne textbook. – Sasha Jan 23 '22 at 14:43

Can effective divisors generate all Cartier divisors?

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