# Can effective divisors generate all Cartier divisors?

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: math.stackexchange.com/questions/3992498/… Jan 22, 2022 at 8:25
• If $X$ has an ample divisor, the answer is positive. Jan 22, 2022 at 8:47
• @Sasha Do you mean $X$ has an ample invertible sheaf? And could you give a proof or share a reference? Thanks! Jan 22, 2022 at 12:52
• @Richard: This can be found in Hartshorne, but you can also see a simple argument below. Jan 22, 2022 at 18:33
• @Richard: this may also help stacks.math.columbia.edu/tag/0AYM Jan 22, 2022 at 20:56

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 $$D_2 - D_1 = (D + nH) - nH = D.$$