Showing that a power of an ample sheaf is equivalent to an effective Cartier divisor

I am trying the following exercise:

Let $X$ be a quasi-projective scheme over a Noetherian ring A. Let $\mathcal{L}$ be an ample sheaf on $X$. Show that there exists an $m \geq 1$ such that $\mathcal{L}^{\otimes m} \cong \mathcal{O}_X(D)$ for some effective Cartier divisor $D$.

I tried to start by using that on $X$, invertible sheaves and Cartier divisors are in bijective correspondence. Now, the obvious thing would be to take a tensor power $\mathcal{L}^{\otimes m}$ of $\mathcal{L}$ that is very ample, and try to argue that if a Cartier divisor $D$ is such that $\mathcal{O}_X(D)$ is isomorphic to a very ample sheaf, it is effective.For this, I thought maybe one should use something about global generation and show that any divisor $D$ such that $\mathcal{O}_X(D)$ is globally generated is actually effective. I have not been able to show this, so, any hints or solutions?

• Dear Tedar, Think about how very ample line bundles correspond to embeddings into projective space. (And note that it is not a particular Cartier divisor that corresponds to an invertible sheaf, but a linear equiv. class of them. Some members of this linear equiv. class might be effective, and others not. So you might want to think more about what extra data you need to attach to the invertible sheaf to actually produce an effective Cartier divisor in the linear equiv. class. This post – Matt E May 4 '14 at 22:07
• @MattE: So, I think I see what you mean ( I actually read that great post before) , namely: Given this invertible sheaf $\mathcal{L}^{\otimes m}$, we want to find a global section $s$ that is not locally a zero-divisor. So, one can show that if we have a line bundle of the form $i^\ast \mathcal{O}_X(1)$ ,it contains a section that is not locally a zero-divisor. My guess would be to use the canonical sections from $\mathbb{P}^n_A$ (for some n). – user101036 May 4 '14 at 22:20
• So one wants to produce such a section. Do you have any hint on how this can be done? Sorry for asking for more! – user101036 May 4 '14 at 22:21
• ... might help. [Added: sorry, this was supposed to be the end of my previous comment, but I only just submitted it for some reason.] Regards, – Matt E May 4 '14 at 23:42