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?
This collects some remarks and hints above:
By passing to a power, we may assume that $\mathcal L$ is ample, and hence is the restriction of $\mathcal O(1)$ to $X$, for some embedding $X \hookrightarrow \mathcal P^n$. Sections of $\mathcal O(1)$ over $\mathcal P^n$ are precisely the linear homogeneous equations, which cut out the hyperplanes in $\mathcal P^n$, and these are the sections of $\mathcal L$ that we can get our hands on.
So we ask: is there a hyperplane $H$ in $\mathcal P^n$ that does not contain any irred. component of $X$? Such a hyperplane will correspond to a section $\ell$ of $\mathcal O(1)$ whose restriction to $X$ is a local non-zero divisor at every point, and hence which will give rise to the desired Cartier divisor.
More geometrically, we will have $D = X \cap H.$
The question shouldn't be too hard to answer; it's a bit like Bertini's theorem, and can be proved in a somewhat similar (but I guess easier) way.