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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?

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  • $\begingroup$ 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 $\endgroup$ – Matt E May 4 '14 at 22:07
  • $\begingroup$ @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). $\endgroup$ – user101036 May 4 '14 at 22:20
  • $\begingroup$ So one wants to produce such a section. Do you have any hint on how this can be done? Sorry for asking for more! $\endgroup$ – user101036 May 4 '14 at 22:21
  • $\begingroup$ ... 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, $\endgroup$ – Matt E May 4 '14 at 23:42
  • $\begingroup$ Dear Tedar, The sections of $\mathcal O(1)$ are hyperplanes. (This is essentially the definition, and the point, of $\mathcal O(1)$.) So your problem essentially amounts to the following: given a quasi-projective variety $X$ embedded in $\mathbb P^n$, can you find a hyperplane whose intersection with $$X$ does not contain any irred. component of $X$. Can you? Regards, $\endgroup$ – Matt E May 4 '14 at 23:44
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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.

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  • $\begingroup$ Dear Matt: yes, your guess is right, this is much easier than Bertini. Choose a point $x_i$ on each irreducible component of $X$. The hyperplanes not going through $x_i$ form an open dense subset $U_i$ of the dual projective space $(\mathbb P^n)^\ast$. The finite intersection $\cap_iU_i\subset (\mathbb P^n)^\ast$ is a dense open subset of $(\mathbb P^n)^\ast$, any element $H\in U_i$ of which represents a hyperplane not containing any irreducible component of $X$. $\endgroup$ – Georges Elencwajg May 5 '14 at 6:11
  • $\begingroup$ Dear Georges, Indeed, that is the argument I had in mind. Best wishes, $\endgroup$ – Matt E May 5 '14 at 11:09
  • $\begingroup$ Dear MattE, I think $H$ must avoid all associated points ($A$ is noetherian) for $D$ to be a Cartier divisor. But the proof is the same as suggested by @GeorgesElencwajg. $\endgroup$ – Cantlog May 5 '14 at 12:56
  • $\begingroup$ @GeorgesElencwajg Sorry for being a bit slow here, but I don't see the proof directly. We want to find a hyperplane avoiding all associated points. So if we look at say $\mathcal{O}(1)$ ,I know that for any finite set of points I can find a non-empty hypersurface that avoids the given points, but I don't know if I can do it for a hyperplane. I am not so good with the geometry (regretfully) in algebraic geometry, so I have trouble translating Georges proof to language I understand fully. $\endgroup$ – user101036 May 5 '14 at 16:03
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    $\begingroup$ Dear Tedar, if a point $p\in \mathbb P^n$ has coordinates $(p_0:p_1:...:p_n)$, then the hyperplanes not containing it are of the form $\sum a_i x_i=0$ with $\sum p_i a_i \neq0$. The last equation must be interpreted as a an open condition on the variables $a_i$ with the $p_i$'s being thought of as constants. This is so tautologous that it seems difficult! You might try to post a new question if you find Matt's or my statements still not clear. $\endgroup$ – Georges Elencwajg May 5 '14 at 18:35

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