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Let $X$ be a smooth projective complex surface and $H$ an ample divisor on $X$.

My main question is whether for any divisor $D$ we can say that eventually $mH+D$ will be globally generated.

Secondly, I wonder if the viceversa holds as well.

Heitor Fontana
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  • You only have to check with the definition of an ample invertible sheaf over any separated algebraic variety (or any quasi-separated noetherian scheme more generally). – Cantlog Feb 09 '14 at 16:10
  • Look in Lazarsfeld, Principles of Algebraic Geometry Volume 1, for all kinds of results of this kind (which hold much more generally, as Cantlog indicates). –  Feb 09 '14 at 16:51
  • @AsalBeagDubh. Thanks for the reference Asal, I look forward to taking a look at that book (which I didn't know) now! however at the moment I am not able to get it unfortunately; could you please just sketch in an answer the idea for proving that? – Heitor Fontana Feb 10 '14 at 09:16

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A precise reference for the main question is Lazarsfeld, Positivity in Algebraic Geometry Volume 1, Theorem 1.2.6. The idea is to use Serre vanishing for the ample line bundle $H$.

Details: fix a point $x \in X$, and consider the coherent sheaf $\mathfrak m_x \cdot O_X(D)$ Then by Serre vanishing there exists a natural number $M$ such that $H^1(X, \mathfrak m_x \cdot O_X(D) \otimes H^{\otimes m})=0$ for $m \geq M$.

Now look at the exact sequence

$$ 0 \rightarrow m_x \cdot O_X(D) \rightarrow O_X(D) \rightarrow O_x(D) \rightarrow 0.$$

Twist this by $H^{\otimes m}$ and use the vanishing about to conclude that there is a section of $O_X(D) \otimes H^{\otimes m}$ that does not vanish at $x$.

To conclude, since this section doesn't vanish at $x$, it doesn't vanish in an open neighbourhood either. Then use quasi-compactness of $X$ to find an appropriate $M$ such that this holds everywhere --- i.e. $mH+D$ is globally generated for $m \geq M$. $\square$

I interpret the second question as asking "if $H$ has the property that for any divisor $D$, it holds that $mH+D$ is eventually globally generated, is it true that $H$ is ample?" The answer is yes, for the following reason:

Suppose $H$ is not ample, and let $D$ be a divisor such that $-D$ is ample. Suppose that there exists $m>0$ such that $mH+D$ is globally generated. Then $mH=(mH+D)-D$ is the sum of a globally generated and an ample divisor, hence is ample, contrary to hypothesis.