# Technique to prove basepoint-freeness

Let $X$ be a smooth projective variety over $\mathbb{C}$. And let $L$ be a big and nef line bundle on $X$. I want to prove $L$ is semi-ample($L^m$ is basepoint-free for some $m > 0$).

The only way I know is using Kawamata basepoint-free theorem:

Theorem. Let $(X, \Delta)$ be a proper klt pair with $\Delta$ effective. Let $D$ be a nef Cartier divisor such that $aD-K_X-\Delta$ is nef and big for some $a > 0$. Then $|bD|$ has no basepoints for all $b >> 0$.

Question. What other kinds of techniques to prove semi-ampleness or basepoint-freeness of given line bundle are?

Maybe I miss some obvious method. Please don't hesitate adding answer although you think your idea on the top of your head is elementary.

Addition : In my situation, $X$ is a moduli space $\overline{M}_{0,n}$. In this case, Kodaira dimension is $-\infty$. More generally, I want to think genus 0 Kontsevich moduli space of stable maps to projective space, too. $L$ is given by a linear combination of boundary divisors. It is well-known that boundary divisors are normal crossing, and we know many curves on the space such that we can calculate intersection numbers with boundary divisors explicitely.

• Are there any other special circumstances around your situation? Can you say anything more specifically about $X$, or $L$ (where they come from, etc.) Sep 3, 2010 at 16:50
• Recently Jian Song gave an analytical approach via Cheeger-Coling-Tian theory to solve base point free theorem arxiv.org/abs/1409.8374
– user21574
Jul 6, 2017 at 21:18

I don't think that your assertion is true; for example, Lazarsfeld gives an example (PAG, 2.3.3) of a big and nef divisor on a surface such that its graded algebra is not finitely generated, so that the divisor can't be semiample.

But there are some close results for nef and big divisors, or even for good divisors (when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed : for example, Wilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $m\geq m_0$, the linear system $|mD-N|$ has no base-point. (PAG, 2.3.9)

• You are right, nefness + bigness does not guarantee semi-ampleness. My question is following: Is there any sufficient condition to get semi-ampleness? With which conditions we get semi-ample property?
– Moon
Sep 4, 2010 at 10:46
• Anyway, thank you for answer. By the way, what is the result of Mourougane and Russo? Do you mean the theorem 2.3.9 in PAG?
– Moon
Sep 4, 2010 at 10:49
• For example, as I mentionned it, a nef and big divisor $D$ is semiample if and only if its graded ring of sections $R(X,D)=\bigoplus_{m\in \mathbb N} H^0(X,mD)$ is finitely generated. Sep 4, 2010 at 10:51
• Yes, you could have a look at it here : www.math.jussieu.fr/~mourouga/note_abondant.pdf (its both in french and in english, don't worry!) Sep 4, 2010 at 10:52

Numerical criteria for base-pointfreeness are known only in specific cases such as the Kawamata basepoint-free theorem and Reider's theorem (for $\dim X=2$).

In the case you mention, $X=\mathcal{M}_{0,n}$, the problem of classifying semi-ample divisors is an important problem. It is slightly easier in positive characteristic, thanks to a theorem of Keel which says that a nef line bundle $L$ is semi-ample if and only if the restriction $L|_E$ is semiample, where $E$ is the exceptional locus of subvarieties $Z$ such that $L^{\dim Z}.Z=0$. If $f:X\to Y$ is a morphism with exceptional locus $E$, then $L$ is semi-ample if and only $L^r$ is the pullback of an ample line bundle on $Y$ for $r>0$. For the precise statements, you might want to take a look at

According to G. Farkas' article, there are currently no known examples of nef divisors which are not semi-ample.