# if the canonical divisor is nef, then a multiple if effective

Let $S$ be a complex projective non-singular surface. Couldd you explain the following implication:

If the canonical divisor $K_S$ is nef then there exists a number $m>>0$ such that $mK_S$ is equivalent to an effective divisor.

I remember that a divisor $D$ is called nef if $D.C\ge0$ for every irrducible curve $C\subset S$.

It is a general fact, let me try to sketch a proof of it.

If a divisor $D$ is nef then $D^2\geq0$.

If $D^2>0$ just apply the Riemann-Roch inequality to $mD$ and get $h^0(mD)>0$ for $m$ big enough.

If $D^2=0$ then $D$, being nef, is numerically trivial. Then it might be that $D$ is itself equivalent to zero (in which case you are done) or it is not. So you only are left with the latter case to deal with. It is a non-trivial fact that any numerically trivial $D$ is such that $mD\sim0$ for $m$ big enough. (See Remark 1.1.20 in Lazarsfeld' book on Positivity). Surely for the special case $D=K$ it has to be simpler but I can't think of a straightforward argument right now.

Edit: as Schemer suggests in the comments the cited Remark of Lazarsfeld is false as stated. The precise content is: a line bundle $L$ is numerically trivial if and only if there is some $m\neq0$ such that $L^{\otimes m}\in\mathrm{Pic}^0(X)$. Therefore the end of the above proof does not apply to general $D$. However when $D=K$ it follows e.g. by the Enriques-Castelnuovo classification of smooth projective surfaces. Indeed, if $K$ is nef then $S$ is not rational. Hence, by a rationality criterion of Enriques you get $h^0(12K)\neq0$. Therefore $12K$ is equivalent to an effective divisor.

• It isn't true that a multiple of a numerically trivial bundle is trivial. (Pic^0 is a torus, with lots of non-torsion points!) I cannot find the remark you refer to in Lazarsfeld; the closest thing I see is Remark 1.1.20. But if you meant that remark, you left out the key word deformation. Nevertheless it is true that if $K$ is nef and $K^2=0$, then $K$ is torsion; the only argument I see is classification, which is unsatsifying. – Schemer Oct 27 '15 at 21:32
• @Schemer: oops! You are right. Let me edit. – Heitor Fontana Oct 27 '15 at 22:01