# Why is the mobile part of a complete linear system big and nef?

I am trying to learn about linear systems of divisors. Let $X$ be a smooth projective complex surface and $L$ a line bundle. Let's write the complete linear system $|L|=F+|M|$ decomposed in its fixed and mobile part.

I don't see why is $M$ big and nef, i.e. $M^2\geq0$ and $M.C\geq0$ for all irreducible curves. Can anybody explain me this apparently obvious fact?

$M$ need not be big and nef: for nef divisors ($M.C \geq 0$ for all curves $C$) the condition to be big is that $M^2 >0$, a strict inequality. To see this more concretely, just take $L$ to be a basepoint-free but not big line bundle.
On the other hand, the inequalities you wrote are true for the mobile part of a linear system on a surface. They are trivial if $|M|$ is empty, so assume it isn't: since $M$ is mobile then for any curve $C$ there is a representative of $M$ that meets $C$ properly, so $M$ must have non-negative intersection with all curves (in particular with any curve representing $M$ itself).
Postscript: In fact we can say a bit more in this situation. Zariski showed that a line bundle $L$ whose base locus is zero-dimensional must actually be semi-ample (i.e. some positive multiple $L^m$ is basepoint-free). Fujita generalised this as follows: if the restriction of $L$ to the base locus is ample, then again $L$ is semi-ample. (Since any line bundle restricted to a point is ample, this implies Zariski's original statement.) As usual, Lazarsfeld's book is a good place to read about these things.
• @Cantlog: no, Zariski--Fujita is true in all dimensions. Unfortunately, in higher dimensions there is a gap between 0 and $n-1$, so movable divisors need not be semi-ample! But that is what makes birational geometry interesting... – user64687 Oct 3 '13 at 16:38