Abundance for algebraic surfaces I am currently teaching a course in algebraic geometry where one of the aims is to give an overview of the Enriques-Kodaira classification of surfaces. I am trying to throw in some modern aspects so I formulated the cone theorem and then used it (together with the existence of extremal contractions) to show that one can blow down to either $\mathbb P^2$, a ruled surface or one with $K_X$ nef. That worked quite well (the right amount of detail and non-detail). However, continuing with the nef case one of the major results is abundance (a positive multiple of $K_X$ is basepoint free). The classical Enriques-Kodaira classification does give abundance but only at the end of an almost complete classification of Kodaira dimension $\leq0$ surfaces (with $K_X$ nef). 
Hence my question is: Using modern ideas is it possible to give a quicker proof of abundance for surfaces?
(Actually I am not quite sure that abundance for Kodaira dimension $2$ can be considered to be part of the E-K classification but let's ignore that).
 A: I am a novice here, but have you looked at Lazarsfeld's chapter "Lectures on linear series" in the book Complex algebraic geometry, IAS Park City math series vol 3?  There he deduces it from Reider's theorem which he deduces from Bogomolov's instability theorem, whose proof he also sketches.  This is a rewrite of some parts of his chapter on application of vector bundles techniques in the book Lectures on Riemann surfaces from Trieste, where however he says his argument has an error.  I am assuming the result you want is that when K is nef then some positive multiple of K is free, (in fact 4K).
yes you are right he does assume general type, but unfortunately for my understanding he does not state that in the theorem itself but only in a paragraph above the theorem, which is then apparently a blanket assumption not restated later.  So since you are presumably dealing with the opposite case this is useless to you.  On p. 81 of Kollar and Mori they merely state that for surfaces base point freeness is a "non trivial result".  This suggests they did not know an easy proof.
edit:  In Miles Reid's chapters on algebraic surfaces, he discusses this point explicitly at the end of his treatment of classification of surfaces with K nef.  See E.9.1. "Abundance as a logical bottleneck".  He states there that he knows only one proof in the literature not using Enriques' argument at a crucial point, namely that in the book of Barth, Peters, and Van de Ven, where they use Ueno's prof of Iitaka's additivity conjecture C(2,1) via moduli of curves.  Does that help?
A: There is a proof of aboundance in the case of surfaces in Section 1.5 of the book "Introduction to the Mori Program" by Kenji Matsuki. 
A: This is more a remark than an answer.
It is perhaps worth noting that in the related classification of foliations on surfaces by McQuillan, Brunella, and Mendes abundance does not hold. The so called Hilbert Modular foliations are examples of foliations with nef canonical bundle but with Kodaira-Iitaka
dimension negative. These turn out to be the only examples, and the proof of this fact is the harder part of the classification.
A: Looking at both Matsuki's "Introduction to the Mori Program"  and Reid's "Chapters on Algebraic Surfaces", it does not seem that there is a uniform and quick proof of abundance conjecture for surfaces. In both books, the Authors separately consider the three cases:


*

*$\textrm{kod}(X)=2$. Then abundance is proven by explicitly constructing the canonical model $X^{can}$ of $X$, by means of contractions of the $(-2)$-cycles. Another approach is using Kawamata-Viehweg vanishing, if one wants to prove base-point freeness in a more general setting. Also, one could apply Bombieri's result, which ensures that $|5K_X|$ is always a birational morphism onto the canonical model for any surface of general type. 

*$\textrm{kod}(X)=1$. Then one must prove the existence of an elliptic pencil on $X$ and the canonical bundle formula for elliptic fibrations. These are both rather subtle results, 
and I do not know any proof avoiding them.

*$\textrm{kod}(X)=0$. Then the proof is obtained by looking at the Albanese map, and the analysis needed is essentially equivalent to the Enriques-Kodaira classification. 
If a quicker proof avoiding this case-by-case analysis actually exists, I really would like to see it.
