# existence of birational morphism and divisors

The following result was metioned in a lecture: A nonsingular (or smooth) projective surface (variety of dimension 2) has a birational morphism to the projective plane, if and only if there exists an effective divisor D such that |D| is base point free (i.e. for every point P there is a member D' of |D| with P not in D') and D has self-intersection number 1. However I didn't find the proof in any book. Dose some research paper contain the proof of the above result?

One direction is easy. If $$S \rightarrow \mathbf P^2$$ is a birational morphism, then let $$D$$ be the pullback of a hyperplane on $$\mathbf P^2$$. This is basepoint-free and has $$D^2=1$$, because both properties are preserved by pullback.
The converse is a little more work. Suppose $$D$$ is a divisor with the given properties. Since $$|D|$$ is basepoint-free, it defines a morphism $$f: S \rightarrow \mathbf P^n$$ for some $$n$$ such that $$D$$ is the preimage of a hyperplane. Note that $$f(S)$$ must have dimension 2, since if it had smaller dimension then we would get $$D^2=0$$. So $$f$$ is generically finite. Then $$D^2$$ equals $$(\operatorname{deg}(f)) \cdot \operatorname{deg}(f(S))$$. So $$D^2=1$$ implies that $$\operatorname{deg}(f)=1$$, which means $$f$$ is birational, and that $$\operatorname{deg}(f(S))=1$$, which means that $$f(S)\cong \mathbf P^2$$.