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?
1 Answer
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I don't know where to find a proof written, but it is not hard to give one here.
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 $f(S)\cong \mathbf P^2$.
