Some naive question on birational equivalence

Question

Recall that a variety $V$ is called rational if there exists a birational map (not necessarily everywhere defined) between $V$ and some projective space. A seemingly stronger condition is that there exists a birational morphism. Are these two conditions equivalent? For curves I known that it is true.

Answer 1

If by morphism you mean regular map (my AG is not very developed), then no, they are not equivalent. A nonsingular quadric $Q\subset \mathbb P^n$ is rational. This can be seen by projecting from a point $x\in Q$ to a hyperplane $L$ which doesn't contain $Q.$ However, $Q$ is not isomorphic to $\mathbb P^{n-1}$ since there exist disjoint curves on $Q.$ For example, if $Q\subset \mathbb P^3$ is a surface given by $F(x:y:z:w)=0,$ then the generating lines obtained by holding some of the variables constant are disjoint. If $Q$ were isomorphic to the projective plane, then this would contradict Bézout's theorem.

Answer 2

No, these two conditions are not equivalent.

The simplest example is $V=\mathbf P^1 \times \mathbf P^1$. This is rational, but I claim there is no morphism $V \rightarrow \mathbf P^2$ or $\mathbf P^2 \rightarrow V$.

To prove that claim, one can use the fact that if $f: Y \rightarrow X$ is a birational morphism of smooth surfaces which is not an isomorphism, then $Y$ must contain a curve of selfintersection $-1$. (A reference is Shafarevich Basic Algebraic Geometry 1, Section IV.3.4.)

Since neither of the two surfaces $\mathbf P^2$ and $\mathbf P^1 \times \mathbf P^1$ contains such a curve, this proves the claim.