Question

Let $X$ be a smooth projective variety and $\phi: X \to \mathbb P^n$ be a map. If $\phi$ is an embedding then $E=\phi^*(O(1))$ is very ample. But can one say something if $\phi$ is birational (but not isomorphism) to its image? Is it ample?

Answer 1

Suppose $\phi$ is a morphism (i.e., defined everywhere) which is birational, but not an embedding. Then there are two cases:

1. $\phi$ is finite. In this case $\phi^*\mathscr L$ is ample for any ample $\mathscr L$ on the target. An example (pretty much the only one) when this happens is if $\phi$ is the normalization of $\phi(X)$. For instance if $Y$ is any projective singular curve, or for a slightly more interesting example, $Y=Z(xy^2=tz^2)\subset \mathbb P^n$ and $\phi:X\to Y$ is its normalization.
2. $\phi$ is not finite. In this case, (since it's projective) $\phi$ must have positive dimensional fibers, so there exists a curve on which $\phi^*\mathscr L$ is trivial and hence cannot be ample.