# What can be said about a pullback of a very ample line bundle w.r.t birational maps?

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?

-

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.
Thank you very much, Can one say something else about $E$? –  Rami Feb 28 '12 at 4:15
For $\phi$ proper and birational onto its image, then $E$ is always big and semi-ample (and in particular, big and nef). –  Karl Schwede Feb 28 '12 at 4:34
In particular, there is always an effective divisor $D$ on $X$ such that $E - \varepsilon D$ is ample, for all sufficiently small $\varepsilon D$... Typically if $\phi$ is the blow-up of some codimension 2 subvariety, then you can take this $D$ to be the exceptional divisor with appropriate coefficients. –  Karl Schwede Feb 28 '12 at 4:36
Sorry, that should have read "for all sufficiently small $\varepsilon$". –  Karl Schwede Feb 28 '12 at 4:48