TITLE: Projectivity of blowups
QUESTION [6 upvotes]: If I have a quasiprojective variety $X$, and a subscheme $Z$, then the blowup 
$$f:Y = Bl(X,Z)\rightarrow X$$ 
is projective over $X$, since it is constructed by a relative Proj construction.  Can I find a relatively ample bundle on $Y$ that is trivial on
$$f^{-1}(X\backslash Z)?$$
At first I thought the construction of $O(1)$ in the Proj construction guarantees this property.  However, if I have a projective small contraction $Y \rightarrow X$, then it is a blowup map, but then certainly an ample bundle can't be trivial on the birational locus, since it would be trivial on $Y$.
So under what conditions can I find such a relatively ample bundle?
Edit:  Karl's answer explains away my confusion:  I have to blow up more than the image of the exceptional locus in case of a small contraction.  It's still reasonable to ask when it's trivial on the birational locus, in which case Sandor's answer applies.

REPLY [2 votes]: anon, is your definition of relatively ample local over the base?  (There are some different definitions out there, but most people seem to use the one that is local over the base now, compare for instance Ravi Vakil's notes with Hartshorne).  
If your definition of relatively ample is local over the base, then there is nothing to show.  If you blow-up $Z$, say with ideal sheaf $I_Z$, then $I_Z \cdot O_Y$ is relatively ample.  
With respect to a small contraction $\pi : Y \to X$ (at least with $X$ normal), you can NEVER get one of these just by blowing up a scheme supported on the locus where $\pi$ is an isomorphism.  In general, the way you get a small resolution, is by blowing up a Weil divisor that is not Cartier.  Of course, on a normal variety a Weil divisor is Cartier except on a small close set, and blowing up a Cartier divisor obviously doesn't do anything.
For fun, I'd suggest that you try blowing up the ideal $(x, u)$ on the scheme $\text{Spec} k[x,y,u,v]/(xy-uv)$.  $(x,u)$ is a non-Cartier divisor, but it's still a divisor (it's also not $\mathbb{Q}$-Cartier, as Sándor effectively points out above).  Blowing it up gives a small resolution.