It is a basic fact that $H^n(X, F) = 0$ if $X$ is noetherian affine, $n > 0$, and $F$ a quasi-coherent sheaf.


If $Y \to X$ is a blow-up of a smooth variety in a smooth center, then then exceptional divisor is a projective bundle over the center, and so $H^n(Y, \mathcal{O}\_Y) = H^n(X, \mathcal{O}\_X)$. (right?)


I have not seen any examples of blow-ups of smooth varieties (with arbitrary center) whose fibres are not unions of projective spaces (sub-question: do such blow-ups exist?).


So I am lead to wonder: is it always true that $H^n(Y, \mathcal{O}\_Y) = 0$ if $f: Y \to X$ is birational, $Y$ is integral, and $X$ is smooth and affine?


Edit: I mean't to include the hypothesis that $f$ is proper, and that $n$ is any integer $> 0$.



Hn(X, OX) = 0 for X birational to a regular affine variety?
Let me give a counter-example, where $f$ is moreover proper (see Piotr Achinger's comment), but $Y$ is not normal. 


Let $X$ be the affine plane over a field $k$ and let $\tilde{Y}$ be the blow-up of the origin in $X$. We choose two $k$-points $P\_1$ and $P\_2$ on the exceptional divisor, and consider the variety $Y$ constructed by gluing $P\_1$ and $P\_2$ transversally. There is an induced proper and birational morphism $f:Y\to X$, and the contraction $c:\tilde{Y}\to Y$ is nothing else than the normalization of $Y$. Moreover, we have the following exact sequence on $Y$:
$$0\to\mathcal{O}\_Y\to c\_\*\mathcal{O}\_{\tilde{Y}}\to\mathcal{O}\_P\to 0,$$ where we denoted by $P$ the common image of $P\_1$ and $P\_2$ in $Y$, and where the last arrow is given by the difference of evaluations at $P\_1$ and $P\_2$.


We get a long exact sequence in cohomology (here we use that, the cohomology of $c\_\*\mathcal{O}\_{\tilde{Y}}$ coincides with the cohomology of $\mathcal{O}\_{\tilde{Y}}$ because $c$ is finite, and that $H^1(\tilde{Y},\mathcal{O}\_\tilde{Y})=0$): 
$$0\to H^0(Y,\mathcal{O}\_Y)\to H^0(\tilde{Y},\mathcal{O}\_\tilde{Y})\to H^0(P,\mathcal{O}\_P)\to H^1(Y,\mathcal{O}\_Y)\to 0.$$
Every global function on $\tilde{Y}$ is constant on the exceptional divisor (as it is proper and integral), and hence descends to $Y$ showing that the first map is an isomorphism. As a consequence, $H^0(P,\mathcal{O}\_P)\to H^1(Y,\mathcal{O}\_Y)$ is an isomorphism, and $H^1(Y,\mathcal{O}\_Y)$ is nonzero.


