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Let $X$ be a smooth projective scheme over an algebraically closed field $k$. I have heard it claimed that $\dim_k H^i(X, \mathcal{O}_X)$ is a birational invariant of $X$ for all $i$. Does anyone know where to find a proof of this?

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    $\begingroup$ You should also assume smoothness. See the answers to this question: mathoverflow.net/questions/25922/… $\endgroup$ Apr 7, 2011 at 22:30
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    $\begingroup$ You did see the accepted answer in the question Dan Petersen referenced I assume. The point is that $X$ has rational singularities, and so if $\pi : X' \to X$ is a birational map with $X'$ smooth, then $R^j \pi_* \mathcal{O}_{X'} = 0$ for $j > 0$. Therefore, the Leray spectral sequence $H^i(X, R^j\pi_* \mathcal{O}_{X'})$ converges to $H^{i+j}(X, \mathcal{O}_X)$. However, all the $R^j \pi_*$ vanish, and so the spectral sequence degenerates giving you the equality that you want. $\endgroup$ Apr 8, 2011 at 0:59
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    $\begingroup$ In addition to the above comments/answers, I'd like to point (as VA had done in the link) that Chatzistamatiou and Rülling have established this in any characteristic. Their argument is entirely different, and bypasses lack of resolution of singularities in a clever way. It should be appearing in Algebra & Number theory. $\endgroup$ Apr 8, 2011 at 12:14

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Karl's nice answer settles the question in general. Let me give here a different proof in the case where $X$ is smooth over $\mathbb{C}$, which involves Dolbeault theorem instead of Leray spectral sequence.

Assume that $X$ is smooth over $\mathbb{C}$ and let $f \colon X \dashrightarrow Y$ be a birational map. By Dolbeault isomorphism, it is sufficient to prove that

$H^0(X, \Omega_X^i)=H^0(Y, \Omega_Y^i) \quad (*)$

for all $i$. Let $\omega \in H^0(Y, \Omega_Y^i)$ be a holomorphic $i$-form on $Y$. Then $f^* \omega$ is a meromorphic $i$-form on $X$, which is holomorphic outside the indeterminacy locus $Z \subset X$ of $f$. Since the codimension of $Z$ is at least $2$ and the poles of a meromorphic form are a divisor, it follows that $f^* \omega$ is actually holomorphic on the whole of $X$. This enables us to define an injective map

$f^* \colon H^0(X, \Omega^i_Y) \to H^0(X, \Omega_X^i)$.

Since $f$ is birational it follows that $f^*$ has an inverse, so we have proven $(*)$.

EDIT. After writing this answer, I realized that it essentially coincides with the one given by Dan Petersen for the question quoted in the comments. Anyway, I think there is no harm in leaving it here.

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