Let $X$ be an affine variety. Let $Y$ be smooth and let the map $f\colon Y\rightarrow X$ be proper birational. We will call $Y$ a smooth resolution of $X$.
Do the cohomology groups $H^i(Y,\mathcal{O}_Y)$ depend on $Y$?
I'm sure that the answer is 'no' if $X$ is normal, but I can't find a reference. Is it true in general (if $X$ is not normal)?
Edit. This follows from the Elkik-Fujita Vanishing Theorem. There is a more general vanishing theorem due to Elkik and Fujita. One version of this theorem (where I read the theorem) is Theorem 1.3.1 of the following article.
MR0946243 (89e:14015)
Kawamata, Yujiro; Matsuda, Katsumi; Matsuki, Kenji
Introduction to the minimal model problem. Algebraic geometry, Sendai, 1985,
pp. 283–360,
Adv. Stud. Pure Math., 10,
North-Holland, Amsterdam, 1987.
The proof in Kawamata-Matsuda-Matsuki is similar in spirit to the proof in my original answer, so I will leave that answer below. In place of the Hodge symmetries, K-M-M use Kawamata-Viehweg vanishing (the original proofs of that theorem use the Hodge decomposition theorem and Kodaira vanishing). In particular, the argument using the Leray spectral sequence and intersecting with general hyperplanes is in the proof in Kawamata-Matsuda-Matsuki (that is probably where I first learned those techniques). There is an extension of Kawamata-Viehweg vanishing to positive characteristic ala Deligne-Illusie (under appropriate hypotheses, of course). So there is probably also an extension of the Elkik-Fujita Vanishing Theorem.
To deduce your result from Theorem 1.3.1, apply the theorem to the morphism $g:Z\to Y$ with $E$ equal to the (scheme-theoretic) support of the cokernel of the natural morphism, $$ f^*: f^*\omega_{Y/k} \to \omega_{Z/k}. $$ Take $L$ to be the invertible sheaf $f^*\omega_{Y/k}$ (this will pull through the derived pushforward by the projection formula). Define $D$ and $\widetilde{D}$ to be the empty divisors. Define $\widetilde{L}$ to be the structure sheaf.
Original answer. Let me expand a bit on my comment. First of all, for every field $k$, for every birational, proper morphism of $k$-schemes, $g:Z\to Y$, each of the following natural $\mathcal{O}_Y$-module homomorphisms is an isomorphism, $$g^q: \Omega^q_{Y/k} \to g_*\Omega^q_{Z/k}.$$ Since $\Omega^q_{Y/k}$ is locally free, hence torsion-free, and since $g^q$ restricts to an isomorphism on the dense, maximal open $U$ over which $g$ is an isomorphism, the morphism $g^q$ is injective. Next, for every open $V$ of $Y$ and for every section $s$ of $g_*\Omega^q_{Z/k}(V) = \Omega^q_{Z/k}(g^{-1}(V))$, the restriction of $s$ to $U\cap V$ is the image of a section of $\Omega^q_{Y/k}$. The complement of the open subset $U$ has codimension $2$ (by Zariski's Main Theorem, or easier arguments). Since the sheaf $\Omega^q_{Y/k}$ is locally free on a regular scheme, in particular it is $S2$. Thus, the section $s$ extends to a section of $\Omega^q_{Y/k}$ on all of $V$. Therefore $g^q$ is an isomorphism.
If $k$ has characteristic $0$, and if $Y$ is projective, then the Hodge symmetries imply that the following maps are also isomorphisms, $$\psi^q:H^q(Y,\mathcal{O}_Y)\to H^q(Z,\mathcal{O}_Z).$$ The claim, to be proved by increasing induction on $m$, is that $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism in degrees $\leq m$, i.e., $g^\#:\mathcal{O}_Y\to g_*\mathcal{O}_Z$ is an isomorphism and every $R^qg_*\mathcal{O}_Z$ is zero for $0<q\leq m$.
The $q=0$ case of the first paragraph proves that $\mathcal{O}_Y\to g_*\mathcal{O}_Z$ is an isomorphism. Thus, by way of induction, assume that $m>0$, and assume that the result is proved for all birational morphisms $g$ of smooth projective varieties for all degrees strictly less than $m$. The claim, to be proved by increasing induction on $d$, is that the support of the sheaf $R^mg_*\mathcal{O}_Z$ has no irreducible component of dimension $d$.
For the base case of the induction, assume that the support of $R^mg_*\mathcal{O}_Z$ has an irreducible component of dimension $0$. If there were such a component, then it would give a summand of $R^mg_*\mathcal{O}_Z$ that is a skyscraper sheaf. Thus, $H^0(Y,R^mg_*\mathcal{O}_Z)$ is nonzero. The Leray spectral sequence for the cohomologies $H^m(Z,\mathcal{O}_Z)$ is a stage $2$, first quadrant, cohomological spectral sequence, $$E_2^{p,q} = H^p(Y,R^qg_*\mathcal{O}_Z)\Rightarrow H^{p+q}(Z,\mathcal{O}_Z).$$ By hypothesis, all of the sheaves $R^qg_*\mathcal{O}_Z$ are zero for $0<q\leq m$. Thus the only possible nonzero differential out of $H^0(Y,R^mg_*\mathcal{O}_Z)$ is the differential to $H^{m+1}(Y,g_*\mathcal{O}_Z)$. If this map were nonzero, then the induced map $$\psi^{m+1}:H^{m+1}(Y,\mathcal{O}_Y)\to H^{m+1}(Z,\mathcal{O}_Z),$$ would have a nonzero kernel. Since that map is an isomorphism, all of the differentials are zero. Thus, the spectral sequence gives that $H^0(Y,R^mg_*\mathcal{O}_Z)$ is the cokernel of $\psi^m$. Since $\psi^m$ is an isomorphism, it follows that the support of $R^mg_*\mathcal{O}_Z$ has no irreducible component of dimension $0$.
Now, by way of induction, let $d>0$ be an integer, assume that the vanishing of $R^qg_*\mathcal{O}_Z$ has been proved for all proper birational morphisms from a smooth, proper $k$-scheme to a smooth, projective $k$-scheme for $0<q<m$, and assume also that no irreducible component of the support of $R^qg_*\mathcal{O}_Z$ has dimension $0,\dots,d-1$. By Bertini's theorem, for a general hyperplane $H$, both the hyperplane section $Y_H$ of $Y$ and $Z_H=g^{-1}(Y_H)$ are smooth. Choosing $H$ to be transverse to the complement of $U$, also $g_H:Z_H\to Y_H$ is birational. Thus, by the induction hypothesis, the support of $R^m(g_H)_* \mathcal{O}_{Z_H}$ has no irreducible component of dimension $m-1$. For $H$ general, so that the Tor sheaves are zero on $Z$, $$\text{Tor}^{\mathcal{O}_Y}_r(\mathcal{O}_{Y_H},\mathcal{O}_Z),\ \ r>0,$$ then $R^m(g_H)_*\mathcal{O}_{Z_H}$ equals $R^mg_*\mathcal{O}_Z\otimes_{\mathcal{O}_Y}\mathcal{O}_{Y_H}$. Choose $H$ to be transverse to every $d$-dimensional irreducible component of the support of $R^mg_*\mathcal{O}_Z$ (if there are any). Then the intersection of that component with $H$ gives a $(d-1)$-dimensional irreducible component of the support of $R^m(g_H)_*\mathcal{O}_{Z_H}$. This is a contradiction. Therefore, there are no $d$-dimensional irreducible components of the support of $R^mg_*\mathcal{O}_Z$. By induction on $d$, the support of $R^mg_*\mathcal{O}_Z$ is empty, i.e., this is the zero sheaf. Thus, by induction on $m$, the map $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism.
Now, let $X^o$ be affine and finite type over a field, let $f^o:Y^o\to X^o$ be a projective, birational morphism with $Y^o$ smooth, and let $g^o:Z^o\to Y^o$ be a proper, birational morphism with $Z^o$ smooth. The closure of $X^o$ in a projective space containing affine space is a projective scheme $X$. By the strong resolution of singularities of Hironaka, there exists a projective, birational morphism $f:Y\to X$ with $Y$ smooth and with $f^{-1}(X^o)$ equal to $Y^o$ as an $X^o$-scheme. Similarly, by Nagata compactification and Hironaka, there exists a proper, birational morphism $g:Z\to Y$ with $Z$ smooth and with $g^{-1}(Y^o)$ equal to $Z^o$ as a $Y^o$-scheme. By the previous paragraph, $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism. By flat base change, also $\mathcal{O}_{Y^o}\to R(g^o)_*\mathcal{O}_{Z^o}$ is a quasi-isomorphism. Thus, applying the derived functor $R(f^o)_*$ also gives a quasi-isomorphism. Since a composition of derived functors of pushforward is the derived functor of pushforward by the composition, taking cohomology modules gives an isomorphism, $$H^q(Y^o,\mathcal{O}_{Y^o})\to H^q(Z^o,\mathcal{O}_{Z^o}).$$
Finally, for any proper, birational morphism $h^o:W^o\to X^o$ with $W^o$ smooth, and for any projective, birational morphism $f^o:Y^o\to X^o$ with $Y^o$ smooth (and such a morphism $f^o$ exists by Hironaka), then again by Hironaka, and by Chow's Lemma, there exists a pair of projective, birational morphisms, $$g^o:Z^o\to X^o, \ \ i^o:Z^o\to W^o,$$ such that $h^o\circ i^o$ equals $f^o\circ g^o$. Covering $W^o$ by open affines and using the previous result with $\text{Id}:W^o\to W^o$ and $i^o:Z^o\to W^o$ in the place of $f^o$ and $g^o$, it follows that $\mathcal{O}_{W^o}\to R(i^o)_*\mathcal{O}_{Z^o}$ is a quasi-isomorphism. Therefore each natural pullback map, $$H^q(W^o,\mathcal{O}_{W^o})\to H^q(Z^o,\mathcal{O}_{Z^o}),$$ is an isomorphism. Therefore every $H^q(W^o,\mathcal{O}_{W^o})$ equals $H^q(Y^o,\mathcal{O}_{Y^o})$.
