This question consists of two parts. I will try to be as short and clear as possible.

Let $S$ be a Dedekind scheme of characteristic zero. The main examples are $\mathbf{P}^1_k$, with $k$ a field of characteristic 0 and $\textrm{Spec} \ \mathbf{Z}$.

A fibered surface is a projective flat morphism $X\longrightarrow S$ with $X$ an integral scheme of dimension 2.

Suppose $S=\mathbf{P}^1_k$ with $k$ a field of characteristic 0. Then $X$ is an algebraic surface over a field of characteristic zero. The theory of rational singularities for $X$ is then explained in Chapter 5 of Kollar and Mori: Birational geometry of algebraic varieties.

Now, I would like to know if there are analogous results for when $S=\textrm{Spec} \ \mathbf{Z}$ as in *loc. cit.*. Let me make this more precise.

Assume that $X$ is normal. (This will suffice for my applications.)

Call a resolution of singularities $\rho:Y\longrightarrow X$ rational if it satisfies

$$R^i\rho_\ast \mathcal{O}_Y =0 \ \textrm{for} \ i>0.$$

We say that $X$ has rational singularities if all resolutions of $X$ are rational.

To have a good theory, we should probably show that $R^i \rho_\ast \omega_{Y/S} = 0$ for $i>0$.

**Question 1:** Is it true that $R^i \rho_\ast \omega_{Y/S} = 0$ for $i>0$ for any resolution of singularities?

Compare my last question with Theorem 5.10 of Kollar-Mori.

**Question 2:** Is it true that the following properties

A. $X$ has a rational resolution

B. $X$ has rational singularities

C. For every resolution of singularities $\rho:Y\longrightarrow X$, we have that $\rho_\ast \omega_{Y/S} = \omega_{X/S}$.

are equivalent?

I guess one could try to see if some of the arguments given in Kollar-Mori carry over to this setting. For example, it probably suffices to show that a resolution is rational if and only if it satisfies C.

Here is a possible application: Assume $X$ to be regular from now on. Let $S$ be $\mathbf{P}^1_{\mathbf{C}}$. Let $\pi:Y\longrightarrow X$ be a finite surjective morphism with branch locus a normal crossings divisor. Then, it is easy to see that the singularities of $Y$ are quotient. (Consider the analytic topology or the etale topology and use Abhyankar's lemma). It is well-known that quotient singularities are rational.

I guess Abhyankar's lemma is still valid and therefore I was hoping to show that $Y$ in case $S= \textrm{Spec} \mathbf{Z}$ still has rational singularities. (The notion of quotient singularities seems only to work over $\mathbf{C}$...)