The following is a theorem of Elkik from 1978 describing how rational singularities behave under flat morphisms.


Let $f:X\rightarrow S$ be a flat morphism of schemes of finite type over $\mathbb{C}$ such that both $S$ and the fibres of $f$ have rational singularities.


I am confused about one step in the proof and I am afraid that it should be obvious.


Let $\varphi:S'\rightarrow S$ be a resolution of singularities and let $X'=X\times\_S S'$ be the fiber product, with projections $\varphi'$ and $f'$ to $X$ and $S'$ respectively. Also, let $\psi:Y\rightarrow X'$ be a resolution of singularities.


$$\begin{array}{ccccc} Y & \stackrel{\psi}{\longrightarrow} & X'=X\times\_S S' & \stackrel{\varphi'}{\longrightarrow} & X \\ & & \downarrow & & \downarrow \\ & & S' & \stackrel{\varphi}{\longrightarrow} & S \end{array} f$$


By assumption, we have that $R^{\bullet}\psi\mathcal{O}\_Y=\mathcal{O}\_{X'}$ and $R^{\bullet}\varphi\mathcal{O}\_{S'}=\mathcal{O}\_S$. I am trying to show that then $X$ also has rational singularities, namely $R^{\bullet}(\varphi'\circ \psi)\mathcal{O}\_Y=\mathcal{O}\_X$.


By Grothendieck spectral sequence, since $X'$ has rational singularities, we have that $R^i(\varphi'\circ \psi)\mathcal{O}\_Y = R^i\varphi'\mathcal{O}\_{X'}$ so we have to show that $R^{\bullet}\varphi'\mathcal{O}\_{X'}=\mathcal{O}\_X$.


I am not sure about how this follows: by flat base change the natural map $$f^{\ast}R^i\varphi\_{\ast}\mathcal{O}\_{S'}\rightarrow R^i\varphi'f'^{\ast}\mathcal{O}\_{S'}$$ is an isomorphism.


Now for instance for $i=0$ flat base change yields $\varphi\_{\ast}'f'^{\ast}\mathcal{O}\_{S'}=f^{\ast}\mathcal{O}\_S$. Pushing forward by $f$ on both sides yields $$f\_{\ast}\varphi'\_{\ast}f'^{\ast}\mathcal{O}\_{S'}=f\_{\ast}f^{\ast}\mathcal{O}\_S=\mathcal{O}\_S\otimes f\_{\ast}\mathcal{O}\_X=f\_{\ast}\mathcal{O}\_X$$


Commutativity on the LHS yields $$f\_{\ast}\varphi'\_{\ast}f'^{\ast}\mathcal{O}\_{S'}=\varphi\_{\ast}f'\_{\ast}f'^{\ast}\mathcal{O}\_{S'}=\varphi\_{\ast}(\mathcal{O}\_{S'}\otimes f'\_{\ast}\mathcal{O}\_{S'})=\varphi\_{\ast}f'\_{\ast}\mathcal{O}\_{X'}$$


So $f\_{\ast}\varphi'\_{\ast}\mathcal{O}\_{X'}=f\_{\ast}\mathcal{O}\_X$.


Does it follow that $\varphi'\_{\ast}\mathcal{O}\_{X'}=\mathcal{O}\_X$? Am I making this more complicated than it actually is?


**Added**: For instance, if $f$ is injective, the claim would follow, since then $f\_{\ast}\mathcal{F}\simeq f\_{\ast}\mathcal{G} \Longrightarrow \mathcal{F}\simeq \mathcal{G}$. (by definition $\mathcal{F}(f^{-1}(U))\simeq \mathcal{G}(f^{-1}(U))$ for every $U\subseteq Y$ open and provided that $f$ is flat and injective, for any open set $V\subseteq X$, we have $V=f^{-1}(f(V))$ (since flat morphisms of finite type between Noetherian schemes are open). But this is more than we need.



Rational singularities under flat morphisms
I think you are. That is, making it more complicated than it is. 


You were done right after saying that "by flat base change the natural map $$f^{\ast}R^i\varphi\_{\ast}\mathcal{O}\_{S'}\rightarrow R^i\varphi'\_\*f'^{\ast}\mathcal{O}\_{S'}$$ is an isomorphism."


*By definition* $f^\*\mathscr O\_S=\mathscr O\_X$ and ${f'}^\*\mathscr O\_{S'}=\mathscr O\_{X'}$. In other words, you get that


$$R^i{\varphi'\_\*}\mathscr O\_{X'}\simeq
R^i\varphi'\_\*f'^{\ast}\mathscr{O}\_{S'}\simeq f^{\ast}R^i\varphi\_{\ast}\mathscr{O}\_{S'}\simeq \begin{cases} 0 \qquad \text{if $i>0$} \\ \mathscr O\_X \quad \text{if $i=0$} \end{cases},$$ 
exactly what you need.


