# Understand the proof that rational resolution is independent of the resolution

EDIT. Karl Schwede has given a different approach to solve this question, which is very clear. However, I still want to figure out how to deal with this problem along Ein's line, and waiting for the answer about my QUSETION A. B. below.

This is a duplication of a question from MSE.

I am currently reading the notes Lectures on singularities and adjoint linear systems written by Ein and encounter some difficulties as follows, see Prop1.12.

Definition (Rational Singularity). A morphism $$f: Y \rightarrow X$$ is said to be a rational resolution if $$Y$$ is smooth and $$f$$ is a proper and birational morphism such that $$R^{i} f_{*} \mathscr{O}_{Y}=0$$ for $$i>0$$.

Proposition. Let $$f: Y \rightarrow X$$ be a rational resolution and $$f^{\prime}: Y^{\prime} \rightarrow X$$ be another resolution. Then $$f^{\prime}: Y^{\prime} \rightarrow X$$ is also a rational resolution.

Proof. We have a birational map $$\varphi: Y -\to Y^{\prime}$$. Successively blowing up the undefined locus of $$\varphi$$, we get a variety $$Z$$ and two proper birational morphisms $$g: Z \rightarrow Y$$ and $$g^{\prime}: Z \rightarrow Y^{\prime}$$ such that $$h:=f \circ g=f^{\prime} \circ g^{\prime}$$. Since $$g$$ is the composition of blowing-ups. Then $$R^{q} g_{*}\left(\mathscr{O}_{Z}\right)=0$$ for $$q>0$$. Apply the Leray spectral sequence $$E_{2}^{p, q}=R^{p} f_{*}\left(R^{q} g_{*}(\mathscr{F})\right) \Rightarrow R^{p+q}(f \circ g)_{*}(\mathscr{F}) .$$ It follows that $$R^{i} h_{*} \mathscr{O}_{Z}=0$$ for $$i>0$$. Apply the Leray spectral sequence to $$f^{\prime} \circ g^{\prime}$$. It is easy to see that $$R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$$. In fact, it fits in the following exact sequence $$0 \rightarrow R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime} \rightarrow R^{1} h_{*} \mathscr{O}_{Z} \rightarrow f_{*}^{\prime} R^{1} g^{\prime} \mathscr{O}_{Z}$$ Since $$Y^{\prime}$$ is smooth hence $$Y^{\prime}$$ has a rational resolution. Now $$Z$$ is another resolution of $$Y^{\prime}$$. By the above argument, we can conclude that $$R^{1} g_{*}^{\prime} \mathscr{O}_{Z}=0$$. Apply the Leray spectral sequence to $$p+q=2$$. We see that $$R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$$. Hence $$R^{2} g_{*}^{\prime} \mathscr{O}_{Z}=0$$. By induction, we conclude that $$R^{p} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$$ for $$p>0$$.

Here is my question:

QUESTION A. Suppose we know that $$R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime}=0$$ and $$R^{1} g_{*}^{\prime} \mathscr{O}_{Z}=0,$$ how to deduce $$R^{2} g_{*}^{\prime} \mathscr{O}_{Z}=0$$ by applying the Leray spectral sequence as Ein said rather than using the five-term sequence, that is $$0 \rightarrow R^{1} f_{*}^{\prime} \mathscr{O}_{Y^\prime} \rightarrow R^{1} h_{*} \mathscr{O}_{Z} \rightarrow f_{*}^{\prime} R^{1} g_*^{\prime} \mathscr{O}_{Z}\rightarrow R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}\rightarrow R^{2} h_{*} \mathscr{O}_{Z}?$$

We can say $$R^{2} f_{*}^{\prime} \mathscr{O}_{Z}=0$$ since its prior one and posterior one both vanish.

I wonder how to use Leray spectral sequence at $$p+q=2$$, since now, we have $$0=R^{2} h_{*} \mathscr{O}_{Z}=R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}\oplus R^{1} f_{*}^{\prime}(R^1 g_{*}^{\prime} \mathscr{O}_{Z})\oplus R^{2} g_{*}^{\prime} \mathscr{O}_{Z}$$ with the middle term vanishing, but we cannot say anything about the vanishing of $$R^{2} f_{*}^{\prime} \mathscr{O}_{Y^\prime}$$ or $$R^{2} g_{*}^{\prime} \mathscr{O}_{Z}$$.

QUESTION B. How to do the induction if we cannot use the five-term sequence for larger $$p$$, $$q$$?

• It seems a lot easier to use the whole derived pushforward (as a complex, not just the cohomology sheaves), so that if $\pi$ is a resolution and $\sigma$ is a smooth blow up, then $R(\pi\sigma)_*(\mathcal{O}) = R\pi_* R\sigma_*(\mathcal{O}) = R\pi_*(\mathcal{O})$. Using Weak factorization (in char. 0) this implies that the complex $R\pi_*(\mathcal{O})$ on a given singular variety $X$ is independent of the choice of the resolution $\pi$. Commented Sep 12, 2022 at 20:31

This is not exactly what is asked for, rather it is a different proof of the fact in question.

I assume you are working in characteristic zero with varieties. I'm also going to assume you want $$X$$ to be normal (for simplicity, your rational resolutions will all factor through the normalization). The proof in other characteristics is hard I think (see relevant papers by Kovács and A. Chatzistamatiou and K. Rülling). The way I prefer to see this is via Grothendieck duality and also utilizing Grauert–Riemenschneider vanishing. If you want to work with Q-Schemes with dualizing complexes, everything below works by the version of Grauert–Riemenschneider vanishing by Murayama (but let's stick with varieties).

Ok, if $$X$$ has a rational resolution $$f : Y \to X$$, then your hypothesis means that $$O_X \to R f_* O_Y$$ is an isomorphism in the derived category. In particular, by Grothendieck duality, you have that $$R f_* \omega_Y^{\bullet} \to \omega_X^{\bullet}$$ is an isomorphism in the derived category as well. But $$Y$$ is smooth and so Cohen-Macaulay, and in particular $$\omega_Y^{\bullet} = \omega_Y[d]$$ (ie, it's a shifted sheaf). Furthermore by Grauert–Riemenschneider vanishing we have that $$R\pi_* \omega_Y[d] = \pi_* \omega_Y[d]$$ is a sheaf. Thus $$\omega_X^{\bullet}$$ is a sheaf too and $$X$$ is Cohen–Macaulay. In particular, we now have that

1. $$X$$ is Cohen-Macaulay.
2. $$\pi_* \omega_Y = \omega_X$$.

Ok, now if $$f' : Y' \to X$$ is another resolution, we can find $$Y''$$ birationally mapping to both $$Y$$ and $$Y'$$. Furthermore we may assume $$Y'$$ a smooth variety (you can resolve indeterminacies and then resolve the singularities, or take the product $$Y \times_X Y'$$, take the irreducible component dominating $$X$$ and resolve that).

In particular, since we need to show that $$O_X \to R \pi_* O_{Y'}$$ is a quasi-isomorphism, by Grothendieck duality it suffices to show that $$g_* \omega_{Y'} = \omega_X$$. To do this, it suffices to show that both $$\pi_* \omega_{Y''} = \omega_{Y}$$ and $$\nu_* \omega_{Y''} = \omega_{Y'}$$ (where $$\pi$$ and $$\nu$$ are the relevant maps) since then $$f_* \pi_* \omega_{Y''} = \omega_X$$ and chasing the diagram the other way we get $$g_* \omega_{Y'} = g_* \nu_* \omega_{Y''} = \omega_X$$.

Ok, so in particular we need to show that any resolution of a smooth variety is a rational resolution. But to do that, it is a computation (that I can show some details if you want) that $$\pi_* \omega_{Y''} = \pi_* \pi^* \omega_Y \otimes O_Y(D)$$ where $$D$$ is an effective divisor. Ie, the relative canonical over a smooth variety is effective (you might have seen this in other places in notes of Ein). It follows now that $$\pi_* \omega_{Y''} = \omega_Y$$ which is what we wanted to show.

• Thanks! Yes, it looks right. Commented Sep 12, 2022 at 1:36