Proper birational morphism from a Gorenstein normal scheme to a normal local domain, with trivial higher direct images, implies Cohen-Macaulay?

Question

Let $k$ be a field of characteristic $0$. Let $R$ be a Noetherian local normal domain containing $k$. Also assume that $R$ is the homomorphic image of a Gorenstein ring of finite dimension, hence $R$ admits a Dualizing complex. Let $Y$ be a Gorenstein normal scheme over Spec$(k)$ of finite Krull-dimension.

If there exists a proper birational map $f: Y \to \text{Spec}(R)$ such that $R^i f_* \mathcal O_Y=0, \forall i>0$, then is it true that $R$ is Cohen-Macaulay? If this is not true in general, what if I also assume $Y$ is regular?

As $R$ is normal, $R$ is of course Cohen-Macaulay when $\dim R \leq 2$. So we only have to think about $\dim R \geq 3$.

Answer 1

I started writing this last night, but didn't finish. In addition to Jason Starr's answer, you can use my new vanishing theorems to remove the assumption that $R$ is essentially of finite type over a field. Note that the special case when $R$ is essentially of finite type over a field of characteristic zero also follows from Kempf's criterion for rational singularities [Kempf 1973, p. 50].

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In short, if $Y$ is assumed to be regular, then $R$ will be Cohen–Macaulay. In fact, we have the following stronger result.

Theorem. Let $R$ be a Noetherian ring and let $f\colon Y \to \operatorname{Spec}(R)$ be a proper birational morphism such that $Y$ is regular and $R^if_*\mathcal{O}_Y = 0$ for all $i > 0$. Suppose $R$ contains $\mathbf{Q}$. Then, every localization of $R$ is pseudo-rational, and in particular $R$ is Cohen–Macaulay.

Proof. Since the conclusion can be checked locally, it suffices to consider the case when $(R,\mathfrak{m})$ is local. Set $\hat{Y} := Y \times_{\operatorname{Spec}(R)} \operatorname{Spec}(\hat{R})$, and consider the Cartesian diagram $$\require{AMScd}\begin{CD} \hat{Y} @>>> Y\\ @V\hat{f}VV @VVfV\\ \operatorname{Spec}(\hat{R}) @>>> \operatorname{Spec}(R) \end{CD}$$ By base change, $\hat{f}$ is proper, and by flat base change, $\hat{f}$ is birational [EGAI$_\text{new}$, Proposition 3.9.9] and $R^i\hat{f}_*\mathcal{O}_{\hat{Y}} = 0$ for all $i > 0$. Note that $\hat{R}$ is normal by [Lipman 1969, Remark 16.2] and that the pseudo-rationality of $\hat{R}$ would imply that $R$ is pseudo-rational [Murayama 2022, Proposition 4.20]. Since $\hat{Y}$ is regular by [EGAIV$_2$, Lemme 7.9.3.1], we may replace $R$ by $\hat{R}$ and $Y$ by $\hat{Y}$ to assume that $R$ is complete local.

We now prove the special case when $(R,\mathfrak{m})$ is complete local. First, by my version of Grauert-Riemenschneider vanishing [Murayama, Theorem B(i)], we know that $R$ has rational singularities (see, e.g., [Kollár 2013, Definition 2.76] for the definition). Since $R$ having rational singularities is equivalent to saying that $R$ is pseudo-rational in our setting [Murayama, Remark 7.4], we are done.

We can also show that $R$ is Cohen–Macaulay directly. Since Cohen–Macaulayness can be checked locally, it suffices to consider the case when $(R,\mathfrak{m})$ is local. Since $R$ is normal, the map $R \overset{\sim}{\to} f_*\mathcal{O}_Y$ is an isomorphism [EGAIII$_1$, Corollaire 4.3.12]. Combined with the assumption that $R^if_*\mathcal{O}_Y = 0$ for all $i > 0$, we have the quasi-isomorphism $R \overset{\sim}{\to} \mathbf{R}f_*\mathcal{O}_Y$. Applying $\mathbf{R}\Gamma_{\mathfrak{m}}(-)$, we obtain $$\mathbf{R}\Gamma_{\mathfrak{m}}(R) \overset{\sim}{\longrightarrow} \mathbf{R}\Gamma_{\mathfrak{m}}(\mathbf{R}f_*\mathcal{O}_Y) \cong \mathbf{R}\Gamma_{f^{-1}(\{\mathfrak{m}\})}(\mathcal{O}_Y).$$ Now taking $i$-th cohomology modules, we obtain the isomorphisms $$H^i_{\mathfrak{m}}(R) \overset{\sim}{\longrightarrow} H^i_{\mathfrak{m}}(\mathbf{R}f_*\mathcal{O}_Y) \cong H^i_{f^{-1}(\{\mathfrak{m}\})}(\mathcal{O}_Y).$$ Since the local cohomology modules on the right vanish for all $i < \dim(R)$ by my version of Grauert–Riemenschneider vanishing (in its dual formulation) [Murayama, Theorem B*(i)], we see that $R$ is Cohen–Macaulay. $\blacksquare$

Answer 2

Edit. Thanks to user @Johan for pointing out the indexing mistake. It is now corrected.

I read more of Elkik's article. I am just expanding my comments into one answer. Edit. Assume that $R$ is essentially of finite type over $\text{Spec}\ k$. By hypothesis, the scheme $X=\text{Spec}\ R$ has a dualizing complex $\omega^\bullet_R$ that has nonzero homology sheaf $\mathcal{H}^{\text{dim}\ R}(\omega^\bullet_R)$, and has vanishing homology sheaves $\mathcal{H}^i(\omega^\bullet_R)$ for $i>0$ and for $i<-\text{dim}\ R$. The ring $R$ is Cohen-Macaulay if and only if the only integer $i$ with $\mathcal{H}^i(\omega^\bullet_R)$ nonzero is $i=-\text{dim}\ R$. It is Gorenstein if, further, this homology sheaf is a projective $R$-module of rank $1$.

Also by hypothesis, the scheme $Y$ has a dualizing complex $\omega_Y^\bullet$ such that the homology sheaf $\mathcal{H}^i(\omega_Y^\bullet)$ is nonzero if and only if $i=-\text{dim}\ Y = -\text{dim}\ R$. By relative duality, we have a natural quasi-isomorphism, $$\alpha:Rf_* \omega_Y^\bullet \cong R\text{Hom}_{\mathcal{O}_X}(Rf_*\mathcal{O}_Y,\omega_X^\bullet).$$ By hypothesis, the natural map $\mathcal{O}_X\to Rf_*\mathcal{O}_Y$ is a quasi-isomorphism. Thus, $\alpha$ is equivalent to a quasi-isomorphism, $$Rf_*\omega^\bullet_Y \cong \omega_X^\bullet.$$

Since the only integer $i$ with a nonvanishing homology sheaf $\mathcal{H}^i(Rf_*\omega^\bullet_Y)$ is $i=-\text{dim}\ Y$, the complex $Rf_*\omega^\bullet_Y$ has nonvanishing homology sheaves $R^if_*\omega^\bullet_Y$ only for $ -\text{dim}\ Y \leq i \leq 0$.

Edit. Since $R$ is essentially of finite type and $Y$ is regular, by Grauert-Riemenschneider Vanishing the derived direct image $R^if_*\omega^\bullet_Y$ is nonzero only for $i=-\text{dim}\ Y = -\text{dim}\ R$, in which degree it equals $f_*\mathcal{H}^{-\text{dim}\ Y}(\omega^\bullet_Y)$.

Thus the complex $\omega_X^\bullet$ has nonvanishing homology sheaves only for $i =- \text{dim}\ R = \text{dim}\ Y$. Therefore $R$ is Cohen-Macaulay.