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 $\dim R \geq 3$.

  • $\begingroup$ I believe that this is a theorem of Elkik when $R$ is the localization of a finitely generated $k$-algebra. Thus, it also holds if $R$ sits "between" such a localization and its completion (since you can check the Cohen-Macaulay property after passage to the localization). So probably you can deduce your case from the Cohen structure theorem. $\endgroup$ Commented Jul 6, 2023 at 11:33
  • $\begingroup$ @JasonStarr: Thank you for your comment. Could you please give a reference for the case of when $R$ is essentially of finite type over $k$? Also, Is Spec$(R)$ necessary, or can we just do it for any normal scheme essentially of finite type over $k$? $\endgroup$
    – Snake Eyes
    Commented Jul 7, 2023 at 0:27
  • $\begingroup$ I learned about this in Koll'ar's paper, "Singularities of pairs," in the Santa Cruz volume, but he references an earlier paper of Elkik. I will try to track down the original article. $\endgroup$ Commented Jul 7, 2023 at 11:09
  • $\begingroup$ Elkik explains the proof in the discussion leading up to Definition 1 in the article, "Rationalit'e des singularit'es canoniques": eudml.org/doc/142810 $\endgroup$ Commented Jul 7, 2023 at 11:16
  • $\begingroup$ I read the proof, and I think it probably directly applies to your situation, without using the completion. I will try to write more soon. $\endgroup$ Commented Jul 7, 2023 at 12:48

2 Answers 2


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].

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], 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. $\blacksquare$

  • $\begingroup$ Thank you for your answer. I will read it carefully, and will surely have many questions, but I quickly want to ask two things first: $Y$ is just a regular scheme with no other assumption? By $\widehat Y$, do you actually mean the fiber-product $Y \times_{Spec(R)} Spec(\widehat R)$? $\endgroup$
    – Snake Eyes
    Commented Jul 8, 2023 at 2:25
  • $\begingroup$ I like your second argument... could you please spell out the Leray spectral sequence isomorphism ? $\endgroup$
    – Snake Eyes
    Commented Jul 8, 2023 at 3:59
  • $\begingroup$ This is definitely better than the original version. $\endgroup$ Commented Jul 8, 2023 at 10:10
  • $\begingroup$ @SnakeEyes I edited my answer. For the regularity of $\hat{Y}$, see [Stacks Project, Tag 0BG6] for another reference. $\endgroup$ Commented Jul 9, 2023 at 15:36

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.

  • $\begingroup$ Sorry, Jason, but there is something wrong with the numbering. Usually, people normalize the dualizing complex of R such that it has nonzero cohomology in degrees $[-\dim R, 0]$. When I use that normalization I have to use Grauert-Riemenschneider vanishing as in the article of Elkik. Cheers! $\endgroup$
    – Johan
    Commented Jul 7, 2023 at 23:14
  • $\begingroup$ You are right: I misread the indexing convention. I will correct it now. $\endgroup$ Commented Jul 8, 2023 at 0:57
  • $\begingroup$ I know that $\omega_Y^{\bullet}$ is only supported at $\dim Y$, but could you please explain why $\mathcal H^i(Rf_* \omega_Y^{\bullet })=0$ for $i\neq -\dim Y$? If we can just say this, we would be done at this point, no? Because $X$ is Cohen-Macaulay if and only if $\omega_X^{\bullet}$ is supported in a single degree ... $\endgroup$
    – Snake Eyes
    Commented Jul 8, 2023 at 1:36
  • $\begingroup$ As pointed out by user @Johan (and explained in the article by Elkik), this follows from Grauert-Riemenschneider Vanishing. $\endgroup$ Commented Jul 8, 2023 at 1:43
  • $\begingroup$ @JasonStarr: Thank you for your edits ... I do not have a reference in front of me right now, but for Grauert-Riemenschneider, don't you need both the schemes to be actually a variety over a characteristic $0$ field? $\endgroup$
    – Snake Eyes
    Commented Jul 8, 2023 at 1:55

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