Suppose that $X$ is a Cohen-Macaulay normal scheme/variety and $\pi : Y \to X$ is a proper birational map with $Y$ normal.

Question: Is $Y$ also Cohen-Macaulay? Are there common conditions which imply it is?

If $Y$ is not normal I know of several ways to show that the answer to the first question is no.

There are obvious spectral sequences but I don't see how to deduce what I want from them, perhaps I'm being dumb (or maybe there is an obvious example).

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    $\begingroup$ Karl, just to clarify, when X is affine, you don't want the Rees algebra to be CM, but just the Proj to be CM, right? There are examples of the Rees algebra not CM. $\endgroup$ Oct 24, 2010 at 6:26
  • $\begingroup$ Long, yes. I'm just interested in the Proj of the Rees algebra. $\endgroup$ Oct 24, 2010 at 13:33

2 Answers 2


An example was given in Section 3 of this paper by Cutkosky: "A new characterization of rational surface singularities." (The scheme $Z$ in the last page, which is a blow up of some $m$-primary ideal of a regular local ring of dimension $3$, is normal but not Cohen–Macaulay.)

The algebraic side of this example has been studied quite a bit, so perhaps more explicit examples are known. I am not an expert here, but you can check out a paper by Huckaba–Huneke (Wayback Machine), or papers by Vasconcelos (he has a book called "Arithmetic of Blowup Algebras" which discussed, among other things, Serre's condition $(S_n)$ on Rees algebras), and the references there.

  • $\begingroup$ Long thanks, that's quite interesting. I've looked at Vasconcelos' book a long time ago. This example seems to show that there is no reasonable condition on $X$ which would imply blow-ups are Cohen-Macaulay. $\endgroup$ Oct 24, 2010 at 19:51
  • $\begingroup$ Actually, another reason this is interesting, is that I've certainly had people tell me several times that normal blow-ups of varieties with rational singularities still have rational singularities. This seems to debunk that assertion. $\endgroup$ Oct 25, 2010 at 14:11

Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

  1. The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

  2. I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

  3. So, let's see how one could construct a counterexample. Perhaps cones will do...

  4. Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

  5. So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

  6. So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. I think this might give you what you want: Grothendieck duality gives that then $$R\pi_*\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$ Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as $$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1} $$ So, if we also knew that $R^i\pi_*\omega_Y=0$ for $i>0$, then it would follow that $R\pi_*\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition:

If $R^i\pi_*\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume.

Any thoughts?

  • $\begingroup$ S\'andor, I agree that it seems unlikely to be true. For 4), I'm slightly confused, because $Y$, the cone over $Z$ is also affine, so I don't think the map $f:Y\to X$ will be proper+birational (is there even a map in general). For the existence of $f$, I suppose it depends on the choice of a particular line bundle on $Y$ to take the sectionf. Start with the pull back of whatever ample on $X$ twisted by something relatively ample? My guess is that the map $Y \to X$ will not be defined everywhere, maybe resolving the indeterminacies will help... then do what you did? I'll think about this. $\endgroup$ Oct 24, 2010 at 14:01
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    $\begingroup$ Karl, yes, regarding the map in #4, these are exactly my worries, but here is what I was thinking: It seems that the morphism is OK outside of the vertices of the cones, so if you resolve the indeterminacies, then you get a morphism extending this on a scheme that you get by blowing up an ideal supported at the vertex of $Y$ and the morphism maps the entire exceptional set to the vertex of $X$. But then, it seems that this morphism should actually factor through $Y$. In fact, this might be a place where $Y$ being normal makes a difference. $\endgroup$ Oct 24, 2010 at 14:39

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