Blowups of Cohen-Macaulay varieties

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).

• 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. Oct 24, 2010 at 6:26
• Long, yes. I'm just interested in the Proj of the Rees algebra. Oct 24, 2010 at 13:33

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.

• 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. Oct 24, 2010 at 19:51
• 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. 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 $$0 and $$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?

• 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. Oct 24, 2010 at 14:01
• 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. Oct 24, 2010 at 14:39