What do we mean by a variety being arithmetically Cohen-Macaulay? Is every such variety also Gorenstein?
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1$\begingroup$ Determinental varieties (whose defining ideal is genenerated by minors of a matrix) are ACM but usually not Gorenstein. $\endgroup$– J.C. OttemNov 28, 2010 at 18:05
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1 Answer
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Arithmetically Cohen-Macaulay means, depending on the source/context, either:
- The homogeneous coordinate ring (with respect to a given embedding into $\mathbb{P}^n$) is Cohen-Macaulay. This seems to be more common.
- The section ring (with respect to a given ample line bundle) of the variety is Cohen-Macaulay.
Of course if you are projectively normal in $\mathbb{P}^n$ and the ample line bundle is a very ample line bundle of that embedding, these two definitions coincide.
It doesn't imply anything about Gorenstein-ness. In fact, any Cohen-Macaulay projective variety with $H^i(X, \mathcal{O}_X) = 0$ for $0 < i < \dim X$ is arithmetically Cohen-Macaulay with respect to some embedding into projective space.
To see this, take a sufficiently ample line bundle $L$ such that $H^i(X, \omega_X \otimes L^n) = 0$ and $H^i(X, L^n) = 0$ for all $n \geq 1$ and all $0 < i < \dim X$. In the previous version of this answer, I forgot the Cohen-Macaulay hypothesis on $X$, in which case the first vanishing can't be forced to hold.
If I recall correctly, these notions appear prominently in the study of Linkage (see Eisenbud's book for an introduction).
A related notion is that of arithmetic Macaulayfication of a ring. This means that there exists an ideal $I$ such that the Rees algebra of $I$ (the ring you blow-up to get the blow-up of $I$) is Cohen-Macaulay. These were shown to exist in the last decade by Kawasaki. If I recall correctly, a corollary of this result is that every ring with a dualizing complex is a quotient of a Gorenstein ring (this was previously a conjecture of Sharp). Someone correct me if I'm wrong on this.
EDIT: Added the CM hypothesis on the variety and added an explanation (thanks to Long). EDIT2: Added the two possible definitions (section ring vs coordinate ring). Thanks to J. C. Ottem.
answered Nov 28, 2010 at 17:03
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$\begingroup$ A stupid question: When $X$ is projectively normal under the embedding given by $L$, the section ring $A=\bigoplus_{n\ge 0}H^0(X,L^n)$ is isomorphic to the coordinate ring $S(X)$ of X under this embedding. In general $A$ can be regarded as a $S(X)$-module. When $X$ is not projectively normal, is there any essential difference between ACM and $S(X)$ being CM? $\endgroup$ Nov 28, 2010 at 18:41
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3$\begingroup$ Karl, don't you need the middle cohomology of all the twists to vanish? $\endgroup$ Nov 28, 2010 at 20:25
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3$\begingroup$ @Ottem. They are different. Take $\mathbb P^1$ embedded in $\mathbb P^4$ by $x^4$, $x^3y$, $xy^3$, $y^4$. This is a closed immersion, but not projectively normal. The section ring is Cohen-Macaulay, but the homogeneous coordinate ring is not, because it is regular in codimension 1, but not normal and therefore not S2. @Pascal. It is the section ring which is losing information. In the example above, the section ring doesn't distinguish between the given embedding and the embedding by all sections of $\mathcal O(4)$. $\endgroup$ Nov 28, 2010 at 22:36
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1$\begingroup$ @Long, all the middle twists vanish by Serre vanishing and Serre duality. Just take a very high veronese subring. $\endgroup$ Nov 28, 2010 at 22:48
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2$\begingroup$ @J. C. Ottem, I think you are right, the coordinate ring definition is much more common. Before posting the original answer, I found some places where it was defined the other way, and went with that. But I probably should not have $\endgroup$ Nov 29, 2010 at 5:40