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(EDIT 07/06/11: although the question has not been settled definitely, Sándor's excellent answer and the comments by Angelo and ulrich have highlighted many potential obstructions to the constructions I wanted. Thank you all! I am still very interested in any leads, so please keep them coming if you have some more.)

I would like to know examples of log-canonical singularities of low codimension which is normal but non-Cohen-Macaulay. A silly way to do it may be just adjoining variables, so here is the precise question:

Fix a number $c$, for what $n$ can one construct an affine variety $X \subseteq \mathbb A^n_{\mathbb C}$ such that: $X$ is indecomposable and normal of codimension $c$, $X$ has at worst log-canonical singularity but $X$ is not Cohen-Macaulay? Given $c$, can we construct such $X$ for all $n$ big enough?

The case $c=1$ is easy, there are no example since hypersurfaces are Cohen-Macaulay, so let's begin with $c=2$.

Motivation/Comments: I am actually looking for $F$-pure rings (i.e., the Frobenius is a pure morphism), but conjecturally my question above is virtually the same. Karl Schwede told me one can also try to look for (projective) Calabi-Yau varieties with some non-vanishing middle cohomolgy and low codimension embedding, then take their cones. But not being a geometer, I do not know how to construct such things.

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There is a conjecture of Hartshorne that says that a smooth projective variety in $\mathbb{P}^n$ of codimension $<n/3$ should be a complete intersection. Very little is known about it but it does suggest that it is unlikely that taking cones over smooth projective varieties will give examples of fixed codimension $c$ for large $n$. –  ulrich Jul 4 '11 at 6:30
    
@Dear ulrich: that is a very good point. Actually, it reminded me that even for normal projective varieties of small codimensions, there are some restrictions: the Picard group is $\mathbb Z$. That probably rules out quite a few candidates. –  Hailong Dao Jul 4 '11 at 13:47
    
Oops, thank you Sándor! –  Hailong Dao Jul 6 '11 at 14:54
    
No problem. (I removed the incriminating evidence, too....) –  Sándor Kovács Jul 6 '11 at 14:55
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1 Answer

1 Examples, smallest $n$

Let $A$ be an abelian variety and $X$ a cone over $A$. Then $X$ is log canonical, but as soon as $\dim A\geq 2$, then $X$ is not Cohen-Macaulay. (This you can see by computing the local cohomology at the vertex).

For the $c=1$ case: those are obviously CM.

As ulrich points out, there exist abelian surfaces in $\mathbb P^4$, so those give you $c=2$ with $3$-dimensional singularities, that is, $n=5=2c+1$. For $c=2$ this is the smallest $n$ you can get. Here is why: For $n\leq 4$ anything of codimension $c=2$ would be of dimension at most $2$ and hence if it is normal, it is $S_2$ and in particular CM.

So, let's assume that $c>2$.

As any quasi-projective variety $A$ of dimension $d$ maybe embedded in $\mathbb P^{2d+1}$ (Embed in some $\mathbb P^N$ and notice that the closure of the secant variety of $A$ is of dimension $2d+1$, so as long as $N>2d+1$, one may find a point and a projection that gives an embedding of $X$ into $\mathbb P^{N-1}$. Repeat.), you can do that with abelian varieties as well. This gives you $c=d+1$, so turning it around, for any $c>2$ you can find an $X$ that you're looking for in $\mathbb A^n$ with $n=2c-1$.

2 Non-indecomposables

(Note: this is here, because I originally did not read the requirements carefully. Then I didn't feel like erasing it. )

If you did not required an indecomposable singularity, then you could take the product of this $X\subset \mathbb A^{2c-1}$ and an arbitrary $\mathbb A^r$. Of course, this is why you made that requirement. Anyway, you'd get $$ X_{c,r}:=X\times \mathbb A^r\subset \mathbb A^{2c-1+r} $$ of codimension $c$ with $n=2c-1+r$. In other words, yes, you can construct such an $X$ for all $n$ big enough. I am not sure whether the bound $2c-1$ is optimal, but I have a feeling that you can't get much better than that. Note that this construction works for $c=2$ as well, so you can get examples in all dimensions $n\geq 5=2\cdot 2+1$.

If you wanted indecomposable singularities, I would expect that the codimension is actually increasing with the dimension. In other words, I would expect low codimensional examples in low dimension and not in (arbitrarily) high dimension.

3 Indecomposable vs. isolated

The previous point shows why Hailong assumed indecomposable. I claim that one may as well also assume isolated. At least to start.

First, let $X$ be an example as required, of dimension $d=\dim X$ and of codimension $c$. Let $s=\mathrm{Sing} X$ the dimension of the singular set of $X$ and assume that $\mathrm{Sing} X$ is irreducible. Take a general complete intersection of codimension $s$. This will have dimension $d-s$, codimension $c$ and it is an isolated non-CM log canonical singularity.

Next, let $X$ be a codimension $c$ and it is an isolated non-CM log canonical singularity and consider a $\mathbb Q$-Gorenstein deformation of $X$ over a base of dimension $s$. (A $\mathbb Q$-Gorenstein deformation means that the relative dualizing sheaf of the family is a $\mathbb Q$-line bundle and its line bundle powers restrict to the appropriate power of the dualizing sheaf of the members of the family). The total space of the deformation will be an example of the kind you want. (For an indecomposable example you need a deformation without a trivial component).

The fact that this gives you an example that you want is non-trivial. It is log canonical by inversion of adjunction see the main result of Kawakita's paper and it is non-CM at every point of the singular set by Corollary 1.3 of this paper.

From this you can conclude that any example you get will be a deformation of an isolated example. A priori you get that at the generic points of the singular set. At non-generic points those are still degenerations (i.e., non-small deformations) of the general ones.

The problem you run into is that you need non-trivial deformations of these singularities and they have finite dimensional versal deformation spaces, so you can't get too far with this idea. (Angelo will correct me if this is wrong, since he is one of the ultimate experts on this. See also Artin's extended work on this topic.)

This suggests that given your restriction of being indecomposable, regarding your "every big enough $n$" question, it seems that in order for that to happen you really need low codimensional isolated examples, which will be hard to construct since you can't get too far with cones (cf. ulrich's and Angelo's comments).

On the other hand we do get new examples out of this: A flat family of polarized abelian/CY/etc varieties will give you a flat family of the cones over them. As long as the family has maximal variation (i.e., the moduli map of the base is generically finite) the resulting singularity will be what you want. So, this way you can get higher dimensional examples, but it seems that all constructions are limited in dimension.

All of this suggests that the answer to your big enough question will be "no".

4 Better examples (manageability over low codimension)

To get an example that you want, you do not need to have an abelian variety for the cone construction. If $A$ is a smooth projective variety of dimension $d$ such that $\omega_A\simeq \mathscr O_A$ and there exist two integers $i,m\in\mathbb Z$ such that $0<i<d$ and $$H^i(A,\mathscr O_A(m))\neq 0,$$ then the cone over $A$ has non-CM log canonical singularities. In other words, to get more examples you just need to find such subvarieties of low codimension.

Of course, complete intersections do not satisfy this, but for example the product of any two CYs do. So, you could take, say, a CY hypersurface $H$ and an elliptic curve $E$ and then $A=H\times E$ satisfies the condition (and $A$ is generally neither CY nor abelian).

The obvious embedding via Segre gives larger codimension than what you get from the above procedure, so this does not give you smaller codimensional examples, but they may be more manageable since you have the product of a hypersurface and a plane curve and even if the codimension is high, you know the embedding pretty well. So, from a practical point of view these are perhaps better examples after all.

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Dear S\'andor, how can we make sure that $X$ has low codimension? –  Hailong Dao Jul 4 '11 at 3:10
    
And how can I type your name correctly in the comments? Thanks. –  Hailong Dao Jul 4 '11 at 3:11
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An abelian variety of dimension $n$ cannot be embedded in $\mathbb{P}^m$ for $m < 2n$. See, for example, Fulton's Intersection Theory, Example 3.2.15. Also, there do exist abelian surfaces in $\mathbb{P}^4$ (constructed by Horrocks and Mumford) but I'm not sure about the higher dimensional case. (Of course one can always embed an $n$ dimensional abelian variety in $\mathbb{P}^{2n+1}$.) –  ulrich Jul 4 '11 at 5:31
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Van De Ven proved that an abelian variety of dimension $n$ can not be embedded in $\mathbb{P}^{2n}$, as soon as $n \geq 3$ (On the embedding of abelian varieties in projective spaces, Annali di Matematica Pura ed Applicata 103, 127-129). –  Angelo Jul 4 '11 at 8:29
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The Barth-Lefschetz theorem implies that for any smooth projective variety $X$ of dimension $m$ in $\mathbb{P}^n$, $H^1(X, \mathbb{Z}) = 0$ if $2m > n$. So one can get an improvement of at most $1$ by using a product of an elliptic curve and a CY hypersurface rather than an abelian variety. –  ulrich Jul 4 '11 at 9:40
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