Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $R = S/I = k[x_1, \dots, x_n]/I$ is a (normal) domain of finite type over a field (or any semi-local ring $k$ with a dualizing complex). In this case, I can define $\omega_R = \textrm{Ext}^{n - \dim R}_{S}(R, S)$.

$R$ is called quasi-Gorenstein if $\omega_R$ is locally free of rank 1. Presumably there are easy examples where $\omega_R$ is locally free but not free, does anyone know one off the top of their head?

EDIT: Of course, It would be even better to have a Gorenstein (ie, quasi-Gorenstein + CM) example too.

Obviously I can choose a canonical module up to tensoring with a locally free at some level, but for many purposes the $\omega_R$ I defined above is better to work with.

share|improve this question
    
Dear Karl, in my experience "quasi-Gorenstein" is a word describing local rings, and it means that the canonical module (unique since the ring is local) is free. The "quasi" in this case means that the ring may not be Cohen-Macaulay. The reference I usually see given for this is Platte & Storch, Math. Z., 1977, but see also Aoyama, J. Math Kyoto 1981. The canonical (har har) example is $R = k[[x,y,z,w]]/(x,y) \cap (z,w)$, with $\omega_R = R/(x,y) \oplus R/(z,w)$. Does your definition come up in the literature? –  Graham Leuschke Feb 9 '11 at 18:44
    
Graham. The definitions are the same. The module $\omega_R$ I gave is the (canonical) canonical module. So I call a non-local ring quasi-Gorenstein if it is quasi-Gorenstein at each point. By the way, how is the example you gave quasi-Gorenstein, $R/(x,y) \oplus R/(z,w)$ is not a free $R$-module at the origin, is it? –  Karl Schwede Feb 9 '11 at 18:57
    
The canonical example of a quasi-Gorenstein ring in my experience is the cone over an Abelian surface. But that's graded and so essentially local. –  Karl Schwede Feb 9 '11 at 19:01
    
Oops, that "canonical example" is an example of something else. Terrible, terrible. Sorry. –  Graham Leuschke Feb 9 '11 at 22:11
    
Here's a related thought. Is it possible to have $R \cong S/I \cong T/J$, where $S$ and $T$ are both regular (say, polynomial rings) of the same dimension $D$, and $\mathrm{Ext}^{D-d}_S(R,S) \not \cong \mathrm{Ext}_T^{D-d}(R,T)$? The idea would be that choosing an affine embedding is more or less the same thing as choosing a canonical canonical module. –  Graham Leuschke Feb 9 '11 at 22:16
show 1 more comment

2 Answers 2

up vote 2 down vote accepted

Karl, I think one can construct a smooth affine curve with non-zero canonical class by removing some general points from a smooth projective curve of genus $>1$. Details can be found in this paper (Theorem 6), which I gave in this answer.

share|improve this answer
    
Thanks, I'll look into this. –  Karl Schwede Feb 9 '11 at 23:05
    
I see, fix a curve $C$, choose a representative of the canonical class $K$ and find some point $x \in C$ not in the support of $K$ such that $K$ is not linearly equivalent to any multiple of $x$. This should be easy for any higher genus curve. On the affine scheme $U = C \setminus \{ x \}$, $K|_U$ can't possibly be linearly equivalent to $0$. –  Karl Schwede Feb 10 '11 at 2:57
    
Hi Karl, glad it worked for you, I wasn't sure if this is what you wanted. –  Hailong Dao Feb 10 '11 at 20:32
add comment

Karl, here is a simple construction in arbitrary dimension $>1$:

Let $X$ be a smooth projective variety with Picard number at least $2$ and non-trivial canonical bundle $\omega_X$. Let $A$ be a smooth (irreducible) very ample divisor on $X$ such that $\mathscr O_X(A)$ and $\omega_X$ are not generating the same ray in ${\rm Pic}\\, X$. Finally let $U=X\setminus {\rm supp}\\, A$. Then $\omega_U={\omega_X}|_{U}$ is locally free but not free. (If it were free, that is, trivial in ${\rm Pic}\\, U$ then $\omega_X$ would be a power of $\mathscr O_X(A)$).

Example: Let $X=\mathbb P^1\times \mathbb P^1$ and $A$ a general divisor of type $(a,b)$ with $a>b\gg 0$.

Remark: This needs $\dim X>1$ in order to get a Picard number larger than $1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.