I read that in dimension $\geq 4$ there are Gorenstein abelian quotient singularities that have no crepant resolutions. What is the simplest example? I immagine that there should be toric examples. Is it the case that the cone does not admit a suitable subdivision in simplicial cones, corresponding to a crepant resolution? In your answers, please consider that I am only an amateur algebraic geometer, but I know some toric geometry.
3 Answers
[EDIT: added proof that $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor.]
Definition A variety is called $\mathbb Q$factorial if every Weil divisor on it is $\mathbb Q$Cartier, i.e., some multiple of it is a Cartier divisor.
The general statement you might want is that
Claim A $\mathbb Q$factorial, terminal singularity does not admit a nontrivial crepant resolution.
Proof $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor, and being terminal implies that all the discrepancies are positive. $\square$
Remark One might think that a Gorenstein terminal singularity does not admit a nontrivial crepant resolution. Here is an example that this is not true. Consider a cone over a smooth quadric surface in $\mathbb P^3$. This is a hypersurface in $\mathbb A^4$, so it is clearly Gorenstein. Blowing up the vertex and a simple calculation using adjunction shows that this is a terminal singularity. However, blowing up a divisor that is a cone over a line on the quadric surface gives a small resolution which will be crepant by being an isomorphism in codimension $1$. This shows that it is necessary to add the $\mathbb Q$factoriality condition for the above Claim.
Example For the example in Jim Bryan's answer, $\mathbb C^4/\pm$, or more generally, $\mathbb C^{m}/\pm$, the point to notice is that this is just the cone over the Veronese embedding of $\mathbb P^{m1}$. Isolated quotient singularities are $\mathbb Q$factorial and an easy computation shows that the discrepancy of the single exceptional divisor of the blow up of the vertex is $\dfrac m21$. This implies that it is terminal as soon as $m>2$, but for $m$ odd it will not be Gorenstein, so the first example of the desired kind is for $m=4$.
Addendum Here is a proof that $\mathbb Q$factoriality implies that the exceptional set of any resolution is a divisor:
Claim Let $X$ be a $\mathbb Q$factorial variety and $f:Y\to X$ a proper birational morphism. Let $E=\mathrm{Exc}(f)$ denote the exceptional set of $f$, i.e., the largest (closed) subset of $Y$ such that $f_{Y\setminus E}:Y\setminus E\to X\setminus f(E)$ is an isomorphism. Then $E$ is of pure codimension $1$ in $Y$.
Proof Let $y\in E$ and suppose that $\mathrm{codim}_YE\geq 2$ in a neighborhood of $y$. Let $C\subseteq E$ be an arbitrary proper curve such that $f(C)$ is a point and $y\in C$ and let $H\subseteq Y$ be an effective divisor such that $y\in H$, but $C\not\subseteq H$. This implies that $H\cdot C>0$. Consider the Weil(!) divisor $f_*H$ on $X$ (the pushforward is meant as a cycle). As $X$ is $\mathbb Q$factorial, some multiple of $f_*H$ will be Cartier, so replacing $H$ with that multiple we may assume that actually $f_*H$ is Cartier. Then it makes sense to pull it back (as a Cartier divisor). So we get a (Cartier) divisor $f^*f_*H$ which agrees with $H$ on $Y\setminus E$. In particular, if $\mathrm{codim}_YE\geq 2$ in a neighborhood $U$ of $y$, then $H_U=(f^*f_*H)_U$. Now by construction $y\in C\cap U\neq\emptyset$, so along $C$, $f^*f_*H=H+F$ where $F$ is an effective (exceptional) divisor that does not contain $C$. Finally, this leads to a contradiction, because we get that $$ 0=f^*f_*H\cdot C \geq H\cdot C>0$$ since $f(C)$ is a point.

$\begingroup$ Sandor, very nice and instructive! Is it hard to prove that the exceptional set of a resolution of a $\mathbb Q$factorial singularity is a divisor? $\endgroup$ Commented Jun 2, 2011 at 9:01

$\begingroup$ Dmitri, I added a proof for that. I think it is really not too hard, although it may seem at first because one has to be careful with the details, but the main idea is clear. (I don't know if this is the "usual" proof. Someone might know a shorter one. This is what I could come up with.) $\endgroup$ Commented Jun 3, 2011 at 5:38

$\begingroup$ thank you very much sandor! my knowledge of algebraic geometry is (sigh!) still too weak to understand all the words you mention (e.g. $\mathbb{Q}$factorial) but i'll keep this as something to learn! $\endgroup$ Commented Jun 3, 2011 at 8:19

$\begingroup$ Sandor, thanks a lot!! Do I understand correctly that in order to get an $f$ample divisor you need to work locally in a neighbourhood of $y$? Otherwise, is it true that an $f$ample divisor $H$ always exists (say when $Y$ is a small nonprojective resolutions of a projective $3$fold with simple double points)? $\endgroup$ Commented Jun 3, 2011 at 9:22

1$\begingroup$ Dmitri: Actually $f$ample is not needed. I edited the answer to show this. However, at least for this proof, you need a divisor that is not a pullback via $f$. (Actually, this is the point!). On the other hand, it only needs to be a Weil divisor, so it seems that a proper map will do as well. $\endgroup$ Commented Jun 3, 2011 at 9:52
The simplest example is $\mathbb{C}^4/\pm1$ where $1$ acts diagonally. I'm sure there is an elementary proof that this does not have a crepant resolution, maybe via toric geometry. Someone else on MO might know a reference. The proof I know uses a result of Yasuda which says that if $X$ is a Gorenstein orbifold and $Y\to X$ is a crepant resolution, then $H^*_{orb}(X) = H^*(Y)$ as graded vector spaces. In the case at hand, this would imply that the exceptional fiber of a crepant resolution $Y\to \mathbb{C}^4/\pm1$ would have cohomology in degree 0 and in degree 4 only, which is impossible for a projective variety.

$\begingroup$ See the proof you asked for in my answer. $\endgroup$ Commented Jun 2, 2011 at 2:43

$\begingroup$ thank you very much for the example, it should be easy enough for me to understand it! $\endgroup$ Commented Jun 3, 2011 at 8:21
There is a nice explicit description, in this paper of Morrison and Stevens, of four dimensional cyclic quotient singularities that are Gorenstein and terminal (cf. this MO questionterminal implies the nonexistence of crepant resolutions).

1$\begingroup$ A smooth point is both Gorenstein and terminal :) $\endgroup$– nafCommented Jun 1, 2011 at 14:31

1$\begingroup$ @Tom: for your last statement you also need the singularity to be $\mathbb Q$factorial. Otherwise it might admit a small resolution in which case there are no discrepancies at all. $\endgroup$ Commented Jun 2, 2011 at 7:30

1$\begingroup$ ps: See an example that shows that this can actually happen in my answer. $\endgroup$ Commented Jun 2, 2011 at 7:43

$\begingroup$ Hi Sandor, Thanks for the commentif I'm not mistaken this is true for cyclic quotient singularities, no? [That's why I referred to your earlier MO answer that spells this out.] Though if not, then I'll correct my answer...in any case, again thanks for the clarification! $\endgroup$ Commented Jun 2, 2011 at 21:57

$\begingroup$ All quotient singularities are $\mathbb{Q}$factorial. $\endgroup$– nafCommented Jun 3, 2011 at 6:14