Quotient of affine space by finite subgroup of SL(V) is Gorenstein I am looking for a proof of the following fact:
If $G$ is a finite subgroup of $SL_n(\mathbb{C})$ acting on $\mathbb{A}_{\mathbb{C}}^n$, then the resulting quotient scheme is Gorenstein.
Thanks.
 A: *

*By the Hochster-Roberts Theorem it is Cohen-Macaulay

*The canonical sheaf of $\mathbb A^n$ is $G$-invariant (because the elements of $G$ have det=1) and hence it descends to the canonical sheaf of the quotient, which is then a line bundle.

A: You probably want to show that the quotient scheme $\mathbb{A}^n_{\mathbb{C}}/G$ is Gorenstein.
A proof can be made along the following line. Since $G \subset SL_n(\mathbb{C})$, the canonical bundle of $\mathbb{A}^n_{\mathbb{C}}$ is $G$-equivariant and satisfy the following : for all $x \in \mathbb{A}^n_{\mathbb{C}}$ which is fixed by some subgroup $H \subset G$, the group $H$ acts trivially on $\omega_{\mathbb{A}^n_{\mathbb{C}}} \otimes \mathbb{C}(x)$. Hence, by a classical criterion for descent, there is a line bundle $L$ on $\mathbb{A}^n_{\mathbb{C}}/G$ such that $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \pi^*L$, where $\pi : X \rightarrow \mathbb{A}^n_{\mathbb{C}}/G$ is the quotient map.
Now we know that $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}}$ and that $\pi_*(\mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}})^G = \mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}/G}$. Using Grothendieck-Serre duality for the projective morphism $\pi$, it's easy to see that $\pi_*(\omega_{\mathbb{A}^n_{\mathbb{C}}})^G$ is a (up to a shift) a dualizing object on $\mathbb{A}^n_{\mathbb{C}}/G$. Using the projection formula and the equality $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \pi^*L$, one finds that $L$ is a dualizing object (up to a shift) on $\mathbb{A}^n_{\mathbb{C}}/G$. Hence $\mathbb{A}^n_{\mathbb{C}}/G$ is Gorenstein.
