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.



Quotient of affine space by finite subgroup of SL(V) is Gorenstein
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.


