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Sándor Kovács
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It probably goes something like this:

  1. By the Hochster-Roberts Theorem it is Cohen-Macaulay
  2. The canonical sheaf of the quotient$\mathbb A^n$ is a line bundle, because$G$-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.

It probably goes something like this:

  1. By the Hochster-Roberts Theorem it is Cohen-Macaulay
  2. The canonical sheaf of the quotient is a line bundle, because the elements of $G$ have det=1.
  1. By the Hochster-Roberts Theorem it is Cohen-Macaulay
  2. 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.
Source Link
Sándor Kovács
  • 42.9k
  • 2
  • 109
  • 155

It probably goes something like this:

  1. By the Hochster-Roberts Theorem it is Cohen-Macaulay
  2. The canonical sheaf of the quotient is a line bundle, because the elements of $G$ have det=1.