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Could you give us an variety X which is Q-Gorenstein variety, but this variety is not Gorenstein variety? What is the canonical divisor K_X on X? How can we compute its canonical divisor? Could you give us an variety which is not Gorenstein variety? What is the canonical divisor K_X on X? How can we compute its canonical divisor?

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    $\begingroup$ Just to quickly answer the other question. The canonical divisor $K_X$ on a normal variety $X$ is any divisor $D$ such that if $U \subseteq X$ is the regular locus, then $$O_U(D|_U) \cong \Omega_{U/k}^{\dim X}.$$ One common way to compute a canonical divisor is by adjunction. For example, if $X \subseteq Y$ is a normal hypersurface in a normal variety $Y$, and we know the canonical divisor on $Y$, then $$(K_Y + X)|_X = K_X.$$ Alternately, for any $X$ of dimension $d$ in $\mathbb{P^n}$, then $$O_X(K_X) \cong \mathcal{E}xt^{n-d}(O_X, O_{\mathbb{P}^n}(-n-1)).$$ $\endgroup$ Commented Apr 28, 2012 at 17:15

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Let me give a couple more examples to complement what J.C.Ottem already said.

First I'll point out a couple discrepancies in terminology for a normal variety:

Gorenstein: This means that both $K_X$ is Cartier and $X$ is Cohen-Macaulay.

$Q$-Gorenstein: This means that $nK_X$ is Cartier for some integer $n > 0$.

Quasi-Gorenstein (or $1$-Gorenstein): This means that $K_X$ is Cartier but that $X$ is not necessarily Cohen-Macaulay.

In particular, $Q$-Gorenstein does not usually imply Cohen-Macaulay (because of this a lot of people don't like the terminology, notably János Kollár). On the other hand, some people like to require the Cohen-Macaulay condition in their $Q$-Gorenstein definition (adding to the confusion). It would also be "reasonable" to require that some canonical cover of $X$ is Cohen-Macaulay!

Ok, let me at least state the general rule for when a cone (or rather a section ring) is $Q$-Gorenstein or Gorenstein.

First note that because $S$ is graded, a module is free of rank one if and only if it is isomorphic to $S$ itself (up to a shift in grading).

Suppose that $X$ is a smooth (or even Gorenstein) projective algebraic variety over a field $k$ and suppose that $L$ is an ample line divisor. Form the ring: $$ S = \bigoplus_{k \in \mathbb{Z}} O_X(kL) $$ and let $m = S_+$ denote the irrelevant ideal. Also form the $S$-module $$ \omega_S = \bigoplus_{k \in \mathbb{Z}} (\omega_X \otimes O_X(kL)) $$ I claim that $\omega_S$ is the canonical module of $S$. This is actually pretty easy to see, $S$ is always normal so $\omega_S$ is determined on $U = \text{Spec} S \setminus m$. Therefore since $f : U \to X$ is a $\mathbb{A}^1$-bundle it follows that $\omega_U$ is just $f^* \omega_X$ which it is easy to see coincides with $\omega_X|_U$.

Then

Theorem (quasi-Gorenstein): The ring $S$ is quasi-Gorenstein if and only if $K_X \sim nL$ for some integer $n$ (possibly $n = 0$).

The proof idea is as follows. Since $K_X \sim nL$, it follows that $\omega_S$ is just $S$ shifted over by $n$. But that's a free module. Conversely, if $\omega_S$ is a free module, since $S$ is graded, it must be a shift of $S$.

Theorem ($Q$-Gorenstein): $S$ is $Q$-Gorenstein if and only if $mK_X \sim nL$ for some integers $n$ and $m \neq 0$.

The proof idea is similar, it follows from the same sorts of arguments as above that $$O_{\text{Spec} S}(mK_S) = \omega_S^{(m)} = \bigoplus_{k \in \mathbb{Z}} O_{X}(kL + mK_S).$$ Therefore for this to be free, we need exactly $mK_S = nL$ for some integer $n$.

Theorem (Cohen-Macaulay): Finally $S$ is Cohen-Macaulay if and only if $H^i(X, O_X(kL)) = 0$ for all $\dim X > i > 0$ and all $k \in \mathbb{Z}$.

Note that the points away from the origin are Cohen-Macaulay since $X$ was smooth (or at least Gorenstein $\Rightarrow$ Cohen-Macaulay). The point is then that $[H^{i+1}_m(S)]_k = H^i(X, O_X(kL))$ where $[\bullet]_k$ is the $k$th graded piece of the module. This follows from computations in Cech cohomology.

The upshot:

This gives us lots of examples of varieties that are $Q$-Gorenstein or not.

Fano: If $X$ is Fano (meaning $-K_X$ is ample) and we take $L = -K_X$, then $S$ is certainly quasi-Gorenstein. On the other hand, if we take $L = -mK_X$, then $mK_X = (-1)L$, $S$ is $Q$-Gorenstein.

Now let's suppose we are in characteristic zero and that $X$ is smooth. If $k > 0$, we have $H^i(X, O_X(kL)) = H^i(X, O_X(K_X + (m-1)K_X + (k-1)L) = 0$ by Kodaira vanishing. If $k < 0$, by Serre-duality, $H^i(X, O_X(kL)) = H^{\dim X - i}(X, O_X(K_X - kmL)) = 0$ by Serre vanishing. On the other hand for $k = 0$, $H^{i}(X, O_X) = H^i(X, O_X(K_X - K_X)) = 0$ by Kodaira vanishing. Thus $S$ is Cohen-Macaulay.

Calabi-Yau: If $X$ is Calabi-Yau (meaning $K_X = 0$) and we choose any ample $L$, then $S$ is quasi-Gorenstein. It may not be that $X$ is NOT Gorenstein though, it depends on the vanishing of $H^i(X, O_X(kL))$. For example, an Abelian surface has $H^1(X, O_X) \neq 0$.

Others: If $K_X$ is neither ample nor anti-ample, nor $Q$-trivial, then you are out of luck. The cone will never be $Q$-Gorenstein.

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  • $\begingroup$ is there a relation between gorenstein and having quotient singularities? $\endgroup$ Commented Apr 28, 2012 at 17:49
  • $\begingroup$ No, there are non-quotient singularities that are Gorenstein (any hypersurface singularity is Gorenstein). There are quotient singularities that are not Gorenstien, for example $k[x^3, x^2y, xy^2, y^3]$. Normal finite quotient singularities are always Cohen-Macaulay and $Q$-Gorenstein though. $\endgroup$ Commented Apr 28, 2012 at 18:15
  • $\begingroup$ thanks for the quick reply! where would I go about learning the normal finite case for example? $\endgroup$ Commented Apr 28, 2012 at 21:19
  • $\begingroup$ For the Cohen-Macaulay bit, this follows since the ring is a direct summand of the ring it is the invariants of. There is a paper of Hochster and J.~Roberts that does this. In fact, Boutot's theorem actually says that the singularity is rational, which amoung other things is Cohen-Macaulay. This stuff works for reductive groups. $$\text{ }$$ For the $Q$-Gorenstein bit, see for example Lemma 5.16 in Kollár-Mori, Birational geometry of algebraic varieties. $\endgroup$ Commented Apr 29, 2012 at 1:40
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Cones are usually a good source of examples when it comes to questions like these. In your case, let $Y\subset \mathbb{P}^5$ be the Veronese embedding of $\mathbb{P}^2$ and let $X\subset \mathbb{P}^6$ be the projective cone of $Y$. Then $X$ is a normal non-Gorenstein variety which is not Gorenstein. Indeed, Let $L\subset Y$ be the image of a line in $\mathbb{P}^2$ and let $D$ be the Weil divisor given by the cone over $L$. Let $H$ be the generator of $Pic(X)$, that is, the hyperplane section from the embedding in $\mathbb{P}^5$. Then $-K_X=3D$ is not Cartier (as you can see by blowing up the vertex of the node and use the adjunction formula). But $2K_X$ is certainly Cartier as $2K_X\sim -6D\sim -2H$.

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It is also quite easy to generate a lot of examples with toric geometry. We have next characterization:

Let $\Sigma$ be a fan in a lattice $N$ and $M = N^*$ dual lattice. Then

  1. $X_{\Sigma}$ is $\mathbb Q$-Gorenstein iff for each cone $\sigma \in \Sigma$ generated by $n_1,...,n_r \in N$ there is $m \in M \otimes \mathbb Q$ such that $(n_i,m) = 1$ for every $i=1,...,r$
  2. $X_{\Sigma}$ is Gorenstein iff for each cone $\sigma \in \Sigma$ generated by $n_1,...,n_r \in N$ there is $m \in M$ such that $(n_i,m) = 1$ for every $i=1,...,r$

For example for a complete fan with generators $(-1,-2),(1,0),(0,1)$ in $\mathbb Z^2$ we can check that the only solution of system $-1x-2y=1, x=1$ is $(1,-\frac{1}{2})$. So $X_{\Sigma}$ is not Gorenstein, but $\mathbb Q$-Gorenstein. In fact, $X_{\Sigma} \cong \mathbb P(1,1,2)$ in this case.

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  • $\begingroup$ +1 WPPs were the first example that came to mind. It should be straightforward to characterize which ones specifically satisfy the desired condition. $\endgroup$
    – user347489
    Commented Aug 3, 2021 at 20:18

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