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Let $P$ be a normal, $\mathbb{Q}$-Gorestein variety with terminal singularities. Let $X \subseteq P$ be a normal, irreducible Weil divisor such that $X \sim_{\mathbb{Q}} - K_P$, that is $\mathbb{Q}$-linearly equivalent to the anticanonical divisor of $P$ (which is $\mathbb{Q}$-Cartier) .

My question is, does $X$ have $\mathbb{Q}$-trivial canonical divisor? In the adjuction formula for Weil divosr, it seems there is a $Diff$ term appears. I was wondering if this term is zero here?

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  • $\begingroup$ You probably mean $\mathbb{Q}$-trivial? $\endgroup$ – abx Jan 28 '15 at 14:55
  • $\begingroup$ Yes, that is want I meant, thank you! $\endgroup$ – Li Yutong Jan 28 '15 at 16:01
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    $\begingroup$ Spelling? "Gorenstein"? $\endgroup$ – paul garrett Jan 29 '15 at 15:42
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How about this: As you assumed, $P$ has terminal singularities, which means that the singular locus of $P$ has codimension at least 3 (see Corollary 5.18 in Kollár-Mori's book). Hence, the singular locus has codimension at most 2 in $X$. We can consider everything outside the singular locus because we work on the level of divisors of $X$. Hence $P$ is just smooth and $X$ is Cartier and everything is true for you.

EDIT: @LiYutong, of course I did not mean that $K_P$ is Cartier (which is not true in general). I mean that there exists a codimension 3 subset $Z$ of $P$ such that $U=P-Z$ is smooth and hence $X|_U$ is Cartier divisor on $U$. Hence $K_{X|_U}=(K_U+X|_U)|_{X|_U}$ is $\mathbb{Q}$-trivial by adjunction. Note that $X-X|_U$ has codimension at least 2 in $X$, so the Weil divisors class $\rm{Cl}(X)$ on $X$ coincides with the Weil divisors class $\rm{Cl}(X|_U)$ on $X|_U$ by restriction. So $K_X\sim_\mathbb{Q}0$.

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  • $\begingroup$ Dear Chen Jiang, thank you for your answer! Since I assume $X$ is normal, $X$ is already smooth in codimensional 1. Do you want to argue: when $P$ smooth in codimensional 2, then the canonical divisor $K_P$ is Cartier (which I don't think is true by considering some toric example)? $\endgroup$ – Li Yutong Jan 29 '15 at 13:59
  • $\begingroup$ @LiYutong, I edited. Please check additional explanation. $\endgroup$ – Chen Jiang Jan 29 '15 at 14:44
  • $\begingroup$ Yes, this makes sense for me, thank you again!! $\endgroup$ – Li Yutong Jan 29 '15 at 15:07

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