In a previous question Moving a Weil divisor on a normal surface away from a finite set of closed points I probably asked for too much. As J.C. Ottem pointed out, it is not always possible to move a Weil divisor on a normal surface away from a given closed set of points. Fortunately in my set-up this generality is not required. Therefore, I propose the following more mild set-up and hope that the answer will be positive.

Let $Y$ be a normal surface and let $K_Y$ be the Weil divisor obtained by taking the closure of a canonical divisor on the nonsingular locus of $Y$.

Question. Is $K_Y$ linearly equivalent to a divisor which does not go through the singular locus of $Y$?

Again, by a normal surface I mean an integral normal excellent separated noetherian 2-dimensional scheme. I will also assume $Y$ to be (locally?) $\mathbf{Q}$-factorial in the motivation below.

Motivation. Given a resolution of singularities $\rho:Y^\prime\longrightarrow Y$, I would like to show that the intersection number $(\psi^\ast K_{Y},E) =0$, where $E$ is an exceptional component of $\psi$ and $\psi^\ast K_Y$ is the pull-back of the $\mathbf{Q}$-Cartier divisor $K_Y$. This will hold if I can move $K_Y$ away from the singular locus.

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    $\begingroup$ In order to prove the result in your motivation, you don't need the answer to your question to be positive (in fact, as it is shown below it is negative in general). Instead, you can use the following (standard) argument: since $Y$ is $\mathbb{Q}$-factorial, $mK_Y$ is Cartier for $m$ large enough. Then the support of $mK_Y$ can be moved away from the singular locus (same proof as in the smooth case). This implies $0=(\psi^*(mK_Y), E)=m \cdot (\psi^*K_Y, E)$, hence $(\psi^*K_Y, E)=0$. $\endgroup$ Jul 31, 2011 at 19:31
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    $\begingroup$ It might be worth mentioning that $\psi^* K_Y \cdot E = 0$ as long as the projection formula $\psi^* K_Y \cdot E = K_Y \cdot \psi_* E$ holds (since $\psi_* E = 0$ because $E$ is exceptional). That projection formula does hold at least when $K_Y$ is $\mathbb{Q}$-Cartier (reference: proposition 2.3 in Fulton's Intersection theory). $\endgroup$
    – cgodfrey
    Jan 24, 2020 at 21:03

2 Answers 2


I assume that you also allow non effective divisors.

Then the answer to your question is a consequence of the following fact (see Kollar-Mori, Birational geometry of algebraic varieties, Proposition 5.75):

Given a normal variety $Y$, its dualizing sheaf $\omega_Y$ is given by $\omega_Y = \mathcal{O}_Y(K_Y)$.

If $K_Y$ is linearly equivalent to a divisor whose support is disjoint from the singular locus of $Y$, then $K_Y$ is Cartier. So your question is equivalent to the following:

when is $\omega_Y$ a line bundle?

The well-known answer is that this happens if and only if $Y$ is a Gorenstein variety (in fact, some people use this as the definition of Gorenstein variety).

In particular, this is true if the surface $Y$ contains only Rational Double Points as singularities.

For instance, let $Y \subset \mathbb{P}^{n+1}$ be the cone over the rational normal curve of degree $n$ in $\mathbb{P}^n$. The unique singular point of $Y$ is its vertex, which is a quotient singularity of type $\frac{1}{n}(1,1)$. Then $Y$ is Gorenstein if and only if $n=2$, i.e. if and only if it is the quadric cone in $\mathbb{P}^3$. In this case $K_Y$ is linearly equivalent to $-2H$, where $H$ is the hyperplane section. Hence you can move $K_Y$ away from the vertex.

For all $n \geq 3$, instead, every divisor linearly equivalent to $K_Y$ must contain the vertex.

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    $\begingroup$ Francesco, just to do a little nitpicking: Your definition of a Gorenstein variety this way only works for normal surfaces. In higher dimensions Gorenstein implies Cohen-Macaulay. Of course, a normal surface is CM, but there are higher dimensional varieties with trivial canonical class that are not Gorenstein. $\endgroup$ Sep 5, 2012 at 22:30

This should be impossible, for basically the reason you describe in your Motivation. Note that, if things worked the way you wanted, not only would the numerical intersection $(\psi^* K_Y, E)$ be zero, but the actual class of $\psi^* K_Y$ in $\mathrm{Pic}(E)$ would also be zero.

Suppose that $Y$ has one singular point, at $x$; that $\rho: Y' \to Y$ is a resolution and that $\rho^{-1}(x)$ is a single curve $E$. Then, by adjunction, $K_E = (K_{Y'} + E)|_E$. Also, $\psi^* K_Y = K_{Y'} + r E$ for some integer $r$. So $\psi^* K_Y |_E = (K_E + (r-1) E)|_E$. If things worked the way you wanted, this class would be zero. So we'll get a counter-example if we can set things up so that $K_E$ is not a multiple of $\mathcal{O}(E)|_E$ in $\mathrm{Pic}(E)$.

Start with $E$ a curve of genus $\geq 2$. Choose $L$ any negative degree line bundle so that $K_E$ is not a multiple of $[L]$ in $\mathrm{Pic}(E)$. Let $Y'$ be the total space of $L$, and let $Y$ be $Y'$ with $E$ blowndown. (We can do this because we used a negative degree line bundle.) If I'm not confused, this should be a counter-example.

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    $\begingroup$ Yes, blowing down $E$ you produced a non-Gorenstein singularity in $Y'$. A similar construction also works over $\mathbb{P}^1$: the cones $Y$ I used in my answer are precisely the blow-down of the unique curve of negative self intersection in the projective bundle $\mathbb{P}(\mathcal{O} \oplus \mathcal{O}(-n)$. $\endgroup$ Jul 31, 2011 at 18:21
  • $\begingroup$ Great. I finally understand the situation. Francesco's comment above makes it clear that what I wanted to be true is still true. Thanks for the counterexample! $\endgroup$ Jul 31, 2011 at 19:41

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