Moving a canonical divisor on a normal surface away from the singular locus

Question

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.

Answer 1

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.

Answer 2

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.