In a previous question [Moving a Weil divisor on a normal surface away from a finite set of closed points](https://mathoverflow.net/questions/71735/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.



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


