Return to Answer

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.

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.

I assume that you also allow non effective divisors.

Then the answer to your question follows from 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 $\omega_Y = \mathcal{O}_Y(K_Y)$.

So your question is equivalent to the following: when is $\omega_Y$ a line bundle? In fact, if $K_Y$ is equivalent to a divisor whose support is disjoint from the singular locus of $Y$, then $K_Y$ is Cartier. 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 $Y$ 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 $\mathbb{P}^n$. 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.

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.

I assume that you also allow non effective divisors.

Then the answer to your question follows from 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 $\omega_Y = \mathcal{O}_Y(K_Y)$.

So your question is equivalent to the following: when is $\omega_Y$ a line bundle? In fact, if $K_Y$ is equivalent to a divisor disjoint from the singular locus of $Y$, then $K_Y$ is Cartier. 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).

For instance, let $Y \subset \mathbb{P}^{n+1}$ be the cone over the rational normal curve of degree $n$ in $\mathbb{P}^n$. 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 $K_Y$ outside the vertex.

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

I assume that you also allow non effective divisors.

Then the answer to your question follows from 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 $\omega_Y = \mathcal{O}_Y(K_Y)$.

So your question is equivalent to the following: when is $\omega_Y$ a line bundle? In fact, if $K_Y$ is equivalent to a divisor whose support is disjoint from the singular locus of $Y$, then $K_Y$ is Cartier. 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 $Y$ 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 $\mathbb{P}^n$. 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.