Locally principal

At least in height-1 ideal case, in a normal domain (or at least G1+S2 domain), the following should let you know whether the ideal is locally principal.

Let $I$ be the ideal in question and let $J$ be another ideal isomorphic to $Hom_R(I, R)$ (which you can do in a number of ways, say by forming a colon or by embedding the module back into R).

Then $I\cdot J$ is an ideal. If $I$ is locally principal, it corresponds to a Cartier divisor $D$. Then $J$ corresponds to $-D$. It's a fact (first pointed out to me by Tommaso de Fernex) that $D$ is Cartier if and only if

$$O(D) \cdot O(-D) = I\cdot J = \langle g \rangle$$

is principal.

That doesn't help...

But you say, this doesn't help us at all (we have another ideal we need to check whether or not it is principal)!

The point however, is that we know that the reflexification of $I\cdot J$ is principal! (Recall reflexification just means applying $Hom(\bullet, R)$ twice, Macaulay2 can do this for instance).

On the other hand $I \cdot J$ agrees in codimension 1 with its reflexification (since reflexification won't change anything in codimension 1 for a normal domain).

It follows that $I\cdot J$ is principal if and only if $I\cdot J$ is reflexive.

Proposition: Hence $I$ is locally principal if and only if $I \cdot J$ is reflexive.

Locally principal

At least in height-1 ideal case, in a normal domain (or at least G1+S2 domain), the following should let you know whether the ideal is locally principal.

Let $J$ be the ideal in question and let $J_2$ be another ideal isomorphic to $Hom_R(J, R)$ (which you can do in a number of ways, say by forming a colon or by embedding the module back into R).

Then $J\cdot J_2$ is an ideal. If $J$ is locally principal, it corresponds to a Cartier divisor $D$. Then $J_2$ corresponds to $-D$. It's a fact (first pointed out to me by Tommaso de Fernex) that $D$ is Cartier if and only if

$$O(D) \cdot O(-D) = J\cdot J_2 = \langle g \rangle$$

is principal.

That doesn't help...

But you say, this doesn't help us at all. All I've done is give you another ideal we need to check whether or not it is principal!

The point however, is that we know that the reflexification of $I\cdot J$ is principal! (Recall reflexification just means applying $Hom(\bullet, R)$ twice, Macaulay2 can do this for instance). Also remember that locally principal ideals are always reflexive.

On the other hand $J \cdot J_2$ agrees in codimension 1 with its reflexification (since reflexification won't change anything in codimension 1 for a normal domain).

It follows that $J \cdot J_2$ is principal if and only if $J\cdot J_2$ is reflexive.

Proposition: Hence $J$ is locally principal if and only if $J \cdot J_2$ is reflexive.

Perhaps its worth noting that you can also use this to identify the locus where $J$ is not locally principal. If $L$ is the reflexification of $J \cdot J_2$, then compute the $$\text{Ann}_R(L/(J \cdot J_2)).$$ That is just the locus where $J$ is not locally principal.

Locally principal

At least in height-1 ideal case, in a normal domain (or at least G1+S2 domain), the following should let you know whether the ideal is locally principal.

Let $I$ be the ideal in question and let $J$ be another ideal isomorphic to $Hom_R(I, R)$.

Then $I\cdot J$ is an ideal. If $I$ is locally principal, it corresponds to a Cartier divisor $D$. Then $J$ corresponds to $-D$. It's a fact (first pointed out to me by Tommaso de Fernex) that $D$ is Cartier if and only if

$$O(D) \cdot O(-D) = I\cdot J$$

is principal.

That doesn't help...

But you say, this doesn't help us at all (we have another ideal we need to check whether or not it is principal)!

The point however, is that we know that the reflexification of $I\cdot J$ is principal! (Recall reflexification just means applying $Hom(\bullet, R)$ twice, Macaulay2 can do this for instance).

On the other hand $I \cdot J$ agrees in codimension 1 with its reflexification (since reflexification won't change anything in codimension 1 for a normal domain).

It follows that $I\cdot J$ is principal if and only if $I\cdot J$ is reflexive.

Proposition: Hence $I$ is locally principal if and only if $I \cdot J$ is reflexive.

Locally principal

At least in height-1 ideal case, in a normal domain (or at least G1+S2 domain), the following should let you know whether the ideal is locally principal.

Let $I$ be the ideal in question and let $J$ be another ideal isomorphic to $Hom_R(I, R)$ (which you can do in a number of ways, say by forming a colon or by embedding the module back into R).

Then $I\cdot J$ is an ideal. If $I$ is locally principal, it corresponds to a Cartier divisor $D$. Then $J$ corresponds to $-D$. It's a fact (first pointed out to me by Tommaso de Fernex) that $D$ is Cartier if and only if

$$O(D) \cdot O(-D) = I\cdot J = \langle g \rangle$$

is principal.

That doesn't help...

But you say, this doesn't help us at all (we have another ideal we need to check whether or not it is principal)!

The point however, is that we know that the reflexification of $I\cdot J$ is principal! (Recall reflexification just means applying $Hom(\bullet, R)$ twice, Macaulay2 can do this for instance).

On the other hand $I \cdot J$ agrees in codimension 1 with its reflexification (since reflexification won't change anything in codimension 1 for a normal domain).

It follows that $I\cdot J$ is principal if and only if $I\cdot J$ is reflexive.

Proposition: Hence $I$ is locally principal if and only if $I \cdot J$ is reflexive.