Can I get help on showing this?
Let $D$ be a Dedekind domain, with non-zero ideals $I, J$. Prove that the multiplication map $I \otimes_D J \to IJ$ is an isomorphism. Show that this can fail when $D$ is not Dedekind: for example, if $k$ is a field, and $R = k[x,y]/(y^2 - x^3)$ show that if $I = (x, y) \subset R$, then $I \otimes_R I \to I^2$ is not an isomorphism.
I feel like this should be relatively straightforward, but I'm struggling to do this. I tried to do it directly using the definition of the tensor product, but was not making progress.
The assertion is local, hence we can assume $D$ to be a PID and in particular $I,J$ are free $D$-modules.
Clearly the map is surjective. To show injectivity, consider the inclusion map $I \hookrightarrow R$ and tensor with $J$. Since $J$ is free, we get an injection $I \otimes_R J \hookrightarrow R \otimes_R J \xrightarrow{\cong} J$, which is given by $a \otimes b \mapsto a \otimes b \mapsto ab$. So this is just the multiplication map. Thus we have shown that the multiplication map is injective.
In the given counterexample, note that $y \otimes y$ and $x^2 \otimes x$ have the same image.