Example of a ring $R$ and ideals $A$ and $B$ of $R$ such that $AB\neq A\cap B$.

Question

Give an example ofa ring $R$ and ideals $A$ and $B$ of $R$ such that $AB\neq A\cap B$. I know that $AB\subset A\cap B$. But I can't find an example such that $AB\neq A\cap B$.

Answer 1

Take $R=\mathbb Z$, $A=(6)$, $B=(10)$. Then $AB=(60)$ but $A\cap B=(30)$.

In general, if $A=(m)$ and $B=(n)$ then $AB=(mn)$ and $A\cap B=(lcm(m,n))$ and so they are equal iff $\gcd(m,n)=1$.