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

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 $AB\neq A\cap B$.

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Take $A=B$. In most of reasonable situation, $AB=A^2\ne A$ (find an example with $R=\mathbb Z$). –  Cantlog Oct 12 '13 at 13:21

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 $\gcd(m,n)=1$.