Let $\mathcal{I}, \mathcal{J}$ be ideals of the commutative ring $\mathcal{A}$. It's a well know fact that:
$\mathcal{I} + \mathcal{J}= A \Rightarrow \mathcal{I} \mathcal{J} = \mathcal{I} \cap \mathcal{J} $.
What can I say about the vice versa?
Let $\mathcal{I}, \mathcal{J}$ be ideals of the commutative ring $\mathcal{A}$. It's a well know fact that:
$\mathcal{I} + \mathcal{J}= A \Rightarrow \mathcal{I} \mathcal{J} = \mathcal{I} \cap \mathcal{J} $.
What can I say about the vice versa?
Consider the case where $\mathcal{I}=(0)$ and $\mathcal{J}$ is a proper ideal of $\mathcal{A}$.
Then $\mathcal{IJ}=\mathcal{I}\cap\mathcal{J}=(0)$, but $\mathcal{I}+\mathcal{J}=\mathcal{J}\neq\mathcal{A}$.
in $\mathbf Z[x_1,x_2,x_3]$ consider the ideals $(x_1)$ and $(x_2)$