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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?

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Could you have something like $\mathcal{I}=(0)$, and $\mathcal{J}$ a proper ideal of $A$? –  Noomi Holloway Dec 22 '12 at 9:09
    
If possible I would accept this comment as answer. –  Ivan Dec 27 '12 at 12:37
    
I've added the comment as an answer now. –  Noomi Holloway Dec 27 '12 at 21:00

2 Answers 2

up vote 3 down vote accepted

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 $\mathcal{I}+\mathcal{J}=\mathcal{J}\neq\mathcal{A}$.

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in $\mathbf Z[x_1,x_2,x_3]$ consider the ideals $(x_1)$ and $(x_2)$

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let me know if you got it –  Koushik Dec 22 '12 at 10:03
    
Noomi's idea was already enough. –  Ivan Dec 22 '12 at 10:56

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