Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $\mathcal{I}+\mathcal{J}=\mathcal{J}\neq\mathcal{A}$.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.