Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Suppose that $R$ is a ring, $I$ and $J$ are ideals in $R,$ and $R/I\cong R/J$ as rings. When does $I\cong J$ as $R$-modules hold?

share|cite|improve this question
And an example where they are not even isomorphic: $R=\mathbb Z_2\times\mathbb Z_4$ with $I=\mathbb (1)\times(2)$ and $J=(0)\times(1)$. – Henning Makholm Dec 15 '12 at 17:39
In the case when $R$ is noetherian, and one of the ideals is contained in the other, we do have an equality. – Hmm. Dec 31 '15 at 18:55
up vote 4 down vote accepted

We can also consider $$\left(\prod_{n=1}^\infty\mathbb{Z}\right)/\mathbb{Z}\cong\left(\prod_{n=1}^\infty\mathbb{Z}\right)/(\mathbb{Z}\times\mathbb{Z}).$$ Clearly $\mathbb{Z}\not\cong\mathbb{Z}\times\mathbb{Z}$ since $\mathbb{Z}$ only has one generator while $\mathbb{Z}\times\mathbb{Z}$ has two generators.

share|cite|improve this answer

$\Bbb R[X,Y]/(X)\simeq\Bbb R[X,Y]/(Y)$ but $(X)\neq(Y)$. The question should be if $I$ and $J$ are isomorphic as $R$-modules.

share|cite|improve this answer
Now the question is as you wished. Unfortunately, the answer has became obsolete. – user26857 May 5 at 14:45

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.