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

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What do you mean by "When"? It does not always hold, but do you have some other assumptions in mind? The ideals $I$ and $J$ need not even be isomorphic. For a quick example where they are not equal, consider $I=\mathbb Z\times \{0\}$, $J=\{0\}\times\mathbb Z$ in $R=\mathbb Z\times\mathbb Z$. – Jonas Meyer Dec 15 '12 at 17:30
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
I meant "What assumption about R,I,J (prime or maximal or others) makes this true?”.I'm especially interested in the case R=A[X_1,...,X_n],A is PID and I and J are maximal ideals.Sorry,I'm not gut at English. – user53216 Dec 15 '12 at 17:50
In the case when $R$ is noetherian, and one of the ideals is contained in the other, we do have an equality. – Soumya Sinha Babu Dec 31 '15 at 18:55

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.

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Thank you.. : ) – user53216 Dec 15 '12 at 18:44

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

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Thanks, I'll correct it. – user53216 Dec 15 '12 at 17:55

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