Isomorphism of quotients implies isomorphism of the denominator

Question

Let $A$ be a commutative ring and $I \subset J$ two ideals of $A$. Let's suppose we have $A/I \simeq A/J$ as $A$-modules. Using the second isomorphism theorem can we deduce that $I \simeq J$ ?

I think it is true but i would like to check it (or find a counterexample).

Answer 1

It is important to clarify whether you are requiring $A/I\simeq A/J$ as $A$-modules, or as just abelian groups.

If you are considering isomorphism as abelian groups, then Angry_Math_Person provides a counter-example.

However, $A/I\simeq A/J$ as $A$-modules implies $I=J$. Indeed, suppose $I\subset J$ and $\varphi\colon A/I\to A/J$ is an $A$-module isomorphism. For any $x\in J$, we have $\varphi(x)=x\varphi(1)=0\in A/J$ so since $\varphi$ is injective, we have $J=I$.

Answer 2

What about $A$ = infinite product of $\mathbb Z$. $I= \mathbb Z$ and $J = \mathbb Z ^2$?