# Isomorphism between quotient modules

Is it true for a commutative domain $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$?

-
$F[x,y]/(x) \cong F[y] \cong F[x] \cong F[x,y]/(y)$. But $(x)\neq(y)$. –  Arturo Magidin Jun 22 '12 at 2:38

No, just take a polynomial ring $k[X]$ for some field $k$ (which is a commutative domain), and consider the ideal generated by $X$ and the ideal generated by $X-1$. The quotient rings are both (isomorphic to) $k$.
Problem 22 of Chapter 4 of Steven Roman's Advanced Linear Algebra asks to prove this question in the affirmative when $R/I\simeq R/J$ are isomorphic as $R$-modules, and then asks about the case when $R/I\simeq R/J$ as rings.
The nice existing answers show it is not necessarily true that $I=J$ when the quotients are isomorphic as rings. However, suppose $R/I\simeq R/J$ as $R$-modules with the standard $R$-module structure.
Let $\tau\colon R/I\to R/J$ be an $R$-module isomorphism. Then for any $j\in J$, $$\tau(j+I)=\tau(j\cdot 1+I)=j\cdot\tau(1+I)=0+J$$ since $\tau(1+I)\in R/J$, and any element of $R/J$ is annihilated by multiplication by elements of $J$. Thus $j+I\in\ker\tau=\{I\}$, the equality following since $\tau$ is injective. So $j+I=I$, thus $j\in I$, and $J\subseteq I$. The reverse containment follows similarly, by looking at $\tau^{-1}$ say. So $I=J$.