If $A/I \cong A/J$ as rings and $I\subseteq J,$ then $I=J.$ [duplicate]

Let $A$ be a commutative Noetherian ring and let $I$ and $J$ be ideals of $A.$

Suppose that $I\subseteq J$ and that $A/I \cong A/J$ as rings.

I want to prove that $I=J.$

Observations so far:

1) If we drop the hypothesis $I\subseteq J,$ then the result is false:

$$\mathbb{Q}[X,Y]/(X)\cong \mathbb{Q}[X,Y]/(Y) \text{ but } (X)\neq (Y).$$

2) If $A/I\cong A/J$ as $A$-modules, then $I=J$:

$$I=\mathrm{Ann}_A(A/I)=\mathrm{Ann}_A(A/J)=J.$$

I'm now a bit stuck as to how to proceed. Any hints would be most helpful!

Many thanks :)

Solution 1:

Assume that $I\subsetneq J$. By induction, we can construct an infinite ascending chain of ideals as follows:

• Put $J_0=I$ and $J_1=J$. Then $J_0\subsetneq J_1$ and $A/J_0 \cong A/J_1$.
• Assume that for a certain integer $n$, we have ideals $J_n$ and $J_{n+1}$ such that $J_n\subsetneq J_{n+1}$ and $A/J_n \cong A/J_{n+1}$. Via this isomorphism, the non-trivial ideal $J_{n+1}/J_n$ of the left-hand side corresponds to a non-trivial ideal of the right-hand side. This ideal has the form $J_{n+2}/J_{n+1}$, where $J_{n+2}$ is an ideal of $A$ with $J_{n+1}\subsetneq J_{n+2}$. Moreover, $$ A/J_{n+2} \cong (A/J_{n+1})/(J_{n+2}/J_{n+1}) \cong (A/J_{n})/(J_{n+1}/J_n)\cong A/J_{n+1}. $$

By induction, we have constructed an infinite sequence of ideals $$J_0\subsetneq J_1\subsetneq J_2\subsetneq \ldots, $$

so $A$ is not Noetherian, a contradiction.