Do we have $R\simeq S$ for two submodules $R,S$ of $A^n$?

Question

Let $A$ be a commutative ring with identity. Given two submodules $R,S$ of $A^n$ (where $n\in\Bbb N$), if there exists an isomorphism of $A$-modules $A^n/R\simeq A^n/S$, then do we have $R\simeq S$?

Note that this is definitely false for quotients of non-free modules: see, e.g., Quotient modules isomorphic $ \Rightarrow$ submodules isomorphic or Isomorphy of quotient modules implies isomorphy of submodules .

Answer 1

No.

Let $A=\Bbb Z^\Bbb N=\{\,f\colon \Bbb N\to\Bbb Z\,\}$, $n=1$, $R=\Bbb Z=\{\,f\in A\mid \forall n>0\colon f(n)=0\,\}$, and $S=0$. Then $A^1/R\cong A^1/S\cong A$, but of coure $R\not\cong S$.

Answer 2

This is false even for direct summands. For example, take $R=\mathbb{R}[x,y,z]/(x^2+y^2+z^2=1)$, the co-ordinate ring of the real sphere. Let $P$ be defined as the kernel of the surjective map $R^3\to R$ given by $(x,y,z)$. ($P$ is the tangent bundle of the sphere). Then $P\oplus R\cong R^3$, but $P$ is not free.