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

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 .

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.
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 $$R\not\cong S$$.
• These are not isomorphic as $A$-modules. Note that $\mathrm{Ann}(A/R)=R$ while $\mathrm{Ann}(A/S)=S$. But, annihilators of isomorphic modules coincide. – David Hill Oct 9 '18 at 17:29