Cyclic modules over local rings

Question

Let $R$ be a local ring. If two cyclic right $R$-modules are epimorphic images of each other, are these modules necessarily isomorphic?

Answer 1

I assume that $R$ is commutative. Let $M=R/I$ and $N=R/J$ be cyclic and let $\phi:M\to N$ and $\psi:N\to M$ be surjective. Then $\tau:=\psi\circ\phi$ is a surjective endomorphism of $R/I$. In particular, $a:=\tau(1)$ is a generator of $R/I$, i.e., $Ra+I=R$. Thus, there is $b\in R$ with $ba\in 1+I$. Since $\tau$ is multiplication by $a$, multiplication by $b$ is a (left) inverse of $\tau$. Thus, $\tau$ and therefore $\phi$ is injective.

Edit: Here is a proof for non-commutative rings which works under the additional hypothesis that $R$ is left Noetherian. So let $I,J,\phi,\psi,\tau,a,b$ as above. Then $I,J$ are only left ideals. Since $ba\in 1+I$ and $R$ is local, $ba$ is invertible. Replace $b$ by $(ba)^{-1}b$. Then $ba=1$. Since $ab$ is a non-zero idempotent and $R$ is local, we have $ab=1$. Thus $a$ is a unit of $R$. Since $x\mapsto xa$ induces an endomorphism of $R/I$ we have $Ia\subseteq I$. I claim $Ia=I$. Otherwise $$ I\subsetneq Ia^{-1}\subsetneq Ia^{-2}\subseteq\ldots $$ would violate left Noetherianity. The claim implies that also $Ia^{-1}\subseteq I$. Thus, $x\mapsto xa^{-1}$ yields an inverse of $\tau$.

Edit 2: I think, I now have a counterexample. From the arguments above it suffices to construct a local ring $R$ and two left ideals $I,J$ such that there are units $a,b$ with $Ia\subseteq J$ and $Jb\subseteq I$ but there is no unit $c$ with $Ic=J$. Then $R/I$ and $R/J$ are epimorphic images of each other without being isomorphic.

For this, let $k$ be any commutative field, $t,u,v$ transcendental elements and $R:=k(t)\oplus k(u,v)$ with multiplication $$ (f_1(t)+g_1(u,v))(f_2(t)+g_2(u,v))=f_1(t)f_2(t)+[f_1(u)g_2(u,v)+g_1(u,v)f_2(v^2)]. $$ In other words, $R$ is the ring of matrices $$ \left(\matrix{f(u)&g(u,v)\\0&f(v^2)}\right) $$ It is clearly local with maximal ideal $\{f=0\}$. Now put $$ I:=k(u)[v]\text{ and }J:=k(u)[v]\cdot v. $$ Then $I\cdot t=k(u)[v]\cdot v^2\subseteq J$ and $J\subseteq I$ but there is no $f(t)$ with $I\cdot f(t)=J$.