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

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    $\begingroup$ If $R$ is implicitly assumed commutative, the tag ac.commutative-algebra should be used. If not, perhaps it could be specified too. $\endgroup$ – YCor Jun 6 '16 at 13:35
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    $\begingroup$ @YCor: The wording "right R-module" usually indicates that one is in the non-commutative setting. $\endgroup$ – Todd Leason Jun 6 '16 at 19:07
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    $\begingroup$ I think this is an interesting and nontrivial question. It would probably get a better reception if you added some context/motivation (why local?, result essentially clear when $R$ is commutative, for what rings do you have a positive answer, ...) $\endgroup$ – Frieder Ladisch Jun 7 '16 at 13:43
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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 $I\cdot f(t)=J$.

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  • $\begingroup$ Thank you. But under commutative condition over the ring $R$, it will be clear and then we do not need local condition. But my question is in non commutative case.Thanks $\endgroup$ – Najmeh Dehghani Jun 7 '16 at 7:58
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    $\begingroup$ Do you know of examples of a local ring $R$ with a right (or left) ideal $I \leq R$ and a unit $u\in R^*$ such that $uI < I$ ($Iu < I$) (strict containment)? I would guess that this is impossible, but I can't show it. $\endgroup$ – Frieder Ladisch Jun 7 '16 at 13:39
  • $\begingroup$ Your proof in the Edit gives more precise results that could be helpful in the local case. But for the record let me note: Every surjective endomorphism of a left noetherian module over any ring (I think not even an identity is needed) is an isomorphism. $\endgroup$ – Todd Leason Jun 8 '16 at 10:33
  • $\begingroup$ @Todd Leason: Thanks, I was not aware of that. After searching I found that modules with that property are called Hopfian and your statement is a very popular homework assignment. I also constructed a counterexample to the original problem. See edit 2. $\endgroup$ – Friedrich Knop Jun 8 '16 at 18:04
  • $\begingroup$ Dear all, Thank you so much for your comments on my question. In fact under right Noetherian condition over the ring R, the result will be clear also. To see this, let R be a right Noetherian ring. Then every finitely generated module is Noetherian. Note that every Noetherian module is Hopfian (An R-module M is called Hopfian if every epimorphism on M is an isomorphism). So this imply that over a right Noetherian ring R, two finitely generated R-modules which are epi-morphic images of each others are isomorphic. $\endgroup$ – Najmeh Dehghani Jun 9 '16 at 10:10

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