Let $R$ be a ring with 1, $M$ be a left $R$-module. Then $M$ is fp-injective if every $R$-homomorphism from a finitely presented left ideal to $M$ extends to a homomorphism of $R$ to $M$ i.e. if $\operatorname{Ext}^1(R/I, M) = 0$ for every finitely presented left ideal $I$. And $M$ is fp-projective if $\operatorname{Ext}^1(M, N) =0$ for any fp-injective left $R$-module $N$. It is known that $R$ is left Noetherian iff every left $R$-module is fp-projective. It is known also that if $R$ is left Noetherian, then every fp-injective left $R$-module is injective.

My conjecture is that if $R$ is a left Noetherian, and every cyclic left $R$-module embeds in projective, then $R$ is a left self-injective. I was trying solve this question by using the concept of fp-projectivity, but I didn't solve it. Could someone give the answer?


There's a clue to a counterexample at the end of Section 5 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra 5, 203-221 (1967). There are rings $R$ that have only three left ideals, $0$, $\text{rad}(R)$ and $R$, with $R/\text{rad}(R)\cong\text{rad}(R)$ as left modules, but which are not right Artinian. Such a ring clearly satisfies the conditions of your conjecture, but can't be quasi-Frobenius as it is not right Artinian.

For an example of such a ring, let $K$ be a field that has an embedding $\alpha:K\to K$ so that $K$ is an infinite degree field extension of $\alpha(K)$ (e.g., a field of rational functions in infinitely many variables), and let $R=K\oplus J$, where $J$ is a square zero ideal isomorphic to $K$ as an abelian group, with $K$ acting on the left in the natural way and on the right via $\alpha$.

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    $\begingroup$ @R. Shhaied after "Thank you" note, you can mark "accept". :) you didnt accept any answer till now $\endgroup$ – user 1 Mar 22 '16 at 16:36

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