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If $R$ is a left noetherian ring, then every finitely generated left $R$-module $M$ is noetherian, and hence every proper submodule of $M$ is contained in some maximal submodule of $M$.

Is it possible to weaken the hyphotesis? I.e. when can we assure that a left (non-zero) $R$-module has a maximal submodule?

Suppose now that $R$ is left artinian ring with identity. I was wondering, for instance, if it was true that (non-zero) left $R$-modules of finite projective dimension have some maximal submodule. If this latter statement is not true, would it be true if the base ring was an Artin algebra?

I am asking this to try to solve the problem I posted in "Artinian rings with zero finitistic dimension" in full generality.

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up vote 2 down vote accepted

I think you would be interested in On rings over which each module has a maximal submodule by Y. Hirano Communications in Algebra Volume 26, Issue 10, 1998.

When H. Bass was investigating perfect rings, he found that for a right perfect ring $R$, every nonzero right $R$ module has a maximal submodule, and that $R$ does not contain infinite sets of orthogonal idempotents. He conjectured that the converse might also be true.

The conjecture turned out to be true for commutative rings (Hamsher, cited in the above paper) but false in general (Cozzens, Koifmann, also cited in the paper above).

Since a right Artinian ring does not admit infinite sets of orthogonal idempotents, and is right (and left) perfect, you're right that the modules of right Artinian rings have maximal submodules.

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