Existence of a maximal submodule Let $R$ be a left Artinian ring and let $M$ be a nonzero left $R$-module. Prove that $M$ has at least one maximal submodule.
I don't really know where to start. Any hint?
Thanks!
 A: Consider the following fact, which can be found in, e.g., Chap. 17 of Lang's Algebra or elsewhere in the internet:
Let $R$ be (left) Artinian and let $J$ be its Jacobson radical. Then $J$ is nilpotent and $R/J$ is semisimple.
In particular $M/JM\neq 0$. (Otherwise contradicts the nilpotency of $J$.) But $M/JM$ can be seen as an $R/J$-module and hence is semisimple. (Any module over a semisimple ring is semisimple.) A semisimple module has a maximal submodule (pick a simple submodule and take a complement). Thus $M/JM$ has a maximal submodule, which in turn implies $M$ has a maximal submodule.
A: Apologies in advance for this answer. The problem with it is that it is too advanced, and also relies on a lemma that is very similar to your question. I will continue to seek a more elementary answer.
Lemma: Every left $R$ module over a left Artinian ring $R$ has a projective cover.
Lemma: Every nonzero projective module has a maximal submodule.
Lemma: Superfluous submodules are contained in all maximal submodules.
Combining these three, you have that $M\cong P/S$ where $P$ is a projective module and $S$ is superfluous submodule. Then $P$ has a maximal submodule, and it necessarily contains $S$. By submodule correspondence, there is a maximal submodule of $M$.
