# On maximal submodules of projective modules

I know that any non-zero projective module has a maximal submodule. But is it true that any proper submodule is contained in a maximal submodule !?

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No. Let $R=\mathbb{Z}$ and consider $M=\mathbb{Q}$. There is a projective module $F$ and surjective homomorphism $f:F\to M$ (for example, take $F=R^{(M)}$ and $f(e_m) = m$). The submodule $\ker(f) \leq F$ is not contained in any maximal submodule, since the lattice of submodules containing $\ker(f)$ is isomorphic to the submodule lattice of $M$, and the submodule lattice of $M$ contains no (proper) maximal elements.

Thanks for the nice question. I was surprised by the answer.

A related question: Well which rings work?

Proposition: The following are equivalent:

• Every proper submodule of a projective module is contained in a maximal submodule
• Every proper submodule of a module is contained in a maximal submodule
• Every module has a maximal submodule

Proof: The second clearly implies both of the others. Let $K < M$. Then there is a free module $F$ and surjective homomorphism $f:F \to M$. Let $L=f^{-1}(K)$. In the first case, we can find some maximal submodule $X$ with $L \leq X \lessdot F$. Then $K \leq f(X) \lessdot M$, and so the first case implies the second. In the third case, the module $M/K$ contains a maximal submodule $L/K \lessdot M/K$. Then $K \leq L \lessdot M$, so the third case implies the second. $\square$

In the commutative case, Hamsher studied these rings, first (1966) in the noetherian case, and then (1967) in the general case. I’m reading through these now, and might read the noncomm results too. Let me know if you are interested.

• Hamsher, Ross M.
“Commutative, noetherian rings over which every module has a maximal submodule.”
Proc. Amer. Math. Soc. 17 (1966) 1471–1472.
MR200303
DOI:10.2307/2035767
• Hamsher, Ross M.
“Commutative rings over which every module has a maximal submodule.”
Proc. Amer. Math. Soc. 18 (1967) 1133–1137.
MR217059
DOI:10.2307/2035815