Let $R$ be a commutative ring with $1$. We know that every finitely generated $R$-module has a maximal proper submodule. Is it true for any free $R$-module? In particular, can we do the following:
Take a basis $\mathfrak{B}$ of the free module $M$ and remove one element $x \in \mathfrak{B}$ from it to get a proper subset $\mathfrak{B}\setminus \lbrace x\rbrace$. Then the claim is that Span$( \mathfrak{B} \setminus \lbrace x \rbrace )$ generates a maximal proper submodule -- however, I cannot show that it is maximal. Any ideas?
Yes, the idea of focusing on what happens with a single basis element is sound, sort of.
Let's say $M\cong \oplus _{i\in I}R$. You can project any of the coordinates (say the first one) onto $R$, yielding a surjective homomorphism $\phi :M\to R$. (This is basically the same as picking a basis element $x$ and making a mapping from $M$ to $R$ by projecting onto the coordinate of $x$.)
We know $R$ has maximal right ideals, so by correspondence of submodules, there's a maximal submodule of $M$ containing $\ker(\phi)$ that is the preimage of a maximal right ideal of $R$. This is a maximal submodule of $M$.
Actually, more is true:
This is not as simple as the free case, so we'll pass on a proof for now.
Added for reference request in comments: In the interest of saving time I did a quick google search and found that Andrew Hubery had found a reference for someone with the same question. He gave:
I also can't help but mention this neat answer by Jack Schmidt concerning the question "Must every proper submodule of a projective module be contained in a maximal submodule?" The surprising answer is apparently, "no!"