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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?

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  • $\begingroup$ Compare $R$ - module and $R-$ module. $\endgroup$ – Pedro Tamaroff Aug 28 '14 at 4:38
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    $\begingroup$ @Pedro: And the spacing-correct, $R$-module. $\endgroup$ – Asaf Karagila Aug 28 '14 at 4:48
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    $\begingroup$ Take the quotient $M/(\mathrm{Span}(\mathfrak{B}\setminus \{x\}))$, this will be isomorphic to $R$, while a submodule $N$ of $M$ is maximal if and only if $M/N$ is simple. Your idea works iff $R$ is simple as an $R$-module. $\endgroup$ – Simone Aug 28 '14 at 8:39
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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:

Every nonzero projective module has a maximal submodule.

This is not as simple as the free case, so we'll pass on a proof for now.

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!"

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  • $\begingroup$ Why wouldn't this idea work for an arbitary $R$-module $M$. Since every $R$-module $M$ has a surjection from some free module $F(\mathfrak{S})$ where $\mathfrak{S}$ is the set of generators of $M$,we have $\pi \colon F(\mathfrak{S}) \to M$,if we can find a maximal submodule in $F(\mathfrak{S})$ using the above method which also contains the $ker(\pi)$ then we get a maximal submodule of $M$. i guess this will work out if the ring was a PID. $\endgroup$ – user2902293 Aug 29 '14 at 12:46
  • $\begingroup$ @user2902293 While every free module has a maximal submodule, we unfortunately do not have enough control to find maximal submodules containing a proper submodule of our choice! I added an answer I remembered by Jack Schmidt that is related. $\endgroup$ – rschwieb Aug 29 '14 at 12:50

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