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Let $R$ be a commutative ring, let $M$, $N$ and $P$ be $R$-modules, and let $N' \subseteq N$ and $P' \subseteq P$ be submodules. Let $\mu:M\times N \to P$ be a surjective bilinear map. Define the module quotient of $P'$ by $N'$ with respect to $\mu$ to be the submodule $(P' : N')_\mu = \{x \in M \mid \mu(x, N') \subseteq P'\} \subseteq M.$

Question: With the notation as above, suppose further that $M$, $N$ and $P$ and the submodules $N'$ and $P'$ are all projective. Is $(P' : N')_\mu$ necessarily projective? If not, how can we strengthen the conditions on $R$ and $\mu$ so that it is?

If $R$ is a PID for example, then projective modules and free modules coincide, and submodules of free modules are free; so the result is clear. This gives an "upper bound" on the strength of conditions needed to impose on $R$ to obtain the result. I would like to know for what more general class of rings $R$ the result holds.

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    Yeah, Kaplansky's theorem is a good point, but local rings can have big dimension (even infinite, I'm pretty sure, say a localization of the entire functions...) and in particular there's no reason a submodule of a projective/free ought to be a direct summand.2012-10-16

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