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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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    Since $(P' :N')_\mu$ is a submodule of $M$, on which no hypothesis of projectivitiy was placed, how exactly do you see the partial result you stated?2012-10-16
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    @KevinCarlson: I forgot to assume that $M$, $N$ and $P$ are also projective. Fixed.2012-10-16
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    The simplest generalization of your result is to Dedekind domains, which are exactly the Noetherian domains for which every submodule of a projective module is projective and also those with Krull dimension 1. I have an inkling that you'll be able to make $(P' :N')_\mu$ non-projective once you pass to Krull dimension at least $2$, say looking at $k[x_1,x_2]$, but I don't see how.2012-10-16
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    @KevinCarlson: Thanks for your comment. I didn't know that Dedekind domains are hereditary, that will be useful. Do you have any inklings about the case of semi-local rings?2012-10-16
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    Well, heck, we haven't even settled local rings, have we? And, yes, Dedekind domains are actually exactly the hereditary domains.2012-10-16
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    @KevinCarlson: Over a local ring, projective modules are free (Kaplansky's Theorem), but I think that not all submodules of free modules over local rings are free. In which case you're right, there's still work to be done to establish the result for local rings.2012-10-16
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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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