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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? – Kevin Carlson Oct 16 '12 at 22:08
@KevinCarlson: I forgot to assume that $M$, $N$ and $P$ are also projective. Fixed. – Hamish Oct 16 '12 at 22:11
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. – Kevin Carlson Oct 16 '12 at 22:33
@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? – Hamish Oct 16 '12 at 22:41
Well, heck, we haven't even settled local rings, have we? And, yes, Dedekind domains are actually exactly the hereditary domains. – Kevin Carlson Oct 16 '12 at 22:48

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