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Let $R$ be a Noetherian integral domain, $M$ and $R$-module and $F_1$, $F_2$ free $R$-modules of rank $n$ for which $$F_1 \subseteq M \subseteq F_2$$ Is it true that $M$ is finitely generated? If so, is there a "slick" proof of this (perhaps using the fact that $F_1, F_2$ are projective)?

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Yes, because $F_2$ is necessarily Noetherian, and then so are all of its submodules.

You can even drop the domain condition and the other free module $F_1$.

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    $\begingroup$ As slick and short as possible...+1] $\endgroup$ – DonAntonio Jan 3 '14 at 22:11

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