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Suppose we are given a finitely generated free module $M$ over a ring $R$. Assume $N \subseteq M$ is a free submodule of $M$.

If $B$ is a basis for $M$, does it follow that there exists a subset $A \subseteq B$ such that $A$ is a basis for $N$?

As stated, I'm pretty sure the question is incorrect: Just take $R=M=\mathbb{Z}$. Then $2 \mathbb{Z}$ is a free submodule of $M$ but the basis $\{1\}$ of $M$ does not contain a basis for $2 \mathbb{Z}$.

Are there any cases in which this is true?

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This is not even true for vector spaces, which is just about the nicest possible case. Take $k$ a field, $M = k^2$ with basis $e_1, e_2$, and $N = \text{span}(e_1 + e_2)$.

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  • $\begingroup$ Oh of course, makes perfect sense $\endgroup$
    – Exit path
    Feb 11, 2016 at 4:55
  • $\begingroup$ Nevertheless, thinking backwards, in the vector space case, you can always extend the subspace basis to a basis of the whole space. This is not possible in the general case of free modules. $\endgroup$
    – MooS
    Feb 11, 2016 at 5:56
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This never holds if $R$ is a nonzero ring and $B$ has more than one element: just take any two distinct elements $a,b\in B$, and let $N$ be the submodule generated by $a+b$.

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