Can we extend a linearly independent list in a f.g. module of a commutative ring? Let's say we're given a commutative ring $A$ with identity. We are given that $M$ is a f.g. $A$-module. For any elements $m_1,\dots,m_n$ of $M$, we have an $A$-module homomorphism $\phi_{m_1,\dots,m_n}$ from $A^n$ to $M$ defined by $\phi_{m_1,\dots,m_n}(a_1,\dots,a_n)=a_1m_1+\dots+a_nm_n$.
Thus $m_1,\dots,m_n$ becomes a linearly independent list if and only if $\phi_{m_1,\dots,m_n}$ is injective, a span list if and only if $\phi_{m_1,\dots,m_n}$ is surjective. Now we say that $m_1,\dots,m_n$ is a basis for $M$ if and only if $\phi_{m_1,\dots,m_n}$ is bijective.
Suppose that we are given a linearly independent list $n_1,\dots,n_k$ of $M$, we say $n_1,\dots,n_k$ can be extended to a basis of $M$ if $n_1,\dots,n_k,m'_1,\dots,m'_r$ is a basis for $M$ for some $m'_1,\dots,m'_r \in M$.
I came up with this question when I was trying to generalize the following proposition on Sheldon Axler's Linear Algebra Done Right:

Every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space.

 A: This is usually not possible.  For one thing, a typical module $M$ will not have any basis at all, let alone one extended from a basis of $N$.  For instance, take $A=\mathbb{Z}$, $M=\mathbb{Z}/2\mathbb{Z}$, and $N=0$.  Then the empty set is a basis for $N$, but there does not exist any basis for $M$.
Even if $M$ is a free module (so it does have a basis), it is still typically not possible to extend a basis of $N$ to a basis of $M$.  For instance, take $A=\mathbb{Z}$, $M=\mathbb{Z}$, and $N=2\mathbb{Z}$.  Then $\{2\}$ is a basis for $N$, but it cannot be extended to a basis of $M$ even though $M$ is free.
In general, if $N$ is a free module which is a submodule of $M$, then a basis of $N$ extends to a basis of $M$ iff the quotient module $M/N$ is free, in which case you can extend any basis of $N$ to a basis of $M$ by just adding elements of $M$ which are preimages of a basis of $M/N$.
(Note that finite generation is pretty much irrelevant to this entire discussion; indeed, it is an unnecessary hypothesis in the case of vector spaces as well.)
