Let $M$ be a finitely generated $R$-module, where $R$ is an integral domain. Let $N$ be a submodule of $M$. Let $B'$ be a maximal linearly independent subset of $M/N$ and define $B=\{x\in M \mid x+N\in B'\}$. I want to show that $|B|=|B'|$.
Consider the following function: $f:B\rightarrow B', f(b)=b+N$. If $b\in B$, then by definition $b+N\in B'$. If $a=b$, then $a-b=0\in N$ so that $a-b+N=N$ and thus $a+N=b+N$. So we have shown that $f$ is well defined and defined everywhere.
Suppose $y\in B'$ then there exists $b\in B$ s.t. $b+N=y$ and so $f$ is surjective.
I am having problems showing that $f$ is injective. I tried to start it but I get nowhere fast. If $f(a)=f(b)$ then $a+N=b+N$ and so $a-b+N=0$. I am not sure what to do next. I am guessing that the linearly independence of $B'$ will come into play but I can't figure out how.
I would love some help.
Your map $f$ is never injective if $N \neq 0$, since its fibres have cardinality $|N|$ by the virtue of $f(b)=f(b+n)$ for all $n \in N$. Thus $|B| = |B'| \cdot |N|$, which is greater than $|B'|$ whenever $N \neq 0$, since $|B'|$ is finite.
However, by picking a set of representatives of $B'$ in $M$, you can achieve the map to become injective without losing the surjectivity.