In an assignment, I was asked to prove that if $M$ is an $R$ module and $N\subseteq M$ is a submodule such that $M/N$ is finitely generated and free, then there exists a submodule $F$ of $M$ which is isomorphic to $M/N$.
I attempted this independently and I was convinced that I had proved it, but after checking through the solutions I feel like I'm missing something.
So here's what I did:
If $M/N$ is free then it has a basis $B=\{[x_1],[x_2],\dots, [x_t]\}$ where $x_i\in M$. The set $F=\langle x_1,x_2,\dots x_t \rangle$ is obviously a submodule of $M$. Let $\phi : M/N \rightarrow F$, where $\phi (\sum_{i=1}^tr_i[x_i])=\sum_{i=1}^tr_ix_i$. This is well defined because any element can be uniquely expressed as such a sum. I had no trouble proving that this is a surjective homomorphism. To show it's injective, suppose that $\phi (\sum_{i=1}^tr_i[x_i])=\sum_{i=1}^tr_ix_i=0$. Then $[0]=[\sum_{i=1}^tr_ix_i]=\sum_{i=1}^tr_i[x_i]$. This shows that if the image of the class of an element of $M$ is $0$ in $F$, then the class of that element is the class of $0$.
I can't see what's wrong with this, but the solutions don't seem to think this is enough. They say that since $B$ is a basis, and $\sum_{i=1}^tr_i[x_i]=[0]$, then each of the $r_i$'s are $0$, therefore the kernel is trivial. I agree that each of the $r_i$'s is $0$, but I don't see why this is necessary, isn't it enough merely that $\sum_{i=1}^tr_i[x_i]=[0]$.
I know the question seems trivial but I might be over-simplifying it, and missing some understanding. Thanks for any help!
Edit: I'm actually more confused that I thought: I can't see why the $F$ I defined isn't dependent on choice of representative of the basis elements!
The sequence $0 \to N \to M \to M/N \to 0$ of $R$-modules splits, since $M/N$ is free. Hence we have the desired property.
To help you with your actual problem: you certainly don't need the freeness for the injectivity. But however: note that you need the freeness to define the map!
Edit (as a response to your edit where you raised the question about the choice of $F$): Of course $F$ is dependent on the choice of the $x_i$. But we don't require $F$ to be unique. Hence different representatives may yield different submodules, but all of those are isomorphic to the desired quotient.
Short story: we want to show existence and for the existence we allow us to choose some elements which exist.