Prove $M/N$ is free implies $M \cong N \oplus M/N$ I am asked to prove that if $M/N$ is free then $M \cong N \oplus M/N$, where M is a module over a ring, R and $N \leq M$.
My first thought is that if $S$ is the basis of $M/N$, then we can consider the function $\phi: S \rightarrow M$ via $\phi(s_i + N) = s_i$ which then extends to some module homomorphism $\Phi: M/N \rightarrow M$. 
However I don't really know what to do beyond this point. I thought maybe I could have a function $\Theta$ that maps from $N \times M/N$ to $M$ and that $\Theta(n, s + N) = \theta(n) + \Phi(s + N)$ for some appropriate function $\theta$ 
If I could then show that this is a bijective function then I'd be done. However, if this is the correct approach I am unsure of how I can construct such a function and how to show it'd be bijective. I am really thinking that there's something simple I've missed but I am unsure how to proceed with this question. 
Thank you for any help you might be able to offer.
 A: If $M/N$ is free, then the exact sequence
$$0 \to N \to M \to M/N \to 0 $$
splits, since you can define manually a map $f:M/N \to M$ on the basis of $M/N$ such that $\pi \circ f=Id$.
A: Hint: If $M/N$ is generated by the basis $(x_1 + N,..., x_n + N)$, then what can you say about $\phi(n, x) = n+ \displaystyle\sum_{i=0}^n \lambda_i x_i$, where $x = \displaystyle\sum_{i=0}^n \lambda_i (x_i + N)$ ? 
A: I try to give a Naive method here.
Lemma 
Suppose $P_0$ is the subset of $M$ such that $P_0/N$ is a basis of $M/N$, we define $P$ as the minimal set of $P_0$, i.e., $P/N=P_0/N$ and
$$p+N=p'+N\iff p=p',\quad \forall p,p'\in P.$$
Then 
$$A:=(\{0\}\times (P/N))\cup(N\times \{N\})$$
is a basis of $B:=N\oplus M/N$.
Proof
Given any $n\in N,m\in M$, 
\begin{align*}
  (n,m+N)&=(n,N)+(0,m+N)\\
  &=(n,N)+(0,\sum_i \lambda_i(p_i+N))\\
  &=(n,N)+\sum_i \lambda_i (0,p_i+N),
 \end{align*}
    thus $A$ generate $B$. Suppose
    \begin{align*}
  (0,N)&=(n,N)+\sum_i \lambda_i (0,p_i+N)\\
  &=(n,\sum_i \lambda_i(p_i+N)),
 \end{align*}
    we know $n=0,\lambda_i=0$ since $P/N$ is a basis of $M/N$.
Now back to the original question.
Proof
Define $\phi:A\to M$ as
$$\phi(0,p+N)=p,\quad \phi(n,N)=n,$$
this is well-defined since $P$ is minimal. Extends it to $\theta: B\to M$ where $\theta$ is a module homomorphism. Easy to check $\theta$ is surjective. Now Suppose $(n,m+N)\in\ker(\theta)$, 
\begin{align*}
  \theta(n,m+N)&=\sum_i \lambda_i\,\theta(0,p_i+N)+\theta(n,N)\\
  &=\sum_i\lambda_i\,p_i+n\\
  &=0.
\end{align*}
Thus $\sum_i \lambda_i\,(p_i+N)=N$ and thus $\lambda_i=0$ and $n=0$. So $m+N=N$ and 
$$\ker(\theta)=\{(0,N)\}$$
is trivial. So done.
