If $R$-Module M has a submodule $N$ such that $N/M$ is free, then there exist a submodule $L$ such that $M = N \oplus L$

Question

I know you can prove this using the splitting lemma but I was thinking about a proof using the snake Lemma.

$0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0$ and $0 \longrightarrow N \longrightarrow N \oplus M/N \longrightarrow M/N \longrightarrow 0$ are both exact with the obvious homomorphisms and its just as obvious that $N \cong N, M/N \cong M/N$.

So if I found a Homomorphism $\varphi :M \longrightarrow N \oplus M/N$ that commutates i.e. $\varphi(n) = (n, 0+N) \forall n \in N$ and $\varphi(m) = (n, m+N)$ for some $n\in N$, then $\varphi$ must be an isomorphism and $L = \varphi^{-1}(\{(0, m+N)|m\in M\})$.

I'm having some trouble actually describing $\varphi$ im guessing the fact that $M/N$ is free, makes sure $\varphi$ is well defined. So my question is, what is $\varphi$?

Answer 1

In fact, any exact sequence $0 \longrightarrow N \stackrel j\longrightarrow M \stackrel r\longrightarrow F \longrightarrow 0$ of $R$-modules, with $F$ free, splits.

Proof: If $B$ is a basis for $F$, note that surjectivity of $r$ implies that for each $b \in B$ there is $m_b \in M$ with $b = r(m_b)$, so that the homomorphism $i : F \to M$ given by $b \mapsto m_b$ (for $b \in B$) is a section of $r$, that is, $r \circ i = \operatorname{id}_F$. Thus, by the splitting lemma, the submodule $\ker r = \operatorname{im} j$ of $M$ is a direct summand.

In your case, take $F = M/N$, $j$ the inclusion $n \mapsto n$, and $r$ the canonical projection $m \mapsto m+N$, so that $\ker r = \operatorname{im} j = N$.

If you're still worried about how is $\varphi$ defined, at the end of the proof consider this:

Note that $j$ and $i$ induce a homomorphism $\varphi : N \oplus F \to M$ that maps $(x,y) \in N \oplus F$ to $j(x) + i(y)$. Now, $\varphi$ is an isomorphism: the function $\psi : M \to N \oplus F$ that maps $m \in M$ to $(n_m,r(m))$, where $n_m$ is the unique element in $N$ such that $j(n_m)=m-i(r(m))$ (why such an element exists?), is a homomorphism with the property that $\psi \circ \varphi = \operatorname{id}_{N \oplus F}$ and $\varphi \circ \psi = \operatorname{id}_M$. Thus $M = \operatorname{im}j \oplus \operatorname{im}i$.