Let $R$ be a commutative ring. For $R-$modules $L,M,N$ show that the following conditions are equivalent.(all functions are $R-$ module homomorphisms.)
a- $M \cong_{R} L \oplus N.$
b- There exists a left-split short exact sequence $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$
Here is my trial:
$a \implies b.$
Let $R$ be a commutative ring. And let $L,M,N$ be $R-$modules.
Assume that $M \cong_{R} L \oplus N.$ We want to show that there exists a left-split short exact sequence $$0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$$ i.e., $\exists$ a retraction $r: M \rightarrow{L}$ s.t.$$r \circ i = id_L \quad (1)$$ Where $i: L \rightarrow M $ and it is injective. So we need to find the functions $r,i$ that satisfies $(1)$ above.
So, since $M \cong_R L \oplus N,$ we can say that we have $0 \rightarrow L \rightarrow L \oplus N \rightarrow N \rightarrow 0,$ with $i: L \rightarrow L \oplus N$ the embedding $i(l) = (l,0)$ and $p: L \oplus N \rightarrow N$ the projection $p(l,n) = n.$
If we define $r: L \oplus N \rightarrow N $ with $r(l,n) = l$ then we have $(r\circ i) (l) = r((l,0)) = l$ as required.
Now, we want to verify that $r$ is an $R-$module homomorphism:
1- $r((l_{1},n_{1}) + (l_{2}, n_{2})) = r((l_{1} + l_{2}, n_{1} + n_{2})) = l_{1} + l_{2} = r((l_{1}, n_{1})) + r((l_{2},n_{2}))$ where the first equality is by definition of addition in direct sum.
2- Let $r' \in R$ and let $m \in M,$ then $r(r'm) = $ but then I do not know how to complete. could anyone help me in this step please? also, is my solution correct?
Also I do not know how to prove the reverse direction, could anyone help me in this please?
Given $r' \in R$ and $(l,n) \in L \oplus N$, $$r(r'(l,n)) = r((r'l,r'n)) = r'l = r'r((l,n)),$$ and this shows that $r(r'm) = r'r(m)$ for every $r' \in R$ and $m \in L \oplus N$. Now, for $b \!\implies\! a$, tag the left-split short exact sequence $0 \to L \to M \to N \to 0$ as $$0 \longrightarrow L \stackrel{i}{\longrightarrow} M \stackrel{p}{\longrightarrow} N \longrightarrow 0$$ and let $r : M \to L$ be the $R$-module homomorphism such that $r \circ i = \operatorname{id}_L$. Show that the induced homomorphism $(r,p) : M \to L \oplus N$ given by $(r,p)(m) := (r(m),p(m))$ is an isomorphism using the fact that $r$ and $p$ are surjective (why?) and that $\ker r \cap \operatorname{im} i = 0$.