I want to prove the following $a \Longleftrightarrow d$ in the following questoin:
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 $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$ (i.e., $\exists$ a retraction )
c- There exists a right-split short exact sequence $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$ (i.e., $\exists$ a section )
d- $L$ and $N$ are $R-$submodules of $M,$ $L + N = M$ and $L \cap N = \{0\}. $
My trial is:
$a \implies d$
This follows directly from the definition of direct sum.
$d \implies a$
We will prove this in the following $2$ steps:
1-$d \implies b.$
Proof:
Assume that $L$ and $N$ are $R-$submodules of $M,$ $L + N = M$ and $L \cap N = \{0\},$ and 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 )
Given $m \in M,$ by hypothesis there exist $l \in L$ and $n \in N$ with $m = l + n$ and $L \cap N = \{0\}.$
If also $m = a + b$ for $a \in L$ and $b \in N,$ then $l + n = a +b,$and so $a - l = n - b$ where $(a - l) \in L$ and $(n - b) \in N$ because $L$ and $N$ are modules. But then $(a-l), (n-b) \in L \cap N = \{0\}.$ Which means that $a - l = 0$ and $n - b = 0,$ hence $l = a$ and $n = b.$ Therefore, $l$ and $n$ are uniquely determined by the choice of $m.$
Let $i: L \rightarrow M $ be the inclusion homomorphism. Define $r: M \rightarrow L$ and $p: M \rightarrow N$ by $r(l + n) = l$ and $p(l+n) = n.$
Proving that $r$ and $p$ are homomorphisms.
First.
a- Given $a,l \in L$ and $b,,n \in N,$ we have:
$r((l + n) + (a + b)) = r((l + a) + (n + b)) = (l + a) = r(l + n) + r(a + b)$
b- Given $r' \in R$ and $m\in M,$ then we have:
$r(r'm) = r(r'(l+n)) = r((r'l + r'n)) = r'l = r'r((l + n)).$
Second.
a- Given $a,l \in L$ and $b,,n \in N,$ we have:
$p((l + n) + (a + b)) = p((l + a) + (n + b)) = (n + b) = p((l + n)) + p((a + b))$
b- Given $r' \in R$ and $m\in M,$ then we have:
$p(r'm) = p(r'(l+n)) = p((r'l + r'n)) = r'n = r'p((l + n)).$
So, $r$ and $p$ are $R-$module homomorphisms.
Proving that $\operatorname{ker} p = \operatorname{im} i.$
$p(m) = p((l+n)) = n = 0_{N}$ iff $m = l \in L.$ So, $\operatorname{ker} p = L + 0_N = i(L).$ Therefore the sequence $$0 \longrightarrow L \stackrel{i}{\longrightarrow} M \stackrel{p}{\longrightarrow} N \longrightarrow 0$$ is exact.
Moreover, $r$ is a retraction because $(r \circ i)(l) = r(i(l)) = r((l + 0_N)) = l$ i.e.$(r \circ i) = id_L$ as required.
2-$b \Longleftrightarrow a.$ (proved previously)
Then from $1$ and $2$ the required is proved.
My questions are:
1- Is my proof correct?
2- Do I have to show that elements of $L$ commutes with elements of $N$ (like in case of direct product) and if so, why?
3- When should I take the element of $M$ as an ordered pair and when should I take it as addition of two elements in $L$ and $N$?
Could anyone help me answer these questions please?
EDIT
My answers to my questions:
1- I guess yes.
2- I guess no because addition is commutative.
3- I do not know the answer to this.
Are my answers to myself correct?
It is not true that $M \cong L \oplus N$ implies that $L$ and $N$ are submodules of $M$. I suggest you to change $(d)$'s statement to something like:
Now, to prove $(a) \!\!\implies\!\! (d)$, consider an isomorphism $\phi : L \oplus N \to M$ and take $L' = \{\phi(l,0) : l \in L\}$ and $N' = \{\phi(0,n) : n \in N\}$. On the other hand, to prove $(d) \!\!\implies\!\! (a)$ consider the map $\phi : L' \oplus N' \to M$ given by $\phi(l,n) = l+n$, and conclude that $M \cong L' \oplus N' \cong L \oplus N$.
So, since you already done $(a) \!\!\iff\!\! (b)$ and $(a) \!\!\iff\!\! (c)$, you already have the difficult (and interesting) parts.