Here is the question I want to prove so far:
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- $M \cong_{R} L \oplus N.$
c- There exists a right-split short exact sequence $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$
My trial is as follows:
$a \implies c.$
Also, using the fact that the direct sum of a finite number of modules coincides with their direct product we will prove the required.
Assume that $M \cong_{R} L \oplus N.$ We want to show that there exists a right-split short exact sequence $$0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$$ i.e., $\exists$ a section $s: N \rightarrow{M}$ s.t.$$p \circ s = id_N \quad (2)$$ Where $p: M \rightarrow N $ and it is onto. So we need to find the functions $s,p$ that satisfies $(2)$ 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 $i(l) = (l,0_N)$ and $p: L \oplus N \rightarrow N$ the projection $p(l,n) = n.$
Also, if we define $s: N \rightarrow L \oplus N $ with $s(n) = (0_L,n)$ then we have $(p\circ s)(n) = p ((0_L,n)) = n $ as required.
Now, we want to verify that $s$ is an $R-$module homomorphism:
Let $n,n_1, n_2 \in N$ and $r\in R,$ then we have the following:
1- $s((n_{1} + n_{2})) = (0_L, n_{1} + n_{2})) = (0_L, n_{1}) + (0_L, n_{2}) = s(h_{1})s(h_{2}).$
Where the second equality is by definition of multiplication in the direct product.
2- $s(rn) = (0_L, rn) = r(0_L, n) = rs(n).$
And hence $s$ is an $R-$module homomorphism as required.
$c \implies a$
I do not know exactly how to define the isomorphism in that case, I am just guessing that it should be $\varphi :=(i,s): L \oplus N \rightarrow M$ for the following ses:
$$0 \longrightarrow L \stackrel{i}{\longrightarrow} M \stackrel{p}{\longrightarrow} N \longrightarrow 0$$
Where $i$ is injective and $p$ is surjective. Is my isomorphism function $\varphi$ defined correctly?
If so, I know that $\varphi $ is an $R-$module homo. (I can easily prove that), but how can I show that it is injective and bijective?
Could anyone help me in answering those questions please?
Define the $R$-module homomorphism $\varphi : L \oplus N \to M$ by $\varphi(l,n) = i(l) + s(n)$. Note that if $\varphi(l,n)=0$, then $$\begin{align} 0 &= p(\varphi(l,n)) \\ &= p(i(l))+p(s(n)) \\ &= p(s(n)) = n \end{align}$$ ($p \circ i = 0$ by exactness) and then $0=\varphi(l,n)=i(l)$ implies $l=0$ (since $i$ is injective). Hence $$\varphi(l,n)=0 \implies (l,n)=(0,0)$$ which means that $\varphi$ is injective. Can you now figure out how to prove the surjectivity of $\varphi$?