This is a question of Commutative Algebra.It was given in my class.I was able to solve it to some extend.But I have some doubts. Please help me. Thnx in advance.
$A$ is a commutative ring with identity.$L,M,N$ are all $A$-modules.Now consider the diagram which gives some $A$-module homomorphisms between them.
$$L\xrightarrow{{\alpha}}M\xrightarrow{{\beta}}N$$$$L\xleftarrow{{\rho}}M\xleftarrow{{\sigma}}N$$
Then prove that The Following Are Equivalent:
(a) $M=L\oplus N,\alpha=i_L,\beta=\pi_N,\sigma=i_N,\rho=\pi_L$
(b) $\beta\alpha=0,\beta\sigma=1_N,\rho\sigma=0,\rho\alpha=1_L,\alpha\rho+\sigma\beta=1_M $
Where $i_L(x)=(x,0)\in L\oplus N,i_N(y)=(0,y)\in L\oplus N,\pi_N(x,y)=y\in N,\pi_L(x,y)=x\in L$ and $1_X$ denotes the identity map on $X$
What I have done so far:
I was able to prove (a)$\Rightarrow$(b)
To prove the reverse I managed to show $M\cong L\oplus N$ using the map $f:M\to L\oplus N$ by $f(m)=(\rho(m),\beta(m))$
But now if I take $M=L\oplus N$ (not isomorphic,but exactly same)then how can I prove $\alpha=i_L$ and other equalities?
I was able to show the said equalities via the isomorphism $f$ like $fo\alpha:L\oplus N$ and $fo\alpha=i_L$.
But I could not show without using the isomorphism.
Is there any way to solve this?
Please help me.Thnx again
When you write $M=L \oplus N$, you are saying that $M$ is the direct sum of $L$ and $N$ in the sense of the universal property of the direct sum in the category of $A$-modules.
In fact, since condition (b) implies that $\sigma$ and $\alpha$ are injective, you should implicitly identify $L$ with $\alpha(L)\subseteq M$ and $N$ with $\sigma (N) \subseteq M$. In this way you will have that $\alpha = i_L$, $\sigma = i_N$.