# Direct summand of modules proof

I'm have problems trying to show this, any help would be appreciated.

Show $i: N \subset M$ is a direct summand iff $\exists$ a module map $r: M \to N$ s.t $ri = 1_{N}$, and that any complement of $N$ is isomorphic to $M/N$

I'm using it as a build up to prove Maschke's Theorem.

Thanks

• Hint for the first part: If $M = N \oplus S$ then write down the obvious map $r\colon M \to N$ and try and prove that $ri = 1_N$. For the other direction if you have $r$ such that $ri = 1_N$, then you should try and prove that $M \simeq N \oplus M/N$. Try using $r$ to define a map $M \to N \oplus M/N$. Then use the condition $ri = 1_N$ to prove that this map is a bijection. – Jim Nov 19 '14 at 19:11

One direction is trivial: if $M=N\oplus N'$, then by definition any element $m\in M$ can be written in one and only one way as $m=x+y$, with $x\in N$ and $y\in N'$; verifying that $r\colon m\mapsto x$ is the homomorphism you need is very easy.

Conversely, let $N'=\ker r$. Try to prove that

1. $M=N+N'$; hint: write $m=r(m)+(m-r(m))$.

2. $N\cap N'=\{0\}$.

If $M=N\oplus N'$, then the homomorphism theorem says $$M/N=(N+N')/N\cong N/(N\cap N')$$ and…

Let $A$ be a commutative ring with identity and $$0 \rightarrow N \xrightarrow{\alpha} M \xrightarrow{\beta} P \rightarrow 0$$ be an exact sequence of $A$-modules and homomorphisms. Then the followings are equivalent:

(i) There is an isomorphism of sequences with the sequence $$0 \rightarrow N \xrightarrow{i} N\oplus P \xrightarrow{\pi} P \rightarrow 0$$ where the maps $i$ and $\pi$ are the natural injection and projection respectively.

(ii) There is a $A$-morphism $P \xrightarrow{f} M$ such that $\beta f = 1_P.$

(iii) There is a $A$-morphism $M \xrightarrow{g} N$ such that $g\alpha = 1_N.$

(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iii) are obvious.

(ii)$\Rightarrow$(i): $M =$ Im$f$ + Ker$\beta$, because any $m \in M$ can be written as $m = m - f\beta(m) + f\beta(m)$ and $m - f\beta(m) \in$ Ker$\beta$. Also $m \in$ Im$f \cap$ ker$\beta \Rightarrow f(n) = m, \beta(m) = 0$ for some $n \in P$ $\Rightarrow n = \beta f(n) = \beta(m) = 0 \Rightarrow m = f(n) = 0.$

(iii)$\Rightarrow$(i): $M =$ Im$\alpha$ + ker$g$ because any $m \in M$ can be written as $m = m - \alpha g(m) + \alpha g(m)$ and $m - \alpha g(m) \in$ Ker$g$. Also $m \in$ Im$\alpha \cap$ ker$g \Rightarrow \alpha(n) = m, g(m) = 0$ for some $n \in N$ $\Rightarrow n = g \alpha(n) = g(m) = 0 \Rightarrow m = \alpha(n) = 0.$

• (i) is not stated correctly. You need that the whole exact sequence is isomorphic to $0 \to N \to N \oplus P \to P \to 0$ (with the obvious maps here) - not just the middle objects. – Martin Brandenburg Nov 19 '14 at 20:04
• @Martin Brandenburg: Sorry, I forget to mention it. I'll edit my answer accordingly. Thanks for pointing it out. – Krish Nov 19 '14 at 20:11