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.


  • 1
    $\begingroup$ 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. $\endgroup$ – 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.$

  • $\begingroup$ (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. $\endgroup$ – Martin Brandenburg Nov 19 '14 at 20:04
  • 1
    $\begingroup$ @Martin Brandenburg: Sorry, I forget to mention it. I'll edit my answer accordingly. Thanks for pointing it out. $\endgroup$ – Krish Nov 19 '14 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.