direct sum

Overview

Direct sum is a way to make a new algebraic structure (like a group, ring, module, etc.) from existing ones such that the projection should respect the structure of the building blocks.

Direct sum of groups

Direct sum of R-modules

Given two R-modules $M$ and $N$, the direct sum $M\oplus N$ is the set $M\times N$ with the addition defined as the abelian group case.

The $R$ action is defined by

$$r\cdot (m,n)=(r\cdot m,r\cdot n)$$

Direct summand

Let $R$ be a ring. A direct summand of an $R$-module $N$ is a submodule $M\subset N$ for which there exists a submodule $P\subset N$ such that $M\oplus P\simeq N$.

Split inclusion

Let $M$ be a submodule of an $R$-module $N$. $M$ is a direct summand if and only if the inclusion $i:M\hookrightarrow N$ is split, that is, there exists a morphism $s:N\to M$ which satisfies $s\circ i={\text{id}}_M$.

Proof

First assume $M$ is a direct summand of $N$, so $N\simeq M\oplus P$. Then let $s:M\oplus P\to M$ be the projection in the first element. Looking at the maps

$$\begin{array}{r}M\hookrightarrow M\oplus P\stackrel{s}{\to }M\\ m⟼(m,0)⟼m\end{array}$$

$s$ is certainly surjective, and $s\circ i={\text{id}}_M$

Next, let there be the morphisms $i,s$ as described in the prompt:

Thus, we know that $i$ must be injective and $s$ must be surjective or else the composition would not be the identity.

From this we claim that $N\simeq \text{im}i\oplus \ker s\simeq M\oplus \ker s$. It is clear that $\text{im}i\cap \ker s=\{0\}$ since if $n\in \text{im}i\cap \ker s$ then $s(n)=0$ and $i(s(n))=0$ where $i$ is injective so $n=0$. Next, let $n\in N$ then we can take $m=s(n)$, using this element, let $n_i=i(m)$. Now we can take the element $n_k=n-n_i$.

$$s(n_k)=s(n-n_i)=s(n)-s(n_i)=m-m=0$$

so $n_k\in \ker s$ and $n=n_i+n_k$. Therefore, $N=M\oplus \ker s$.