Semi Simplie Ring and Module

Let $R$ be a ring with unity $1\ne 0$.

Module Case

Equivalent Statements

Let $E$ be an $R$-module. Then the following are equivalent

1. $E$ is a sum of simple $R$-module. (SS1)
2. $E$ is a direct sum of simple $R$-module. (SS2)
3. For all submodule $F$ of $E$, there exists $F^{\prime }$ such that $E=F\oplus F^{\prime }$. (SS3)

Proof

SS1 implies SS2 by the Lemma.

SS2 $\Rightarrow$ SS3

Suppose $F⊲E={\oplus }_{i\in I}{E}_i$ where $E_i$‘s are simple. Let $J\subset I$ be maximal such that $F+\oplus E_{j\in J}{E}_j=F\oplus {\oplus }_{j\in J}{E}_j$. Then for all $i\in I\setminus J$, by the maximality, $E_i\cap (F\oplus {\oplus }_{j\in J}{E}_j)\ne 0$. And do the similar argument with SS1 to SS2.

SS3 $\Rightarrow$ SS1

Let $E_0={\sum }_{\text{simple }S⊲E}S$. Then by SS3, $E=E_0\oplus E_0^{\prime }$ for some submodule $E_0^{\prime }$. Suppose $E_0\ne E$. Then there is a simple submodule $S⊲E^{\prime }$ by the lemma2, so it contradicts to the direct sum. There $E=E_0$ which implies SS1 holds.

Remark

For SS3, for any monomorphism $i:E\to F$ splits, i.e. there exists $s:F\to E$ such that $s\circ i=i{d}_E$. (SS3`)

So also equivalent to a short exact sequence $0\to F\to E\to H\to 0$ splits.(SS3“)

Lemma

1. If $E={\sum }_i{E}_i$ where $E_i$‘s are Simple Module, the there exist $J\subset I$ such that $E={\oplus }_{j\in J}{E}_j$

Proof

Let $J\subset I$ be maximal such that ${\sigma }_{j\in J}{E}_j={\oplus }_{j\in J}{E}_j$. By the maximality, for all $i\in I\setminus J$ satisfies $E_i\cap {\oplus }_{j\in J}{E}_j\ne 0$. Since $E_i$ is simple, $E_i\subset {\oplus }_{j\in J}{E}_j$ so that $E={\oplus }_{j\in J}{E}_j$.

2. Let $0\ne F⊲E$ aIf $E$ satisfies SS3, then there exists simple $S⊲F$.

Proof

Simple Module

Definition of Semi-simple module

A $R$-module $E$ is semi simple module if it satisfies one of the Equivalent Statements.

Example

For ring $R=\mathbb{R}\times \mathbb{R}$ and $E$ be a $R$-module $\mathbb{R}\times \mathbb{R}=(\mathbb{R}\times 0)\oplus (0\times \mathbb{R})$.

Proposition

Every quotient and submodule of semi simple module is semi simple.

Proof

Quotient

Let $f:E\twoheadrightarrow M$ with $E$ is a semi simple, i.e. $E={\sum }_i{S}_i$ where $S_i$‘s are simple. Then $M={\sum }_{i\in I}f(S_i)$. Since $kerf{|}_{S_i}⊲S_i$, $f(S_i)=0$ or simple. Therefore $M$ satisfies SS1.

Submodule

Let $F⊲E$ where $E$ is semi simple. Then $E=F\oplus F^{\prime }$ by SS3, and the result from the quotient, $F{\cong }_{Mod}E/F^{\prime }$ is semi simple.

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Ring

Definition

Ring $R$ is semi simple if $R$ is a semi simple $R$-module as itself.

Proposition

If $R$ is a semi simple ring, then all $R$-modules are semi simple.

Proof

All modules are quotient of the free module. Since, quotient module is semi simple, all modules are semi simple.

Characterization

Artin-Weddenburn Theorem

Application

Maschke’s Theorem