Artin-Weddenburn Theorem

Statement

A ring $R$ is ring isomorphic to a finite product of matrix rings over division rings $D_i$.

$$M_{n_1}(D_1)\times \cdots M_{n_k}(D_k)$$

if and only if $R$ is a semi simple ring.

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To prove this big theorem we need a lemma.

Lemma

Ring can be written as a finite direct sum of left ideals.

Let $R$ be a ring, and as a $R$-module $R={\oplus }_{j\in J}{L}_j$ where $L_j$‘s are non-zero ideals. Then $\left|J\right|<\infty$.

Proof

As a $R$-module, $1={\sum }_{j\in J}{e}_j={\sum }_{i\in I}{e}_i$ where $i\ne 0$ and $\left|I\right|<\infty$. So that for each $r\in R$,

$r=r{\sum }_{i\in I}{e}_i={\sum }_{i\in I}r{e}_i\in {\sum }_{i\in I}{L}_i.$

Simple Ideal

Let $L$ be a simple left ideal of ring $R$, and let $E$ be a simple $R$-module. If $L$ is not isomorphic to $E$, then $LE=0$, i.e. $L\subset An{n}_R(E)$.

Proof

We have $RLE=LE$, and $LE$ is a submodule of $E$, hence equal to 0 or $E$. Suppose there exists $y\in E$ such that $Ly\ne 0$. Then $Ly=E$ so that it is an isomorphism between $L$ and $E$ which contradicts to the hypothesis.

Simple Ring

If $R$ is a simple ring, then $k=1$, i.e. $R{\cong }_{Ring}{M}_n(D)$ where $D$ is a skew field. Also it holds in converse direction.

Proof

Suppose $R$ is a simple ring such that $R=L^n$ where $L$ is a simple $R$-module. We know that $R^{op}{\cong }_{Ring}En{d}_R(R)$ from the Differences between Ring, Matrix, and Endomorphism notes and $En{d}_R(R)$ is a Division Ring by Schur’s lemma. So

$$R^{op}{\cong }_{Ring}En{d}_R(L^n){\cong }_{Ring}{M}_n(En{d}_R(L))$$

Also, through the exercise, we know that $M_n(K{)}^{op}=M_n(En{d}_R(L{)}^{op})$ by the transpose map. Let denote $D=En{d}_R(L{)}^{op}$. Then $R{\cong }_{Ring}{M}_n(D)$ as desired.

On the other hand, suppose $R{\cong }_{Ring}{M}_n(D)$ for some division ring $D$. We can break down $M_n(D)$ as a direct sum with $n$ summands where each direct summands are $i$th column of $M_n(D)$. Each summand is isomorphic to simple $R$-module $D^n$. Therefore $R$ is a simple ring.

Semi simple

Assume $R$ is a semi simple ring. By the [[Artin-Weddenburn Theorem#Lemma#Ring can be written as a finite direct sum of left ideals.|lemma 1]], we can choose finitely many($n$) simple left ideals $L_i$ where not isomorphic to each other. Let

$$R_i=\sum _{L{\cong }_{R-mod}{L}_i}L.$$

Then, by the [[Artin-Weddenburn Theorem#Lemma#Simple Ideal|lemma 2]], $R_i$ is two-sided ideal because

$$R_i=R_{i}R=R_i{R}_i\subset R{R}_i\subset R_i.$$

Claim : $R=R_1\oplus \cdots \oplus R_n$ which is Block Decomposition of the ring. Proof) Clearly, $R={\sum }_{i=1}^n{R}_i$ from the construction. And again using [[Artin-Weddenburn Theorem#Lemma#Simple Ideal|lemma 2]], we can see that sum is a direct sum.

By the claim, we can write $1_R=e_1+\cdots +e_n$. Hence $R_i$ is a ring with unity $e_i$ and denote as $(R_i,e_i)$. Since $R_i{R}_j=0$ for $i\ne j$,

$$R{\cong }_{Ring}(R_1,e_1)\times \cdots \times (R_n,e_n).$$

On the other hand, suppose $R{\cong }_{Ring}{M}_1(D_1)\times \cdots M_n(D_n)$. Apply the simple case and mathematical induction.¹

Different Point of View

In $1_R=e_1+\cdots +e_n$, $e_i=e_i^2$ that $e_i$ is idempotent, and

$$r\cdot 1_R=1_{R}r=\sum _{i=1}^{n}r{e}_i=\sum _{i=1}^n{e}_{i}r.$$

that every $e_i$‘s is central. Therefore, Block Decomposition is equivalent to finding orthonormal non trivial central idempotents.

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We can determine $D$

Lemma

Let $k$ be a field and $D$ skew field which is a $k$-algebra and finite dimension over $k$. Then

1. For all $\alpha \in D$, $k[\alpha ]$ is field.
2. If $K$ is algebraically closed, $D=k$

Proof

Basic Field Theory and Weddenburn’s Little Theorem.

Example

For example $k=\mathbb{R}$, $D=\mathbb{H}$ and $\alpha \in \mathbb{H}-\mathbb{R}$, then $k[\alpha ]\cong \mathbb{C}$.

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Therefore, for the case $\mathbb{C}G$ where $G$ is a finite group, then $\mathbb{C}G{\cong }_{Ring}{\prod }_{i=1}^s{M}_{n_i}(\mathbb{C}).$

Exercise

Block decomposition and Simple Ring

Footnotes

1. Exercise. However I think it is quite routine. ↩