Semisimple algebra

The Jacobson radical of an algebra over a field is the ideal consisting of all elements that annihilate every simple left-module. The radical contains all nilpotent ideals, and if the algebra is finite-dimensional, the radical itself is a nilpotent ideal. A finite-dimensional algebra is then said to be semisimple if its radical contains only the zero element.

An algebra ${\displaystyle A}$ is called simple if it has no proper ideals and ${\displaystyle A^{2}=\{ab|a,b\in A\}\neq \{0\}}$. As the terminology suggests, simple algebras are semisimple. The only possible ideals of a simple algebra ${\displaystyle A}$ are ${\displaystyle A}$ and ${\displaystyle \{0\}}$. Thus if ${\displaystyle A}$ is simple, then ${\displaystyle A}$ is not nilpotent. Because ${\displaystyle A^{2}}$ is an ideal of ${\displaystyle A}$ and ${\displaystyle A}$ is simple, ${\displaystyle A^{2}=A}$. By induction, ${\displaystyle A^{n}=A}$ for every positive integer ${\displaystyle n}$, i.e. ${\displaystyle A}$ is not nilpotent.

Any self-adjoint subalgebra ${\displaystyle A}$ of ${\displaystyle n\times n}$ matrices with complex entries is semisimple. Let ${\displaystyle \operatorname {Rad} (A)}$ be the radical of ${\displaystyle A}$. Suppose a matrix ${\displaystyle M}$ is in ${\displaystyle \operatorname {Rad} (A)}$. Then ${\displaystyle M^{*}M}$ lies in some nilpotent ideals of ${\displaystyle A}$, therefore ${\displaystyle (M^{*}M)^{k}=0}$ for some positive integer ${\displaystyle k}$. By positive-semidefiniteness of ${\displaystyle M^{*}M}$, this implies ${\displaystyle M^{*}M=0}$. So ${\displaystyle Mx}$ is the zero vector for all ${\displaystyle x}$, i.e. ${\displaystyle M=0}$.

If ${\displaystyle {A_{i}}}$ is a finite collection of simple algebras, then their Cartesian product ${\displaystyle A=\prod A_{i}}$ is semisimple. If ${\displaystyle (A_{i})}$ is an element of ${\displaystyle \operatorname {Rad} (A)}$ and ${\displaystyle e_{1}}$ is the multiplicative identity in ${\displaystyle A_{1}}$ (all simple algebras possess a multiplicative identity), then ${\displaystyle (a_{1},a_{2},\dots )\cdot (e_{1},0,\dots )=(a_{1},0,\dots )}$ lies in some nilpotent ideal of ${\displaystyle \prod A_{i}}$. This implies, for all ${\displaystyle b}$ in ${\displaystyle A_{1}}$, ${\displaystyle a_{1}b}$ is nilpotent in ${\displaystyle A_{1}}$, i.e. ${\displaystyle a_{1}\in \operatorname {Rad} (A_{1})}$. So ${\displaystyle a_{1}=0}$. Similarly, ${\displaystyle a_{i}=0}$ for all other ${\displaystyle i}$.

It is less apparent from the definition that the converse of the above is also true, that is, any finite-dimensional semisimple algebra is isomorphic to a Cartesian product of a finite number of simple algebras.

Let ${\displaystyle A}$ be a finite-dimensional semisimple algebra, and

${\displaystyle \{0\}=J_{0}\subset \cdots \subset J_{n}\subset A}$

be a composition series of ${\displaystyle A}$ then ${\displaystyle A}$ is isomorphic to the following Cartesian product:

${\displaystyle A\simeq J_{1}\times J_{2}/J_{1}\times J_{3}/J_{2}\times ...\times J_{n}/J_{n-1}\times A/J_{n}}$

where each

${\displaystyle J_{i+1}/J_{i}\,}$

is a simple algebra.

The proof can be sketched as follows. First, invoking the assumption that ${\displaystyle A}$ is semisimple, one can show that the ${\displaystyle J_{1}}$ is a simple algebra (therefore unital). So ${\displaystyle J_{1}}$ is a unital subalgebra and an ideal of ${\displaystyle J_{2}}$. Therefore, one can decompose

${\displaystyle J_{2}\simeq J_{1}\times J_{2}/J_{1}.}$

By maximality of ${\displaystyle J_{1}}$ as an ideal in ${\displaystyle J_{2}}$ and also the semisimplicity of ${\displaystyle A}$ the algebra

${\displaystyle J_{2}/J_{1}\,}$

is simple. Proceed by induction in similar fashion proves the claim. For example, ${\displaystyle J_{3}}$ is the Cartesian product of simple algebras

${\displaystyle J_{3}\simeq J_{2}\times J_{3}/J_{2}\simeq J_{1}\times J_{2}/J_{1}\times J_{3}/J_{2}.}$

The above result can be restated in a different way. For a semisimple algebra ${\displaystyle A=A_{1}\times \cdots \times A_{n}}$ expressed in terms of its simple factors, consider the units ${\displaystyle e_{i}\in A_{i}}$. The elements ${\displaystyle E_{i}=(0,\dots ,e_{i},\dots ,0)}$ are idempotent elements in ${\displaystyle A}$ and they lie in the center of ${\displaystyle A}$ Furthermore, ${\displaystyle E_{i}A=A_{i},E_{i}E_{j}=0}$ for ${\displaystyle i\neq j}$, and ${\displaystyle \sum E_{i}=1}$, the multiplicative identity in ${\displaystyle A}$.

Therefore, for every semisimple algebra ${\displaystyle A}$, there exists idempotents ${\displaystyle \{E_{i}\}}$ in the center of ${\displaystyle A}$, such that

1. ${\displaystyle E_{i}E_{j}=0}$ for ${\displaystyle i\neq j}$ (such a set of idempotents is called central orthogonal),
2. ${\displaystyle \sum E_{i}=1}$,
3. ${\displaystyle A}$ is isomorphic to the Cartesian product of simple algebras ${\displaystyle E_{1}A\times \dots \times E_{n}A}$.

A theorem due to Joseph Wedderburn completely classifies finite-dimensional semisimple algebras over a field ${\displaystyle k}$. Any such algebra is isomorphic to a finite product ${\displaystyle \prod M_{n_{i}}(D_{i})}$ where the ${\displaystyle n_{i}}$ are natural numbers, the ${\displaystyle D_{i}}$ are division algebras over ${\displaystyle k}$, and ${\displaystyle M_{n_{i}}(D_{i})}$ is the algebra of ${\displaystyle n_{i}\times n_{i}}$ matrices over ${\displaystyle D_{i}}$. This product is unique up to permutation of the factors.^[1]

This theorem was later generalized by Emil Artin to semisimple rings. This more general result is called the Wedderburn–Artin theorem.