Associative Artinian algebra with a trivial Jacobson radical

Semisimple algebra

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In ring theory, a branch of mathematics, a semisimple algebra is an associative Artinian algebra over a field which has trivial Jacobson radical (only the zero element of the algebra is in the Jacobson radical). If the algebra is finite-dimensional this is equivalent to saying that it can be expressed as a Cartesian product of simple subalgebras.

Definition

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

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

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

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.

Characterization

Let $A$ be a finite-dimensional semisimple algebra, and

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

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

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

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

is a simple algebra.

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

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

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

J_{2}/J_{1}\,

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

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 $A=A_{1}\times \cdots \times A_{n}$ expressed in terms of its simple factors, consider the units $e_{i}\in A_{i}$ . The elements $E_{i}=(0,\dots,e_{i},\dots,0)$ are idempotent elements in $A$ and they lie in the center of $A$ Furthermore, $E_{i}A=A_{i},E_{i}E_{j}=0$ for $i\neq j$ , and $\sum E_{i}=1$ , the multiplicative identity in $A$ .

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

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

Classification

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