Semisimple algebra

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In ring theory, a semisimple algebra is a finite dimensional algebra that can be expressed as a Cartesian product of simple subalgebras.

[edit] Definition

Given an algebra, its radical is the (unique) nilpotent ideal that contains all nilpotent ideals in the algebra. A finite dimensional algebra is then said to be semi-simple if its radical is {0}, where 0 denotes the zero element of the algebra.

A algebra A is called simple if it has no proper ideals and A² = {ab | a, b ∈ A} ≠ {0}. As the terminology implies, simple algebras are semi-simple. Only possible ideals in a simple algebra are A and {0}. Thus if A is not nilpotent, then A is semisimple. Because A² is an ideal of A and A is simple, A² = A. By induction, Aⁿ = A for every positive integer n, i.e. A is not nilpotent.

Any self-adjoint subalgebra A of n × n matrices with complex entries is semisimple. Let Rad(A) be the radical of A. Suppose a matrix M is in 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 M x 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_i is semi-simple. If (a_i) is an element of Rad(A). Let e₁ be the multiplicative identity in A₁ (all simple algebras possess a multiplicative identity). Then (a₁, a₂, ...) · (e₁, 0, ...) = (a₁, 0..., 0) lies in some nilpotent ideal of ∏ A_i. This implies, for all b in A₁, a₁b is nilpotent in A₁, i.e. a₁ ∈ Rad(A₁). So a₁ = 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 semisimple algebra is isomorphic to a Cartesian product of simple algebras. The following is a semisimple algebra that appears not to be of this form. Let A be an algebra with Rad(A) ≠ A. The quotient algebra B = A ⁄ Rad(A) is semisimple: If J is a nonzero nilpotent ideal in B, then its preimage under the natural projection map is a nilpotent ideal in A which is strictly larger than Rad(A), a contradiction.

[edit] Characterization

Let A be a finite dimensional semisimple algebra, and

be a composition series of A, then A is isomporphic to the following Cartesian product:

where each

is a simple algebra.

The proof can be sketched as follows. First, invoking the assumption that A is semisimple, one can show that the J₁ is a simple algebra (therefore unital). So J₁ is a unital subalgebra and an ideal of J₂. Therefore one can decompose

By maximality of J₁ as an ideal in J₂ and also the semisimplicity of A, the algebra

is simple. Proceed by induction in similar fashion proves the claim. For example, J₃ is the Cartesian product of simple algebras

The above result can be restated in a different way. For a semisimple algebra A = A₁ ×...× A_n expressed in terms of its simple factors, consider the units e_i ∈ A_i. The elements E_i = (0,...,e_i,...,0) are idempotents in A and they lie in the center of A. Furthermore, E_i A = A_i, E_iE_j = 0 for i ≠ j, and Σ 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

1. E_iE_j = 0 for i ≠ j (such a set of idempotents is called orthogonal),
2. Σ E_i = 1,
3. A is isomorphic to the Cartesian product of simple algebras E₁ A ×...× E_n A.