- Semisimple algebra
In
ring theory , a semisimple algebra is anassociative algebra which has trivialJacobson radical (that is 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 that can be expressed as a Cartesian product of simple subalgebras.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 "A"2 = {"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 "A"2 is an ideal of "A" and "A" is simple, "A"2 = "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 {"Ai"} is a finite collection of simple algebras, then their Cartesian product ∏ "Ai" is semi-simple. If ("ai") is an element of Rad("A"). Let "e"1 be the multiplicative identity in "A"1 (all simple algebras possess a multiplicative identity). Then ("a"1, "a"2, ...) · ("e"1, 0, ...) = ("a"1, 0..., 0) lies in some nilpotent ideal of ∏ "Ai". This implies, for all "b" in "A"1, "a"1"b" is nilpotent in "A"1, i.e. "a"1 ∈ Rad("A"1). So "a"1 = 0. Similarly, "ai" = 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.
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 "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
:
By maximality of "J"1 as an ideal in "J"2 and also the semisimplicity of "A", the algebra
:
is simple. Proceed by induction in similar fashion proves the claim. For example, "J"3 is the Cartesian product of simple algebras
:
The above result can be restated in a different way. For a semisimple algebra "A" = "A"1 ×...× "An" expressed in terms of its simple factors, consider the units "ei" ∈ "Ai". The elements "Ei" = (0,...,"ei",...,0) are
idempotent s in "A" and they lie in the center of "A". Furthermore, "Ei A" = "Ai", "EiEj" = 0 for "i" ≠ "j", and Σ "Ei" = 1, the multiplicative identity in "A".Therefore, for every semisimple algebra "A", there exists idempotents {"Ei"} in the center of "A", such that
#"EiEj" = 0 for "i" ≠ "j" (such a set of idempotents is called "orthogonal"),
#Σ "Ei" = 1,
#"A" is isomorphic to the Cartesian product of simple algebras "E"1 "A" ×...× "En A".Classification
The
Artin–Wedderburn theorem completely classifies semisimple algebras: they are isomorphic to a product where the are some integers, the aredivision ring s, and means the ring of matrices over . This product is unique up to permutation of the factors.
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