- Simple ring
In
abstract algebra , a simple ring is a non-zero ring that has no ideal besides thezero ideal and itself. A simple ring can always be considered as asimple algebra .According to the
Artin-Wedderburn theorem , every simple ring that is left or right Artinian is amatrix ring over adivision ring . In particular, the only simple rings that are a finite-dimensionalvector space over thereal number s are rings of matrices over either the real numbers, thecomplex number s, or thequaternion s.Any quotient of a ring by a
maximal ideal is a simple ring. In particular, afield is a simple ring. A ring "R" is simple if and only itsopposite ring "R"o is simple.An example of a simple ring that is not a matrix ring over a division ring is the
Weyl algebra .Wedderburn's theorem
Wedderburn's theorem characterizes simple rings with a unit and a minimal left ideal. (The left Artinian condition is a generalization of the second assumption.) Namely it says that every such ring is, up to isomorphism, a ring of "n" × "n" matrices over a division ring.
Let "D" be a division ring and M(n,"D") be the ring of matrices with entries in "D". It is not hard to show that every left ideal in M(n,"D") takes the following form:
:{"M" ∈ M(n,"D") | The "n"1..."nk"-th columns of "M" have zero entries},
for some fixed {"n"1,...,"nk"} ⊂ {1, ..., "n"}. So a minimal ideal in M(n,"D") is of the form
:{"M" ∈ M(n,"D") | All but the "k"-th columns have zero entries},
for a given "k". In other words, if "I" is a minimal left ideal, then "I" = (M(n,"D")) "e" where "e" is the idempotent matrix with 1 in the ("k", "k") entry and zero elsewhere. Also, "D" is isomorphic to "e"(M(n,"D"))"e". The left ideal "I" can be viewed as a right-module over "e"(M(n,"D"))"e", and the ring M(n,"D") is clearly isomorphic to the algebra of homorphisms on this module.
The above example suggests the following lemma:
Lemma. "A" is a ring with identity 1 and an idempotent element "e" where "AeA = A". Let "I" be the left ideal "Ae", considered as a right module over "eAe". Then "A" is isomorphic to the algebra of homomorphisms on "I", denoted by "Hom"("I").
Proof: We define the "left regular representation" Φ : "A" → "Hom"("I") by Φ("a")"m" = "am" for "m" ∈ "I". Φ is injective because if "a · I" = "aAe" = 0, then "aA" = "aAeA" = 0, which implies "a" = "a" · 1 = 0.
For surjectivity, let "T" ∈ "Hom"("I"). Since "AeA" = "A", the unit 1 can be expresses as 1 = ∑"aiebi". So
:"T"("m") = "T"(1·"m") = "T"(∑"aiebim") = ∑ "T"("aieebim") = ∑ "T"("aie) ebim" = [ ∑"T"("aie")"ebi"] "m".
Since the expression [∑"T"("aie")"ebi"] does not depend on "m", Φ is surjective. This proves the lemma.
Wedderburn's theorem follows readily from the lemma.
Theorem (Wedderburn). If "A" is a simple ring with unit 1 and a minimal left ideal "I", then "A" is isomorphic to the ring of "n" × "n" matrices over a division ring.
One simply has to verify the assumptions of the lemma hold, i.e. find an idempotent "e" such that "I = Ae" and "A = AeA"."A" being simple also implies "eAe" is a division ring.
See also
*
simple (algebra) References
*D.W. Henderson, A short proof of Wedderburn's theorem, "Amer." "Math." "Monthly" 72 (1965), 385-386.
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