Simple ring

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In abstract algebra, a simple ring is a non-zero ring that has no ideal besides the zero ideal and itself. A simple ring can always be considered as a simple algebra.

According to the Artin-Wedderburn theorem, every simple ring that is left or right Artinian is a matrix ring over a division ring. In particular, the only simple rings that are a finite-dimensional vector space over the real numbers are rings of matrices over either the real numbers, the complex numbers, or the quaternions.

Any quotient of a ring by a maximal ideal is a simple ring. In particular, a field is a simple ring. A ring R is simple if and only its opposite 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.

[edit] 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 n₁...n_k-th columns of M have zero entries},

for some fixed {n₁,...,n_k} ⊂ {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 = ∑a_ieb_i. So

T(m) = T(1·m) = T(∑a_ieb_im) = ∑ T(a_ieeb_im) = ∑ T(a_ie) eb_im = [ ∑T(a_ie)eb_i]m.

Since the expression [∑T(a_ie)eb_i] 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 semisimple 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.