In abstract algebra, a simple ring is a non-zero ring that has no (two-sided) ideal besides the zero ideal and itself. A simple ring can always be considered as a simple algebra. This notion must not be confused with the related one of a ring being simple as a left (or right) module over itself (although both notions coincide in the commutative setting). Rings which are simple as rings but not as modules do exist: the full matrix ring over a field does not have any nontrivial ideals (since any ideal of M(n,R) is of the form M(n,I) with I an ideal of R), but has nontrivial left ideals (namely, the sets of matrices which have some fixed zero columns).
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 Ro 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 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:
for some fixed {n1,...,nk} ⊂ {1, ..., n}. So a minimal ideal in M(n,D) is of the form
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 homomorphisms 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
Since the expression [∑T(aie)ebi] does not depend on m, Φ is surjective. This proves the lemma.
- T(m) = T(1·m) = T(∑aiebim) = ∑ T(aieebim) = ∑ T(aie) ebim = [ ∑T(aie)ebi]m.
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 then show that eAe is a division ring. The assumption A = AeA follows from A being simple.