# Simple ring

In abstract algebra, a simple ring is a non-zero ring that has no two-sided ideal besides the zero ideal and itself.

One should notice that several references (e.g., Lang (2002) or Bourbaki (2012)) require in addition that a simple ring be left or right Artinian (or equivalently semi-simple). Under such terminology a non-zero ring with no non-trivial two-sided ideals is called quasi-simple.

A simple ring can always be considered as a simple algebra over its center. 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 Mn(R) is of the form Mn(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 (possibly one-sided) maximal ideal is a simple ring. In particular, a field is a simple ring. In fact a division ring is also a simple ring. A ring 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.

Furthermore, a ring ${\displaystyle R}$ is a simple commutative ring if and only if ${\displaystyle R}$ is a field. This is because if ${\displaystyle R}$ is a commutative ring, then you can pick a nonzero element ${\displaystyle x\in R}$ and consider the ideal ${\displaystyle \{xr:r\in R\}}$. Then since ${\displaystyle R}$ is simple, this ideal is the entire ring, and so it contains 1, and therefore there is some element ${\displaystyle y\in R}$ such that ${\displaystyle xy=1}$, and so ${\displaystyle R}$ is a field. Conversely, if ${\displaystyle R}$ is known to be a field, then any nonzero ideal ${\displaystyle I\subseteq R}$ will have a nonzero element ${\displaystyle i\in I}$. But since ${\displaystyle R}$ is a field, then ${\displaystyle i^{-1}\in R}$ and so ${\displaystyle i^{-1}i=1\in I}$, and so ${\displaystyle I=R}$.

## 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 Mn(D) be the ring of matrices with entries in D. It is not hard to show that every left ideal in Mn(D) takes the following form:

{M ∈ Mn(D) | The n1...nk-th columns of M have zero entries},

for some fixed {n1, ..., nk} ⊆ {1, ..., n}. So a minimal ideal in Mn(D) is of the form

{M ∈ Mn(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 = Mn(D)e, where e is the idempotent matrix with 1 in the (k, k) entry and zero elsewhere. Also, D is isomorphic to eMn(D)e. The left ideal I can be viewed as a right module over eMn(D)e, and the ring Mn(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 : AHom(I) by Φ(a)m = am for mI. Φ is injective because if aI = aAe = 0, then aA = aAeA = 0, which implies that a = a ⋅ 1 = 0.

For surjectivity, let THom(I). Since AeA = A, the unit 1 can be expressed 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 then show that eAe is a division ring. The assumption A = AeA follows from A being simple.