Noncommutative algebra requires explicit control of multiplication order, one-sided ideals and module conventions. Representation theory then encodes group or algebra structure through linear actions and characters. This page consolidates the relevant material from the supplied algebra source into a stand-alone handbook chapter.
Learning pathNoncommutative Algebra and Representations
LevelAdvanced
FormatHandbook guide
Read time11 min
Executive summary
This chapter develops division algebras and central simple algebras as part of a connected advanced-algebra learning sequence. The emphasis is on definitions, hypotheses, structural results and repeatable methods rather than historical narrative.
The source material is theorem-rich. Accordingly, the handbook presentation separates vocabulary from results and then adds a verification workflow so that each statement can be applied safely. Mathematical examples in the source are treated as examples, not as universal rules.
Problem-solving workflow
Fix left-versus-right conventions before beginning.
Keep multiplication order unchanged unless commutativity has been proved.
Use one-sided ideals, radicals and chain conditions appropriate to the setting.
Decompose modules into simple pieces only when the semisimplicity hypotheses apply.
For representations, identify the field, dimension, kernel and invariant subspaces.
Use characters as class functions only after verifying the representation-theoretic assumptions.
Core definitions
Definition
A division algebra over a field k is a division ring regarded as an algebra over its center k. Let us begin by considering the wider class of simple algebras.
Definition
A k-algebra A over a field k is central simple if it is finite-dimensional,18 simple (no two-sided ideals other than A and {0}), and its center Z(A) = k. Notation. If A is an algebra over a field k, then we write [A : k] = dimk(A). 18Some authors do not assume finite-dimensionality.
Definition
A splitting field for a central simple k-algebra A is a field extension E/k for which there exists an integer n such that E ⊗k A ∼= Matn(E).
Definition
If A is a k-algebra and X ⊆A is a subset, then its centralizer, CA(X), is defined by CA(X) = {a ∈A : ax = xa for every x ∈X}. It is easy to check that centralizers are always subalgebras.
Definition
Two central simple k-algebras A and B are similar, denoted by A ∼B, if there are integers n and m with A ⊗k Matn(k) ∼= B ⊗k Matm(k).
Definition
If [A] denotes the equivalence class of a central simple k-algebra A under similarity, define the central-simple algebra class groups Br(k) to be the set Br(k) = { [A] : A is a central simple k-algebra } with binary operation [A][B] = [A ⊗k B].
Definition
If E/k is a field extension, then the relative central-simple algebra class groups, Br(E/k), is the kernel of homomorphism fE/k : Br(k) →Br(E): Br(E/k) = ker fE/k = { [A] ∈Br(k) : A is split by E } .
Principal results and structural facts
Key result
Let A be a central simple k-algebra. If B is a simple k-algebra, then A ⊗k B is a central simple Z(B)-algebra. In particular, if B is a central simple k-algebra, then A ⊗k B is a central simple k-algebra.
Key result
Let k be a field and let A be a central simple k-algebra. (i) If k is the algebraic closure of k, then there is an integer n with k ⊗k A ∼= Matn(k). (ii) If A is a central simple k-algebra, then there is an integer n with [A : k] = n2.
Key result
says that the algebraic closure k of a field k is a splitting field for every central simple k-algebra A.
Key result
Let A be a central simple algebra over a field k and let B be a simple subalgebra of A. (i) CA(B) is a simple k-algebra. (ii) B ⊗k Aop ∼= Mats(Δ) and CA(B) ∼= Matr(Δ) for some division algebra Δ, where r | s. (iii) [B : k][CA(B) : k] = [A : k]. (iv) CA(CA(B)) = B.
Key result
If Δ is a division algebra over a field k, then a subfield E of Δ is a maximal subfield if and only if CΔ(E) = E.
Key result
If D is a division algebra over a field k and E is a maximal subfield of D, then E is a splitting field for D; that is, E ⊗k D ∼= Mats(E), where s = [D : E] = [E : k].
Key result
Let k be a field, let B be a simple k-algebra, and let A be a central simple k-algebra. If there are algebra maps f, g : B →A, then there exists a unit u ∈A with g(b) = u f (b)u−1 for all b ∈B.
Key result
Let A be a central simple k-algebra over a field k, and let B and B′ be isomorphic simple k-subalgebras of A. If ψ : B →B′ is an isomorphism, then there exists a unit u ∈A with ψ(b) = ubu−1 for all b ∈B.
Key result
Therefore, D = ! x x Ex−1, and so D× = ≠ x x E×x−1. Therefore, D = E is commutative. •
Key result
Let A be a finite-dimensional algebra over a field k. If S and T are k-subalgebras of A such that (i) st = ts for all s ∈S and t ∈T ; (ii) A = ST ; (iii) [A : k] = [S : k][T : k], then A ∼= S ⊗k T .
Key result
gives the desired isomorphism. (ii) If V and W are vector spaces over k of dimensions n and m, respectively, it suffices to prove that Endk(V ) ⊗k Endk(W) ∼= Endk(V ⊗k W). Define S to be all f ⊗1W, where f ∈Endk(V ), and define T to be all 1V ⊗g, where g ∈Endk(W). It is routine to check that the three conditions in Lemma 9.125 hold. Division Algebras (iii) Since k = Mat1(k), we have A ∼= A ⊗k k ∼= A ⊗k Mat1(k), so that ∼is reflexive. Symmetry is obvious; for transitivity, suppose that A ∼B and B ∼C; that is, A ⊗k Matn(k) ∼= B ⊗k Matm(k) and B ⊗k Matr(k) ∼= C ⊗k Mats(k). Then A ⊗k Matn(k) ⊗k Matr(k) ∼= A ⊗k Matnr(A), by part (ii). On the other hand, A ⊗k Matn(k) ⊗k Matr(k) ∼= B ⊗k Matm(k) ⊗k Matr(k) ∼= C ⊗k Matm(k) ⊗k Mats(k) ∼= C ⊗k Matms(k). Therefore, A ∼C, and so ∼is an equivalence relation. (iv) Define f : A×Aop →Endk(A) by f (a, c) = λa◦ρc, where λa : x ↦ax and ρc : x ↦ xc; it is routine to check that λa and ρc are k-maps (so their composite is also a k-map), and that f is k-biadditive. Hence, there is a k-map f : A ⊗k Aop →Endk(A) with f (a ⊗c) = λa ◦ρc. Associativity a(xc) = (ax)c in A says that λa ◦ρc = ρc ◦λa, from which it easily follows that f is a k-algebra map. As A ⊗k Aop is a simple k-algebra and ker f is a proper two-sided ideal, we have f injective. Now dimk(Endk(A)) = dimk(Homk(A, A)) = n2, where n = [A : k]. Since dimk(im f ) = dimk(A ⊗k Aop) = n2, it follows that f is a k-algebra isomorphism: A ⊗k Aop ∼= Endk(A). • We now extend Theorem 9.118 from division algebras to central simple algebras.
Key result
Let A be a central simple k-algebra over a field k, so that A ∼= Matr(Δ), where Δ is a division algebra over k. If E is a maximal subfield of Δ, then E splits A; that is, there is an integer n and an isomorphism E ⊗k A ∼= Matn(E). More precisely, if [Δ : E] = s, then n = rs and [A : k] = (rs)2.
Key result
If k is a field, then there is a bijection from Br(k) to the family D of all isomorphism classes of finite-dimensional division algebras over k, and so | Br(k)| = |D|. Therefore, there exists a noncommutative division ring, finite-dimensional over its center k, if and only if Br(k) ̸= {0}.
Key result
If E/k is a field extension, then there is a homomorphism fE/k : Br(k) →Br(E) given by [A] ↦[E ⊗k A].
Source-grounded examples
Worked source example
(i) Every division algebra Δ that is finite-dimensional over its center k is a central simple k-algebra. The quaternions H is a central simple R-algebra, and every field is a central simple algebra over itself. (ii) If k is a field, then Matn(k) is a central simple k-algebra. (iii) If A is a central simple k-algebra, then its opposite algebra Aop is also a central simple k-algebra. ◀
Worked source example
(i) If k is an algebraically closed field, then Theorem 9.114 shows that Br(k) = {0}. (iii) If k = R, then fixed-point-action’s Theorem 9.124 shows that Br(R) ∼= I2. (iv) It is proved, using class field theory, that Br(Qp) ∼= Q/Z, where Qp is the field of p-adic numbers. Moreover, there is an exact sequence 0 →Br(Q) →Br(R) ⊕ p Br(Qp) ϕ −→Q/Z →0. Division Algebras If we write Br(R) = ⟨1 2 + Z⟩⊆Q/Z, then ϕ is the “sum of coordinates” map. In a series of deep papers, Br(k) was computed for the most interesting fields k arising in algebraic number theory (local fields, one of which is Qp, and global fields) by A. central-simple class, H. ◀
How to reason with these results
Most advanced-algebra problems become manageable when the representation is separated from the invariant structure. Begin with the definition, then decide whether the problem is asking for an elementwise calculation, a statement about a morphism, or a classification up to isomorphism. That choice determines the correct proof language.
When a theorem gives a structural conclusion, do not jump directly to the conclusion. Write the hypotheses next to the object you are studying and check them one by one. If a hypothesis fails, either strengthen the object, pass to a quotient or localisation where the theorem applies, or use a more elementary argument.
For computational work, record each transformation together with the equivalence relation it preserves. In algebra, row operations, similarity, quotienting, localisation and isomorphism preserve different kinds of information. A calculation is useful only when the preserved structure matches the question.
Common failure modes
Failure mode
Control
Silently commuting factors.
Return to the definition or theorem hypotheses and verify the missing condition before continuing.
Confusing left modules with right modules.
Return to the definition or theorem hypotheses and verify the missing condition before continuing.
Assuming every module decomposes into simples.
Return to the definition or theorem hypotheses and verify the missing condition before continuing.
Using character identities outside the required field or finiteness conditions.
Return to the definition or theorem hypotheses and verify the missing condition before continuing.
Treating a matrix representation as faithful without checking its kernel.
Return to the definition or theorem hypotheses and verify the missing condition before continuing.
Verification checklist
The ambient set, ring, field, group, module or category has been stated.
Every operation and map used is well-defined in that setting.
The hypotheses of each structural result have been checked before use.
Representatives, coordinates or generators have not been confused with the underlying object.
Existence and uniqueness have been separated where both matter.
The final result has been checked against the original defining relation or universal property.
Quick questions
What should I identify first in a problem about division algebras and central simple algebras?
Start with the ambient algebraic structure, its operation or maps, and the exact hypotheses. Most incorrect solutions begin by using a familiar rule that is not valid in the stated structure.
How should definitions be used in proofs?
Expand the definition at the point where it becomes useful. Definitions are not background prose; they are the conditions that determine what must be proved and which implications are available.
When is a structural theorem safer than direct calculation?
Use a structural theorem when its hypotheses are satisfied and the calculation would otherwise depend on arbitrary coordinates, representatives or generators. The theorem usually identifies an invariant that survives those choices.
How can a final answer be checked?
Substitute the result back into the defining relation, verify any required closure or map property, and check edge cases such as zero, the identity, the empty object or degenerate quotients where relevant.
Connections within the handbook
Source basis: supplied advanced algebra reference. Source-identifying authorship, publisher information, acknowledgements and biographical material are intentionally omitted. Mathematical terminology and results are retained in handbook form.
