Brauer Groups — Mathematics Knowledge System

Overview

The Brauer group Br(k) of a field k is the group of equivalence classes of central simple algebras over k, under the operation of tensor product. Central simple algebras generalise matrix algebras and division algebras. The Brauer group measures the 'arithmetic complexity' of k: Br(R) = 0 (all CSAs are matrix algebras), Br(C) = 0, Br(R) = Z/2 (the only non-trivial class is the Hamilton quaternions). For number fields, the Brauer group is computed by class field theory and plays a key role in the arithmetic of quadratic forms and representation theory in characteristic 0.

Intuition

A central simple algebra (CSA) over k is an algebra A that is simple (no two-sided ideals) and has centre exactly k. The simplest examples are matrix algebras M_n(k) (these are 'trivial' in the Brauer group). Non-trivial elements come from division algebras like the Hamilton quaternions H over R. Two CSAs A and B are Brauer equivalent if A tensored with some matrix algebra = B tensored with some matrix algebra. The Brauer group is then the set of equivalence classes, with group operation = tensor product over k.

Formal Definition

Definition

An algebra A over a field k is a central simple algebra (CSA) if A is finite-dimensional, simple, and Z(A) = k. By Wedderburn's theorem, every CSA is isomorphic to M_n(D) for a unique (up to isomorphism) division algebra D over k. Two CSAs A and B are Brauer equivalent (A ~ B) if A tensor M_m(k) is isomorphic to B tensor M_n(k) for some m, n. The Brauer group Br(k) = {CSAs over k}/~ with group operation [A]*[B] = [A tensor_k B]. The identity is [M_1(k)] = [k], and [A]^{-1} = [A^op] (the opposite algebra).

\mathrm{Br}(k) = \{\text{CSAs over }k\}/\!\sim,\quad [A]\cdot[B] = [A\otimes_k B]

Brauer group definition

\mathrm{Br}(k) \cong H^2(\mathrm{Gal}(k^s/k),\, (k^s)^\times)

Brauer group = second Galois cohomology

\mathrm{Br}(\mathbb{R}) \cong \mathbb{Z}/2,\quad \mathrm{Br}(\mathbb{C}) = 0,\quad \mathrm{Br}(\mathbb{F}_q) = 0

Examples

\mathrm{Br}(k) \hookrightarrow \bigoplus_v \mathrm{Br}(k_v) \xrightarrow{\sum \mathrm{inv}_v} \mathbb{Q}/\mathbb{Z} \to 0

Exact sequence for number fields (Albert-Brauer-Hasse-Noether)

Notation

Notation	Meaning
$\mathrm{Br}(k)$	Brauer group of field k
$A \sim B$	A and B are Brauer equivalent CSAs
$A^{\mathrm{op}}$	Opposite algebra (inverse in Brauer group)
$\mathrm{inv}_v$	Local invariant map at place v

Theorems

Theorem 1: Wedderburn's Structure Theorem

Every central simple algebra A over a field k is isomorphic to M_n(D) for some n >= 1 and some division algebra D over k (both n and D up to isomorphism are unique). In particular, any CSA is a matrix algebra over a division algebra. The Brauer class of A = [D].

Theorem 2: Brauer = Second Galois Cohomology

For a field k with separable closure k_s and absolute Galois group G_k = Gal(k_s/k), there is a natural isomorphism Br(k) = H^2(G_k, (k_s)^times) where (k_s)^times is the multiplicative group with the natural Galois action. This identifies the Brauer group with the 'twisted' G_k-cohomology.

Theorem 3: Albert-Brauer-Hasse-Noether Theorem

For a number field k, a central simple algebra A over k is trivial (isomorphic to M_n(k)) if and only if it is trivial locally at every place v (i.e., A tensor_k k_v is isomorphic to M_n(k_v) for all completions k_v). This is the Hasse principle for CSAs and gives an exact sequence 0 -> Br(k) -> direct sum_v Br(k_v) -> Q/Z -> 0.

Worked Examples

1
H = {a + bi + cj + dk : a,b,c,d in R} with i^2=j^2=k^2=-1 and ij=k, jk=i, ki=j.
2
H is simple (no proper two-sided ideals) with centre Z(H) = R. So H is a CSA over R.
$Z(\mathbb{H}) = \mathbb{R},\quad \dim_{\mathbb{R}} \mathbb{H} = 4$
3
H is a division algebra (every non-zero element has an inverse: (a+bi+cj+dk)^{-1} = (a-bi-cj-dk)/(a^2+b^2+c^2+d^2)).
$x^{-1} = \bar{x}/|x|^2$
4
H is not isomorphic to M_2(R) as M_2(R) has zero divisors. So [H] is a non-trivial element of Br(R).
5
[H]^2 = [H tensor_R H] = [M_4(R)] = 0 by a standard computation. So Br(R) has an element of order 2; in fact Br(R) = Z/2 = {[R], [H]}.
$\mathrm{Br}(\mathbb{R}) = \{[\mathbb{R}],[\mathbb{H}]\} \cong \mathbb{Z}/2$

✓ Answer

H is a division CSA over R of dimension 4, giving the unique non-trivial element of Br(R) = Z/2.

Practice Problems

Hardfree response

Explain why Br(C) = 0 and Br(F_q) = 0 for finite fields F_q.

Common Mistakes

Common Mistake

Thinking Br(k) consists of all division algebras over k.

Br(k) consists of Brauer equivalence classes of central simple algebras. Each class is represented by a unique division algebra D (by Wedderburn), so there is a bijection between Br(k) and isomorphism classes of central division algebras over k. But the group operation is [A]*[B] = [A tensor B], not direct product of division algebras.

Quiz

Two central simple algebras A and B over k are Brauer equivalent if:

A is isomorphic to BA tensor M_m(k) is isomorphic to B tensor M_n(k) for some m,nA and B have the same dimensionA tensor B is isomorphic to a matrix algebra

Historical Background

Richard Brauer introduced the group now bearing his name in 1929 as part of his study of division algebras over number fields. The Brauer group was computed for number fields by the Albert-Brauer-Hasse-Noether theorem (1932), a triumph of class field theory. Artin and Tate showed the Brauer group equals the second Galois cohomology group H^2(Gal(k_s/k), k_s^*). Grothendieck extended the theory to schemes in the 1960s, defining the cohomological Brauer group.

1929
Brauer introduces the group of division algebras over a field
Richard Brauer
1932
Albert-Brauer-Hasse-Noether theorem: Brauer group of number fields computed
Abraham Adrian Albert, Richard Brauer, Helmut Hasse, Emmy Noether
1950s
Brauer group identified with H^2(Gal(k_s/k), k_s^*) via Galois cohomology
Emil Artin, John Tate
1968
Grothendieck extends Brauer group to algebraic geometry
Alexander Grothendieck

Summary

Br(k) = equivalence classes of central simple algebras (CSAs) over k, under [A]*[B]=[A tensor_k B].
Every CSA is M_n(D) for a unique division algebra D; [A]=[D] depends only on D.
Br(R) = Z/2 (represented by H), Br(C) = Br(F_q) = 0.
Br(k) = H^2(Gal(k_s/k), k_s^*) -- the second Galois cohomology group.

References

BookGille, P. and Szamuely, T. Central Simple Algebras and Galois Cohomology. Cambridge, 2006.
BookJacobson, N. Basic Algebra II. Freeman, 1980.
WebsiteWikipedia -- Brauer group