## 58.60 Brauer groups

Brauer groups of fields are defined using finite central simple algebras. In this section we review the relevant facts about Brauer groups, most of which are discussed in the chapter Brauer Groups, Section 11.1. For other references, see , or [Weil].

Theorem 58.60.1. Let $K$ be a field. For a unital, associative (not necessarily commutative) $K$-algebra $A$ the following are equivalent

1. $A$ is finite central simple $K$-algebra,

2. $A$ is a finite dimensional $K$-vector space, $K$ is the center of $A$, and $A$ has no nontrivial two-sided ideal,

3. there exists $d \geq 1$ such that $A \otimes _ K \bar K \cong \text{Mat}(d \times d, \bar K)$,

4. there exists $d \geq 1$ such that $A \otimes _ K K^{sep} \cong \text{Mat}(d \times d, K^{sep})$,

5. there exist $d \geq 1$ and a finite Galois extension $K \subset K'$ such that $A \otimes _{K'} K' \cong \text{Mat}(d \times d, K')$,

6. there exist $n \geq 1$ and a finite central skew field $D$ over $K$ such that $A \cong \text{Mat}(n \times n, D)$.

The integer $d$ is called the degree of $A$.

Proof. This is a copy of Brauer Groups, Lemma 11.8.6. $\square$

Lemma 58.60.2. Let $A$ be a finite central simple algebra over $K$. Then

$\begin{matrix} A \otimes _ K A^{opp} & \longrightarrow & \text{End}_ K(A) \\ \ a \otimes a' & \longmapsto & (x \mapsto a x a') \end{matrix}$

is an isomorphism of algebras over $K$.

Proof. See Brauer Groups, Lemma 11.4.10. $\square$

Definition 58.60.3. Two finite central simple algebras $A_1$ and $A_2$ over $K$ are called similar, or equivalent if there exist $m, n \geq 1$ such that $\text{Mat}(n \times n, A_1) \cong \text{Mat}(m \times m, A_2)$. We write $A_1 \sim A_2$.

By Brauer Groups, Lemma 11.5.1 this is an equivalence relation.

Definition 58.60.4. Let $K$ be a field. The Brauer group of $K$ is the set $\text{Br} (K)$ of similarity classes of finite central simple algebras over $K$, endowed with the group law induced by tensor product (over $K$). The class of $A$ in $\text{Br}(K)$ is denoted by $[A]$. The neutral element is $[K] = [\text{Mat}(d \times d, K)]$ for any $d \geq 1$.

The previous lemma implies that inverses exist and that $-[A] = [A^{opp}]$. The Brauer group of a field is always torsion. In fact, we will see that $[A]$ has order dividing $\deg (A)$ for any finite central simple algebra $A$ (see Lemma 58.61.2). In general the Brauer group is not finitely generated, for example the Brauer group of a non-Archimedean local field is $\mathbf{Q}/\mathbf{Z}$. The Brauer group of $\mathbf{C}(x, y)$ is uncountable.

Lemma 58.60.5. Let $K$ be a field and let $K^{sep}$ be a separable algebraic closure. Then the set of isomorphism classes of central simple algebras of degree $d$ over $K$ is in bijection with the non-abelian cohomology $H^1(\text{Gal}(K^{sep}/K), \text{PGL}_ d(K^{sep}))$.

Sketch of proof.. The Skolem-Noether theorem (see Brauer Groups, Theorem 11.6.1) implies that for any field $L$ the group $\text{Aut}_{L\text{-Algebras}}(\text{Mat}_ d(L))$ equals $\text{PGL}_ d(L)$. By Theorem 58.60.1, we see that central simple algebras of degree $d$ correspond to forms of the $K$-algebra $\text{Mat}_ d(K)$. Combined we see that isomorphism classes of degree $d$ central simple algebras correspond to elements of $H^1(\text{Gal}(K^{sep}/K), \text{PGL}_ d(K^{sep}))$. For more details on twisting, see for example . $\square$

If $A$ is a finite central simple algebra of degree $d$ over a field $K$, we denote $\xi _ A$ the corresponding cohomology class in $H^1(\text{Gal}(K^{sep}/K), \text{PGL}_ d(K^{sep}))$. Consider the short exact sequence

$1 \to (K^{sep})^* \to \text{GL}_ d(K^{sep}) \to \text{PGL}_ d(K^{sep}) \to 1,$

which gives rise to a long exact cohomology sequence (up to degree 2) with coboundary map

$\delta _ d : H ^1(\text{Gal}(K^{sep}/K), \text{PGL}_ d(K^{sep})) \longrightarrow H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*).$

Explicitly, this is given as follows: if $\xi$ is a cohomology class represented by the 1-cocycle $(g_\sigma )$, then $\delta _ d(\xi )$ is the class of the 2-cocycle

58.60.5.1
\begin{equation} \label{etale-cohomology-equation-two-cocycle} (\sigma , \tau ) \longmapsto \tilde g_\sigma ^{-1} \tilde g_{\sigma \tau } \sigma (\tilde g_\tau ^{-1}) \in (K^{sep})^* \end{equation}

where $\tilde g_\sigma \in \text{GL}_ d(K^{sep})$ is a lift of $g_\sigma$. Using this we can make explicit the map

$\delta : \text{Br}(K) \longrightarrow H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*), \quad [A] \longmapsto \delta _{\deg A} (\xi _ A)$

as follows. Assume $A$ has degree $d$ over $K$. Choose an isomorphism $\varphi : \text{Mat}_ d(K^{sep}) \to A \otimes _ K K^{sep}$. For $\sigma \in \text{Gal}(K^{sep}/K)$ choose an element $\tilde g_\sigma \in \text{Gl}_ d(K^{sep})$ such that $\varphi ^{-1} \circ \sigma (\varphi )$ is equal to the map $x \mapsto \tilde g_\sigma x \tilde g_\sigma ^{-1}$. The class in $H^2$ is defined by the two cocycle (58.60.5.1).

Theorem 58.60.6. Let $K$ be a field with separable algebraic closure $K^{sep}$. The map $\delta : \text{Br}(K) \to H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*)$ defined above is a group isomorphism.

Sketch of proof. In the abelian case ($d = 1$), one has the identification

$H^1(\text{Gal}(K^{sep}/K), \text{GL}_ d(K^{sep})) = H_{\acute{e}tale}^1(\mathop{\mathrm{Spec}}(K), \text{GL}_ d(\mathcal{O}))$

the latter of which is trivial by fpqc descent. If this were true in the non-abelian case, this would readily imply injectivity of $\delta$. (See [SGA4.5].) Rather, to prove this, one can reinterpret $\delta ([A])$ as the obstruction to the existence of a $K$-vector space $V$ with a left $A$-module structure and such that $\dim _ K V = \deg A$. In the case where $V$ exists, one has $A \cong \text{End}_ K(V)$. For surjectivity, pick a cohomology class $\xi \in H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*)$, then there exists a finite Galois extension $K \subset K' \subset K^{sep}$ such that $\xi$ is the image of some $\xi ' \in H^2(\text{Gal}(K'|K), (K')^*)$. Then write down an explicit central simple algebra over $K$ using the data $K', \xi '$. $\square$

Comment #4087 by cristian d.gonzalez-aviles on

Please, where can I find a proof of the statement

The Brauer group of C(x,y) is uncountable?

Many thanks

Comment #4092 by on

This would follow from the Artin-Mumford sequence: for every 2-torsion point on a plane elliptic curve we get an element of $\mathrm{Br}(\mathbb{C}(x,y))$ (and they are different), but there are uncountably many of such curves because they are parametrised by the $j$-line.

Comment #4093 by on

Thanks Pieter! Another way to see this is to show that the quaternion algebras $Q_\lambda = \mathbf{C}(x, y)\langle u, v \rangle/(u^2 - x, v^2 - y + \lambda, uv + vu)$ are all pairwise not isomorphic for $\lambda \in \mathbf{C}$. To do this you can try to show that the quadratic extension of $\mathbf{C}(x, y)$ gotten by adjoining the square root of $y - \mu$ is contained in $Q_\lambda$ if and only if $\lambda = \mu$.

Comment #5066 by Tom Graber on

The statement of Thm 03R7 here claims that the map is a group isomorphism, but the sketch of proof gives no hint of how to compare the group law on the two sides.

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