Brauer groups and the reciprocity map

Section 7.6 Brauer groups and the reciprocity map

Reference.

[36] IV (for the general theory of Brauer groups); VII.7 and VII.8 (for the application to reciprocity). For the general theory, see also [24], Chapter 4.

We discuss Brauer groups of fields, especially number fields. On one hand these can be used to give an alternate construction of the global reciprocity map, not based on abstract class field theory; on the other hand, they carry important information from class field theory which is useful in numerous applications.

In this lecture, we reprise a bit of shorthand from Section 4.2, writing $H^i(L/K)$ to mean $H^i(\Gal(L/K), L^*)\text{.}$

Subsection The Brauer group of a field

Definition 7.6.1.

Recall from Definition 4.1.15 that we have defined the Brauer group of a field $K$ as the group

\begin{equation*} \Br(K) = H^2(\overline{K}/K) = \varinjlim_{L/K} H^2(L/K) \end{equation*}

where $L$ runs over finite Galois extensions of $K$ and the transition maps in the direct limit are inflation maps. By Lemma 1.2.3 and Proposition 4.2.14, these maps are all injective, so the direct limit is actually a union.

This definition of Brauer groups is not the original one; we give that next.

Lemma 7.6.2.

For any field $K\text{,}$ there is a natural bijection between $\Br(K)$ and the isomorphism classes of division algebras which are finite-dimensional $K$-algebras with center $K\text{.}$

Proof.

See [36], Corollary IV.3.16; [46], Chapter X, Proposition 9; or [24], Theorem 8.11.

Example 7.6.3.

For $K$ an algebraically closed field, every division algebra which is finite dimensional over $K$ is trivial. Namely, if $D$ is such an algebra, then for each $x \in D\text{,}$ multiplication by $x$ defines a $K$-linear endomorphism of $D\text{,}$ which necessarily has at least one eigenvalue $y \in K\text{.}$ Then $x-y$ is an element of $D$ which cannot be invertible (since multiplication by this element is a $K$-linear endomorphism of $D$ with 0 as an eigenvalue), so it must be zero; hence $x \in K\text{.}$

Example 7.6.4.

For $K$ a finite field, $\Br(K)$ is trivial. In the classical interpretation, this is Wedderburn’s theorem that every finite division algebra is commutative. In the cohomological interpretation, it follows from Proposition 4.2.4 via the periodicity of Tate groups Theorem 3.4.1.

Remark 7.6.5.

While Lemma 7.6.2 only characterizes the Brauer group as a set, the original construction of Brauer included the group structure. Namely, for any two central simple algebras $D_1, D_2$ over $K\text{,}$ we have an isomorphism of $K$-algebras

\begin{equation*} D_1 \otimes_K D_2 \cong M_n(D) \end{equation*}

for some positive integer $n$ and some division algebra $D$ with center $K\text{,}$ and $D$ is the product of $D_1$ and $D_2$ in $\Br(K)$ (in particular, it is characterized by this construction up to isomorphism).

In this construction, the identity element in $\Br(K)$ is $K$ viewed as a division algebra with itself as the center. The inverse element of a division algebra $D$ is the opposite algebra in which multiplication is reversed.

Remark 7.6.6.

The property of a field $K$ of characteristic 0 having trivial Brauer group is useful in the theory of finite group representations: for such a field, any $K$-valued character of a finite group arises from a representation defined over $K\text{.}$ (This follows from Schur’s lemma: the character in question appears within some irreducible $K$-linear representation, whose endomorphism ring is a division algebra; the triviality of the Brauer group forces this to split without any base extension.)

By contrast, for $G = \{\pm 1, \pm i, \pm j, \pm k\}$ the unit quaternion group, the standard 2-dimensional representation of $G$ has a $\QQ$-valued character but cannot be realized as a representation over $\QQ\text{.}$ In other words, this representation has nontrivial Schur index.

Remark 7.6.7.

One can also associate Brauer groups to arbitrary rings and even to schemes in algebraic geometry, by replacing division algebras (or more precisely, central simple algebras) with Azumaya algebras and Galois cohomology with étale cohomology. See [15].

Subsection The Brauer group of a number field

We state the formula for the Brauer group of a number field, and prove it modulo one key step.

Lemma 7.6.8.

Let $L/K$ be a cyclic extension of number fields of degree $n\text{.}$ Then there is a commutative diagram as in Figure 7.6.9 in which the vertical arrows are isomorphisms.

Proof.

The left square comes from applying Theorem 3.4.1 to the morphism $L^* \to I_L$ of $\Gal(L/K)$-modules. Since $r_{L/K}$ is defined in terms of local reciprocity maps, the right square comes from Lemma 4.2.21.

Theorem 7.6.10.

For any number field $K\text{,}$ the group $\Br(K)$ fits into an exact sequence of the form

\begin{equation*} 0 \to \Br(K) \to \Br(\AA_K) = \bigoplus_v \Br(K_v) \to \QQ/\ZZ \to 0 \end{equation*}

in which

\begin{equation*} \Br(K_v) = \begin{cases} \QQ/\ZZ & \mbox{for $v$ finite} \\ \frac{1}{2} \ZZ/\ZZ & \mbox{for $v$ real} \\ 0 & \mbox{for $v$ complex} \end{cases} \end{equation*}

and the map on the right is summation.

Proof.

The map $\Br(K) \to \bigoplus_v \Br(K_v)$ is the injection from Theorem 7.2.14. The value of $\Br(K_v)$ for $v$ finite is given by Lemma 4.2.21. For $v$ complex, it is evident that $\Br(K_v) = 0\text{.}$ For $v$ real, by Theorem 3.4.1 we have

\begin{align*} \Br(\RR) &= H^2(\Gal(\CC/\RR), \CC^*)\\ &\cong H^0_T(\Gal(\CC/\RR), \CC^*)\\ &\cong \RR^* / \Norm_{\CC/\RR} \CC^* = \RR^*/\RR^+ \cong \frac{1}{2} \ZZ/\ZZ. \end{align*}

Since the values of $\Br(K_v)$ are the ones given, the surjectivity of the map $\bigoplus_v \Br(K_v) \to \QQ/\ZZ$ is evident.

It remains to establish exactness at the middle of the sequence. For any finite Galois extension $L/K\text{,}$ we have the exact sequence

\begin{equation*} H^2(\Gal(L/K), L^*) \to H^2(\Gal(L/K), I_L) \to H^2(\Gal(L/K), C_L). \end{equation*}

If $L/K$ is cyclic, then by reciprocity (Theorem 7.3.8), the top row of the commutative diagram in Lemma 7.6.8 is exact, as then is the bottom row. Consequently, we could conclude the proof if we knew that every class in $\Br(K)$ is the image of a class in $H^2(\Gal(L/K), L^*)$ for some finite cyclic extension $L$ of $K\text{.}$ In fact, something even stronger is true; see Proposition 7.6.13.

Definition 7.6.11.

For $K$ a number field and $\alpha \in \Br(K)\text{,}$ the image of $\alpha$ in $\Br(K_v)$ is often called the local invariant of $\alpha$ at $v\text{.}$ The exact sequence appearing in Theorem 7.6.10 is sometimes called the fundamental exact sequence associated to $K\text{;}$ it can be viewed as another source of “reciprocity” in class field theory. For example, applying the fundamental exact sequence to a quaternion algebra over $\QQ$ (see Exercise 5) gives rise to Hilbert’s reformulation of the law of quadratic reciprocity using Hilbert symbols.

The fundamental exact sequence also plays a key role in various applications of Brauer groups in number theory. One of these is the detection of obstructions to the existence of rational points on algebraic varieties over number fields, called Brauer-Manin obstructions. This construction is based on the following observation: for $X$ an algebraic variety over a number field $K\text{,}$ each class in $\Br(X)$ gives rise to a commutative diagram

in which the vertical maps are evaluation maps and the bottom row is the fundamental exact sequence.

Subsection All Brauer classes are (cyclic) cyclotomic

Recall that in the cohomological approach to local class field theory, the crucial computation was that of the Brauer groups of local fields, which involved first studying unramified extensions and then transferring the knowledge to general extensions (see the proof of Proposition 4.2.18). The missing step in Theorem 7.6.10 is of a very similar nature, except that we have to vary the extension based on the class.

Proposition 7.6.13.

Let $L/K$ be a Galois extension of number fields. Then for any element $x$ of $H^2(L/K)\text{,}$ there exist a cyclic cyclotomic extension $M$ of $K$ and an element $y$ of $H^2(M/K)$ such that $x$ and $y$ map to the same element of $H^2(ML/K)\text{.}$

Proof.

By Theorem 7.2.14, any class in $H^2(L/K)$ is determined by its images in $H^2(L_w/K_v)$ for all places $v$ in $K$ (where $w$ denotes any place of $L$ above $v$), with only finitely many of these being nonzero. Moreover, a class in $H^2(L_w/K_v)$ of some order $m$ is killed by replacing $K_v$ by any extension of degree $m$ (by Remark 4.2.22 and Proposition 4.2.23; see also [36], Theorem III.2.1). It thus suffices to find a cyclic cyclotomic extension for which, for some fixed finite set of finite places $S$ of $K\text{,}$ the degrees $[L_w:K_v]$ for all $v \in S$ are conveniently large; for this, see Exercise 1. (Compare also [36], Proposition VII.7.2.)

Remark 7.6.14.

By Proposition 7.6.13, the field $\QQ^{\ab}$ has trivial Brauer group. Since in addition every complex character of a finite group has values in $\QQ^{\ab}\text{,}$ it follows that every irreducible complex representation of a finite group can be realized over $\QQ^{\ab}\text{;}$ for a more direct proof of this, see [45], Chapter 12, Theorem 24.

Subsection Local-global compatibility via Brauer groups

To conclude, we turn things around and show that Proposition 7.6.13 can be used to recover local-global compatibility for the reciprocity map (Proposition 6.4.5). This makes no use of abstract class field theory, although it does use the same inputs (notably the First and Second Inequality).

Proposition 7.6.15.

Let $K$ be a number field and put $L = K(\zeta_n)\text{.}$ Then $r_{L/K}: I_K \to \Gal(L/K)$ maps all principal idèles to the identity.

Proof.

For $K = \QQ\text{,}$ this follows from the explicit description of the Artin map given in Definition 1.1.7 (or from Lemma 7.5.3). In general, we have a commutative diagram

and we know the bottom row kills principal idèles and the right column is injective. Thus the top row kills principal idèles too.

Proposition 7.6.17.

For any cyclic extension $L/K$ of number fields, the map $r_{L/K}: I_K \to \Gal(L/K)$ kills principal idèles.

Proof.

To begin with, Proposition 7.6.15 implies that $r_{L/K}$ kills principal idèles whenever $L/K$ is a cyclotomic extension, and Lemma 7.6.8 implies that in this case the composite $H^2(L/K) \to \QQ/\ZZ$ along the bottom row of Figure 7.6.9 vanishes. By Proposition 7.6.13, the same then holds for any cyclic extension $L/K\text{.}$ By Lemma 7.6.8 again, the composition along the top row of Figure 7.6.9 vanishes, proving the claim.

Remark 7.6.18.

Let $L/K$ be a cyclic extension of number fields. At this point, $r_{L/K}$ kills both principal idèles (by Proposition 6.4.5) and norms (since it does so locally), so it induces a map $C_K/\Norm_{L/K} C_L \to \Gal(L/K)\text{.}$ By the surjectivity of the Artin map, as deduced from the First Inequality (Proposition 7.3.7), this map is surjective; by comparing orders using the Second Inequality (Theorem 7.2.10), we see that the map is also an isomorphism. This establishes local-global compatibility (Proposition 6.4.5) for cyclic extensions, from which it directly follows also for abelian extensions. Hooray again!

Remark 7.6.19.

Note that for a cyclic extension $L/K$ of number fields, Remark 7.6.18 establishes not just local-global compatbility, but the entire reciprocity isomorphism

\begin{equation*} C_K/\Norm_{L/K} C_L \cong \Gal(L/K)^{\ab} \end{equation*}

without use of abstract class field theory. One can say the same for an abelian extension: in this case, local reciprocity (Theorem 4.1.2) and Remark 7.6.18 together imply that we have a well-defined map. Using the cyclic case, we may see that this map is surjective; by Corollary 7.2.8 (a side effect of our proof of the Second Inequality), the map is forced to be an isomorphism.

It is less clear how to recover the norm limitation theorem, which is needed to prove the existence theorem. The difficulty is that if $L/K$ is not abelian and $M/K$ is its maximal abelian subextension, then the maximal abelian subextension of a completion of $L$ can be strictly larger than the corresponding completion of $M\text{;}$ so we cannot simply apply the local norm limitation theorem. Instead, one first uses the fundamental exact sequence (Theorem 7.6.10, whose proof depended on reciprocity only for cyclic extensions) to argue that $C_L$ satisfies the hypotheses of Tate’s theorem (Theorem 4.3.1), which yields an isomorphism

\begin{equation*} \Gal(L/K)^{\ab} \cong H^{-2}_T(\Gal(L/K), \ZZ) \to H^0_T(\Gal(L/K), C_L) = C_K/\Norm_{L/K} C_L. \end{equation*}

By comparing with the construction of the local reciprocity map, we see that the inverse of this isomorphism is exactly $r_{L/K}\text{,}$ which yields the norm limitation theorem. See [36], Theorem VIII.4.8 for more details.

Exercises Exercises

1.

Let $K$ be a number field, let $S$ be a finite set of finite places of $K\text{,}$ and let $m$ be a positive integer. Prove that there exists a subextension $L$ of $K^{\smcy}/K$ (which is necessarily cyclic) such that for all $v \in S\text{,}$ for some place $w$ of $L$ above $K\text{,}$ $[L_w:K_v]$ is divisible by $m\text{.}$

Hint.

See [36], Lemma VII.7.3.

2.

Let $D$ be a quaternion algebra over a field $K$ (see Exercise 5). Prove the following statements directly (without using Lemma 7.6.2).

$D$ is isomorphic to its opposite algebra.
There is an isomorphism $D \otimes_K D \cong M_4(K)$ of $K$-algebras. Consequently, if $D$ is not split, then it represents an element of $\Br(K)$ of order 2.

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