A computation a day: the Brauer group of a number ring

In this post we compute the group \mathrm{Br}(\mathcal{O}_K) where K is a number field.

A reminder on some definitions

The Brauer group of a scheme X has several equivalent definitions. The first, the one most similar to the case one usually encounters (the Brauer groups of fields), involves so-called Azumaya algebras. Namely, let us call a quasicoherent sheaf of \mathcal{O}_X-algebras \mathcal{A} an Azumaya algebra of rank n if there is a covering \{U_i\} of X in the étale topology such that \mathcal{A}_{U_i}\cong \mathrm{Mat}_n(\mathcal{O}_{U_i}).

As an example, if X=\mathrm{Spec}(k), where k is a field, one can show that a k-algebra A satisfying the property that A\otimes_k\overline{k}=\mathrm{Mat}_n(\overline{k}) must also satisfy the property A\otimes_k K\cong \mathrm{Mat}_n(K) for some finite Galois extension K/k. In particular, the Azumaya algebras on \mathrm{Spec}(k) are just the central simple algebras on k.

We shall call two Azumaya algebras equivalent if they are Morita equivalent. Specifically, we say \mathcal{A} is equivalent to \mathcal{B} if there are vector bundles \mathcal{E},\mathcal{E}' on X such that

\mathcal{A}\otimes_{\mathcal{O}_X}\mathrm{End}_{\mathcal{O}_X}(\mathcal{E})\cong \mathcal{B}\otimes_{\mathcal{O}_X}\mathrm{End}_{\mathcal{O}_X}(\mathcal{E}')

as sheaves of algebras. One can show that the set of Azumaya algebras, up to equivalence, is a group under tensor product. We denote this group as \mathrm{Br}(X).

Again, as an example, we see that if X=\mathrm{Spec}(k) then the above becomes equivalent to the statement that two central simple algebras A,B are equivalent if, as k-algebras,


for some m,n. Thus, \mathrm{Br}(\mathrm{Spec}(k))=\mathrm{Br}(k).

Finally, if X=\mathrm{Spec}(R), we will often times denote \mathrm{Br}(X) as \mathrm{Br}(R). This will cause no confusion in the classical case of fields, as mentioned in the last paragraph.

Some useful tools

Brauer group equals cohomological Brauer group (for affine schemes)

Before we go galavanting into the computation of \mathrm{Br}(\mathcal{O}_K) we recall some important facts which will make the computation possible.

First, we have the following theorem:

Theorem: Let X=\mathrm{Spec}(R). Then, there is an isomorphism:

\mathrm{Br}(R)\cong H^2(X_\mathrm{\acute{e}t},\mathbf{G}_m)

I am not positive to who the attribution of this theorem should go. It is certainly proven, in greater generality than this, in Gabber’s thesis (the results of which, as well as a generalization of, can be found here). While this theorem is difficult to prove in general, one can fairly easily describe the map \mathrm{Br}(X)\to H^2(X_\mathrm{\acute{e}t},\mathbf{G}_m) which ends up being the isomorphism.

Note first that since rank n Azumaya algebras on X are étale locally \mathrm{Mat}_n, whose automorphism group is \mathrm{PGL}_n, we get a bijection (admittedly with some detail checking)

\left\{\begin{matrix}\text{rank }n\text{ Azumaya}\\ \text{algebras on }X\end{matrix}\right\}\leftrightarrow H^1(X_{\mathrm{\acute{e}t}},\mathrm{PGL}_n)

Thus, we get a bijection

\mathrm{Br}(X)\xrightarrow{\approx}H^1(X_{\mathrm{\acute{e}t}},\mathrm{PGL}_\infty)=\varinjlim H^1(X_\mathrm{\acute{e}t},\mathrm{PGL}_n)

One can show that this is even an isomorphism of groups when one gives the right hand side the group structure coming from the natural map


which endows the right hand side (together with the cup-product) with a group structure.

Now, the map \mathrm{Br}(X)\to H^2(X_{\mathrm{\acute{e}t}},\mathbf{G}_m) is just the natural map one obtains from the above equivalence and the natural maps

H^1(X_{\mathrm{\acute{e}t}},\mathrm{PGL}_n)\to H^2(X_{\mathrm{\acute{e}t}},\mathbf{G}_m)

one obtains from short exact sequence of sheaves

1\to \mathbf{G}_m\to\mathrm{GL}_n\to\mathrm{PGL}_n\to 1

which is still exact on the étale site.

A little class field theory

The key result for this computation will be the following ‘fundamental exact sequence of (global) class field theory’

Theorem: Let K be a global field. Then, the following sequence is short exact

\displaystyle 0\to\mathrm{Br}(K)\to\bigoplus_{v}\mathrm{Br}(K_v)\to\mathbb{Q}/\mathbb{Z}\to 0

where the first map is A\mapsto (A\otimes_K K_v), v ranges over the places of K, and the last map is the so-called ‘invariant’ map. To be more specific, if one identifies each \mathrm{Br}(K_v) with \mathbb{Q}/\mathbb{Z} (which one can do), the invariant map is just summing the coordinates.

Note that this is indeed a deep theorem. Contained in its fold (specifically the injectivity of the first map) is the Brauer-Hasse-Arf theorem that a central simple algebra A/K is split (i.e. is a matrix algebra over K) if and only if it is split locally (i.e. A\otimes_K K_v is split for all v). Applying this to the case of quaternion algebras over K yields the famous Hasse pricinple for quadratic forms: that a quadratic form has a solution in K if and only if it has a solution in K_v for all v.

Now, using the fact that

\mathrm{Br}(K_v)=\begin{cases}\mathbb{Q}/\mathbb{Z} & \mbox{if}\quad v\text{ is finite}\\ \mathbb{Z}/2\mathbb{Z} & \mbox{if}\quad v\text{ is real}\\ 0 & \mbox{if}\quad v\text{ is complex}\end{cases}

we can non-canonically write the above sequence as

\displaystyle 0\to\mathrm{Br}(K)\to \left(\bigoplus_{\mathfrak{p}\in\mathrm{Spec}(\mathcal{O}_K)}\mathbb{Q}/\mathbb{Z}\right)\oplus \left(\mathbb{Z}/2\mathbb{Z}\right)^r \to \mathbb{Q}/\mathbb{Z}\to 0

where r is the number of real places of K.

A short exact sequence

The last important tool for us will be the existence of a certain short exact sequence of sheaves on X_{\mathrm{\acute{e}t}}. This is a sequence that anyone familiar with the computation of the étale cohomology of curves (cf. SGA 4.5) will likely remember.

Theorem: Let X be an integral, separated, locally factorial scheme with function field K. Let \eta:\mathrm{Spec}(K)\to X denote the inclusion of its generic point. Then, the following sequence is exact:

0\to \mathbf{G}_{m,X}\to \eta_\ast \mathbf{G}_{m,\mathrm{Spec}(K)}\to \mathrm{Div}_X\to 0

where the first map is the obvious one, and the second map is the ‘divisor’ map.

Here \mathrm{Div}_X is the sheaf

\displaystyle \bigoplus_{\stackrel{x\in X}{\mathrm{codim}_X(x)=1}}(i_x)_\ast \underline{\mathbb{Z}}

is the ‘sheaf of Weil divisors’ on X.

Proving this theorem is not difficult, and is a good exercise in working in the étale topology.


So, we can finally get to the computation of \mathrm{Br}(\mathcal{O}_K)=H^2\left(\left(\mathrm{Spec}(\mathcal{O}_K)\right)_{\mathrm{\acute{e}t}},\mathbf{G}_m\right).

So, we begin, as we usually do, with the relevant exact sequence. Thus, if X=\mathrm{Spec}(\mathcal{O}_K), and everything else is in the last section, then we have the following exact sequence

\displaystyle 0\to \mathcal{O}_K^\times\to K^\times\to \bigoplus_{\mathfrak{p}\in\mathrm{MaxSpec}(\mathcal{O}_K)}\mathbb{Z}\to H^1(X_\mathrm{\acute{e}t},\mathbf{G}_{m,X})\to H^1(X,\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)})\to H^1(X_{\mathrm{\acute{e}t}},\mathrm{Div}_X)\to H^2(X_\mathrm{\acute{e}t},\mathbf{G}_{m,X})\to H^2(X_{\mathrm{\acute{e}t}},\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)})\to H^2(X_{\mathrm{\acute{e}t}},\mathrm{Div}_X)

Gussying this up a little bit gives:

\displaystyle 0\to \mathcal{O}_K^\times\to K^\times\to \mathbb{Z}^{\mathrm{MaxSpec}(\mathcal{O}_K)}\to \mathrm{Cl}(K)\to H^1(X_{\mathrm{\acute{e}t}},\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)})\to H^1(X_{\mathrm{\acute{e}t}},\mathrm{Div}_X)\to \mathrm{Br}(\mathcal{O}_K)\to H^2(X_{\mathrm{\acute{e}t}},\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)})\to H^2(X,\mathrm{Div}_X)

Now, some of these terms are easy to compute. For example,

\displaystyle \begin{aligned}H^1(X_{\mathrm{\acute{e}t}},\mathrm{Div}_X) &=\bigoplus_{\mathfrak{p}\in\mathrm{MaxSpec}(\mathcal{O}_K)}H^1\left(X_{\mathrm{\acute{e}t}},(i_\mathfrak{p})_\ast\underline{\mathbb{Z}}\right)\\ &=\bigoplus_{\mathfrak{p}\in\mathrm{MaxSpec}(\mathcal{O}_K)}H^1\left(\left(\mathrm{Spec}(\mathcal{O}_K/\mathfrak{p})\right)_{\mathrm{\acute{e}t}},\underline{\mathbb{Z}}\right)\end{aligned}

where the first isomorphism is because cohomology commutes with direct sums in this case, and the second since i_\mathfrak{p} is a finite map.

But, using the fact that étale cohomology on the spectrum of a field is just Galois cohomology, and under this correspondence constant sheaves go to trivial modules we see that

\displaystyle \begin{aligned}H^1(\mathrm{Spec}(\mathcal{O}_k/\mathfrak{p}),\underline{\mathbb{Z}}) &= H^1(G_{\mathcal{O}_K/\mathfrak{p}},\mathbb{Z})\\ &= \mathrm{Hom}_{\mathrm{cont.}}(\widehat{\mathbb{Z}},\mathbb{Z})\\ &=0\end{aligned}

The computation of the second cohomology of \mathrm{Div}_X succumbs to similar methods. Namely, the same reasoning shows that

\displaystyle H^2(X_{\mathrm{\acute{e}t}},\mathrm{Div}_X)=\bigoplus_{\mathfrak{p}\in\mathrm{MaxSpec}(\mathcal{O}_K)}H^2(\widehat{\mathbb{Z}},\mathbb{Z})

where the cohomology on the right hand side is continous cohomology with \mathbb{Z} a trivial module.

The last relevant calculation we need to do is the computation of H^2(X_{\mathrm{\acute{e}t}},\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)}). To do this we appeal to the Leray spectral sequence which gives us

H^p\left(X_{\mathrm{\acute{e}t}},R^q\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)}\right)\Rightarrow H^{p+q}((\mathrm{Spec}(K))_{\mathrm{\acute{e}t}},\mathbf{G}_{m,\mathrm{Spec}(K)})

But, we claim that R^q\eta_\ast\mathbf{G}_{m,\mathrm{Spec}(K)}=0 for q>0. To do this, it suffices to check that it vanishes on geometric points centered at closed points. So, let \overline{x} be a geometric point of X. Then, by standard theory we know that


where K_{\overline{x}} is the fraction field of strict henselization of X at \overline{x}. One can verify that if \overline{x} hits the point \mathfrak{p} of X then

\displaystyle K_{\overline{x}}=W\left(\overline{\mathcal{O}_K/\mathfrak{p}}\right)\left[\frac{1}{p}\right]

So, we need to prove that the Galois cohomology of \mathbf{G}_m with respect to K_{\overline{x}} is zero in positive degree. This is non-trivial. For q=1 this is Hilbert’s theorem 90. For q=2 we’re asking about the Brauer group of the fraction field of a strictly Henselian ring–this is zero (cf. Milne’s book). Then, in general, one can use the difficult fact that the cohomological dimension of K_{\overline{x}} is 1, together with the Kummer sequence to obtain the result.

Regardless, we see that the Leray spectral sequence degenerates on the first page and so we have that


Thus, we have proven that there is an exact sequence

\displaystyle 0\to \mathrm{Br}(\mathcal{O}_K)\to\mathrm{Br}(K)\to \bigoplus_{\mathfrak{p}}H^2(\widehat{\mathbb{Z}},\mathbb{Z})

This seems relatively unhelpful since we don’t have a handle on this second map. The key observation though is that


and that the diagram

\begin{matrix}\mathrm{Br}(K) & \to & \displaystyle \bigoplus_{\mathfrak{p}}H^2(\widehat{\mathbb{Z}},\mathbb{Z})\\ {^\text{id}}\downarrow & & \downarrow^{\approx}\\ \mathrm{Br}(K) & \to & \displaystyle \bigoplus_{\mathfrak{p}}\mathrm{Br}(K_\mathfrak{p})\end{matrix}

commutes. If this seems unbelievable, it’s because you’ve forgotten that this is precisely how one proves that \mathrm{Br}(K_\mathfrak{p})=\mathbb{Q}/\mathbb{Z} (see, for example, Serre’s relevant article in Cassels and Frolich)!

Thus, we see that we can rewrite our sequence as

\displaystyle 0\to\mathrm{Br}(\mathcal{O}_K)\to\mathrm{Br}(K)\to\bigoplus_{\mathfrak{p}}\mathrm{Br}(K_\mathfrak{p})

But, combining this with the fundamental sequence of class field theory we obtain the following short exact sequence

0\to \mathrm{Br}(\mathcal{O}_K)\to \left(\mathbb{Z}/2\mathbb{Z}\right)^{r}\to \left(\mathbb{Z}/2\mathbb{Z}\right)^\delta\to 0

where \delta is zero or 1 depending on if K has a real place or not. But, this sequence splits (it’s a sequence of \mathbb{F}_2-spaces!) and so we can finally conclude that

\mathrm{Br}\left(\mathcal{O}_K\right)=\begin{cases}(\mathbb{Z}/2\mathbb{Z})^{r-1} & \mbox{if}\quad K\text{ has a real place}\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}

One comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s