# Brauer group of a field of power series in two variables.

Let $k$ be the field $F_2((X,Y))$, where $F_2$ is the field with two elements and $X$ and $Y$ are two indeterminates. Can we describe the Brauer group of $k$, or at least its $2$-torsion?

(My motivation is as one could expect: I have an irreducible representation of a group of dimension 2 over the separable closure of $k$, whose trace is in k, and I am trying to determine whether or not it is realizable over $k$)

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The Brauer group of a field of characteristic $p > 0$ is $p$-divisible. Perhaps you can use the (characteristic $p$ version) of the Milnor conjecture. There is also an exact sequence which may help: [Milnor, 1970], p. 327, Lemma 2.6 und [Gille-Szamuely 2006], p. 189, Corollary 7.1.10. – Timo Keller Oct 15 '11 at 19:11
Another idea: If $(R,\mathfrak{m})$ is Henselian (e.g. complete) local, one has $\mathrm{Br}(R) = \mathrm{Br}(R/\mathfrak{m})$. Also $\mathrm{Br}(R) \hookrightarrow \mathrm{Br}(\eta)$. Perhaps this helps, setting $R = \mathbf{F}_2[[X,Y]]$? – Timo Keller Oct 15 '11 at 19:39
If you could relate $\mathrm{Br}(R)$ to the Brauer group of its completion, you could apply Theorem 3.1 of Grothendieck, Dix Exposes, p. 98 to the morphism $\mathrm{Spec}(\mathbf{F}_2[X,Y]) \to \mathrm{Spec}(\mathbf{F}_2[X])$ (if it were proper). – Timo Keller Oct 15 '11 at 19:53

You can get a fairly good picture of the elements of order $2$ of the Brauer group in the following way. There is no reason to fixate on characteristic $2$ so I assume that we are dealing with $K:=\mathbb F_p((X,Y))$ and in fact the only reason to stick to the prime field is notational convenience as the Frobenius map on $\mathbb F_p$ is the identity so the relative Frobenius is equal to the absolute one which means that I won't have to distinguish between $Z$ and $Z^{(p)}$. Technically it may be that proper references would require $Z$ to be of finite type over the base but everything I say will be clearly true also for $Z=\text{Spec}\mathbb F_p((X,Y))$. (On the other hand the only thing I will need of $Z$ except for a smoothness/regularity assumption is that it is affine and with trivial Picard group).

We define the sheaf (on the small étale site of $Z$) $\nu$ by the exact sequence $$0\rightarrow\mathcal O_Z^\ast\xrightarrow{p}O_Z^\ast\rightarrow\nu\rightarrow0.$$ We have a map $\text{dlog}\colon\mathcal O_Z^\ast\rightarrow\Omega^1_Z$, where $\text{dlog}(f):=df/f$ and it factors to give an injection $\nu\subseteq \Omega^1_Z$. More precisely, it lands in the subsheaf $Z^1$ of closed forms and we have an exact sequence (again on the small étale site): $$0\rightarrow\nu\rightarrow Z^1\xrightarrow{C-\iota}\Omega^1_Z\rightarrow0,$$ where $C\colon Z^1/B^1\rightarrow\Omega^1_Z$, is the Cartier isomorphism ($B^1$ being the exact $1$-forms) and $\iota\colon Z^1\subseteq \Omega^1_Z$ is the inclusion.

Now, the first sequence (and the fact that $\text{Pic}(Z)=0$) gives that $H^1(Z,\nu)$ is the kernel of multiplication by $p$ on the Brauer group. The fact that $Z$ is affine gives that $H^1(Z,Z^1)=0$ and hence the second sequence gives that $H^1(Z,\nu)$ is the cokernel of $C-\iota\colon H^0(Z,Z^1)\rightarrow H^0(Z,\Omega^1_Z)$. This cokernel can be made very explicit (and to make it very explicit we temporaritly assume $p=2$):

$H^0(Z,Z^1)$ is a module over $K$, where scalar multiplication is given by the square map $f\cdot\omega=f^2\omega$, and has a basis given by $dX$, $dX/X$, $d(XY)$, $dY$ and $dY/Y$. We have that $C$ is $0$ on $dX$, $d(XY)$ and $dY$ and $C(dX/X)=dX/X$ and $C(dY/Y)=dY/Y$. Furthermore, $C$ is linear in the sense that $C(f^2\omega)=fC(\omega)$. This implies that the relations in the cokernel are given by $f^2dX=0$, $f^2dY=0$, $f^2XY(dX/X+dY/Y)=0$, $(f^2-f)dX/X=0$ and $(f^2-f)dY/Y=0$ (where $dX$, $dY$, $dX/X$, $dY/Y$, $XdY$ and $YdX$ is a $K$-basis for $H^0(Z,\Omega^1_Z)$). This allows for a fairly transparent normal form for elements in $H^1(Z,\nu)$.

If one wants a direct description of the central simple algebra associated to an element $\omega\in H^0(Z,\Omega^1_Z)$ one can apply the What Else Can It Be-principle (a very useful though somewhat dangerous principle, in this case it is probably OK). Recall that we have the algebra $\mathcal D$ of differential operators of order $<p$. It is the ring generated by derivations and elements of $K$ with the relations $DE-ED=[D,E]$, $Df-fD=D(f)$ and $D^p=D^{[p]}$, where $D^{[p]}$ is the $p$'th power as derivation. Note that the center consists of $K^p$, the subfield of $p$'th powers of $K$ and $K$ is in a natural fashion a $\mathcal D$-module giving an isomorphism $\mathcal D\rightarrow\text{End}_{K^p}(K)$ so that it is a central simple algebra (over $K^p$ which however is isomorphic to $K$).

We can then twist this by $\omega$ by replacing the last relation with $D^p=D^{[p]}+\omega(D)^p$. This still gives a central simple algebra and it should (as I said according to the WECIB-principle) be the associated element of ${}_p\text{Br}(K)$.

There is however a different way of getting explicit representatives. For this we realise instead ${}_p\text{Br}(K)$ as $H^2(Z,\mu_p)$ (now in the flat topology). We then have the usual cup product map $H^1(Z,\mathbb Z/p)\bigotimes H^1(Z,\mu_p)\rightarrow H^2(Z,\mu_p)$. We can represent elements of $H^1(Z,\mathbb Z/p)$ by Artin-Schreier extensions $b^p-b=a$, where $a\in K$ and elements of $H^1(Z,\mu_p)$ by $p$'th root extensions $g^p=f$, where $f\in K^\ast$. The central simple algebra associated to the cup product of these two classes is the algebra generated by $K(b)$ and $g$ and relations $gbg^{-1}=b+1$ and $g^p=f$. On the other hand a straightforward computation shows that the class of the cup product in $H^1(Z,\nu)$ is the residue of $adf/f\in H^0(Z,\Omega^1_Z)$. As $H^0(Z,\Omega^1_Z)$ is generated as a group by such elements we get a different description of the class of ${}_p\text{Br}(K)$ associated to elements of $H^0(Z,\Omega^1_Z)$. Note however, that they will not be isomorphic as algebras, the algebra associated by the first procedure to $adf/f$ has $K$-dimension $p^4$ whereas the second construction has dimension $p^2$.

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See also [Gille-Szamuely, 9.2]. – Timo Keller Oct 18 '11 at 19:49
Now that I have looked at it Gille-Szamuely is indeed a proper reference. I kind of expected there to be a reference but just enjoyed myself with figuring things out and felt that I might as well share... – Torsten Ekedahl Oct 19 '11 at 8:00