Brauer group of a field of power series in two variables. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:01:24Z http://mathoverflow.net/feeds/question/78218 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78218/brauer-group-of-a-field-of-power-series-in-two-variables Brauer group of a field of power series in two variables. Joël 2011-10-15T18:32:40Z 2011-10-18T19:28:03Z <p>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? </p> <p>(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 $k$)</p> http://mathoverflow.net/questions/78218/brauer-group-of-a-field-of-power-series-in-two-variables/78478#78478 Answer by Torsten Ekedahl for Brauer group of a field of power series in two variables. Torsten Ekedahl 2011-10-18T19:28:03Z 2011-10-18T19:28:03Z <p>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).</p> <p>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.</p> <p>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 $p=2$):</p> <p>$H^0(Z,Z^1)$ 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 $H^1(Z,\nu)$.</p> <p>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 $<p>We can then twist this by$\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)$.</p> <p>There is however a different way of getting explicit representatives. For this we realise instead${}_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\$.</p>