most recent 30 from http://mathoverflow.net 2013-05-18T22:03:38Z http://mathoverflow.net/feeds/question/3007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3007/division-algebras-as-algebraic-groups Joel Dodge 2009-10-28T06:02:44Z 2009-10-28T11:38:00Z

<p>If I'm given a division algebra D with Z(D)=F, then how can I view D^x as an algebraic group defined over F? I'd like to see first how D^x can be given the structure of a variety defined over F, and then to see how the group law on D^x is defined over F. </p>

http://mathoverflow.net/questions/3007/division-algebras-as-algebraic-groups/3011#3011 S. Carnahan 2009-10-28T07:05:32Z 2009-10-28T07:05:32Z

<p>Suppose D splits over a finite extension K/F, i.e., the tensor product of D with K over F is isomorphic to M_n(K). Then D^x is the group of F-points of an algebraic group over F that exists as a direct factor (along with all other F-division algebras that split over K, and GL_n,F) in the restriction of scalars Res^K_F GL_n,K.</p> <p>I don't know an explicit presentation in general (say, starting from a Brauer class), although if K/F is a cyclic Galois extension, there is a nice cyclic algebra construction. I think more details can be found in Serre's <i>Local Fields</i> and Cornell-Silverman.</p>

http://mathoverflow.net/questions/3007/division-algebras-as-algebraic-groups/3035#3035 moonface 2009-10-28T11:38:00Z 2009-10-28T11:38:00Z

<p>Choose an F-basis of D. The multiplication is described by certain quadratic functions, with respect to this basis; D* is given by the nonvanishing of a polynomial function (the norm). So the multiplication can be understood as defining an algebraic group structure on the complement of a hypersurface in an affine space. </p>