If I'm given a division algebra D with Z(D)=F, then how can I view Dx as an algebraic group defined over F? I'd like to see first how Dx can be given the structure of a variety defined over F, and then to see how the group law on Dx is defined over F.
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2 Answers
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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.
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1$\begingroup$ And for the record, this is the exact same way that you show GL_n is an algebraic group. $\endgroup$ Commented Oct 28, 2009 at 13:01
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$\begingroup$ Is it part of the general theory of division algebras that the norm is a polynomial function? $\endgroup$ Commented Oct 28, 2009 at 19:27
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2$\begingroup$ @Joel: After base change to a matrix ring, the norm becomes the determinant. $\endgroup$– S. Carnahan ♦Commented Oct 29, 2009 at 2:58
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Suppose D splits over a finite extension K/F, i.e., the tensor product of D with K over F is isomorphic to Mn(K). Then Dx 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 GLn,F) in the restriction of scalars ResKF GLn,K.
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 Local Fields and Cornell-Silverman.
answered Oct 28, 2009 at 7:05