From: t Subject: Re: Division algebras, Milnor conjecture etc Date: Thu, 15 Apr 1999 23:46:01 -0400 Newsgroups: [missing] To: Dave Rusin Dave Rusin wrote: > In article <3715CC0D.60CD4F86@sympatico.ca> you write: > > > >with the recent work of Voevodsky on the Milnor conjecture (on > >homomorphisms from mod 2 K-theory into cohomology) and older work of > >Suslin and Merkur'ev, does anyone know if the following question has > >been resolved or what progress has been made? > > > >Does there exist a non-cyclic division algebra of prime degree? > > The first paragraph means almost nothing to me but the second prompts > this response: by "degree" do you mean dimension over the underlying > field? A finite-dimensional division algebra has a degree over its > center which is a perfect square, so if the dimension is prime, the > division algebra is actually an extension _field_, right? Hi, The degree is actually the square root of the dimension of the algebra over it's centre, where dimension is the dimension of the algebra considered as vector space over the field. By a theorem of Wedderburn (also proved in a clever way by the Unibomber!) if the division algebra is finite then it is a field. Here's a bit of the classical theory: for a field F, an F algebra is called central simple if it is simple and its centre is F. If C(F) denotes the class of finite dimensional central simple algebras, then there is the following theorem: For A and B in F the following are equivalent: 1) There is a division algebra D in C(F) such that A ~ M(D, n) (isomorphic to the nXn matrix algebra over D) and B~M(D, m) for some m, n 2) there are integers r, s such that A (X) M(F, r) ~ B (X) M(F, s) ( (X) = tensor over F ). This defines an equivalence relation on C(F), and the equivalence classes form an abelian group, the so called Brauer group, B(F) (which can be shown to be the second galois cohomology group of F, which is where the K-theory enters) with the identity element being the class of F (ie the matrix algebras over F) and multiplication as tensoring.. So basically, equivalence classes in B(F) are division algebras over F. As a concrete example, the 2-torsion subgroup in B(F) is that generated by the quaternion algebras over F (char(F) <> 2) In some sense, division algebras of prime degree are building blocks: it can be shown that a division algebra is uniquely the tensor product of division algebras of prime degree. The heavy K-theory and algebraic geometry (and associated heavy duty tools such as etale cohomology etc) come into this very classical and old subject from a very beautiul and readable paper paper by John Tate: "Relations between K2 and Galois Cohomology" appearing in Inv. Math in 1976, and Milnor's paper K-theory and Quadratic forms. A very readable account of this stuff is in "Associative Algebras" by Pierce, Springer GTM 88 or the two volume set "Ring Theory" by Rowen, Academic Press. I hope I didn't bore you or waste your time with stuff you already know. Regards, t [MIME formatting deleted -- djr] ============================================================================== From: Dave Rusin Subject: Re: Division algebras, Milnor conjecture etc Date: Fri, 16 Apr 1999 00:57:28 -0500 (CDT) Newsgroups: [missing] To: t@sympatico.ca Thanks for your response. I'm familiar with what you wrote, but I thought it was always true that there was a splitting field K whose degree over F was sqrt([D:F]) (i.e., [K:F]=your degree) for which D\tensor K becomes M(K, d); indeed this K is any maximal (commutative) subfield of the division ring D. From this K one describes D itself by viewing D as a d-dimensional vector space with a basis { e_i }; then D is completely described by defining the products e_i e_j as K-linear combinations of the e_k, right? So the construction of division algebras has to do with these "factor-sets" -- 2-cocyles mapping Gal(K/F) to K or something, isn't it? (It's been 20 yr since I studied this and it's 1am here so I can't reconstruct this). If these half-remembered results are right, then you should have what you want since a Galois group of prime order is necessarily cyclic, eh? What am I missing? dave