Last time we have seen that in order to classify all non-commutative $l$-points one needs to control the finite dimensional simple algebras having as their center a finite dimensional field-extension of $l$. We have seen that the equivalence classes of simple algebras with the same center $L$ form an Abelian group, the Brauer group. The calculation of Brauer groups is best done using Galois-cohomology. As an aside : Evariste Galois was one of the more tragic figures in the history of mathematics, he died at the age of 20 as a result of a duel. There is a whole site the Evariste Galois archive dedicated to his work.

But let us return to a simple algebra $T$ over the field $L$ which we have seen to be of the form $M(k,S)$, full matrices over a division algebra $S$. We know that the dimension of $S$ over $L$ is a square, say $n^2$, and it can be shown that all maximal commutative subfields of $S$ have dimension n over $L$. In this way one can view a simple algebra as a bag containing all sorts of degree n extensions of its center. All these maximal subfields are also splitting fields for $S$, meaning that if you tensor $S$ with one of them, say $M$, one obtains full nxn matrices $M(n,M)$. Among this collection there is at least one separable field but for a long time it was an open question whether the collection of all maximal commutative subfields also contains a Galois-extension of $L$. If this is the case, then one could describe the division algebra $S$ as a crossed product. It was known for some time that there is always a simple algebra $S’$ equivalent to $S$ which is a crossed product (usually corresponding to a different number n’), that is, all elements of the Brauer group can be represented by crossed products. It came as a surprise when S.A. Amitsur in 1972 came up with examples of non-crossed product division algebras, that is, division algebras $D$ such that none of its maximal commutative subfields is a Galois extension of the center. His examples were generic division algebras $D(n)$. To define $D(n)$ take two generic nxn matrices, that is, nxn matrices A and B such that all its entries are algebraically independent over $L$ and consider the $L$-subalgebra generated by A and B in the full nxn matrixring over the field $F$ generated by all entries of A and B. Somewhat surprisingly, one can show that this subalgebra is a domain and inverting all its central elements (which, again, is somewhat of a surprise that there are lots of them apart from elements of $L$, the so called central polynomials) one obtains the division algebra $D(n)$ with center $F(n)$ which has trancendence degree n^2 1 over $L$. By the way, it is still unknown (apart from some low n cases) whether $F(n)$ is purely trancendental over $L$. Now, utilising the generic nature of $D(n)$, Amitsur was able to prove that when $L=Q$, the field of rational numbers, $D(n)$ cannot be a crossed product unless $n=2^s p1…pk$ with the p_i prime numbers and s at most 2. So, for example $D(8)$ is not a crossed product.

One can then ask whether any division algebra $S$, of dimension n^2 over $L$, is a crossed whenever n is squarefree. Even teh simplest case, when n is a prime number is not known unless p=2 or 3. This shows how little we do know about finite dimensional division algebras : nobody knows whether a division algebra of dimension 25 contains a maximal cyclic subfield (the main problem in deciding this type of problems is that we know so few methods to construct division algebras; either they are constructed quite explicitly as a crossed product or otherwise they are constructed by some generic construction but then it is very hard to make explicit calculations with them).

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Posted in geometry, groups
Written on Mon, 09 February 2004 at 8:10 pm
Tags: Brauer, Galois, groups, non-commutative, simples
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