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

$n=2^s p_1…p_k$ 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).