proof of Wedderburn’s theorem
Lemma. The centralizer is a subring.
and are obviously elements of and if and are, then , and , so , and are also elements of . Moreover, for , implies , so is also an element of .
Now we consider the center of which we’ll call . This is also a subring and is in fact the intersection of all centralizers.
is an abelian subring of and is thus a field. We can consider and every as vector spaces over of dimension and respectively. Since can be viewed as a module over we find that divides . If we put , we see that since , and that and .
It suffices to show that to prove that multiplication is abelian, since then and so .
We now consider and apply the conjugacy class formula.
By Zsigmondy’s theorem, there exists a prime that divides but doesn’t divide any of the for , except in 2 exceptional cases which will be dealt with separately. Such a prime will divide and each of the . So it will also divide which can only happen if .
We now deal with the 2 exceptional cases. In the first case equals , which would is a vector space of dimension 2 over , with elements of the form where . Such elements clearly commute so which contradicts our assumption that . In the second case, and . The class equation reduces to where divides 6. This gives with and integers, which is impossible since the right hand side is divisible by 3 and the left hand side isn’t.
|Title||proof of Wedderburn’s theorem|
|Date of creation||2013-03-22 13:10:50|
|Last modified on||2013-03-22 13:10:50|
|Last modified by||lieven (1075)|