6
$\begingroup$

I feel silly asking this, but what is the importance of the Artin-Wedderburn theorem in Algebra?

In A First Course in Noncommutative Rings, T.Y. Lam calls it "the cornerstone of non-commutative ring theory", and he goes on to list some very remarkable properties of modules over semisimple rings, some of which I list below :

  • all modules are projective and injective,
  • the ring itself is isomorphic to a product of matrix rings over division rings, and both these division rings and the size of these matrices are well defined (this is of course the Artin-Wedderburn theorem),
  • group rings are semisimple (except for some special cases).

As someone who knows little to nothing about algebra, even I understand that these are terrific properties to have. However, when do semisimple rings ever occur? Do they only occur as group rings (not that this isn't very important)? Also, is it possible to know the division rings $D_i$ that appear in the theorem (apart from their characterisation as the endomporhism rings of the simple submodules of $R$) and the sizes $n_i$ of the matrices, say if

$$ R\simeq\prod_{i=1}^r\mathrm{Mat}_{n_i}(D_i)~? $$

  • 0
    I think it's cool that most of the basic results in Galois theory and the representation theory of finite groups can be proved using Artin-Wedderburn. See the lecture notes of Mitya Boyarchenko: [one](http://www.math.lsa.umich.edu/seminars/kottwitz/Galois.pdf), [two](http://www.math.lsa.umich.edu/seminars/kottwitz/Lecture2final.pdf).2012-02-28
  • 1
    You ask «is it possible to know the division rings that appear in the theorem?»: that of course depends on what you know about $R$!2012-02-28
  • 0
    @MarianoSuárez-Alvarez Let's say $R$ is a group algebra $\mathbb{C}[G]$ for a finite group $G$. EDIT: I guess this is an easy example, if one already knows the irreducible representations of $G$. My intent with this question is more along the lines of "what parts of the theorem are actually used"?2012-02-28
  • 0
    In that case all the $D_i$ are simply isomorphic to $\mathbb C$ :) Since $\mathbb C$ is an algebraically closed field, there are no non-trivial finite dimensional division algebras over $\mathbb C$.2012-02-28
  • 0
    @MarianoSuárez-Alvarez .......... -_- yes2012-02-28
  • 1
    And there are [only so many](http://en.wikipedia.org/wiki/Division_algebra#Associative_division_algebras) division algebras over $\mathbf R$.2012-02-28
  • 0
    @DylanMoreland I am unaware of the approach to Galois Theory you link to, I'll have a look at it (although I'm no good at Galois theory either ^^). Back to my original question, what are common semisimple rings other than group rings?2012-02-28
  • 0
    For other fields, the set of division algebras that appear in the decompositions of group algebras of finite groups generates a subgroup of the *Brauer group* (Wikipedia surely explains what this is; basically, a group which encodes division algebras) called the *Schur group* of the field. This has been studied for ages now, and a whole lot is known.2012-02-28
  • 1
    You ask «what are common semisimple rings other than group rings?» There are lots of constructions: twisted group algebras, cross-products, cyclic algebras, &c. The study and specially the construction of finite dimensional division algebras is an extremely important part of the classical theory of algebras—and an immensely beautiful subject. There is a nice peek into the theory in Pierce's book *Associative Algebras*, in a modern language. Already the study of quaternion algebras over fields is a huge field, (...)2012-02-28
  • 1
    (...) intimately related to pretty things like the Hasse-Minkowski principle and even very fashionable work connected to a couple of recent Fields medals and all.2012-02-28
  • 0
    So many nice, instructive comments. But why aren't they answers?2012-02-28
  • 0
    @DylanMoreland The links to Boyarchenko's lecture notes seem to be broken. Do you know another way to access these notes?2015-08-25

1 Answers 1