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Fields of finite order are well classified, and classification of groups of finite order has taken some depth in research. Why classification of finite rings and modules is not well studied in research?

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Finite rings are studied; you just need to know where to look. However, most basic algebra texts do not study finite rings deeply simply due to lack of space. –  Amitesh Datta Jun 9 '11 at 8:18
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why a downvote, i think this question is a decent one –  user9413 Jun 9 '11 at 8:36
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The downvote is not mine, but the question has two faulty premises: (i) there is no classification of groups of finite order (only finite simple groups) and (ii) finite rings are very well-studied in research. If the OP had done a quick internet search, s/he would find many papers on the latter. So the question does not look so well thought out. –  Pete L. Clark Jun 9 '11 at 12:31

1 Answer 1

The following basic results can be deduced about finite rings:

  • A finite simple ring is isomorphic to a matrix ring over a finite field. (Artin-Wedderburn theorem.)
  • A finite ring is Artinian. Hence all the known properties of Artinian rings also hold for finite rings.
  • See http://www.csc.villanova.edu/~pcesarz/
  • A finite division ring is a field. (Wedderburn's little theorem.) In fact, many important basic theorems in finite ring theory concern determining conditions under which a finite ring is commutative.
  • Enumerating finite rings of a given order is an important (unsolved) problem.
  • In fact, a finite semisimple ring is isomorphic to a direct product of matrix rings over finite fields. (A ring is semisimple if its Jacobson radical, $J(A)$, is trivial.) Therefore, if $A$ is a ring, then $A/J(A)$ is isomorphic to a direct product of matrix rings over finite fields. Note also that $J(A)$ is a nilpotent ideal.

Of course, there are many more important results about finite rings.

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what about wedderburns theorem on division rings: en.wikipedia.org/wiki/Wedderburn's_little_theorem & the artin-zorn theorem. –  user9413 Jun 9 '11 at 8:37
    
I have already stated Wedderburn's little theorem in my answer. What about it? –  Amitesh Datta Jun 9 '11 at 8:44
    
Oh, it missed my eye. –  user9413 Jun 9 '11 at 9:03

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