2
$\begingroup$

I am studying Artin-Wedderburn structural decomposition theorem for semisimple rings. I understand that it says that any semisimple ring, $R$ is isomorphic (as rings) to $M_{n_1}(D_1) \times M_{n_2}(D_2)\times\cdots\times M_{n_k}(D_k)$ for some $n_i$ and division rings $D_i$. Is this decomposition unique, i.e., suppose $R$ is also isomorphic to $M_{n'_1}(D'_1) \times M_{n'_2}(D'_2)\times\cdots\times M_{n'_{k'}}(D'_{k'})$, then is $k=k'$ and $n_i$ equal to $n'_i$ and $D_i \approx D'_i$ up to some permutation? If so, how?

$\endgroup$
1
4
$\begingroup$

Yes, the number of factors, dimensions of the factors, and the division rings for each factor are unique.

I will outline the general idea.

If two semisimple rings are isomorphic, you know that the isotypes of their minimal right ideals match, so they will have the same number of Wedderburn components. This means the number of simple components ($k$ in your writeup) will be the same for both. Furthermore you know the composition lengths of each component will match, and that determines $n_k$ for each $k$.

Finally, the division rings are just endomorphism rings of the minimal right ideals in each component, and since you know isomorphic simple modules have isomorphic endomorphism rings, the division rings match.


I just scraped up the first reference I could find with a proof.

Passman, Donald S. A course in ring theory. American Mathematical Soc., 2004. (Theorem 4.5 p 36)

I'm also pretty sure it appears in Lectures on modules and rings by Lam. I thought it also appeared in Algebra: a graduate course by Isaacs, but I haven't had time to track them down.

$\endgroup$
4
  • $\begingroup$ Could you provide more details/ a reference to the complete proof? There seem to be subtleties here: like wouldn't a unique semisimple isotype decomposition imply $\sum_i n_i = \sum_i n_i'$? $\endgroup$
    – Noel
    Apr 28 '20 at 11:05
  • $\begingroup$ @Noel I don’t understand the last question... I’m not sure where you’re getting the phrase “unique semisimple isotype decomposition” or what $n_i$ and $n’_i$ denote $\endgroup$
    – rschwieb
    Apr 28 '20 at 11:28
  • $\begingroup$ So, $n_i$ and $n_i'$ are the dimensions of the matrix rings as defined in the question. By the uniqueness of semisimple decomposition, I meant that a R-module $M$ has a unique (up to isomorphism and reordering) representation as a direct sum of simple modules. I believe this is essentially what you have used and I think it would imply $\sum_i n_i = \sum_i n_i'$ $\endgroup$
    – Noel
    Apr 29 '20 at 8:59
  • $\begingroup$ Sorry, yes, I should have remembered the $n_i$ and $n'_i$. I was a little muddled when I read it. Yes, it does imply that, because that would be the composition length of both rings compared. However, knowing the two sums are equal doesn't help a lot proving that the corresponding isotypic components have matching composition lengths, which is what you need. $\endgroup$
    – rschwieb
    Apr 29 '20 at 11:50
3
$\begingroup$

Yes. The number $k$ is equal to the number of isomorphism classes of irreducible $R$-modules, and the $n_i$s are equal to their dimensions, and the $D_i$s are the endomorphism rings of the irreducible $R$-modules.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.