Uniqueness of Artin-Wedderburn decomposition

Question

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?

Answer 1

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.

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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.

Answer 2

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.