## Structure of Semisimple Rings

It turns out there is a nice classification for semisimple rings.

Theorem. Any semisimple ring R is a finite product: $R \cong \prod_{i=1}^k M_{n_i}(D_i),$

where each $D_i$ is a division ring and $M_n(D)$ is the ring of n × n matrices with entries in D. Furthermore, the list of $(n_i, D_i)$ is unique up to permutation, and isomorphism class of $D_i.$

We saw that a semisimple ring R is a finite direct sum of simple submodules (left ideals): $R = M_1^{m_1} \oplus M_2^{m_2} \oplus \ldots \oplus M_k^{m_k},$

where the simple modules $M_1, \ldots, M_k$ are pairwise non-isomorphic. Schur’s lemma says that for simple modules M and M’, $\text{Hom}_R(M, M')$ is zero if M and M’ are not isomorphic, and is a division ring otherwise.

More generally, for a simple module M, we have: $\text{Hom}(M^m, M^n) \cong \text{Hom}(M, M)^{mn} \cong D^{mn}$

where D is a division ring.

[ Note: in general, we have $\text{Hom}(\oplus_i M_i, N) \cong \prod_i \text{Hom}(M_i, N)$ and $\text{Hom}(M, \prod_i N_i) \cong \prod_i \text{Hom}(M, N_i)$ by the universal properties of direct sum and product. When we have finitely many terms, the direct sum is the direct product. ]

Next we have the isomorphism $\text{Hom}_R(R,R) \cong R^{op}$, where $x\in R^{op} \quad \leftrightarrow \quad f_x : R \to R, z \mapsto zx.$

[ We need to take the opposite ring since $f_y\circ f_x = f_{xy}.$ ]

Piecing all these together, we have: $R^{op} \cong \text{Hom}(M_1^{n_1}\oplus \ldots \oplus M_k^{n_k}, M_1^{n_1} \oplus \ldots \oplus M_k^{n_k})\cong \oplus_{i,j} \text{Hom}(M_i^{n_i}, M_j^{n_j}).$

By the above discussion, each term $\text{Hom}(M_i^{n_i}, M_j^{n_j})$ is either 0 (if i ≠ j) or isomorphic to $M_{n_i}(D_i)$, where $D_i := \text{Hom}(M_i, M_i)$ is a division ring. In conclusion: $R\cong \prod_{i=1}^k M_{n_i}(D_i)$ for division rings $D_i.$

[ We leave it to the reader to prove that matrix product in $M_n(\text{Hom}(M, M))$ corresponds to composition of endomorphisms $\text{Hom}(M^n, M^n)$. ]

To show that the set of $(n_i, D_i)$ is unique up to isomorphism, note:

Proposition. The ring $R = M_n(D)$, for division ring D, has a unique simple module up to isomorphism, namely $D^n$.

Proof

Indeed, R is a direct sum of column vectors. E.g. for n = 2, we have: $M_2(D) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix}\right\} = \left\{ \begin{pmatrix} a & 0 \\ b & 0\end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & c\\ 0 & d\end{pmatrix}\right\}.$

It is easy to see that the space of column vectors $D^n$ is simple as an R-module. On the other hand, since every simple R-module occurs as a simple left ideal of R, $D^n$ is the only possible simple R-module. ♦

Hence, for a ring $R = \prod_{i=1}^k M_{n_i}(D_i)$, there are exactly k simple modules up to isomorphism, each simple module occurs $n_i$ times and its endomorphism ring is isomorphic to $D_i^{op}$. This shows that we can recover $(n_i, D_i)$ from the ring R itself, so it must be unique.

The theorem also shows:

Corollary. R is semisimple iff its opposite ring $R^{op}$ is. Another of saying this is: R is “left semisimple” iff it is “right semisimple”.

Coming up next, the most interesting example of semisimple rings: group rings.

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