Morita equivalence of a ring with its matrix rings

Remarks

A quasi-inverse functor sends a left $M_n(R)$-module $L$ to $e_{11} L$, where $e_{11}$ is the matrix with $1$ in position $(1, 1)$ and $0$ elsewhere. It is a left $R$-module through the scalar matrices. The matrix units $e_{ij}$ give mutually inverse isomorphisms between $e_{11} L$ and $e_{jj} L$, and $L = e_{11} L \oplus \cdots \oplus e_{nn} L$, which shows that the two functors are inverse up to natural isomorphism. In the language of progenerators, $P = R^n$ is a progenerator of right $R$-modules and its endomorphism ring is $M_n(R)$. Example: a module over $M_2(\Z)$ is the same as a pair of copies of one abelian group $N$, glued by the matrix units, and the module is $N^2$. So the ideal theory, the Artinian and Noetherian conditions and the homological dimensions of $M_n(R)$ match those of $R$. Being commutative is not shared: $M_2(R)$ is not commutative for $R \ne 0$.

Going deeper

For a student

This is the first example of Morita equivalence and the model for the general theory. For a division ring $D$, it shows that $M_n(D)$, like $D$, has exactly one simple module up to isomorphism, namely $D^n$. Morita's theorem says that every ring Morita equivalent to $R$ is of the form $e M_n(R) e$ for some $n$ and some idempotent $e$ with $M_n(R) e M_n(R) = M_n(R)$. So matrix rings and full corners of them give all rings with the same module category as $R$.