Posted by Jason Polak on 20. October 2016 · Write a comment · Categories: math

Let $R$ be a commutative ring, and $M_n(R)$ be the ring of $n\times n$ matrices with coefficients in $R$. Did you know that the center of $M_n(R)$ is the set of scalar matrices? How can we prove this? Should we use matrix multiplication? No way! Let’s use Morita theory instead!

Morita theory works even if $R$ is not commutative. I’ll state the gist of Morita theory, more of which can be found in the post “The Double Dual and Morita Duality”. Let $M$ a left $R$-module. Write $M^* = {\rm Hom}(M,R)$ for the dual of $M$. We have an evaluation homomorphism $M^*\otimes_R M\to R$ given on pure tensors by $f\otimes m\mapsto f(m)$. The image of this homomorphism is an ideal of $R$ called the trace ideal of $M$. If this ideal is all of $R$, then $M$ is called a generator. If in addition to being a generator, $M$ is finitely generated and projective, then $M$ is called a progenerator.

Progenerators are the main players in Morita theory: the distilled pure essence of this theory is that if $M$ is a progenerator the functor $-\otimes_R M$ from right $R$-modules to left $S = {\rm Hom}_R(M,M)$-modules is an equivalence of categories.

Back to the center of $M_n(R)$ when $R$ is commutative. The module $M = R^n$ is a progenerator. Then, Morita theory says that the category of right $R$-modules is equivalent to the category of left $M_n(R)$-modules via the functor $-\otimes_R M$. From this there is the following consequence: $R\cong {\rm Hom}_R(R,R)\cong{\rm Hom}_{M_n(R)}(R^n,R^n)$, where the composition of these isomorphisms is the map that sends an element $r\in R$ to multiplication by $r$, giving an endomorphism of $R^n\to R^n$.

Now, an $M_n(R)$-endomorphism $R^n\to R^n$ is in particular an $R$-module endomorphism $R^n\to R^n$. Therefore it must be given by multiplication by a matrix in $M_n(R)$. But, being an $M_n(R)$-module endomorphism it must also commute with all the elements of $M_n(R)$. Morita theory gave an isomorphism ${\rm Hom}_R(R,R)\cong {\rm Hom}_{M_n(R)}(R^n,R^n)$, which shows that every such matrix must be a scalar matrix. Great, no matrix multiplication required!

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