# How to prove that a ring $R$ is Morita-equivalent to $M_n(R)$?

Let $R$ be a ring and $M_n(R)$ the ring of $n \times n$ matrices with elements in $R$. Let $\textrm{mod} R$ denote the category of left modules over $R$.

How to define a equivalent functor $F: \textrm{mod} R \to \textrm{mod} M_n(R)$?

In Example, one defines $F(X) = X^n \in \textrm{mod} M_n(R)$ for a left $R$-module $X$.

I do not know how to define the inverse functor $F^{-1}$. In case that $n=2$, $(M_1, M_2)^{T} \in \textrm{mod} M_2(R)$, then what is $F^{-1}((M_1, M_2)^{T})$?

Let me explain the general picture. Let $$R,S$$ be two rings. The cocontinuous functors $$\mathsf{Mod}(S) \longrightarrow \mathsf{Mod}(R)$$ are given by $$P \otimes_R -$$ for some $$(R,S)$$-bimodule $$P$$ - this the Eilenberg-Watts Theorem. It follows that pairs of pseudoinverse functors $$\mathsf{Mod}(R) \longleftrightarrow \mathsf{Mod}(S)$$ are given by pairs $$(P,Q)$$, where $$P$$ is a $$(R,S)$$-bimodule and $$Q$$ is a $$(S,R)$$-bimodule such that $$P \otimes_S Q \cong R$$ as $$(R,R)$$-bimodules and $$Q \otimes_R P \cong S$$ as $$(S,S)$$-bimodules. These pairs (together with the isomorphisms actually) are called Morita contexts, since they classify Morita equivalences between $$R$$ and $$S$$. If you want to know more details, you can find them for instance in Lam's book Lectures on modules and rings in the chapter on Morita theory. There you can also find some details for the example below.

In the standard example $$S=M_n(R)$$, a possible choice is the $$(R,M_n(R))$$-bimodule $$R^{1 \times n}=R^n$$ (row vectors) and likewise the $$(M_n(R),R)$$-bimodule $$R^{n \times 1}=R^n$$ (column vectors). One has to check that $$R^{n \times 1} \otimes_R R^{1 \times n} \cong M_n(R)$$ as $$(M_n(R),M_n(R))$$-bimodules and $$R^{1 \times n} \otimes_{M_n(R)} R^{n \times 1} \cong R$$ as $$(R,R)$$-bimodules.

The corresponding pseudoinverse functors are $$\mathsf{Mod}(R) \to \mathsf{Mod}(M_n(R)),~X \mapsto R^{n \times 1} \otimes_R X = X^{n \times 1}$$ and $$\mathsf{Mod}(M_n(R)) \to \mathsf{Mod}(R),~ X \mapsto R^{1 \times n} \otimes_{M_n(R)} X \cong E_{11} X.$$

• Thank you. But I have a problem: what is $E_{11}$? Whether $E_{11}$ is the matrix whose $(1,1)$-th entry is 1 and 0 elsewhere.
– bing
Apr 13, 2017 at 8:34
• Yes. $E_{ij}$ are the standard matrix units. Apr 13, 2017 at 21:29

Given any $$M_n(R)$$-module $$Y$$, we can treat it like an $$R$$-module (since $$R\hookrightarrow M_n(R)$$ diagonally), in which case $$Y=\bigoplus_i e_{ii}Y$$ as $$R$$-modules. To see why $$Y=\sum_i e_{ii}Y$$, simply note $$I_n=\sum_i e_{ii}$$ and to see the sum is direct, suppose $$y$$ is in both $$e_{ii}Y$$ and $$\bigoplus_{j\ne i}e_{jj}y$$, then apply $$e_{ii}$$ to $$y$$ to get $$y=0$$ (since $$e_{ii}$$ acts as the identity on $$e_{ii}Y$$ and as $$0$$ on all other $$e_{jj}Y$$s). Moreover, observe $$e_{ii}Y\cong e_{jj}Y$$ as $$R$$-modules, with multiplications by $$e_{ij}$$ and $$e_{ji}$$ as mutually inverse homomorphisms (hence isomorphisms), from which we conclude $$Y\cong X^n$$ as $$M_n(R)$$-modules with $$X=e_{ii}Y$$.

So the inverse of $$X\mapsto X^n$$ is $$Y\mapsto e_{11}Y$$.

• I think there's a problem in your proof of $e_{ii}Y \simeq e_{jj}Y$, since multiplying $e_{ii}$ by the permutation matrix associated to the transpositions $(ij)$ can only give you $e_{ji}$, not $e_{jj}$. Nov 10, 2021 at 14:17
• @penseur_32 True. Fixed. Mar 29 at 19:49