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# Morita equivalence

Let $R$ be a ring. Write $\mathcal{M}_{R}$ for the category of right modules over $R$. Two rings $R$ and $S$ are said to be *Morita equivalent* if $\mathcal{M}_{R}$ and $\mathcal{M}_{S}$ are equivalent as categories. What this means is: we have two functors

$F:\mathcal{M}_{R}\to\mathcal{M}_{S}\qquad\mbox{ and }\qquad G:\mathcal{M}_{S}% \to\mathcal{M}_{R}$ |

such that for any right $R$-module $M$ and any right $S$-module $N$, we have

$GF(M)\cong_{R}M\qquad\mbox{ and }\qquad FG(N)\cong_{S}N,$ |

where $A\cong_{R}B$ means that there is an $R$-module isomorphism between $A$ and $B$.

Example. Any ring $R$ with $1$ is Morita equivalent to any matrix ring $M_{n}(R)$ over it.

###### Proof.

Assume $n>1$. For convenience, we will also say a module to mean a right module.

Let $M$ be an $R$-module. Set $F(M)=\{(m_{1},\ldots,m_{n})\mid m_{i}\in M\}$. Then $F(M)$ becomes a module over $M_{n}(R)$ if we adopt the standard matrix multiplication $mA$, where $m\in F(M)$ and $A\in M_{n}(R)$. If $f:M_{1}\to M_{2}$ is an $R$-module homomorphism. Set $F(f):F(M_{1})\to F(M_{2})$ by $F(f)(m_{1},\ldots,m_{n})=(f(m_{1}),\ldots,f(m_{n}))\in F(M_{2})$. Then $F$ is a covariant functor by inspection.

Next, let $N$ be an $M_{n}(R)$-module. Write $e(r)$ as the $n\times n$ matrix whose cell $(1,1)$ is $r\in R$ and $0$ everywhere else. For simplicity we write $e:=e(1)$. Note that $e$ is an idempotent in $M_{n}(R)$: $e=ee$, and $e$ commutes with $e(r)$ for any $r\in R$: $ee(r)=e(r)e$.

Set $G(N)=\{se\mid s\in N\}$. For any $r\in R$, define $se\cdot r:=see(r)=se(r)e$. Since $se(r)\in N$, this multiplication turns $G(N)$ into an $R$-module. If $g:N_{1}\to N_{2}$ is an $M_{n}(R)$-module homomorphism, define $G(g):G(N_{1})\to G(N_{2})$ by $G(g)(se)=g(s)e$. If $N_{1}\stackrel{g}{\longrightarrow}N_{2}\stackrel{h}{\longrightarrow}N_{3}$ are $M_{n}(R)$-module homomorphisms, then

$G(h\circ g)(se)=(h\circ g)(s)e=h(g(s))e=G(h)[g(s)e]=G(h)[G(g)se]=G(h)\circ G(g% )(se)$ |

so that $G$ is a covariant functor.

If $M$ is any $R$-module, then $GF(M)=\{(m_{1},\ldots,m_{n})e\mid m\in M\}=\{(m_{1},0,\ldots,0)^{T}\mid m\in M% \}\cong M$, where $m^{T}$ stands for the transpose of the row vector $m\in M$ into a column vector.

On the other hand, if $N$ is any $M_{n}(R)$-module, then $FG(N)=\{(s_{1}e,\ldots,s_{n}e)\mid s_{i}\in N\}$. Before proving that $FG(N)\cong N$, let’s do some preliminary work.

Denote $e_{{ii}}$ by the $n\times n$ matrix whose cell $(i,i)$ is 1 and $0$ everywhere else. Then each $e_{{ii}}$ is idempotent, $e_{{ii}}e_{{jj}}=0$ for $i\neq j$, and $e_{{11}}+\cdots+e_{{nn}}=1$. From this, we see that $N=N_{1}\oplus\cdots\oplus N_{n}$, where $N_{i}=Ne_{{ii}}$, and $N_{i}\cong N_{j}$ as $M_{n}(R)$-modules. Since $N_{1}=Ne$ has an $R$-module structure as we had shown earlier, $N_{i}$ are all $R$-modules. Let $\pi_{i}:N\to N_{i}$ be the projection map, $\psi_{i}:N_{i}\to N$ be the embedding of $N_{i}$ into $N$, and $\phi_{{ij}}:N_{i}\to N_{j}$ be the isomorphism from $N_{i}$ to $N_{j}$ given by $\phi_{{ij}}(se_{{ii}})=se_{{jj}}$. All these are $M_{n}(R)$-module homomorphisms since $e_{{ii}}A=Ae_{{ii}}$.

Now, take any $s\in N$, then $s\mapsto(\pi_{1}(s),\ldots,\pi_{n}(s))\mapsto(\phi_{{11}}\pi_{1}(s),\ldots,% \phi_{{n1}}\pi_{n}(s))\in FG(N)$ is a homomorphism $\alpha:N\to FG(N)$. Conversely, $(s_{1}e,\ldots,s_{n}e)\mapsto(\phi_{{11}}(s_{1}e),\ldots,\phi_{{1n}}(s_{n}e))% \mapsto\psi_{1}(\phi_{{11}}(s_{1}e))+\cdots+\psi_{n}(\phi_{{1n}}(s_{n}e))\in N$ is also a homomorphism $\beta:FG(N)\to N$. By inspection, $\alpha$ and $\beta$ are inverses of each other, and hence $FG(N)\cong N$. ∎

Remark. A property $P$ in the class of all rings is said to be *Morita invariant* if, whenever $R$ has property $P$ and $S$ is Morita equivalent to $R$, then $S$ has property $P$ as well. By the example above, it is clear that commutativity is not a Morita invariant property.

## Mathematics Subject Classification

16D90*no label found*

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