PlanetMath: Morita equivalence

Let

be a ring. Write $\mathcal{M}_R$ for the category of right modules over

. Two rings

and

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

Proof. Assume

. For convenience, we will also say a module to mean a right module.

Let be an -module. Set $F(M)=\lbrace (m_1,\ldots, m_n)\mid m_i\in M\rbrace$ . Then becomes a module over if we adopt the standard matrix multiplication , where $m\in F(M)$ and $A\in M_n(R)$ . If $f: M_1\to M_2$ is an -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 is a covariant functor by inspection.

Next, let be an -module. Write as the $n\times n$ matrix whose cell is $r\in R$ and 0 everywhere else. For simplicity we write . Note that is an idempotent in : , and commutes with for any $r\in R$ : .

Set $G(N)=\lbrace se\mid s\in N \rbrace$ . For any $r\in R$ , define $se\cdot r:= see(r)=se(r)e$ . Since $se(r)\in N$ , this multiplication turns into an -module. If $g:N_1\to N_2$ is an -module homomorphism, define $G(g): G(N_1)\to G(N_2)$ by . If $N_1\stackrel{g}{\longrightarrow} N_2\stackrel{h}{\longrightarrow} N_3$ are -module homomorphisms, then

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

so that

is a covariant functor.

If is any -module, then $GF(M)=\lbrace (m_1,\ldots,m_n)e\mid m\in M \rbrace = \lbrace (m_1,0,\ldots,0)^T\mid m\in M \rbrace \cong M$ , where stands for the transpose of the row vector $m\in M$ into a column vector.

On the other hand, if is any -module, then $FG(N)=\lbrace (s_1e, \ldots, s_ne)\mid s_i\in N\rbrace$ . Before proving that $FG(N)\cong N$ , let's do some preliminary work.

Denote $e_{ii}$ by the $n\times n$ matrix whose cell is 1 and 0 everywhere else. Then each $e_{ii}$ is idempotent, $e_{ii}e_{jj}=0$ for $i\ne j$ , 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 -modules. Since has an -module structure as we had shown earlier, are all -modules. Let $\pi_i:N\to N_i$ be the projection map, $\psi_i: N_i\to N$ be the embedding of into , and $\phi_{ij}:N_i\to N_j$ be the isomorphism from to given by $\phi_{ij}(se_{ii})=se_{jj}$ . All these are -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, $(s_1e,\ldots,s_n e)\mapsto (\phi_{11}(s_1e),\ldots, \phi_{1n}(s_ne)) \mapsto \psi_1(\phi_{11}(s_1e))+\cdots + \psi_n(\phi_{1n}(s_ne)) \in N$ is also a homomorphism $\beta: FG(N)\to N$ . By inspection, $\alpha$ and $\beta$ are inverses of each other, and hence $FG(N)\cong N$ . $\qedsymbol$