# 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 (http://planetmath.org/EquivalenceOfCategories). What this means is: we have two functors^{}

$$F:{\mathcal{M}}_{R}\to {\mathcal{M}}_{S}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}G:{\mathcal{M}}_{S}\to {\mathcal{M}}_{R}$$ |

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

$$GF(M){\cong}_{R}M\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}FG(N){\cong}_{S}N,$$ |

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 $F(M)=\{({m}_{1},\mathrm{\dots},{m}_{n})\mid {m}_{i}\in M\}$. 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 $F(f)({m}_{1},\mathrm{\dots},{m}_{n})=(f({m}_{1}),\mathrm{\dots},f({m}_{n}))\in F({M}_{2})$. 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 ${N}_{1}\stackrel{g}{\u27f6}{N}_{2}\stackrel{h}{\u27f6}{N}_{3}$ 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 $GF(M)=\{({m}_{1},\mathrm{\dots},{m}_{n})e\mid m\in M\}=\{{({m}_{1},0,\mathrm{\dots},0)}^{T}\mid m\in M\}\cong M$, 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 $FG(N)=\{({s}_{1}e,\mathrm{\dots},{s}_{n}e)\mid {s}_{i}\in N\}$. 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 $i\ne j$, and ${e}_{11}+\mathrm{\cdots}+{e}_{nn}=1$. From this, we see that $N={N}_{1}\oplus \mathrm{\cdots}\oplus {N}_{n}$, where ${N}_{i}=N{e}_{ii}$, 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 ${\varphi}_{ij}:{N}_{i}\to {N}_{j}$ be the isomorphism from ${N}_{i}$ to ${N}_{j}$ given by ${\varphi}_{ij}(s{e}_{ii})=s{e}_{jj}$. All these are ${M}_{n}(R)$-module homomorphisms since ${e}_{ii}A=A{e}_{ii}$.

Now, take any $s\in N$, then $s\mapsto ({\pi}_{1}(s),\mathrm{\dots},{\pi}_{n}(s))\mapsto ({\varphi}_{11}{\pi}_{1}(s),\mathrm{\dots},{\varphi}_{n1}{\pi}_{n}(s))\in FG(N)$ is a homomorphism^{} $\alpha :N\to FG(N)$. Conversely, $({s}_{1}e,\mathrm{\dots},{s}_{n}e)\mapsto ({\varphi}_{11}({s}_{1}e),\mathrm{\dots},{\varphi}_{1n}({s}_{n}e))\mapsto {\psi}_{1}({\varphi}_{11}({s}_{1}e))+\mathrm{\cdots}+{\psi}_{n}({\varphi}_{1n}({s}_{n}e))\in N$ 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.

Title | Morita equivalence |
---|---|

Canonical name | MoritaEquivalence |

Date of creation | 2013-03-22 16:38:49 |

Last modified on | 2013-03-22 16:38:49 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 6 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 16D90 |

Defines | Morita equivalent |

Defines | Morita invariance |

Defines | Morita invariant |