# Classic Morita Theory

I want to write some short introduction of Morita theory, since it seems that not all books introduce this classic thing in the following easy and brief way.

## The quotient relation.

$$\Hom_B^C(Y,\Hom_A(X,Z))=\Hom_A^C(X\otimes_BY,Z)=\Hom_A^B(X,\Hom^C(Y,Z)).$$

There is other isomorphisms

$$\begin{array}{c}\Hom_A(X,Y_B)\otimes_B Z\to \Hom_A(X,Y\otimes_B Z),\\X\otimes_A \Hom_B(Y_A,Z) \to \Hom_B(\Hom^A(X,Y_A),Z). \end{array}$$

For the first map. When $X=A$, it is isomorphic trivially. When $X=A^n$, it is isomorphic trivially. When $X$ is f.g. projective, then as a summand of some $A^n$ it is isomorphic. When $X$ is f.pre. and $Z$ is flat, then it is isomorphic by five lemma.

For the second map. When $X=A$, it is isomorphic trivially. When $X=A^n$, it is isomorphic trivially. When $X$ is f.g. projective, then as a summand of some $A^n$ it is isomorphic.When $X$ is f.pre. and $Z$ is injective, then it is isomorphic by five lemma.

And the dual for right modules

$$\begin{array}{c}X\otimes_A \Hom^B(Y,{}_{A}Z)\to \Hom^B(X,Z\otimes_AY),\\\Hom^A(X_B,Y)\otimes_B Z \to \Hom^A(\Hom_B(X,Z),Y).\end{array}$$

## Left and right adjoint functor of tensor.

For a $P\in B\textrm{-}\mathsf{mod}\textrm{-} A$, and projective as $B$-module, we can consider $Q=\Hom_B(P,B)$. Now

$$Q\otimes_B - = \Hom_B(P,B)\otimes_B -=\Hom_B(P,-)$$

is an adjoint of $P\otimes_A -$.

We can also consider $Q'=\Hom^A(P,A)$, now

$$\begin{array}{rl}\Hom_A(Q'\otimes_B U, V) & = \Hom_B(U,\Hom_A(Q', V))\\& = \Hom_B(U,\Hom_A(\Hom^A(P,A),V))\\& =\Hom_B(U,P\otimes_A \Hom_A(A,V))\\& = \Hom_B(U,P\otimes_A V). \end{array}$$

But we don't use it.

## Equivalence $\Rightarrow$ Projective generator.

Assume $A\textrm{-}\mathsf{mod} \cong B\textrm{-}\mathsf{mod}$. Consider the image of $A$ in $B\textrm{-}\mathsf{mod}$, say $P$. Then $P$ is a projective generator (any $M\in B\textrm{-}\mathsf{mod}$ is a quotient of some $P^n$). Furthermore, $P$ is also f.g. since $\Hom_B(P,-)$ commutes with direct sum. Now, $\Hom_B(P,P)=\Hom_A(A,A)=A^{\op}$, so $P\in B\textrm{-}\mathsf{mod}\textrm{-} A$. Then the functor $A\textrm{-}\mathsf{mod} \to B\textrm{-}\mathsf{mod}$ is given by $P\otimes_A-$.

Since the quasi-equivalence is adjoint, and adjoint is unique, so $Q\otimes-$ is the desired inverse, where$Q=\Hom_B(P,B)$ or $Q=\Hom^A(P,A)$ works (but they are isomorphic by uniqueness).

## Projective generators $\Rightarrow$ Equivalence.

Next, if we have a projective generator $P$ in $B\textrm{-}\mathsf{mod}$, denote $A=\End_B(P,P)^{\op}$, and consider $Q=\Hom_B(P,B)$, then we have two maps

$$Q\otimes_B P=\Hom_B(P,B)\otimes_B P=\Hom_B(P,P)= A\qquad : \eta$$

and

$$\epsilon: \qquad P\otimes_A Q=P\otimes_B \Hom_B(P,B)\to B.$$

Note that

$$P=P\otimes A\cong P\otimes_A Q\otimes_B P\to B\otimes P=P$$

is identity, so $\epsilon\otimes P$ is an isomorphism, but because $P$ is projective generator, so $P\otimes_AQ\otimes_B B\to B$ is isomorphism. So $P\otimes-$ and $Q\otimes-$ induces equivalence.

## Morita context.

We will say a four tuple $(A,B,P,Q)$ with $A,B$ algebras and $P,Q$ bimodules is a Morita context if

$$\epsilon: P\otimes_A Q \to B\qquad \eta:A\to Q\otimes_B P$$

are isomorphisms as bimodule, and satisfy the

$$(P\otimes \eta)\circ (\epsilon\otimes P)=\operatorname{id}_p,\qquad (\eta\otimes Q)\circ (Q\otimes \epsilon)=\operatorname{id}_Q.$$

Of course, this defines an adjoint. So our discussion above shows that the equivalences between $A\textrm{-}\mathsf{mod}$ and $B\textrm{-}\mathsf{mod}$ are all from Morita type.

Note that Morita context is symmetric, not only left modules but also right modules (even the bimodules)!As a result, $A\textrm{-}\mathsf{mod}\cong B\textrm{-}\mathsf{mod} \iff \mathsf{mod}\textrm{-} A\cong \mathsf{mod}\textrm{-} B$.

Then $P$ are generators of both sides. Because we can take some surjective $A^n\rightarrow Q$, then $P^n=P\otimes_AA^n\rightarrow P\otimes Q\cong B$. So

$$\begin{array}{rl}Q & = \Hom_A(A,Q) = \Hom_A(Q\otimes_B P,Q)=\Hom_B(P,P\otimes_A Q)\\& =\Hom_B(P,B). \end{array}$$

The same reason,

$$\begin{array}{rl}Q & = \Hom^B(B,Q) = \Hom^B(P\otimes_A Q,Q)=\Hom^A(Q,Q\otimes_B P)\\& = \Hom^A(Q,A).\end{array}$$

And as algebras

$$\begin{array}{rl}\End_B(P) & =\Hom_B(P,P)=\Hom_A(A,Q\otimes P) =A^{\op}\\\End^A(P) & =\Hom^A(P,P)=\Hom^B(B,P\otimes Q) =B \end{array}$$

posted @ 2020-01-13 04:51  XiongRui  阅读(...)  评论(...编辑  收藏