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.

We know the adjointness

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

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