Duality between Morita contexts.

abstract-algebracategory-theoryduality-theoremshomological-algebramodules

Definition: Let $A,B$ be rings. A Morita context connecting $A$ and $B$ is a sixtuple $(A,B,P,Q,f,g)$ where $P$ is an $(A,B)$-bimodule, $Q$ is a $(B,A)$-bimodule, $f: P \otimes_B Q \to A$ is an $(A,A)$-bimodule morphism and $g: Q \otimes_A P \to B$ is a $(B,B)$-module morphism such that $$q f(p\otimes _B q') = g(q \otimes_A p) q'; \quad pg(q \otimes_A p') = f(p \otimes_B q)p'$$
for all $p,p'\in P'$, $q,q' \in Q$.

Theorem: If $(A,B,P,Q,f,g)$ is a Morita context and $f$ is surjective, then we have the following properties:

(1) $f$ is bijective

(2) $P$ (seen as left $A$-module) and $Q$ (seen as right $A$)-module are finitely generated and projective.

(3) $P \cong _{B}\operatorname{Hom}(Q,B)$ as $(A,B)$-bimodules.

(4) $A\cong \operatorname{End}_B(P)$ as rings

and even more properties hold but I won't write them all down.

Now, if $g$ is surjective, then I expect the following properties to hold:

(1) $g$ is bijective

(2) $Q$ (seen as left $B$-module) and $P$ (seen as right $B$)-module are finitely generated and projective.

(3) $Q \cong _{A}\operatorname{Hom}(P,A)$ as $(B,A)$-bimodules.

(4) $B\cong \operatorname{End}_A(Q)$ as rings

As mathematicians, we are encouraged to work smart and not too hard. Is there a way to prove these results from the corresponding result about $f$? Is it as simple as observing that if $(A,B,P,Q,f,g)$ is a Morita context, then also $(B,A,Q,P,g,f)$ is a Morita context and we can apply the statement there, or are there subtleties involved that I am not seeing?

Best Answer

Yes, it is as simple as that: Morita context is a symmetric notion.

Just some further notes:

  1. The ingredients of a Morita context give rise to a single ring operation, namely we can consider the generalized matrix ring $$\pmatrix{A&P\\Q&B}$$ with multiplication given by multiplications in $A$ and $B$, the module actions, and $f$ and $g$.

  2. Moreover, the category of Morita contexts of rings with $1$ is equivalent to the category of 'idempointed' rings $(R,e)$ where $R$ is a ring and $e\in R$ is an idempotent.
    The correspondence is given by $$A=eRe,\ P=eRe',\ Q=e'Re,\ B=e'Re'$$ where $e'=1-e$.

    The reverse of $(R,e)$ in this setting is $(R,\,1-e)$.

Related Question