Morita Equivalence but not Equivalence

Morita Equivalence but not Equivalence

Non-Commutative Algebra with an Example

This is technically a counterexample, it is not one in the traditional sense. Normally, a counterexample is an object for which some intuition says it should not exist. Nobody thinks all Morita equivalent algebras and equivalent. But, just as nobody thinks all exact sequences split, it is helpful to have examples of non-splitting exact sequences at one’s elbow.

Morita equivalence is a property of (associative) algebras A and B. It means they have the same representation theory.

Definition 1. Let A,B be associative algebras over a field k. Then,  A is said to be Morita equivalent to B, written A \sim B, if there is an equivalence of categories A\textbf{-mod} \simeq B\textbf{-mod}.

To paraphrase: being acted on by A is essentially the same as being acted on by B.

Notation: Throughout, let A,B denote associative algebras over a fixed field \mathbb{k}

For commutative algebras, Morita equivalence coincides with isomorphism.

Lemma 2. Z(A\textbf{-mod}) \simeq Z(A) where Z(A) denotes the center of A and  Z(A\textbf{-mod})=\text{End}(\text{Id}_{A\textbf{-mod}}).

Proof.  Let z \in A be central. Then, define a natural transformation whose component \varphi_M : M \to M at a left A-module M is the action of z on M. By centrality of z\varphi_M is a module homomorphism. Conversely, given a natural transformation \varphi : \text{Id}_{A\textbf{-mod}} \to \text{Id}_{A\textbf{-mod}} define z:= \varphi_A(1_A) \in A. This z is central by naturality of \varphi. Q.E.D.

Proposition 3. Let A,B be commutative. Then, A \sim B if and only if A is isomorphic to B.

Proof. Clearly, if A \cong B then A \sim B. Conversely, suppose A \sim B. Then, Z(A\textbf{-mod}) \simeq Z(B\textbf{-mod}). By Lemma 2, we are done. Q.E.D.

To show this result is “sharp” we construct non-isomorphic algebras that are Morita equivalent. By Proposition 3. (at least) one algebra must be non-commutative.

Theorem 4. For n \geq 1, A \sim \text{Mat}_n(A).

Proof. Let e \in \text{Mat}_n(A) denote the matrix with a 1 in the 1 \times 1 spot and zeros everywhere else. Then, A \simeq e \cdot \text{Mat}_n(A) \cdot e and \text{Mat}_n(A) =\text{Mat}_n(A) \cdot e \cdot\text{Mat}_n(A). So by Morita’s Theorem [1, Corollary 2.3.2] we’re done. Q.E.D.

For n \geq 1, Mat_n(A) is non-commutative. So, for a commutative algebra A we cannot have an isomorphism between A and Mat_n(A).


[1] Ginzburg, V. Lectures on Noncommutative Geometry.


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