Product of rings are Morita equivalent implies the rings are Morita equivalent.

Question

Let $A$ and $B$ be two rings. Then we say $A$ is Morita equivalent to $B$, denoted as $A\sim B$, if the category of left $A$-modules is equivalent to the category of left $B$-modules. Let $A^{n}:=A\times \dotsc \times A$, $B^{n}:= B\times \dotsc \times B $. I am curious about the following question:

If $A^{n}$ is Morita equivalent to $B^{n}$, then is $A$ Morita equivalent to $B$?

I think the answer for the question is yes. But I can not find a reference.

@rschwieb. I think my first question is not so clear. Maybe I just wonder whether $A$ and $B$ are necessarily Morita equivalent. — Mike, Aug 29 '24 at 18:37

score 7 · Accepted Answer · answered Aug 29 '24 at 22:22

This answer to another question gives an example of two commutative rings $A$ and $B$ such that $A\not\cong B$ but $A\times A\cong B\times B$. Since commutative rings are isomorphic if and only if they are Morita equivalent, this also gives a negative answer to this question.

Product of rings are Morita equivalent implies the rings are Morita equivalent.

1 Answers1