Let $A$ be a unital Banach algebra. Is it true that $A$ is Morita equivalent with $M_I(A)$, where $I$ is an arbitrary index set ($M_I(A)$ is the space of $I*I$ matrices with entries in $A$. Let $a,b\in M_I(A)$ and $P$ be an invertible $I*I$ matrix with entries in $A$. The product of $a,b$ is defined by $a.b=aPb$).
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2$\begingroup$ How is $M_I(A)$ a Banach algebra if $I$ is infinite? How do you define multiplication? $\endgroup$– Qiaochu YuanCommented Jan 13, 2018 at 11:47
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1$\begingroup$ As @QiaochuYuan has pointed out, in order for $aPb$ to be well-defined, you must specify some extra conditions on $M_I(A)$, such as convergence with respect to some norm. I have downvoted the question until this is clarified. $\endgroup$– Yemon ChoiCommented Jan 13, 2018 at 14:24
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1$\begingroup$ If P is invertible then your Munn algebra is isomorphic to a usual matrix algebras on I. $\endgroup$– Benjamin SteinbergCommented Jan 13, 2018 at 17:31
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1$\begingroup$ @Mare I think it is up to the OP to clarify what he means, since the question seems currently ill-posed $\endgroup$– Yemon ChoiCommented Jan 13, 2018 at 18:35
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2$\begingroup$ You have now changed the question completely, this is not fair on the previous commenters. Please ask your new question as a separate question. I am reverting this to the original version, since your original question about Morita equivalence does not seem to be directly related to your second question about being self-induced $\endgroup$– Yemon ChoiCommented Jan 15, 2018 at 17:57
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1 Answer
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Taking $A= \mathbb{C}$ , $A$ is not Morita equivalent to $B=M_I (A)$ when $I$ is infinite, since $A$ is artinian but $M_I (A)$ is not (here I assume that $M_I(A)$ is defined as the endomorphism ring of a vector space of $\mathbb{C}$ with a basis of cardinality $I$). One way to see that $B$ is not artinian is for example to note that $B \cong B^2$ as $B$-modules. For finite $I$, and general $A$ and $M_I(A)$ are of course Morita equivalent as is well known (see for example the book "Lectures on modules and rings" chapter 17 by Lam).
