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So I was trying to make an excercise which has multiple parts. The first part is the following.

Show that there exists a Morita context between the rings $eRe$ and $e'Re'$ with $e$ an idempotent and $e' = 1-e$.

This I was able to do seeing it is not very hard to come up with a Morita context.

The part I am having difficulties with is the following:

Show that any Morita context between rings $A$ and $B$ is of the form above. i.e. There exists a ring $R$ with idempotents s.t. $A = eRe$ and $B = e'Re'$

I have done some reading in the past on quivers and more specifically on how they can be used to study the representation theory of finite dimensional algebra's over algebraically closed fields. So my first idea was to perhaps construct a quiver where the nodes are the idempotents and do something with that but I think that would be a rather silly aproach (also in that context one works with algebras where we work with general rings).

Additional comment: I think the given that $e' =1-e$ is not necessary, Just having idempotents is sufficient (they needn't be orthogonal) but that's just a minor detail.

Any hints in the right direction would be appreciated.

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  • $\begingroup$ I think I have found a solution given that the context is a strict one. I think this is an assumption that is missing from the notes of my professor. $\endgroup$ Jan 27, 2018 at 17:37

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Think about it like this: in the first part, you decomposed your ring $R$ as a set into $$R = eRe \times eRe' \times e'Re \times e'Re'$$ and then did some magic. With respect to this decomposition, the multiplication of $R$ looks like the multiplication in the matrix ring $$ \begin{pmatrix} eRe & eRe' \\ e'Re & e'Re' \end{pmatrix} . $$

So, if you have a Morita context between rings $A, B$, with $A-B$-bimodule $M$ and $B-A$-bimodule $N$, you can consider the matrix ring $$ T = \begin{pmatrix} A & M \\ N & B \end{pmatrix}. $$ That this is indeed an associative unital ring boils down exactly to the fact that you have a Morita context: You have "multiplications" $M \times N \to A$ and $N \times M \to B$, which are bimodule maps and satisfy a nice "mixed associativity".

And clearly, the idempotent $e = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ satisfies $eTe \cong A$, $e'Te' \cong B$, and so on.

Finally, strictness is not necessary for this to work.

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