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Wedderburn's theorem

Let $R$ be a ring and $M$ a simple, faithful module over $R$. Let $E = {_R\text{End}(M)}$ and assume that $M$ is finite dimensional over $E$. Then $R\cong {_E\text{End}(M)}$

Apparently this can be proven using a Morita context.

If the following is a Morita context $(R,E,M,M^*,\mu,\tau)$, and if $\mu$ is surjective. I will get my result. (This is some result from Morita Theory). However this requires that $M$ be a $R,E$-bimodule. But I don't see how it is a right $E$-module.

Any help?

What I tried:

I tried defining $m\cdot f:=f(m)$ but that doesn't work for the product. Ie, $$m(fg) = (fg)(m) = f(g(m))\\(mf)g = g(mf) = g(f(m))$$

Another result says that if $(R,E^{op},M,M^*,\mu,\tau)$ is Morita Context and if the context is strict, then $R\cong {_{E^{op}}\text{End}(M)}$

Since $_{E^{op}}\text{End}(M) \cong{_E\text{End}(M)}$, this only leaves me to show that the context is strict which I don't see how I can show.

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  • $\begingroup$ I don't know how relevant this still is but have you tried finding a progenerator? That might be easier $\endgroup$ – M.v.Roozendaal Jan 27 '18 at 16:42

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