Let R be a semisimple ring with identity.

If $e$ and $e'$ are idempotents in R such that $Re \simeq Re'$ then there exists $a \in R^\times$ such that $e' = aea^{-1}$


I know from a previous exercise that if we have $$1=e_1+....e_n=e'_1+...+e'_n$$ where $e_1,...e_n$ and $e'_1,...e'_n$ are orthogonal idempontents, respectively then we have, for every $i$

$Re_i \simeq R'e_i$ if and only if there exists $a \in R^\times$ such that $e_i' = ae_ia^{-1}$

I have $R= Re \oplus L = Re' \oplus L'$ for some semisimple submodules $L$ and $L'$. I think that by Krull Schmidt we have(??) $L \simeq L´$.

From there I can conclude $1=re+l=r'e'+l'$, where $L=Rl$ and $L'=Rl'$ for some idempotents $l$ and $l'$ and $r,r' \in R$

If I could show $r=r'=1$ I would be done but I have no idea on how to do it.

Another approach was to consider the Wedderburn decomposition of $R$ $$R=Re_1+...+Re_t$$ with the $e_i$'s orthogonal idempotents, but I can't relate those idempotents to the ones I am given, $e$ and $e'$. Any help?


You're on the right track. There is really no need for $L$ and $L'$ since you have obvious candidates to complete the idempotent to an orthogonal set.

What I mean is this: from $R=Re\oplus R(1-e)=Re'\oplus R(1-e')$ and $Re\cong Re'$, you can apply the Krull-Schmidt theorem to conclude $R(1-e)\cong R(1-e')$.

Then using your previous exercise, there exists a unit $a$ such that $aea^{-1}=e'$ (and $a(1-e)a^{-1}=1-e'$ too.)


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