I have been trying to prove the statement in the title, however I seem to get stuck at a certain point. Let $M$ be the maximal left-ideal of $R$. Then consider $M.r$ for $r \in R$. If $M.r \neq R$, we have that $M.r \subseteq M$. Now if this assumption would be true, we would have that $M$ is a right ideal as well. However when trying to prove that $M.r \neq R$ for all $r \in R$ I don't get very far. I know that if we assume equality, one would get $m.r=1$ for one $m \in M$. However I don't know how to go from here. Thank you in advance!
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1Does this answer your question? Prove if a non-trivial ring $R$ has a unique maximal left ideal $J$ , then $J$ is two-sided and is also the unique maximal right ideal in $R$. – Rick May 26 '20 at 13:49
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@Rick Look at your proposed dupe: the poster says "i can prove it is two sided but I can't prove uniqueness" and the solver posts an answer that proves uniqueness. That is not helpful for the question asked here. But nevertheless it's a good thing to link as a related question. – rschwieb May 26 '20 at 13:55
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This is also related but interpreting it a a duplicate requires a bit of mental contortion that seems excessive. – rschwieb May 26 '20 at 13:58
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If $R$ has one maximal left ideal, then it is the Jacobson radical. Do you know the Jacobson radical is two-sided?
Different idea:
Lemma: a local ring has only trivial idempotents. (Proof: if $e$ were a nontrivial idempotent, then how could $Re$ and $R(1-e)$ both be contained in the maximal left ideal?)
Hint: Now if $mr=1$ for some $m\in M$, $r\in R$, then $rm$ is an idempotent...
rschwieb
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Thank you for your comment. Would I then do the following: since $rm$ is idempotent it has to be trivial and therefore $rm=1$, which implies $1 \in M$ and is therefore a contradiction? – May 26 '20 at 14:03
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@HenryMeyer correct. Then your original reasoning that $Mr\subseteq M$ holds together than you're done. – rschwieb May 26 '20 at 14:04