If R is a local ring, then its maximal left-ideal is a left and right ideal

Question

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!

Answer 1

If $R$ has one maximal left ideal, then it is the Jacobson radical. Do you know the Jacobson radical is two-sided?

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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...