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Let $A$ be a local unital algebra, that is, it contains a unique maximal left ideal $J$ ($J \neq A$). I want to show that $J$ is also a right ideal. For that, I want to use the fact that an element $y \in A$ is left invertible if and only if $y \not \in J$.

So, let $x \in J$ and $a \in A$, and we want to show that $xa \in J$. If this is not the case, then $xa$ is left invertible: there exists $b \in A$ such that $b(xa) = 1$. How do I conclude?

Eduardo Longa
  • 5,226
  • In a local ring the maximal left ideal is the radical of the ring and this should coincide with the maximal right ideal.You can look this up in "A first course in noncommutative rings" by Lam – mathma Jun 23 '20 at 10:26
  • From where you stopped, $abx$ is idempotent. But the only idempotents in a local ring are $0,1$. Therefore $abx=1=bxa$ shows $a$ is a unit, a contradiction. – rschwieb Jun 23 '20 at 14:13

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