In a (not necessarily commutative) ring, a nilpotent maximal ideal is unique.

Question

Let $R$ be a (not necessarily commutative) ring and suppose $R$ has a maximal left ideal $I$ consisting of nilpotent elements, that is for $x\in R$ there exists some $n\in\mathbb{Z}_{\geq 0}$ such that $x^{m}=1$. I want to show that in this case, $I$ is the unique maximal left ideal.

To attempt to prove this, assume $J$ is some other maximal left ideal of $R$. Since $x^{m}=0$, we have $x^{m}\in J$. This is where I get stuck.

In the commutative case we can note that $I$ is necessarily a prime ideal, so either $x\in J$ or $x^{m-1}\in J$. Iterating this process shows that $x\in J$, then $I=J$ by maximality.

In the non-commutative case I can't seem to proceed. This is probably really easy but I'd appreciate a hint.

Answer 1

It's well-known that a nilpotent element plus $1$ is a unit.

It follows that for every $x\in I$, $x+1$ is a unit, and this proves $I\subseteq J(R)$, the Jacobson radical. Since $J(R)\subseteq K$ for every maximal left ideal $K$, it must be that $I\subseteq K$ for every maximal left ideal $K$. Obviously by maximality $I=K$ for all such $K$, then.

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Note, you called $I$ nilpotent but that is not the normal meaning of "nilpotent ideal". One would say $I$ is a nilpotent (left/right/two-sided) ideal if $I^n=\{0\}$ for some $n$. If merely all its elements are nilpotent, one says it is a nil (right/left/two-sided) ideal, or sometimes a locally nilpotent ideal.