Question

Let J be a nilpotent right ideal in a ring R. If I is a subgroup of J such that I • I ⊆ I and J = I + J2, show that I = J.

Answer 1

Question 1.

Let $J$ be a nilpotent right ideal in a ring $R$. If $I$ is a subgroup of $J$ such that $I\cdot I\subseteq I$ and $J=I+J^{2}$, show that $I=J$.

Solution.

Prove by induction that $J=I+J^{n}$ for all $n\geq 2$. The base of induction is given: $J=I+J^{2}$. The inductive step: suppose $J=I+J^{n}$ and prove that $J=I+J^{n+1}$. We have

$J^{2}=J\cdot J=J(I+J^{n})=JI+J^{n+1}=(I+J^{n})I+J^{n+1}=I\cdot I+J^{n}I+J^{n+1}.$

It is given that $I\cdot I\subseteq I$. Furthermore, $J^{n}I\subset J^{n}\cdot J=J^{n+1}$, because $I\subseteq J$. Finally, $J^{n+1}+J^{n+1}\subseteq J^{n+1}$, as any right ideal is closed under addition. Thus,

$J^{2}=I\cdot I+J^{n}I+J^{n+1}\subseteq I+J^{n+1}+J^{n+1}=I+J^{n+1}.$

Therefore,

$J=I+J^{2}\subseteq I+(I+J^{n+1})=(I+I)+J^{n+1}=I+J^{n+1},$

because $I$ is an additive subgroup of $J$. The converse inclusion $I+J^{n+1}\subseteq J$ is obvious, because $J$ is closed under addition and multiplication and $I\subseteq J$. So, $J=I+J^{n+1}$.

Since $J$ is nilpotent, there is $n\geq 2$ such that $J^{n}=0$. Taking this $n$, we get

$J=I+J^{n}=I+0=I.$

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