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Question: Let $R$ be a ring and let $J$ be a left ideal of $R$.

Assume that $J$ is nilpotent. Prove that $J$ is contained in a nilpotent 2-sided ideal of $R$.

Comments: I have found lots of solutions online which go through this proof but can't fully understand any of them. I want to try and use the fact that $J +JR ⊆ JR$.

One proof online stated the following: "It is easy to see that $J^n = 0$ implies $(J + JR)^n = 0$"

I can't seem to figure out why this would be easy to see.

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You need to show that the $2$-sided ideal $J'$ generated by $J$ is nilpotent. For this, it is enough to show that the product of $n$ elements in of the form $jr$, with $j\in J$ and $r\in R$, is zero. This follows from the fact that $J$ is a nilpotent left ideal: if $j_1,\dots,j_n\in J$ and $r_1,\dots,r_n\in R$, then $$j_1r_1j_2r_2\cdots j_{n-1}r_{n-1}j_nr_n=\underbrace{j_1j'_2\cdots j'_{n-1}j'_n}_{=0}r_n=0$$ with $j_k'=r_{k-1}j_k\in J$ for $k=2,\dots,n$.

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  • $\begingroup$ In this answer I was just wondering where j′2...j′n−1j′n have come from as they haven't been mentioned in the solution until the equation, in the final line did you mean to say that j′k=rk−1jk∈J for k=2,…,n. $\endgroup$ – user225311 Mar 25 '15 at 11:31
  • $\begingroup$ @user225311 yes, fixed it. $\endgroup$ – Olivier Bégassat Mar 25 '15 at 14:21

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