I am trying to understand a proof I am reading but it doesn’t make much sense to me.
If $R$ is a ring with $I$ a nilpotent left ideal then $I$ is contained in a nilpotent two-sided ideal.
The proof goes like this.
If $I^n=0$ then $$(IR)^n=(IR)(IR)…(IR)=I(RI)(RI)…(RI)I \subseteq I^n (\ast) .$$ Apparently I need to use induction to show this. Firstly, I know that $IR$ is a two-sided ideal containing $I$. However I don’t get how $(\ast ) $ works. I get that $RI \subseteq I $ since $I$ is a left ideal in $R$ but then $(IR)^n=I(RI)^{n-1}R \subseteq I^nR $ using some inductive argument but I don’t get why then $I^n R \subseteq I^n $ since surely we would need $I^n $ to be a right ideal of $R$.
Yes, I think you are correct to be suspicious that there is a misprint, or perhaps we should say a misleadingprint.
$(IR)^n$ is certainly going to expand to something with an $R$ on the right. We can say $(IR)^n=I(RI)^{n-1}R\subseteq II^{n-1}R=I^nR=\{0\}R=I^n$
If it were up to me I'd just reprint the tail end $\ldots (RI)I=I^{n}$ as $\ldots (RI)R\subseteq I^{n}R=\{0\}$