A ring $R$ is called with bounded index (of nilpotency) $n$ if $n$ is the smallest natural number such that $a^n=0$ for all nilpotent $a \in R$. Now let $R$ be a commutatitve ring with bounded index $n$. how can I show that $R[x]$ is also with bounded index $m$, where $m \leq 1+ \frac{n(n-1)}{2}$ ? Probably it was proved by A. A. Klein. But I am unable to collect that paper. Please help me solving this problem.
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(I would have made this a comment, but I don't yet have the 50 reps needed.) You can find Klein's 4 page paper by googling "Abraham Klein bounded index nilpotence" and clicking the first link (which should be given by this link to a google book).
The result you want is theorem 4 in his paper "Rings with bounded index of nilpotence".
answered Nov 16, 2014 at 4:08