(January 2014) A nonzero element \(a\) of a ring is said to be nilpotent if there is a positive integer \(n\ge 1\) such that \(a^n=0\). (The element \(0\) itself is not said to be nilpotent.) Let \(R\) be a commutative ring, and let \(I\subseteq R\) be an ideal. Prove that the following two statements are equivalent. (a) The quotient ring \(R/I\) contains no nilpotents. (b) For every element \(b\in R\) such that \(b^m\in I\) for some positive integer \(m\ge 1\), we have \(b\in I\).
Solution
Proof. (=⇒) Given b∈R and m≥ 1 with bm∈I, we have (I +b)m =I +bm =I + 0. SinceI + 0 is the zero element of R/I, it follows that I +b is either nilpotent or zero in R/I. But by assumption (a), I +b is not nilpotent. Therefore, I +b =I + 0, which means b =b− 0∈I. (⇐=) SupposeI +b∈R/I is a nilpotent. Then there is some integerm≥ 1 such that (I +b)m =I +0. That is, I +bm = (I +b)m =I + 0, and hence bm =bm− 0∈I. By property (b), we have b∈I. Thus, I +b =I + 0, contradicting the assumption that I +b is nilpotent, and proving that R/I has no nilpotents. QED