Show \(\nilrad{R} \normal R\) is an ideal and \(A/\nilrad{R}\) is reduced.
Solution
Let \(R\) be a commutative ring with identity.
1. \(R\,\nilrad{R}\subseteq\nilrad{R}\).
If \(r^n=0\) and \(x\in R\), then \((xr)^n=x^nr^n=0\).
2. \(\nilrad{R}+\nilrad{R}\subseteq\nilrad{R}\).
Let \(r^n=s^m=0\) and \(N=n+m-1\). In \((r+s)^N=\sum_{k=0}^N\binom Nk r^ks^{N-k}\), each term has \(k\ge n\) or \(N-k\ge m\), so each term is \(0\). Together with \(0\in\nilrad{R}\) and \(-r\in\nilrad{R}\), steps 1 and 2 show that \(\nilrad{R}\) is an ideal.
3. \(R/\nilrad{R}\) is reduced.
Let \(\bar r=r+\nilrad{R}\) with \(\bar r^n=0\). Then \(r^n\in\nilrad{R}\), so \((r^n)^m=0\) for some \(m\), hence \(r^{nm}=0\) and \(r\in\nilrad{R}\), that is, \(\bar r=0\).