Question

Give an example of a (necessarily noncommutative) ring R in which N1(R) is strictly contained Nil*R.

Answer 1

Fix a commutative ring $S$ with a nilradical $J$ that is not nilpotent, and consider $R_0 = \begin{pmatrix} J & S \\ J & J \end{pmatrix} \subseteq \mathrm{M}_2(S)$ . This $R_0$ is a "subring" of $\mathrm{M}_2(S)$ , except for the fact that it may not possess an identity element.

Then $\mathbf{R}_0$ is a "nil rng" (i.e. every element of $\mathbf{R}_0$ is nilpotent).

We adjoin an identity to $R_0$ to form a ring $R := R_0 \oplus Z$ . Previous claim clearly implies that $\mathrm{Nil}^*R = R_0$ . Therefore, $\mathrm{Nil}^*R \subseteq R_0$ . Let $N = N1(R) \subseteq \mathrm{Nil}^*R$ be the sum of all nilpotent ideals of $R$ . Then the matrix $\beta = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ does not belong to $N$ . Moreover $\beta \in \mathrm{Nil}^*R$ , which implies that $N$ is strictly contained in $\mathrm{Nil}^*R$ .