Question

It is well known that, for any commutative noetherian R, (intersection) (rad R)^n = 0. Show that this need not be true for noncommutative right noetherian rings.

Answer 1

Let $A$ be a commutative discrete valuation ring with a uniformizer $\pi(\text{that is nonzero})$ and quotient field $K$. Consider the ring $R = \begin{pmatrix} A & K \\ 0 & K \end{pmatrix}$, which is right noetherian (but not left noetherian). It is easy to check that

$$J := \left( \begin{array}{cc} \pi A & K \\ 0 & 0 \end{array} \right)$$

is an ideal of $R$, and that $R / J \sim (A / \pi A) \times K$. Since the latter is a semisimple ring, we have $\operatorname{rad} R \subseteq J$. On the other hand, $1 + J$ consists of matrices of the form

$$\left( \begin{array}{cc} 1 + \pi a & b \\ 0 & 1 \end{array} \right) (a \in A, b \in K),$$

which are clearly units of $R$. Therefore, $J \subseteq \operatorname{rad} R$. We have $\operatorname{now} J = \operatorname{rad} R$, from which it is easy to see that

$$(\operatorname{rad} R)^\wedge n = \left( \begin{array}{cc} \pi^n A & K \\ 0 & 0 \end{array} \right),$$

for any $n \geq 1$. It follows that $\bigcap_{n \geq 1} (\operatorname{rad} R)^n = \begin{pmatrix} 0 & K \\ 0 & 0 \end{pmatrix} \neq 0$.