Question

For a triangular ring T =
R M
0 S
(where M is an (R, S) - bimodule), show that rad(T) =
rad(R) M
0 rad(S)

Answer 1

Let $J = \operatorname{rad}(R)$, $J' = \operatorname{rad}(S)$, and $I = \begin{pmatrix} J & M \\ 0 & J' \end{pmatrix}$. It is routine to check that $I$ is an ideal of $T$, with $T/I \sim R/J \times S/J'$. The latter ring is $J$-semisimple, so we have $\operatorname{rad}(T) \subseteq I$. This will be an equality, as asserted, if we can show that $1 + I \subseteq \mathrm{U}(T)$. Now any element of $1 + I$ has the form $\begin{pmatrix} u & m \\ 0 & v \end{pmatrix}$, where $m \in M$, $u \in \mathrm{U}(R)$, and $v \in \mathrm{U}(S)$. This is indeed in $\mathrm{U}(T)$, since it has inverse $\begin{pmatrix} u^{-1} & -u^{-1}mv^{-1} \\ 0 & v^{-1} \end{pmatrix}$.