Jacobson radical of the ring of lower triangular $n \times n$ matrices over $\mathbb{Z}$ I'm trying to solve an exercise, 6. section 3.2, of the book Introductory Lectures on Rings and Modules by John A. Beachy.
It reads:

Find Jacobson radical of the ring of lower triangular $n \times n$ matrices over $ \mathbb{Z}$.

This would not be that difficult given a small $n$, knowing that the intersection of all maximal ideals of $ \mathbb{Z}$ is $0$.
But I need a more general technique to solve for $ n\times n $ matrices.
 A: Fix a positive integer $n$.

Let $R$ be the ring of lower triangular $n \times n$ matrices over $ \mathbb{Z}$, and let $J$ be the Jacobson radical of $R$.

Let $S$ be the set of elements of $R$ which are strictly lower triangular.

It's easily verified that $S$ is an ideal of $R$.

Claim:$\;J=S$

Proof:

Let $A\in S$.

Since $A$ is strictly lower triangular, it follows that $A$ is nilpotent, hence $I_n-A$ is a unit of $R$.

Since $S$ is an ideal such that $I_n-A$ is a unit of $R$ for all $A\in S$, it follows that $S\subseteq J$ (Beacher proves this in the text).

Next suppose $S\subset J$ (proper inclusion).

Our goal is to derive a contradiction.

Let $A\in J{\setminus}S$.

Since $A\in J$, it follows that $I_n-A$ is a unit of $R$, hence each diagonal element of $I_n-A$ is equal to either $1$ or $-1$. But they can't all be equal to $1$, else $A\in S$, hence at least one diagonal element of $I_n-A$ is equal to $-1$, so the corresponding diagonal element of $A$ is equal to $2$.

But $A\in J$ implies $-A\in J$, so $I_n+A$ must be a unit of $R$, contradiction, since $I_n+A$ has a diagonal element equal to $3$.
