Order the simple modules as in part (i), so that $S_r$ records the $r$th diagonal character. The quotient $N^i/N^{i+1}$ has basis
$$
e_{i+1,1},e_{i+2,2},\ldots,e_{n,n-i}.
$$
For $a\in A$, multiplication gives
$$
ae_{r,r-i}\equiv a_{rr}e_{r,r-i}\pmod{N^{i+1}},
$$
because every term below row $r$ lies on a lower subdiagonal. Hence the line generated by $e_{r,r-i}$ is $S_r$, and the lines are independent. Therefore the radical layers of the <lower triangular matrix algebra> are
$$
\boxed{J^i(A)/J^{i+1}(A)
\cong S_{i+1}\oplus S_{i+2}\oplus\cdots\oplus S_n.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 1 / d / iii / Solution

{title2=$T_n(k)$}

The algebra $T_n(k)$ consists of the lower triangular $n$ by $n$ matrices over $k$. Its <Jacobson radical> $N$ consists of the strictly lower triangular matrices, and
$$
N^i=\operatorname{span}\{e_{rs}:r-s\geq i\}.
$$
The $n$ simple left modules $S_r$ are one-dimensional, with a matrix acting through its $r$th diagonal entry. For the left regular module,
$$
N^i/N^{i+1}\cong S_{i+1}\oplus\cdots\oplus S_n.
$$

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