For the left regular module, the <radical series of a module> is obtained by multiplying by powers of the <Jacobson radical>:
$$
J^i({}_AA)=J(A)^i=N^i.
$$
Products of <matrix units> show that
$$
\boxed{N^i=\operatorname{span}_k\{e_{rs}:r-s\geq i\},
\qquad0\leq i\leq n,}
$$
with $N^0=A$ and $N^n=0$. Thus each step removes one subdiagonal, and the <Loewy length> of ${}_AA$ is $n$.

Solution

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

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The Loewy length of a finite-length module is both the least $m$ for which $J^m(M)=0$ and the least $m$ for which $\operatorname{Soc}^m(M)=M$.


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 Incoming links: Loewy length