The <radical of a module> is the smallest submodule $N\subseteq M$ for which $M/N$ is a <semisimple module>. Consequently, if
$$
M=M_0\supseteq M_1\supseteq M_2\supseteq\cdots
$$
has semisimple successive quotients, then $J(M_i)\subseteq M_{i+1}$. Induction gives
$$
J^i(M)\subseteq M_i,
$$
so the <radical series of a module> descends at least as fast as every such series.

Dually, the <socle> is the largest semisimple submodule. If
$$
0=N_0\subseteq N_1\subseteq N_2\subseteq\cdots
$$
has semisimple successive quotients, induction in $M/N_i$ gives
$$
N_i\subseteq\operatorname{Soc}^i(M),
$$
so the <socle series of a module> ascends at least as fast as every such series.

Both series terminate because $M$ has finite <composition length>. More precisely,
$$
J^i(M)=J(A)^iM,
\qquad
\operatorname{Soc}^i(M)=\{x\in M:J(A)^ix=0\}.
$$
Thus $J^r(M)=0$ exactly when $J(A)^r$ annihilates all of $M$, which is exactly when $\operatorname{Soc}^r(M)=M$. The two least terminating indices therefore coincide:
$$
\boxed{m=n=\text{the Loewy length of }M.}
$$

Solution

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

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 $n$.

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

{title2=$J^i(M)$}

The radical series is the descending chain
$$
M\supseteq J(M)\supseteq J^2(M)\supseteq\cdots,
$$
where $J^{i+1}(M)=J(J^i(M))$. For finite-dimensional $A$ and $M$, one has $J^i(M)=J(A)^iM$.


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 Incoming links: Radical series of a module