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

An $A$-module is a semisimple module when it is a direct sum of simple modules, equivalently when every submodule has a complementary submodule. The finite-dimensional algebra $A$ is a semisimple algebra when its left regular module ${}_{A}A$ is semisimple.

Let $C_{p^n}=⟨g⟩$, put $N=p^n$, and suppose $\mathrm{char}k=p$. The freshman's dream givesso, with $u=g-1$,The indecomposable modules of a cyclic p-group in characteristic p are preciselyIndeed, a module is a vector space with a nilpotent operator $u$, and its decomposition into Nilpotent Jordan blocks gives these modules. Each $M_r$ is a uniserial module, with unique chainTherefore

For a finite-dimensional module over $A$, the radical of a module satisfies $J(M)=J(A)M$. Henceand induction gives $J^r(U\oplus V)=J^r(U)\oplus J^r(V)$ for every $r$. A simple submodule of a direct sum projects into semisimple submodules of each summand, and equivalentlyThus the same argument, or induction through the defining quotients, givesThis is radical and socle series of a direct sum.

Now let $G$ be a finite $p$-group and let $k$ have characteristic $p$. The group algebra of a p-group in characteristic p is local, with unique simple module $k$. The socle of its regular module iswhich is one-dimensional. If $kG=U\oplus V$ with both summands nonzero, finite length gives nonzero socles for $U$ and $V$, and the direct-sum identity would make $\mathrm{Soc}(kG)$ at least two-dimensional. Therefore

The radical of a module is the smallest submodule $N\subseteq M$ for which $M/N$ is a semisimple module. Consequently, ifhas semisimple successive quotients, then $J(M_i)\subseteq M_{i+1}$. Induction givesso the radical series of a module descends at least as fast as every such series.

Dually, the socle is the largest semisimple submodule. Ifhas semisimple successive quotients, induction in $M/N_i$ givesso 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,Thus $J^r(M)=0$ exactly when $J(A{)}^r$ annihilates all of $M$, which is exactly when ${\mathrm{Soc}}^r(M)=M$. The two least terminating indices therefore coincide:

For $1\le r\le n$, let $S_r=k$ with a lower triangular matrix $a=(a_{ij})$ acting by the scalar $a_{rr}$. These are pairwise nonisomorphic simple modules, since the diagonal matrix units $e_{ii}$ distinguish them.

Let $N$ be the ideal of strictly lower triangular matrices. It is nilpotent, andis a semisimple algebra. Hence $N$ is the Jacobson radical; explicitly,

For the left regular module, the radical series of a module is obtained by multiplying by powers of the Jacobson radical:Products of matrix units show thatwith $N^0=A$ and $N^n=0$. Thus each step removes one subdiagonal, and the Loewy length of ${}_{A}A$ is $n$.

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 basisFor $a\in A$, multiplication givesbecause 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