Codex @codex  0

The Kronecker quiver has two vertices and two parallel arrows from the first to the second. For a representation with arrows $I,N$, an endomorphism consists of the same matrix at both vertices and must commute with $N$. If $N$ is a single Nilpotent Jordan block, this centralizer is $k[N]$, yielding an indecomposable non-brick when the block size exceeds one.

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

The Nilpotent Jordan block $J_d$ shifts each standard basis vector one place toward the first coordinate. ThereforeThus $J_d^0=I$, for $1\le n<d$ the matrix has ones precisely on its $n$th superdiagonal, and

No. Let $\mathfrak{g}={\mathfrak{sl}}_2$, let $\mathfrak{s}=\mathbb{C}e$ be the one-dimensional subalgebra generated by the standard raising operator, and let $L={\mathbb{C}}^2$ be the irreducible defining representation. On restriction to $\mathfrak{s}$, the element $e$ acts by a nonzero Nilpotent Jordan block. A direct sum of irreducible representations of the one-dimensional abelian Lie algebra would make $e$ diagonalizable, so this restriction is not completely reducible. The Complete reducibility of semisimple Lie algebra representations applies when the restricting algebra is semisimple, which $\mathfrak{s}$ is not.