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For a finitely generated associative algebra $A$, the representation variety of an associative algebra consists of generator matrices satisfying all defining polynomial relations. With fixed orthogonal idempotents $1={\sum }_{i=1}^m{e}_i$, require a representation on ${⨁}_i{k}^{n_i}$ to send $e_i$ to the standard vertex projector. This is the meaning of ${\mathrm{Rep}}_A(\mathbf{n})$; without specified idempotents, use the usual single dimension $r$ and ${\mathrm{Rep}}_A(r)={\mathrm{Hom}}_{k\text{-alg}}(A,M_r(k))$. Polynomial relations cut out a closed affine variety in the space of generator matrices.

The group $G={\prod }_i{\mathrm{GL}}_{n_i}(k)$ acts by conjugation, preserving the prescribed projectors. Its stabilizer is ${\mathrm{Aut}}_A(X)$, a nonempty open subset of ${\mathrm{End}}_A(X)$. The orbit dimension formula consequently givesA degeneration of a module $X$ to $Y$ means that ${\mathcal{O}}_Y\subseteq \overline{{\mathcal{O}}_X}$, equivalently that one representative of $Y$ lies in this closure.

For $0\to X^{\prime }\to X\to X^{\prime \prime }\to 0$, choose a vector-space splitting, so every generator has block matrix $x_g=\left(\begin{array}{cc}{x}_g^{\prime }& b_g\\ 0& x_g^{\prime \prime }\end{array}\right)$. Conjugation by $\mathrm{diag}(tI,I)$, $t\ne 0$, givesThis polynomial family extends to $t=0$, retains all algebra relations, and at zero represents $X^{\prime }\oplus X^{\prime \prime }$. Thus splitting an extension gives a module degeneration. Iterating along a composition series gives $X⇝\mathrm{gr}X={⨁}_j{X}_j/X_{j-1}$. Degenerations are transitive because an orbit closure is closed and invariant under base change.

For the one-arrow quiver with dimension vector $(n_1,n_2)$,The action is $A\mapsto g_{2}A{g}_1^{-1}$. Its rank orbits of a matrix under left-right multiplication are indexed by $r=0,\dots ,\min (n_1,n_2)$, with a representative containing an $r\times r$ identity block and zeros elsewhere. Their dimensions are $r(n_1+n_2-r)$; their closures contain exactly matrices of rank at most $r$.

Put $q=p^n$. A finitely generated module over $A=k[X]/(X^q)$ is finite-dimensional over $k$, and multiplication by $X$ is a nilpotent linear map $T$ with $T^q=0$. Its Jordan normal form exists over $k$ because its minimal polynomial is a power of $X$. Each Jordan block has size $r\le q$ and is the cyclic module $M_r=k[X]/(X^r)$. HenceA submodule of $M_r$ is an ideal of $k[X]/(X^r)$, and its inverse image is an ideal of the principal ideal domain $k[X]$ containing $(X^r)$. The only possibilities are $(X^j)$ with $0\le j\le r$. Thus all submodules form the chainEvery successive quotient is the one-dimensional simple module on which $X$ acts as zero. This is its unique composition series, so $M_r$ is a uniserial module and is an indecomposable representation: two nonzero direct summands would be incomparable submodules. Only $M_1$ is simple.

For $G=⟨g⟩$ of order $q$, the group algebra satisfiessince $(T-1{)}^q=T^q-1$ in characteristic $p$. Consequently, up to isomorphism, $\boxed{M_1,\dots ,M_{p^n}}$ are precisely the finite-dimensional indecomposable modules, and $\boxed{M_1\text{ is the unique simple module}}$. The finite-dimensional convention follows from the finite-generation hypothesis; “exactly” counts isomorphism classes.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.

If a finite group is not simple, choose a maximal proper normal subgroup and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.

For $S_4$ one composition series isThe successive quotients have orders $2,3,2,2$, hence are simple. As subgroups of $S_4$, the groups $S_4,A_4,V_4$, and $1$ are normal. The selected order-two subgroup is not normal, since conjugation permutes the three double transpositions.