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An abelian variety of dimension $g$ has complex multiplication when its rational endomorphism ring contains a commutative semisimple $\mathbb{Q}$-algebra of dimension $2g$. For an elliptic curve in characteristic zero, this means its geometric endomorphism ring of an elliptic curve is an order in an imaginary quadratic field. The resulting rank-one Tate module representations connect torsion points to class field theory.

Take sources $k,k,k$, sink $k^2$, and arrows sending $1$ to $e_1,e_2,e_1+e_2$. An endomorphism preserving these three lines must be scalar, so the representation has endomorphism ring $k$ and is a brick module. It is the only indecomposable of the three-subspace quiver with a two-dimensional sink.

A nonzero module is indecomposable if it cannot be expressed as a direct sum of two nonzero submodules. The zero module is excluded. A module of finite composition length is indecomposable exactly when its endomorphism ring has no idempotents other than zero and one: an idempotent splits the module into its image and kernel.

An endomorphism ring is local when its nonunits form a proper two-sided ideal; commutativity is not required. The Fitting lemma gives this property for an indecomposable finite-length module. If two nonunits had an invertible sum, multiplication by its inverse would give $I=f+g$ with noninvertible $f,g$. Fitting makes $f$ nilpotent, so $I-f=g$ would be invertible, a contradiction. Products with a nonunit are nonunits by finite-length injectivity/surjectivity. The quotient by the nonunit ideal is a division ring.

A module endomorphism is a module homomorphism from a module to itself. These maps form the endomorphism ring ${\mathrm{End}}_R(M)$ under pointwise addition and composition. An invertible module endomorphism is a module automorphism.

The infinitesimal base change action on quiver representations is ${\xi }_x(u{)}_{\rho }=u_{t(\rho )}{x}_{\rho }-x_{\rho }{u}_{s(\rho )}$. Its image is the Zariski tangent space to the orbit, because the stabilizer is a smooth open subset of the endomorphism ring. The extension complex of quiver representations identifies the quotient of the ambient tangent space by this image with ${\mathrm{Ext}}_Q^1(X,X)$. Consequently rigid quiver representations have open orbits.

For an algebraic group action, the orbit dimension is the dimension of the group minus that of its stabilizer. For a finite-dimensional representation, the stabilizer is an automorphism group, open in its endomorphism ring, so their dimensions agree. In a quiver representation space, the Ringel form turns the orbit codimension into the dimension of the self-extension group.

Nagao module theorem states that if $e\in Z(kG)$ is a central idempotent, $eM=M$, and $C_G(Q)\le K\le N_G(Q)$, then ${\mathrm{Br}}_Q(e)$ is central in $kK$ andEvery indecomposable summand of the second term is relatively projective for some subgroup of $K$ not containing $Q$. In particular, if $Q\le K$, the error term has no summand with module vertex containing $Q$. This is the Nagao version of the second main theorem; it identifies the modular block of a module vertex-$Q$ Green correspondent.

Let $e_B$ be the identity of $B$, $N=N_G(D)$, and $b$ its Brauer correspondent of a block in $kN$. Because $D◃N$, the ideal $J(kD)kN$ is nilpotent: normality allows moving $kN$ past $J(kD)$ and the augmentation ideal of the p-group is nilpotent. Hence $b$ has nonzero image in $k(N/D)$. Choose an indecomposable summand $U$ of $bk[N/D]$. It is projective over $k(N/D)$ and belongs to modular block $b$ after inflation.

Since $k[N/D]={\mathrm{Ind}}_D^{N}k$ is a permutation module, its summand $U$ has trivial module source and module vertex contained in $D$. As $D$ acts trivially, all its proper-subgroup traces vanish and its Brauer quotient of a module is $U(D)=U\ne 0$. The module vertex must also contain $D$, so it is exactly $D$. This is inflated quotient projectives have trivial source.

Let $M$ be the Green correspondent of $U$ in $G$, using the theorem proved in the next solution. It has module vertex $D$ and the same trivial module source. For a normal p-subgroup, each central block idempotent of $kN$ equals its Brauer image: the two central idempotents agree modulo the nilpotent ideal $J(kD)kN$, because nonfixed conjugation-orbit sums map to zero in $k(N/D)$; central idempotents congruent modulo a nilpotent ideal coincide. The first main theorem therefore gives ${\mathrm{Br}}_D(e_B)={\mathrm{Br}}_D(b)=b$.

Let $e_C$ be the block identity acting on $M$. Nagao's theorem forces ${\mathrm{Br}}_D(e_C)U=U$, since $U$ has module vertex $D$. But $bU=U$, and for $C\ne B$ orthogonality would give ${\mathrm{Br}}_D(e_C)b={\mathrm{Br}}_D(e_C{e}_B)=0$. Therefore $M$ belongs to $B$, has module vertex $D$ and has trivial module source. For $D=1$, this supplies an indecomposable projective in the defect-zero modular block.

For the representation-type deduction, modules in a block are projective relative to its defect group: the block trace identity acts as a relative trace equal to the module identity, so D. Higman criterion gives split induction from $D$. If $D$ is cyclic of order $d$, then $kD\cong k[t]/(t^d)$, whose indecomposable modules are the nilpotent Jordan chains $k[t]/(t^j)$, $1\le j\le d$. Each indecomposable in $B$ is a summand of the module induction of its module restriction to $D$, and hence occurs among the summands of these finitely many fixed induced modules. Their finite Krull-Schmidt decompositions give finite representation type.

Conversely, the regular defect-group bimodule inside a block givesIndeed this module restriction is a permutation summand. Its Brauer quotient at $\Delta D$ is nonzero, because on the fixed group-basis elements $C_G(D)$ the block projection becomes multiplication by ${\mathrm{Br}}_D(e_B)\ne 0$. Every transitive $D\times D$ orbit in $G$ has stabilizer of order at most $|D|$. A summand with a nonzero quotient at $\Delta D$ must therefore have stabilizer exactly $\Delta D$ up to conjugacy, and is the regular $kD$ bimodule. Tensoring this split inclusion on the right with an arbitrary $kD$-module $V$ yieldsIf $B$ has only finitely many indecomposable modules, their restrictions contain only finitely many indecomposable summands. The displayed inclusion puts every indecomposable $V$ on that finite list, so $kD$ has finite representation type too.

A noncyclic finite p-group has a quotient $C_p\times C_p$, by the Burnside basis theorem for its Frattini quotient. Here is an explicit infinite family over any field, including a finite field. For $s\ge 1$, take basis $v_0,\dots ,v_s,w_1,\dots ,w_s$ and operators $x=g-1$, $y=h-1$ defined byand $x{w}_i=y{w}_i=0$. All products of $x,y$ vanish, so $1+x,1+y$ satisfy the group relations in characteristic $p$. The radical is $W=⟨w_1,\dots ,w_s⟩$. An endomorphism preserves $W$, and commuting with the two shift maps forces its induced maps on $V/W$ and $W$ to be the same scalar. To see this, write the shift matrices as $X=(0I_s)$ and $Y=(I_{s}0)$: the equations $CX=XA$, $CY=YA$ force the off-diagonal entries of $A,C$ to vanish and all diagonal entries to agree, by comparing successive rows, including the zero first column of $X$ and zero last column of $Y$. The remaining linear map sends the top into $W$ and has square zero. Thus the endomorphism ring is a scalar algebra with square-zero radical, hence local; the module is indecomposable. Inflation to $D$ preserves that ring. Its dimensions $2s+1$ are unbounded, contradicting finite representation type. Therefore $B$ has finite representation type if and only if $D$ is cyclic, including trivial defect.

Use finite-dimensional $k$-modules or finite $R$-free lattices. We use the basic relative-projectivity facts that Krull-Schmidt decomposition holds with local endomorphism rings, module vertices are conjugate p-subgroups, and a module source $S$ at a module vertex $D$ satisfies $S\mid M{\downarrow }_D$ and $M\mid S{\uparrow }^G$. Module induction is transitive and Mackey restriction formula gives module vertex containment for restrictions and inductions. In particular, a summand induced from $D$ has module vertex conjugate into $D$, and module restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex. The latter follows from the module source conditions and minimality: restricting further to $D$ must still contain its module source, whose module vertex is $D$.

The module vertex-$D$ form of Green correspondence states that for $N_G(D)\le H\le G$ there is a bijection between indecomposable $RG$-modules with module vertex $D$ and indecomposable $RH$-modules with module vertex $D$. Its maps select the unique module vertex-$D$ summands of module restriction and induction, using $H$-conjugacy for restricted module vertices and $G$-conjugacy for induced module vertices. They occur with multiplicity one and have the same module sources. Other induction summands have module vertices conjugate to proper subgroups of $D$. The module restriction remainder is relatively projective forRelative projectivity for a family means being a summand of a finite sum of inductions from its members. We prove this form; no stronger description of the induction remainder is needed for the application.

Let $U$ have module vertex $D$ over $H$. The identity double coset in Mackey restriction formula givesSince $U$ is relatively $D$-projective, a further Mackey decomposition puts every other double coset term in the relatively-$\mathcal{Y}$ family. No member of $\mathcal{Y}$ contains an $H$-conjugate of $D$. Otherwise, conjugation within $H$ would give $D\le H\cap (ha)D(ha{)}^{-1}$ with $h\in H$, $a\notin H$. Equality of orders then makes $ha\in N_G(D)\le H$, a contradiction. Therefore the module restriction above contains exactly one summand with module vertex $H$-conjugate to $D$, namely $U$ once.

Decompose $U{\uparrow }^G$ into indecomposable modules. All their module vertices are conjugate into $D$. Each one with a module vertex $G$-conjugate to $D$ can take $D$ itself as module vertex and contributes a module vertex-$D$ summand on module restriction to $H$. There can thus be at most one such summand and its multiplicity is one. There must be at least one, because smaller module vertices cannot restrict to produce $U$, with module vertex $D$, in the displayed decomposition. Define $g(U)$ to be that summand. Its module restriction contains $U$ once, with remainder a summand of $T$. All other induction summands have strictly smaller module vertices. This is Green correspondence from Mackey multiplicity.

Conversely take $M$ of module vertex $D$ over $G$ with module source $S$. Decompose $S{\uparrow }^H$ into indecomposable modules. Since $M\mid S{\uparrow }^G=(S{\uparrow }^H){\uparrow }^G$, the local endomorphism ring and Krull-Schmidt decomposition give $M\mid U{\uparrow }^G$ for one indecomposable summand $U$ of $S{\uparrow }^H$. Its module vertex is $H$-conjugate to $D$: all module vertices are conjugate into $D$, and a smaller one could not induce $M$. The preceding uniqueness makes $g(U)=M$ and identifies $U$ as the unique module vertex-$D$ summand of $M{\downarrow }_H$. Define $f(M)=U$. This proves $f(g(U))=U$ and $g(f(M))=M$.

If $S$ is a module source of $U$, then $S\mid U{\downarrow }_D\mid M{\downarrow }_D$ and $M\mid U{\uparrow }^G\mid S{\uparrow }^G$. These two conditions and the common module vertex show that $S$ is also a module source of $M$. Thus the correspondence preserves module sources, including trivial source modules, and the proof is complete under the stated coefficient conventions.

For the required expression, a p-local module is a finite sum of inductions from $N_G(Q)$ for nontrivial p-subgroups $Q$. Induct on the order of a module vertex of $M$. Vertex 1 makes $M$ projective, so choose $L_1=L_2=P_1=0$, $P_2=M$.

For module vertex $D\ne 1$, take $U=f(M)$ for $H=N_G(D)$. Its module induction is p-local andwith smaller module vertices for the $X_i$. By induction, writewith $A_i,B_i$ p-local and $P_i,Q_i$ projective. Adding these identities and substituting the induction decomposition givesHence the explicit choiceswork. All sums are finite and module vertex orders strictly decrease, so the induction terminates. If $N_G(D)=G$, induction from that normalizer is still allowed by the definition of p-local.

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

Put $q=p^n$ and choose a generator $h$ of $H$. In characteristic $p$, $(h-1{)}^q=h^q-1=0$, soA module is a vector space with a nilpotent operator $t$ whose nilpotence index is at most $q$. The nilpotent Jordan normal form, valid over every field, splits it into Jordan blocks. One block of size $i$ isIt is indecomposable: its endomorphism ring is $k[t]/(t^i)$, a local ring with no nontrivial idempotents. Distinct $i$ have distinct dimensions. Conversely a direct sum of at least two Jordan blocks is decomposable. Thus there are exactly $q$ indecomposable isomorphism classes.

A homomorphism $V_i\to V_j$ is determined by the image of $1$, which can be any element of $V_j$ annihilated by $t^i$. If $i\ge j$ this is all of $V_j$, of dimension $j$; if $i<j$ it is the span of $t^{j-i},\dots ,t^{j-1}$, of dimension $i$. HenceThis realizes the indecomposable modules for a cyclic p-group.

Let $B=kGe$ have defect group $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every indecomposable module in $B$ has a module vertex conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=ga{h}^{-1}$, is an indecomposable permutation-module summand with module vertex $\Delta D=\{(d,d):d\in D\}$ and trivial module source. The second fact is the bimodule characterization of a defect group of a block; the first follows by applying its relative-$\Delta D$ induction splitting to $B{\otimes }_{kG}M=M$.

If $D$ is cyclic, every module source for every module vertex $Q\le D$ occurs among the finitely many indecomposable modules for a cyclic p-group. Each module in the block is a summand of an induction of one of these finitely many module sources. There are finitely many subgroups $Q$ and each induced representation has finitely many indecomposable summands by Krull–Schmidt theorem. Thus cyclic defect implies finite representation type of the block.

For the converse, the restriction $B{\downarrow }_{D\times D}$ has the regular $kD$-bimodule as a direct summand. To see this from module vertices and module sources, its Brauer quotient of a module at $\Delta D$ is nonzero. Its summands are permutation modules for the $p$-group $D\times D$ and have module vertices of order at most $|D|$. A summand detected at $\Delta D$ therefore has module vertex $\Delta D$ and trivial module source, hence isHere a transitive permutation module for a $p$-group is indecomposable: it has one-dimensional head of a module over the local group algebra and cannot split into two nonzero modules. This also explains the Brauer-quotient detection used in the argument.

For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It giveswhere $\mid$ denotes being a direct summand. If $B$ has only finitely many indecomposable modules $M_1,\dots ,M_s$, each $B{\otimes }_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i{\downarrow }_D$. Thus $kD$ has finite representation type.

A noncyclic $p$-group has a quotient $C_p\times C_p$: its Frattini quotient has rank at least two, since a one-generator Frattini quotient forces the group itself to be cyclic. Inflation along this quotient embeds all the indecomposable modules of that elementary abelian group, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.

The conclusion remains valid if the splitting field $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ useAn endomorphism commuting with these has form $\left(\begin{array}{cc}A& 0\\ C& A\end{array}\right)$ with $A{T}_f=T_{f}A$. Its endomorphism ring has a square-zero ideal of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a field, so it is local and the module is indecomposable. Finite fields have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,

Put $b={\mathrm{Br}}_D(e)$. By equivariance of the Brauer homomorphism, $b$ is a central idempotent of $kK$, since $C_G(D)\le K\le N_G(D)$. The Nagao module theorem asserts that, if $eM=M$, thenand every indecomposable summand of $(1-b)M$ is relatively projective for a $p$-subgroup of $K$ not containing $D$. Because $D$ is normal in $K$, its module vertices cannot contain $D$ even after $K$-conjugation.

Here is a direct proof. As an element of $(kG{)}^K$, $e-b$ is a linear combination of $K$-conjugation orbit sums supported outside $C_G(D)$. An orbit represented by $x$ has sumIts stabilizer $C_K(x)$ does not contain $D$. On $T=(1-b)M$, the element $e-b$ acts as the identity. Map the orbit-sum expression into ${\mathrm{End}}_k(M)$ by left multiplication, and compose on both sides with the $K$-linear projection onto $T$. It follows thatReplace each centralizer by a Sylow $p$-subgroup of it using trace transitivity and its invertible prime-to-$p$ index. Project this identity further onto an indecomposable summand $U$ of $T$. Its endomorphism ring is local by Krull–Schmidt theorem. A sum of nonunits cannot be the identity in a local ring, so one of the trace terms is a unit $u$ in ${\mathrm{End}}_{kK}(U)$. Multiplying the endomorphism inside that trace by $u^{-1}$ gives ${\mathrm{id}}_U$ as a single relative trace. The D. Higman criterion makes $U$ relatively projective for that Sylow subgroup $Q\le C_K(x)$, and $D\nleqq Q$. This proves the assertion rather than just invoking the theorem's name.

A uniform module is a nonzero module in which any two nonzero submodules intersect nontrivially. An essential extension $V\subseteq E$ means that $V$ is an essential submodule of $E$: every nonzero submodule of $E$ meets $V$. Since $R$ is a left Noetherian ring, the finitely generated $M$ is a Noetherian module.

Every nonzero Noetherian module contains a uniform submodule. Otherwise split a nonuniform submodule into two disjoint nonzero submodules, retain one for further splitting and set the other aside. If no uniform submodule is reached, the pieces set aside form an infinite direct sum, whose partial sums contradict the ascending chain condition. Now successively choose uniform submodules disjoint from the current direct sum. The process again must stop by Noetherianity, giving $V={⨁}_{i=1}^r{V}_i\subseteq M$. At termination no nonzero submodule is disjoint from $V$, soThis proves the finite uniform decomposition of a Noetherian module.

An injective hull $E(M)$ is an injective module containing $M$ essentially. It is equivalently a minimal injective extension; we use the general existence theorem for injective hulls. To prove uniqueness, let $E_1,E_2$ be two hulls of $M$. Injectivity of $E_2$ extends the identity on $M$ to $f:E_1\to E_2$. Its kernel is disjoint from $M$ and therefore zero by essentiality. Its image is injective, so its inclusion splits in $E_2$. The complementary summand is disjoint from $M$, hence zero. Thus $f$ is an isomorphism fixing $M$.

A finite direct sum of injective modules is injective, and a finite direct sum of essential inclusions is essential. For the latter assertion, start with a nonzero vector and process its coordinates in turn: if a coordinate lies outside its specified essential submodule, multiply by a ring element taking that coordinate nontrivially into that submodule. Already processed coordinates stay inside their submodules; the chosen coordinate keeps the vector nonzero. Essentiality is also transitive. Consequently both $E(M)$ and ${⨁}_{i}E(V_i)$ are injective hulls of $V$, so uniqueness givesThe essential extension of a uniform module is uniform, since intersecting any two nonzero submodules with the original uniform module already gives a nonzero common intersection.

Write $E=E(V_i)$ and $S={\mathrm{End}}_R(E)$. If $f\in S$ is injective, its image is isomorphic to the injective module $E$, hence is a direct summand. Uniformity leaves no nonzero complementary summand, so $f$ is an automorphism. Therefore the nonunits are exactly the noninjective endomorphisms. Let $J$ be their set. Two such maps have nonzero kernels, whose intersection is nonzero by uniformity; their sum kills this intersection. For a product, composition on the left preserves the original kernel. Composition on the right either has a noninjective right factor, or that factor is an automorphism and transports the original nonzero kernel. Thus $J$ is a proper two-sided ideal. Every element outside $J$ is a unit, so every nonzero residue class is invertible and every proper left or right ideal is contained in $J$. This proves the local endomorphism ring of a uniform injective module assertion:For the whole hull, let $E_i=E(V_i)$. Its endomorphism ring is the ring of finite block matriceswith multiplication given by summing compositions. The diagonal entries have the local rings just proved. Off-diagonal entries are homomorphism groups, not necessarily zero; in particular this ring is not generally the product of its diagonal rings. Repeated isomorphic summands give matrix blocks over their common local endomorphism ring.

For the left regular module of $A_1(\mathbb{C})$, the Bernstein filtration has graded ring $\mathbb{C}[x,\xi ]$, so finite leading-term cancellation proves that $A_1$ is left and right Noetherian and a domain. Let $Q$ be its classical division ring of fractions. It is an essential extension of $A_1$: a nonzero fraction $q=s^{-1}a$ satisfies $sq=a\ne 0$ in $A_1$.

To prove that $Q$ is injective, use the Baer criterion. For a homomorphism $f:I\to Q$ on a nonzero left ideal, fix $a\in I\setminus \{0\}$. For any nonzero $x\in I$, take common left multiples $sx=ta$ with $s,t\ne 0$. Then $sf(x)=tf(a)$, so in the division ringThus right multiplication by $a^{-1}f(a)$ extends $f$ from $I$ to $A_1$. The zero ideal is harmless. Hence $Q$ is injective and $E({}_{A_1}{A}_1)=Q$. For any $A_1$-linear endomorphism $F$ of $Q$, clearing the left denominator in $q$ gives $F(q)=qF(1)$. These are all right multiplications, and their composition reverses the multiplying elements. ThereforeThe opposite ring in the injective hull of an Ore domain computation is essential for left-module conventions.

Fix a separable closure $k^s$. The Tate module and rational Tate module arewhere the transition maps are multiplication by $\ell$. Since $\ell$ is different from the characteristic, $E[{\ell }^n](k^s)\simeq (\mathbb{Z}/{\ell }^n{)}^2$ and these transition maps are surjective. Choosing compatible bases gives $T_{\ell }(E)\simeq {\mathbb{Z}}_{\ell }^2$, and $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. Both spaces carry the natural continuous action of the absolute Galois group of $k$ and of the base-defined endomorphism ring of an elliptic curve.

For (i), choose a nonrational $a\in R$. Such an element exists because the intersection of the given two-dimensional rational subalgebra with the integral endomorphism ring contains a full lattice. The characteristic equation of an elliptic endomorphism isSince $a\in K$ is nonrational, its minimal polynomial over $\mathbb{Q}$ has degree two. Comparing the two monic equations gives $\mathrm{tr}(a)=a+\bar{a}$ and $\mathrm{deg}(a)=a\bar{a}$. The same is the characteristic polynomial on the rational Tate module: the Weil pairing gives determinant $\mathrm{deg}(a)$, and applying it to $1-a$ gives trace $1+\mathrm{deg}(a)-\mathrm{deg}(1-a)$. Thus on $V_{\ell }(E)$ the characteristic polynomial isIf $K_{\ell }$ is a quadratic field over ${\mathbb{Q}}_{\ell }$, a unital $K_{\ell }$-action on this two-dimensional ${\mathbb{Q}}_{\ell }$-space makes it a one-dimensional $K_{\ell }$-space. If $K_{\ell }\simeq {\mathbb{Q}}_{\ell }\times {\mathbb{Q}}_{\ell }$, the two roots above are distinct in ${\mathbb{Q}}_{\ell }$ and their eigenspaces each have dimension one. The two idempotent factors therefore have equal rank one, so the module is again free of rank one. This also proves faithfulness in the split case. Hence

For (ii), first prove an integral divisibility statement. If $a\in R$ maps $T_{\ell }(E)$ into ${\ell }^n{T}_{\ell }(E)$, then its induced action on $E[{\ell }^n]$ is zero. The kernel of an isogeny of $[{\ell }^n]$ is therefore contained in the kernel of $a$. The quotient property of this isogeny of elliptic curves gives a unique endomorphism $b$ withThis factorization is defined over $k$: applying any descent automorphism preserves it, and uniqueness fixes $b$. Inside the rational endomorphism ring, $b=a/{\ell }^n$ belongs to $K$, so $b\in R$. ConsequentlyHere the prime-to-characteristic torsion is étale, so vanishing on its geometric points is vanishing on the entire kernel scheme.

Every element of $R_{\ell }$ preserves the Tate module by continuity. Conversely suppose $\varphi \in K_{\ell }$ preserves it. Choose $n\ge 0$ with ${\ell }^n\varphi \in R_{\ell }$, and approximate it by $a\in R$ modulo ${\ell }^n{R}_{\ell }$. Write ${\ell }^n\varphi =a+{\ell }^{n}c$ with $c\in R_{\ell }$. Then $a{T}_{\ell }(E)\subseteq {\ell }^n{T}_{\ell }(E)$, so the divisibility statement gives $a/{\ell }^n\in R$. It follows that $\varphi =a/{\ell }^n+c\in R_{\ell }$. Thus the CM Tate module stabilizer isThis does not presume freeness of the integral Tate module over a nonmaximal order.

For (iii), every element of $R$ is defined over $k$, so the Galois representation commutes with $R$ and therefore with $K_{\ell }$. By (i), its action on $V_{\ell }(E)$ is multiplication by an element of $K_{\ell }^{\times }$. The action and its inverse preserve $T_{\ell }(E)$; by (ii), this scalar and its inverse belong to $R_{\ell }$. Hence

Take the Kronecker quiver, with two arrows $\alpha ,\beta :1\to 2$. There are no paths of length two, so its path algebra has basis $e_1,e_2,\alpha ,\beta$ and dimension four.

For each $\lambda \in k$, take a quiver representation with one-dimensional spaces at both vertices and arrow maps $(1,\lambda )$. An endomorphism consists of scalars $a,b$ with $b=a$ from the first arrow; the second imposes no further condition. Its endomorphism ring is therefore $k$, which has no nontrivial idempotent. Thus the representation is indecomposable. An isomorphism to the representation $(1,\mu )$ again requires $b=a\ne 0$, and the second arrow then gives $\lambda =\mu$. Because an algebraically closed field is infinite, this gives infinitely many pairwise nonisomorphic indecomposables. Hence

Write a finite quiver as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A representation of a quiver assigns a vector space $X_i$ to each vertex and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow. A quiver representation morphism $h:X\to Y$ is a family satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$.

The path algebra $A=kQ$ has every directed path, including each length-zero path $e_i$, as a basis. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The orthogonal idempotents satisfy $1={\sum }_i{e}_i$.

The path-algebra module equivalence is explicit. From a representation, form $M={⨁}_i{X}_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-module gives $X_i=e_{i}M$ and $f_{\rho }(x)=\rho x$. An $A$-module homomorphism restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.

$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle. For a finite acyclic quiver, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.

Choose only the orientation $1\to 2\to 3$. The interval representations of an equioriented three-vertex quiver $I[a,b]$ have $k$ at vertices $a,\dots ,b$, zero elsewhere, and identity arrows within that interval. The complete list isHere is an elementary proof, without the Gabriel theorem. For $X_1\stackrel{f}{\to }{X}_2\stackrel{g}{\to }{X}_3$, set $K=\mathrm{im}f\cap \ker g$. Choose $F$ complementing $K$ in $\mathrm{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\mathrm{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has endomorphism ring $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.