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If $S_i$ are the simple modules of a split group algebra and $P_i$ are their projective covers, then

For a finite-dimensional associative algebra, form a graph on its simple modules, joining $S,T$ when ${\mathrm{Ext}}^1(S,T)$ or ${\mathrm{Ext}}^1(T,S)$ is nonzero. Its connected components are exactly the simple modules belonging to each block of a finite-dimensional algebra. They are also the components generated by sharing a Jordan–Hölder factor occurrence in an indecomposable representation that is a projective module.

Indeed, Ext separation of finite-length modules would split the regular module of a block into canonical summands if that block had two graph components. Right multiplication preserves these summands, so its projection supplies a nontrivial central idempotent, contradicting the definition of a block. An indecomposable projective belongs to one block. Finally, a nonsplit extension $0\to U\to V\to W\to 0$ of simple modules is a quotient of the projective cover of $W$: a lift onto $W$ must contain $U$, since otherwise the extension splits. Thus its two endpoints share factors in an indecomposable projective.

Choose a maximal Brauer pair $(D,b_D)$ for $B$. Define its inertial quotient of a block and inertial index of a block byConjugacy of maximal pairs makes the index well-defined. The standard maximal-pair property used here is that $E(B)$ is a $p^{\prime }$-group. Since $D$ is cyclic, its faithful conjugation action embeds it into $\mathrm{Aut}(D)$, so $e\mid p-1$. Write $d=p^n$ and $m=(d-1)/e$.

We use two precise structural results. Over an algebraically closed field, the basic algebra of a block with cyclic defect group is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m$. For a Brauer tree algebra with $t$ edges and exceptional multiplicity $m$, all nonprojective indecomposable modules form the stable Auslander–Reiten quiverThese are the cyclic-defect structure theorem and the stable-quiver classification for tree algebras; we are not taking the requested numerical count as an assumption. Morita equivalence preserves simple, projective and indecomposable isomorphism classes.

There is one simple module per edge, so $l(B)=e$. The vertices of $\mathbb{Z}{A}_s$ can be indexed by $(j,i)$, $j\in \mathbb{Z}$, $1\le i\le s$. The quotient by ${\tau }^e$ leaves $e$ choices of $j$ for each $i$. With $s=em=d-1$, there are $e(d-1)$ nonprojective indecomposable modules. There are also $e$ indecomposable projective modules, one projective cover per simple. ThereforeThis counts isomorphism classes, not multiplicities in the regular module. In the one-edge case it agrees with the $d$ Jordan chains for $k[t]/(t^d)$.

For the definition, take a finite connected tree with a cyclic order of the incident edges at every vertex and positive integer multiplicities equal to one except possibly at a distinguished vertex. Its Brauer tree path presentation has a quiver vertex for each tree edge, and successor arrows around each tree vertex, including loops for valency one. If edge $i$ has endpoints $v,w$, let $C_{i,v}$ and $C_{i,w}$ be the cycles around these endpoints. The relations identifykill $C_{i,v}^{m_v}{\alpha }_{i,v}$, and kill paths switching between successor cycles belonging to different tree vertices. Redundant valency-one loops may be eliminated. This quotient of the path algebra is the basic Brauer tree algebra; allowing an arbitrary algebra with Morita equivalence to it does not change the module assertions. In particular, the cyclic orders and multiplicity, not merely the underlying graph, are part of the data.

For the construction, use right modules with paths multiplied left to right; equivalently use the opposite quiver for left modules, choosing orientation so radical factors follow the specified order. The indecomposable projective $P_1$ with top $S_1$ has simple socle $S_1$ and two uniserial radical branches $U_v,U_w$ meeting in that socle. The relations show that $U_v$ has length $mr$ and factors $S_2,\dots ,S_r,S_1$ repeated $m$ times. The other branch has length $m_w\mathrm{val}(w)$. Thus $W_v=P_1/U_w$ is a uniserial module of length $mr$, with factors $S_1,S_2,\dots ,S_r$ cyclically repeated. For $J$ the Jacobson radical,is the desired uniserial branch of a Brauer tree algebra. If an endpoint is a nonexceptional leaf, its branch is just the socle; the formula still applies.

For uniqueness, length one gives just $S_1$. If $q>1$ and $r>1$, the next factor $S_2$ chooses the successor arrow at $v$ uniquely, because two distinct edges cannot meet at both endpoints in a tree. Subsequent factors force the same cyclic succession. Any arrow into an edge not incident with $v$ acts as zero, because that simple does not occur. An opposite nonexceptional leaf loop is the full socle path around $v$, of length $mr$, hence zero on modules of length at most $mr$. If an opposite leaf is exceptional, then $v$ has multiplicity one and each incident simple occurs at most once; its nilpotent self-loop is zero on that one-dimensional vertex space as well.

The remaining action is one nilpotent successor chain. Choose a top generator $z_0$ and let $z_j$ be its successive path images for $0\le j<q$. Uniseriality makes each image nonzero in its respective radical layer; these images form a basis in the basic algebra description. Their labels are $S_{1+(j\mathrm{mod}r)}$, the successor arrow sends $z_j$ to $z_{j+1}$, and all other actions vanish. Any nonzero scalar choices are absorbed by rescaling this basis. This determines the module solely by its top and length.

If $r=1$ and $q>1$, then $m>1$ and $v$ is the exceptional leaf. The successor is a single loop and its action is the unique nilpotent Jordan chain of length $q$. The other branch either enters a different edge or is a redundant socle loop and contributes nothing. Hence the prescribed uniserial module is unique up to isomorphism for every positive integer $q\le mr$.

Choose a maximal Brauer pair $(D,b_D)$ for $B$, and let $N_G(D,b_D)$ be the subgroup normalizing $D$ and stabilizing the local block $b_D$. The inertial quotient of a block and its order, the inertial index of a block, areThe Brauer extended first main theorem says that maximal local blocks of defect $D$ are parametrized by the normalizer orbits of defect-$D$ blocks of $k(D{C}_G(D))$, with stabilizer quotient of order prime to $p$. For abelian $D$, conjugation embeds $E(B)$ in $\mathrm{Aut}(D)$. When $D=C_{p^n}$, its prime-to-$p$ subgroups are cyclic of order dividing $p-1$, so $e\mid p-1$.

We use two precise structural results for a block with cyclic defect group, with $k$ an algebraically closed field. First, the basic algebra is a Brauer tree algebra with $e$ edges and exceptional multiplicity $m=(p^n-1)/e$; its simple modules are indexed by its edges. Second, its stable Auslander–Reiten quiver has shapeThe vertices are the isomorphism classes of nonprojective indecomposable modules. These are the cyclic-defect structure and indecomposable-classification results, rather than a claim that all Brauer trees are Morita equivalent to a star. The tube description is also recorded in Hiss–Lassueur, Section 1.

The first result gives $l(B)=e$. In the second, there are $p^n-1$ positions in each slice and $e$ slices modulo the translation. Thus there are $e(p^n-1)$ nonprojective indecomposables. A finite-dimensional split algebra has one principal indecomposable module that is a projective cover for each simple module, so there are another $e$ projective indecomposables. ThereforeThe total includes projectives; it is different from the count of ordinary irreducible characters. This is the indecomposable count for a cyclic defect block.

Now let $p=2$ and $D=C_2\times C_2$, the Klein four-group. The extended theorem makes $E(B)$ a subgroup of order prime to $2$ of $\mathrm{Aut}(D)\cong {\mathrm{GL}}_2({\mathbb{F}}_2)\cong S_3$. Its order divides $6$ and is odd, soFor the remaining conclusion we use the central-defect quotient fact: if $Z\le Z(L)$ is a central $p$-subgroup, blocks of $kL$ correspond to blocks of $k(L/Z)$ under the quotient, and their defect groups quotient by $Z$. Central block idempotents lift uniquely here because the kernel is generated by central nilpotent elements; an off-diagonal corner lies in its own product with that nilpotent ideal and is therefore zero. Defect groups quotient by $Z$ by the relative-trace, or equivalently Brauer-pair, characterization. In particular a block with central defect group $D$ has a defect-zero block as its quotient by $J(kD)$.

First suppose $D\le Z(G)$, and write $A=B$. Since $k$ is an algebraically closed field, the defect-zero quotient is $\overline{A}\cong {\mathrm{Mat}}_r(k)$ for some $r$. The algebra $A$ is a direct summand of $kG$ as a $kD$-module, hence is projective and therefore free over the local ring $kD$. Its rank is $r^2$, since reduction modulo $J(kD)$ gives the $r^2$-dimensional quotient. Lift a primitive idempotent $\bar{f}$ of $\overline{A}$ to $f\in A$. The corner $fAf$ is a $kD$-linear direct summand of $A$ and is free over $kD$. Its quotient by $J(kD)$ is $\bar{f}\overline{A}\bar{f}\cong k$, so its rank is one. The central mapis surjective by Nakayama's lemma, since $f$ generates its reduction, and is injective because both sides have dimension $|D|$. Thus $fAf\cong kD$ as algebras. Orthogonal primitive idempotents and their equivalences in the matrix quotient lift modulo the nilpotent ideal; lifting the equivalence maps and correcting their products by units lifts the matrix units. ConsequentlyThis proves the actual algebra isomorphism of central defect blocks are matrix algebras, as an algebra.

Finally suppose $D◃G$ but is not central. Put $C=C_G(D)$, which is normal in $G$ and contains $D$ since $D$ is abelian. The normal-subgroup argument in Question 5 puts the block idempotent $e_B$ in $kC$. It is therefore the sum of the $G$-conjugates of a block idempotent $c$ of $kC$. The extended first main theorem chooses such a $c$ with defect group $D$; since $D$ is normal, its pair stabilizer is exactly $I=\{g\in G:{}^{g}c=c\}$. Nowso $I=C$. Within $C$, the defect group $D$ is central, and the first case gives $kCc\cong {\mathrm{Mat}}_r(kD)$.

We can see the Fong–Reynolds matrix decomposition explicitly in this case. Choose representatives $g_1,\dots ,g_t$ for $G/C$, where $t=[G:C]$, and putDifferent conjugates of $c$ are orthogonal, so $u_{ij}{u}_{\ell m}={\delta }_{j\ell }{u}_{im}$ and ${\sum }_i{u}_{ii}=e_B$. Moreover $cBc=ckGc=kCc$: a term $ckCgc$ vanishes unless $g$ stabilizes $c$, and that stabilizer is $C$. These matrix units therefore identify $B$ with ${\mathrm{Mat}}_t(kCc)$, whenceThe inertial index condition concerns the stabilizer of the local block; it does not require every element of $G$ to centralize $D$. That distinction is why the second case is necessary.

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 $3I$.

A finite-dimensional group algebra has finite representation type of a group algebra when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. The definition concerns all dimensions, not merely finitely many modules in each fixed dimension.

Let $e_B$ be the block idempotent and choose a maximal Brauer pair $(D,b_D)$: $b_D$ is a block idempotent of $k{C}_G(D)$ with $b_D{\mathrm{Br}}_D(e_B)\ne 0$. The inertial index of a block iswhere $N_G(D,b_D)$ is the stabilizer of the chosen pair. Changing the pair by conjugation does not alter this number. For cyclic $D$, it is a prime-to-$p$ divisor of $p-1$. Here $n\ge 1$, as required for the order-$p$ subgroup $Q$ to exist.

Put $H=N_G(Q)$. Since $Q$ is characteristic in $D$, $N_G(D)\le H$, so Green correspondence applies. In fact the nonprojective modules in $B$ have a particularly simple correspondence. If $g\notin H$, a nontrivial intersection $D\cap {}^{g}D$ would contain the unique order-$p$ subgroup of both cyclic groups, giving $Q={}^{g}Q$ and hence $g\in H$, a contradiction. Thus every such intersection is trivial. The error modules in induction are projective. In restriction, an intersection $H\cap {}^{g}D$ cannot contain a conjugate of a nontrivial subgroup of $D$ by an element of $H$: such a conjugate contains $Q$, and that would again force ${}^{g}Q=Q$. The general Green correspondence therefore pairs all indecomposable modules with nontrivial module vertex contained in a conjugate of $D$. In particular it pairs the nonprojective modules of $B$ with the nonprojective modules of its Brauer correspondent $b$ in $H$; the block-compatible form of the correspondence selects $b$. Explicitly,where $P$ is projective and the restriction error family is $\mathcal{Y}=\{H\cap {}^{g}D:g\notin H\}$. One should not replace this restriction error term by projectives without an additional argument about $H$.

For the final structure calculation we use the general cyclic-defect structure theorem, which is permitted here as a general fact: the basic algebra of a split block with cyclic defect group is a Brauer tree algebra. Its tree has $e(B)$ edges, one simple module per edge, and exceptional multiplicityFor a one-edge tree the basic algebra presentation is $k[t]/(t^{m+1})$. This is the local presentation of the tree algebra, not an assertion that an arbitrary block with one simple module is automatically uniserial.

With $e(B)=1$ the tree has one edge and $m=p^n-1$. Thus the basic algebra of $B$ isIt has exactly one simple module, $k=A_0/(t)$. Its ideals are precisely $(t^i)$: for any nonzero element, factor out its lowest power of $t$, after which the remaining factor is a unit. The regular projective therefore has the unique submodule chainwith each consecutive factor one-dimensional and simple. A Morita equivalence from $A_0$ to $B$ preserves simple modules, projective covers, submodule lattices and composition length. HenceIn the split setting one may also write $B\cong M_{{\dim }_{k}S}(A_0)$. The projective cover has $p^n$ composition factors all isomorphic to $S$, so its vector-space dimension is $p^n{\dim }_{k}S$, not generally $p^n$.

There is also an elementary reason why the uniserial conclusion follows once the one-simple-module conclusion and finite type are known. If the basic local algebra had $\dim J/J^2\ge 2$, quotienting its radical square and all but two generators would give $k[x,y]/(x,y{)}^2$. The two-dimensional parameter family from Question 3(b) would then give infinitely many indecomposables. Thus $J/J^2$ is one-dimensional. Nakayama lemma makes $J$ generated by one element $t$, so its powers yield a chain and the basic algebra is $k[t]/(t^r)$. The cyclic-defect calculation above identifies $r=p^n$. This explains why finite representation type rules out branching in this projective cover.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

An integral form of a group representation $V$ is a $G$-stable finite free $\mathcal{O}$-submodule $W\subset V$ with $K{\otimes }_{\mathcal{O}}W\cong V$. Such a form exists: take a basis lattice $L$ and replace it by ${\sum }_{g\in G}gL$, which is finite and torsion-free and therefore free over the discrete valuation ring $\mathcal{O}$.

Choose ordinary simple modules $V_i$ with forms $W_i$, and modular simple modules $S_j$ with projective covers $P_j$. With $\pi$ a uniformizer, put ${\overline{W}}_i=W_i/\pi W_i$. The decomposition matrix and Cartan matrix of a group algebra have entriesThe first numbers do not depend on the integral form: the Brauer character of its reduction is the ordinary character restricted to p-regular elements, and part 2(a) determines all composition multiplicities from this restriction.

For the assertion in (i), set $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. It is a submodule of the finite free module ${\mathrm{Hom}}_{\mathcal{O}}(W,W^{\prime })$, hence is finite free over $\mathcal{O}$. There is a natural injective mapFor any homomorphism $f$ in the target, clearing the finitely many denominators of its values on an $\mathcal{O}$-basis of $W$ gives ${\pi }^{N}f(W)\subset W^{\prime }$. Thus ${\pi }^{N}f\in I$, and the map is surjective. Therefore $\boxed{I\text{ is an }\mathcal{O}\text{-form of }{\mathrm{Hom}}_{KG}(M,M^{\prime })}$, regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.

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 $\boxed{C=D^{T}D}$.

We prove $(iii)\Rightarrow (ii)$. Consider one nonsplit short exact sequencewith $U,W$ simple. Lift the canonical surjection $P(W)\to W$ from the projective cover through $q$, obtaining $f:P(W)\to V$. Its image maps onto $W$. The intersection $\mathrm{im}f\cap U$ is either zero or $U$, since $U$ is simple. If it were zero, $q$ would identify $\mathrm{im}f$ with $W$, splitting the extension. Hence $U\subset \mathrm{im}f$, and surjectivity onto $W$ then implies $\mathrm{im}f=V$.

Therefore both $U$ and $W$ occur as Jordan–Hölder factors of the indecomposable projective $P(W)$. Replacing every edge in an Ext chain by this observation proves (ii). Together with the preceding two implications, this provesThe repeated word “projective” in the printed version of (ii) has no mathematical effect.

Let $g$ be a p-regular element. Its eigenvalues on $M$ are roots of unity of order prime to $p$. If ${\hat{\lambda }}_1,\dots ,{\hat{\lambda }}_d$ are their Teichmuller lifts to characteristic-zero roots of unity, the Brauer character isIt depends only on the conjugacy class of $g$, is additive in short exact sequences, and equals the restriction of an ordinary character whenever the representation lifts.

For linear independence, choose a splitting p-modular system and let $P_i$ be the projective cover of the simple module $S_i$. A projective lattice lifting $P_i$ has an ordinary character ${\Phi }_i$ that vanishes on p-singular elements. Reduction and ordinary character orthogonality givebecause $S_i$ is the head of $P_i$. Pairing a relation ${\sum }_j{c}_j{\chi }_{S_j}=0$ with every ${\Phi }_i$ yields $c_i=0$ for every $i$. HenceThis is the linear-independence part of the Brauer–Nesbitt theorem.

Decompose the projective module aswhere $T$ ranges over the simple modules. Since $P_T/J(P_T)\cong T$, the multiplicity of $S$ in the head of a module $P/J(P)$ is $m_S$. Part (b) gives $\mathrm{Soc}(P_T)\cong T$, so the multiplicity in $\mathrm{Soc}(P)$ is the same $m_S$.

The invariant submodule and coinvariant module satisfyThus $\dim P^G$ is the multiplicity of the trivial module in $\mathrm{Soc}(P)$, while $\dim P_G=\dim {\mathrm{Hom}}_{kG}(P,k)$ is its multiplicity in the head. Applying the same argument to the projective module $P^{\ast }$ yields

The dual $P_S^{\ast }$ is indecomposable projective. Its head is dual to $\mathrm{Soc}(P_S)\cong S$, hence is $S^{\ast }$. Uniqueness of projective covers proves the dual of a projective cover over a group algebra:

For a finite-dimensional algebra $A$, a principal indecomposable module is an indecomposable direct summand of the left regular module, equivalently $Ae$ for a primitive idempotent $e$. Its head of a module is simple, and it is the projective cover of that simple module.