Modular Representations and Projective Covers

By [F1], choose an $\mathcal{O}$-basis of $L$ of size $r<\mathrm{\infty }$. Tensoring with $k=\mathcal{O}\mathrm{/}\mathfrak{m}$ sends that basis to a $k$-basis of $k{\otimes }_{\mathcal{O}}L\cong L\mathrm{/}\mathfrak{m}L$, so ${\dim }_k\bar{L}=r$.

The $G$-action on $L$ is $\mathcal{O}$-linear, so $\mathfrak{m}L$ is $G$-stable and the quotient action on $L\mathrm{/}\mathfrak{m}L$ is well defined. Thus $\bar{L}$ is a finite-dimensional $kG$-module.

The left regular module ${}_{A}A$ is finite-dimensional, so [L1] makes the descending chain $A\supseteq J\supseteq J^2\supseteq \cdots$ stabilize: choose $n$ with $J^n=J^{n+1}$. If $J^n\mathrm{\ne }0$, [L2] gives a maximal submodule $N$ of the finite-dimensional left $A$-module $J^n$, so $S=J^n\mathrm{/}N$ is simple by [L3]. For every nonzero $s\in S$, the map $A\to S$, $a\mapsto as$, is onto; its kernel is a maximal left ideal, hence contains $J$ by [F1]. Thus $Js=0$ for every $s$, so $JS=0$ and $J^{n+1}=J{J}^n\subseteq N$, contradicting $J^{n+1}=J^n$. Therefore $J^n=0$, so $J$ is nilpotent.

Maximal left ideals of $A\mathrm{/}J$ are exactly the quotients $L\mathrm{/}J$ with $L$ a maximal left ideal of $A$ containing $J$, so their intersection is $0$ by [F1]. Let $M={}_{A\mathrm{/}J}(A\mathrm{/}J)$ be the left regular module. It is finite-dimensional. We prove by induction on vector-space dimension that any finite-dimensional module whose maximal submodules intersect trivially is semisimple. If $M=0$ there is nothing to prove. Otherwise choose a nonzero submodule $S\le M$ of least positive dimension, so $S$ is simple. Since the intersection of maximal submodules is $0$, some maximal submodule $N$ does not contain $S$. Then $S\cap N=0$, and maximality makes $(S+N)\mathrm{/}N$ a nonzero submodule of the simple quotient $M\mathrm{/}N$, so $M=S\oplus N$. For any nonzero $x\in N$, choose a maximal submodule $L$ of $M$ avoiding $x$. The image of $N$ in the simple module $M\mathrm{/}L$ is nonzero and hence all of $M\mathrm{/}L$, so $N\cap L$ is a maximal submodule of $N$ avoiding $x$. Thus the maximal submodules of $N$ also intersect trivially. Its dimension is smaller than that of $M$, so induction makes $N$ semisimple, hence so is $M$.

By step 1.2, the left regular module of $A\mathrm{/}J$ is semisimple. Thus [L4] makes the quotient algebra $A\mathrm{/}J$ semisimple. Together with step 1.1, this proves the theorem.

If $M=0$, then $J(A)M=0$ and $\mathrm{rad}(M)=0$, so the claim is immediate. Assume from now on that $M\mathrm{\ne }0$. Let $N$ be a maximal submodule of $M$. Then $M\mathrm{/}N$ has no proper nonzero submodule, so it is simple by [L2]. Fix $m\in M$. If $m\in N$, then every $j\in J$ already satisfies $jm\in N$. If $m\mathrm{\notin }N$, then the nonzero coset $m+N$ generates the simple module $M\mathrm{/}N$, so the $A$-linear map ${\theta }_m:A\to M\mathrm{/}N,a\mapsto am+N$ is surjective. Its kernel $I_m:=\{a\in A:am\in N\}$ is therefore a maximal left ideal of $A$. By [F1], $J$ is contained in every maximal left ideal, so $J\subseteq I_m$ and hence $jm\in N$ for every $j\in J$. Thus $JM\subseteq N$. Since $N$ was arbitrary, $JM\subseteq \mathrm{rad}(M)$.

Consider the quotient module $M\mathrm{/}JM$. Because $J$ acts trivially on it, $M\mathrm{/}JM$ is naturally a module over the semisimple ring $A\mathrm{/}J$ from [L1]. Therefore [L3] makes $M\mathrm{/}JM$ semisimple, so the intersection of its maximal submodules is $0$. Maximal submodules of $M\mathrm{/}JM$ correspond exactly to maximal submodules of $M$ containing $JM$, and their intersection is $\mathrm{rad}(M)\mathrm{/}JM$. Hence $\mathrm{rad}(M)\mathrm{/}JM=0$.

Step 1.1 gives $JM\subseteq \mathrm{rad}(M)$ and step 2.1 gives the reverse inclusion. Therefore $\mathrm{rad}(M)=J(A)M$.

If $M=JM$, then iterating the equality gives $M=J^{n}M$ for every $n\ge 1$. Choose $n$ with $J^n=0$ from [L2]. Then $M=J^{n}M=0$.

Let $N:=A{m}_1+\cdots +A{m}_r$. By hypothesis, the images of the $m_i$ span $M\mathrm{/}JM$, so $M=N+JM$. Hence $M\mathrm{/}N=J(M\mathrm{/}N)$. Applying part 1 to the module $M\mathrm{/}N$ and using [L1], we obtain $M\mathrm{/}N=0$. Therefore $M=N$, so $m_1,\dots ,m_r$ generate $M$.

Steps 1.1 and 2.1 are exactly the two asserted conclusions.

We argue by induction on $\mathrm{\mid }P\mathrm{\mid }$. If $P=1$, then $I_P=0$ by [F1], so the claim is trivial. Assume $P\mathrm{\ne }1$. By [L3], choose a central element $z\in Z(P)$ of order $p$. Then $([z]-1{)}^p=[z^p]-1=0$ in characteristic $p$, using [L1] and the binomial theorem with vanishing intermediate coefficients. Hence the principal ideal generated by $[z]-1$ is nilpotent.

Because $z$ is central, the quotient map $P\to P\mathrm{/}⟨z⟩$ induces an algebra surjection $k[P]\to k[P\mathrm{/}⟨z⟩]$ whose kernel is the ideal $([z]-1)k[P]$. Under this map the augmentation ideal of $k[P]$ maps onto the augmentation ideal of $k[P\mathrm{/}⟨z⟩]$. By the induction hypothesis, some power of $I_P$ lands inside $([z]-1)k[P]$. Since step 1.1 makes that ideal nilpotent, a further power of $I_P$ is $0$. Therefore $I_P$ is nilpotent.

On basis elements one has $⟨[g],[h]⟩=\{\begin{array}{ll}1,& h=g^{-1},\\ 0,& h\mathrm{\ne }{g}^{-1}\mathrm{.}\end{array}$ Thus the matrix of the form in the basis from [F1] is a permutation matrix, so the form is nondegenerate.

Associativity is immediate from the definition, because $⟨xy,z⟩$ and $⟨x,yz⟩$ are both the coefficient of $[e]$ in $xyz$. Symmetry holds because the coefficient of $[e]$ in $[g][h]$ is $1$ exactly when $h=g^{-1}$, which is equivalent to $g=h^{-1}$ and hence to the coefficient of $[e]$ in $[h][g]$ being $1$. Bilinearity is clear from the coefficient functional. Therefore the form is associative, symmetric, and nondegenerate.

A finite-dimensional algebra equipped with such a bilinear form is symmetric Frobenius. Hence $k[G]$ has the stated structure.

Let $A^{\vee }={\mathrm{Hom}}_k(A,k)$. The symmetric Frobenius form from [L1] identifies $A$ with $A^{\vee }$ as left $A$-modules. An $A$-map $U\to A^{\vee }$ is determined by the scalar function $u\mapsto f(u)(1)$, and any $k$-linear extension of that scalar function from a submodule $U\subseteq V$ to $V$ induces an $A$-linear extension $V\to A^{\vee }$. Thus $A^{\vee }$, and hence $A$, is injective.

If $P$ is projective, [F1] makes it a direct summand of a finite free module $A^n$. Step 1.1 makes $A^n$ injective, and a direct summand of an injective module is injective by [F2]. So every finite-dimensional projective module is injective.

Conversely, let $I$ be injective. Dualizing a split monomorphism into $I$ turns the extension property of [F2] into the lifting property for the right $A$-module $I^{\vee }$, so $I^{\vee }$ is projective over $A^{\mathrm{o}\mathrm{p}}$. By the projective characterization in [F1], $I^{\vee }$ is a direct summand of a finite free right $A$-module. Dualizing back and using the symmetric Frobenius identification $(A^n{)}^{\vee }\cong A^n$, the bidual $I^{\vee \vee }\cong I$ becomes a direct summand of a finite free left $A$-module. Hence [F1] makes $I$ projective.

Steps 2.1 and 2.2 prove the equivalence.

Assume for contradiction that $N\mathrm{\ne }M$. Since $M$ has finite length by [L1], it is finitely generated. Therefore the nonzero quotient $M\mathrm{/}N$ has a maximal proper submodule by [L2], and its inverse image in $M$ is a maximal submodule $P$ containing $N$.

By [F1], the radical lies in every maximal submodule, so $\mathrm{rad}(M)\subseteq P$. Hence $M=N+\mathrm{rad}(M)\subseteq P<M$, a contradiction. Therefore $N=M$, and the radical is superfluous.

Choose a finite $k$-basis $m_1,\dots ,m_r$ of $M$. It also generates $M$ as an $A$-module, so sending the standard generators to the $m_i$ gives a surjection $\epsilon :A^r\to M$. Among the direct summands $P$ of the finite-dimensional module $A^r$ for which $\epsilon {\mathrm{\mid }}_P:P\to M$ is surjective, choose one of minimal $k$-dimension; the family is nonempty because it contains $A^r$. Put $\pi :=\epsilon {\mathrm{\mid }}_P$, and write $j:P\to A^r$ and $s:A^r\to P$ for the inclusion and projection, with $sj=1_P$. For any surjection $q:E\to X$ and map $u:P\to X$, choose $y_i\in E$ with $q(y_i)=u(s(e_i))$ for the finitely many standard basis vectors $e_i$. These finite choices need no AC by [L1]. The formula $v({\sum }_i{a}_i{e}_i)={\sum }_i{a}_i{y}_i$ defines an $A$-linear map $v:A^r\to E$ with $qv=us$, so $q(vj)=usj=u$. Thus $P$ is projective by [F2], including when $r=0$.

Put $K=\ker \pi$, and suppose $N\le P$ satisfies $N+K=P$. Then $\pi {\mathrm{\mid }}_N:N\to M$ is surjective. Projectivity of $P$ supplies a map $h:P\to N$ with $(\pi {\mathrm{\mid }}_N)h=\pi$; write $f:P\to P$ for $h$ followed by the inclusion. Thus $\pi f=\pi$, and hence $\pi f^n=\pi$ for every $n\ge 1$.

Since $P$ is finite-dimensional, the kernels and images of the powers of $f$ stabilize. For a sufficiently large $n$, $P=\ker (f^n)\oplus \mathrm{im}(f^n):$ the intersection is zero because $f^n(x)=0$ for $x=f^n(y)$ implies $f^{2n}(y)=0$ and stabilization gives $f^n(y)=0$, while rank-nullity gives that the two dimensions sum to ${\dim }_{k}P$. The restriction of $\pi$ to $\mathrm{im}(f^n)$ is surjective because $\pi f^n=\pi$. Minimality of $P$ in step 1.1 therefore forces $\ker (f^n)=0$. Hence $f$ is injective and thus bijective, so $P=f(P)\subseteq N$. Therefore $N=P$, $K$ is superfluous, and $\pi$ is a projective cover.

Now let $\pi :P\to M$ and $\rho :Q\to M$ be projective covers. Both sources are finite-dimensional: lift a finite $k$-basis of $M$ to $P$, let $P_0$ be the submodule generated by those lifts, and observe that $P=P_0+\ker \pi$; superfluity gives $P=P_0$, which is finite-dimensional because $A$ is. The same argument applies to $Q$. Projectivity yields maps $f:P\to Q$ and $g:Q\to P$ with $\rho f=\pi$ and $\pi g=\rho$. Then $\pi (1_P-gf)=0$, so $(1_P-gf)(P)\subseteq \ker \pi$. Hence $P=gf(P)+\ker \pi$, and the superfluity of $\ker \pi$ gives $gf(P)=P$. Thus $gf$ is surjective, hence bijective on the finite-dimensional module $P$. The same argument shows that $fg$ is bijective on $Q$. Since $gf$ is bijective, $f$ is injective; since $fg$ is surjective, $f$ is surjective. Therefore $f$ is an isomorphism, and it still satisfies $\rho \circ f=\pi$.

Steps 1.1, 2.1, 3.1, and 4.1 prove existence and uniqueness up to isomorphism over the target.

Let $\pi :P\to S$ be the projective cover of a simple module $S$. If $P=P_1\oplus P_2$, then $S=\pi (P_1)+\pi (P_2)$, so simplicity forces one summand, say $\pi (P_1)$, to equal $S$. Then $P=P_1+\ker \pi$, and because $\ker \pi$ is superfluous in a projective cover, $P=P_1$. Hence $P_2=0$, so the projective cover of a simple module is indecomposable.

Now let $P$ be an indecomposable finite-dimensional projective module, and put $J=J(A)$. The quotient map $P\to \mathrm{hd}(P)=P\mathrm{/}JP$ is a projective cover: its kernel is $\mathrm{rad}(P)=JP$ by [F1], and that kernel is superfluous by [L2]. The head is nonzero, since otherwise the superfluity condition applied with the zero submodule would force $P=0$. It is a finite-dimensional module over the semisimple algebra $B=A\mathrm{/}J$ by [L3]. Choose finitely many vector-space generators to make it a quotient of a finite sum of copies of ${}_{B}B$. By [L3] and [F2], that sum is a finite direct sum of simple modules. Its quotient is likewise a finite direct sum of simples: process the summands in order; the image of each is zero or simple, and either lies in the sum of earlier images or meets that sum trivially, so retain exactly the latter images. Thus $\mathrm{hd}(P)\cong S_1\oplus \cdots \oplus S_r$ for some $r\ge 1$.

For each $S_i$, let ${\pi }_i:Q_i\to S_i$ be its finite-dimensional projective cover from [L1], with kernel $K_i$. The finite direct sum $Q={⨁}_i{Q}_i$ is projective by lifting one component at a time, and $\pi ={⨁}_i{\pi }_i:Q\to {⨁}_i{S}_i$ is surjective with kernel $K={⨁}_i{K}_i$. Each $K_i$ is superfluous. If a maximal submodule of $Q_i$ omitted $K_i$, their sum would equal $Q_i$, a contradiction; hence $K_i\subseteq \mathrm{rad}(Q_i)=J{Q}_i$ by [F1]. Therefore $K\subseteq {⨁}_{i}J{Q}_i=JQ=\mathrm{rad}(Q)$, and [L2] makes $K$ superfluous because $Q$ is finite-dimensional. Thus $\pi$ is a projective cover of the head. Uniqueness in [L1] gives $P\cong Q$. Each $Q_i\mathrm{\ne }0$ because it surjects onto $S_i\mathrm{\ne }0$; indecomposability of $P$ therefore forces $r=1$. Thus the head is simple.

Step 1.1 constructs an indecomposable projective from each simple module, and steps 1.2–2.1 show that taking the head of an indecomposable projective returns a simple module. The two constructions are inverse up to isomorphism by uniqueness of projective covers.

Prove existence by induction on $d={\dim }_{k}M$. If $M=0$ or $M$ is indecomposable there is nothing to split. Otherwise $M=M_1\oplus M_2$ with both summands nonzero. By dimension additivity, $0<{\dim }_k{M}_i<d$ for each $i$, so the induction hypothesis decomposes each $M_i$ into finitely many indecomposables. Their combined list decomposes $M$.

Let $X$ be a nonzero indecomposable finite-dimensional module and $f\in \mathrm{End}(X)$. The dimensions of $\ker f^n$ increase and are bounded by ${\dim }_{k}X$, so for some $n\ge 1$, $\ker f^n=\ker f^{2n}$. If $y=f^n(x)$ lies in $\ker f^n$, then $x\in \ker f^{2n}=\ker f^n$ and $y=0$. Rank-nullity therefore gives $X=\ker f^n\oplus \mathrm{im}{f}^n$. Both are submodules. Indecomposability makes one summand zero: in the first case $f$ is injective and hence invertible in finite dimension; in the second $f^n=0$. Thus every endomorphism of $X$ is invertible or nilpotent.

A nilpotent $f$ has inverse $1+f+\cdots +f^{n-1}$ for $1-f$. If nonunits $a,b$ of $\mathrm{End}(X)$ had unit sum $u=a+b$, then $u^{-1}a$ and $u^{-1}b$ would still be nonunits and would sum to $1$; the first is nilpotent by step 1.2, so the second $1-u^{-1}a$ would be a unit, a contradiction. Thus the nonunits are closed under finite sums. This is the only endomorphism-ring property needed below; no commutativity is assumed.

For uniqueness induct on ${\dim }_{k}M$. Suppose $M=X\oplus A=Y_1\oplus \cdots \oplus Y_s$, with $X$ and all $Y_j$ nonzero indecomposable. The identity of $X$ is the sum of composites $e_j:X\hookrightarrow M\twoheadrightarrow Y_j\hookrightarrow M\twoheadrightarrow X$. By step 2.1 at least one $e_j$ is a unit. Write its two factors as $a:X\to Y_j$ and $b:Y_j\to X$, so $ba$ is invertible. Then $a(ba{)}^{-1}$ splits the surjection $b$, giving $Y_j=a(X)\oplus \ker b$. Indecomposability of $Y_j$ forces $\ker b=0$, hence $X\cong Y_j$ and $b$ is an isomorphism.

Put $B={⨁}_{i\mathrm{\ne }j}{Y}_i$. Since the projection $M\to X$ restricts to the isomorphism $b:Y_j\to X$, the projection $M\to B$ restricts to an isomorphism $A\to B$: its kernel is $A\cap Y_j=0$, and for every $z\in B$ subtract from $z$ the unique $y\in Y_j$ with the same $X$-projection to obtain a preimage in $A$. Thus $A\cong B$. Both have dimension smaller than $M$; induction matches their indecomposable decompositions. Adding $X\cong Y_j$ proves uniqueness up to permutation and isomorphism.

By [L1], the regular module decomposes as a finite direct sum of indecomposable projective modules. By [L2], each summand is the projective cover $P(S)$ of a unique simple module $S$, so ${}_{A}A\cong {⨁}_{S}P(S{)}^{\oplus m_S}$ for uniquely determined multiplicities $m_S\ge 0$.

Modding out by the radical preserves direct sums and sends each $P(S)$ to its simple head $S$. Hence $A\mathrm{/}J(A)\cong {⨁}_S{S}^{\oplus m_S}$. By [L3], the quotient is semisimple. Since $k$ is a splitting field by [F1], Wedderburn-Artin writes the semisimple algebra as a product of matrix algebras $M_{n_S}(k)$, and the left regular module of $M_{n_S}(k)$ is the simple column module repeated $n_S$ times. That simple module has $k$-dimension $n_S$, so $m_S={\dim }_{k}S$.

Substituting the multiplicities from step 2.1 gives the displayed decomposition.

Suppose first that $G=P$ is a finite $p$-group. The quotient $k[P]\mathrm{/}{I}_P$ is the field $k$, so $I_P$ is a maximal ideal by [F1]. By [L1], $I_P$ is nilpotent. Every maximal left ideal of $k[P]$ therefore contains $I_P$, so the Jacobson radical from The Jacobson radical of a finite-dimensional algebra is the intersection of its maximal left ideals is exactly $I_P$. Hence $k[P]\mathrm{/}J(k[P])\cong k$ is simple, so $k[P]$ is local.

Conversely, assume $k[G]$ is local and suppose that $G$ is not a $p$-group. Choose a prime $q\mathrm{\ne }p$ dividing $\mathrm{\mid }G\mathrm{\mid }$. By [L3], $G$ contains a subgroup $C$ of order $q$. Since $q$ is invertible in $k$, the element $e_C:=q^{-1}{\sum }_{c\in C}[c]\in k[G]$ satisfies $e_C^2=e_C$ by [L4]. It is neither $0$ nor $1$: its coefficient at every $c\in C$ is the nonzero scalar $q^{-1}$, and either $C\mathrm{\ne }G$ or the coefficients at the nonidentity elements still differ from those of $1=[1_G]$.

A finite-dimensional local algebra has no nontrivial idempotent. Indeed, if $e^2=e$ with $e\mathrm{\ne }0,1$, then the nonzero proper left ideals $Ae$ and $A(1-e)$ are contained in maximal left ideals. Locality puts both inside the unique maximal left ideal, but $e+(1-e)=1$ then lies in that proper ideal, a contradiction. Applying this to $e_C$ contradicts step 1.2. Therefore every prime divisor of $\mathrm{\mid }G\mathrm{\mid }$ is $p$, and $G$ is a $p$-group.

Steps 1.1, 1.2, and 2.1 prove the equivalence. The trivial group is included in the forward direction with $I_1=0$.

By [L1], the algebra $kP$ has a unique simple module up to isomorphism.

The augmentation map $kP\to k$ makes the ground field $k$ into a simple $kP$-module on which every element of $P$ acts trivially. Since step 1.1 says there is only one simple module, it must be this trivial module.

Restrict $S$ from $G$ to the normal subgroup $N$. By [L2], the restricted module has finite length, so it contains a minimal nonzero $N$-submodule $T$. Then $T$ is simple as an $N$-module by [F1].

Since $N$ is a finite $p$-group, [L1] makes $kN$ local. Its augmentation quotient is the trivial simple module $k$, and every simple module over a local finite-dimensional algebra is its unique simple quotient. Hence $T$ is the trivial $kN$-module. Therefore the fixed-point space $S^N:=\{s\in S:ns=s\text{ for every }n\in N\}$ contains $T$ and is nonzero. Because $N$ is normal in $G$, the subspace $S^N$ is $G$-stable: for $g\in G$, $n\in N$, and $s\in S^N$, one has $n(gs)=g(g^{-1}ng)s=gs$.