p-modular System

Definition

Let $p$ be a prime. A $p$-modular system is a triple $(F,R,k)$ where $F$ is a field of character $0$ equipped with a discrete valuation, $R$ is the valuation ring of $F$ with maximal ideal $(\pi )$, and $k=R/(\pi )$ is the residue field of $R$ with $\mathrm{char}k=p$.

Given a finite group $G$, if both $F$ and $k$ are splitting for $G$, then we say that $(F,R,k)$ is a splitting $p$-modular system for $G$. This happens when $F$ contains a primitive $\exp (G)$th root of unity.

Preliminary

RG-module & kG-module

Lemma

Let $R$ be a DVR with maximal ideal $(\pi )$ and $k=R/(\pi )$. Let $G$ be a finite group.

• If $S$ is a simple $RG$-module, then $\pi S=0$.
• The simple $RG$-modules are exactly the simple $kG$-modules made into $RG$-module via the surjection $RG\to kG$.
• For each $RG$-module $U$, $\pi U\subseteq \mathrm{Rad}U$. In particular, $\pi RG\subseteq J(RG)$.
• For each $RG$-module, we have $\mathrm{Rad}U/\pi U=\mathrm{Rad}(U/\pi U)$.

\begin{proof} i) Since $\pi S$ is a $RG$-submodule of $S$ and $S$ is simple, we know $\pi S=0$ or $S$. As $R$-modules, $\mathrm{Rad}(S)=J(R)S=(\pi )S\ne S$ by Nakayama lemma, so $\pi S=0$.

ii) Note that $kG=RG/(\pi G)$ and $\pi G$ annihilates simple $RG$-modules.

iii) If $V$ is a maximal submodule of $U$, then $U/V$ is simple. So $\pi (U/V)=0$ and $\pi U\subseteq V$. Hence $\pi U\subseteq \mathrm{Rad}U$.

iv) Recall that $\mathrm{Rad}U$ is the intersection of kernels of all morphisms from $U$ to simple modules. \end{proof}

Corollary

Let $R$ be a DVR with maximal ideal $(\pi )$ and $k=R/(\pi )$. Let $G$ be a finite group. Let $P,Q$ be finitely generated projective $RG$-modules. Then $P\simeq Q$ as $RG$-modules iff $P/\pi P\simeq Q/\pi Q$ as $kG$-modules.

\begin{proof} One direction is easy. On the other hand, note that $P/\pi P\simeq Q/\pi Q$ yields $P/\mathrm{Rad}P\simeq Q/\mathrm{Rad}Q$. Since projective cover is unique, we have $P\simeq Q$. \end{proof}

Projective kG-module & RG-lattice

Definition

An $RG$-module $L$ is called an $RG$-lattice if it is finitely generated and free as $R$-module.

The key point of the following proposition is, every finitely generated projective $kG$-module can be lifted to be an $RG$-lattice, and such a lift is unique up to isomorphism and projective.

Proposition

Let $R$ be a complete DVR with maximal idea $(\pi )$ and $k=R/(\pi )$. Let $G$ be a finite group.

• For each simple $RG$-module $S$, there is a unique indecomposable projective $RG$-module ${\hat{P}}_S$ that is a projective cover of $S$. It has the form ${\hat{P}}_S=RG{\hat{e}}_S$ where ${\hat{e}}_S$ is a primitive idempotent in $RG$ with ${\hat{e}}_{S}S\ne 0$.

• The $kG$-module $P_S\simeq {\hat{P}}_S/\pi {\hat{P}}_S$ is the projective cover of $S$ as $kG$-modules and ${\hat{P}}_S$ is the projective cover of $P_S$ as $RG$-modules. Furthermore, $S$ is the unique simple quotient of ${\hat{P}}_S$.

  Thus for simple $kG$-modules $S$ and $T$, ${\hat{P}}_S\simeq {\hat{P}}_T$ $⟺$ $S\simeq T$.

• Every finitely generated $RG$-module has a projective cover.

• Every finitely generated indecomposable projective $RG$-module is isomorphic to ${\hat{P}}_S$ for some simple $S$.

• Every finitely generated projective $kG$-module can be lifted to be an $RG$-lattice such a lift is unique up to isomorphism and projective.

  An $RG$-lattice $L$ is a projective $RG$-module iff $L/\pi L$ is a projective $kG$-module.

  Projective $RG$-module $L_1$ and $L_2$ are isomorphic iff $L_1/\pi L_1\simeq L_2/\pi L_2$.

\begin{proof} i) Let $e_S\in kG$ be a primitive idempotent with $e_S\cdot S\ne 0$. Let ${\hat{e}}_S\in RG$ be a primitive idempotent that lifts $e_S$, then ${\hat{e}}_{S}S=e_{S}S\ne 0$. Set ${\hat{P}}_S=RG{\hat{e}}_S$. Since ${\hat{e}}_S$ is a lift of $e_S$, it is indecomposable and projective.

ii) Note that ${\hat{P}}_S/\pi {\hat{P}}_S\simeq kG{e}_S=P_S$. Thus $P_S$ is the projective cover of $S$. Consider the composition ${\hat{P}}_S\to P_S\to S$, then we know ${\hat{P}}_S$ is a projective cover of $S$.

iii) Let $U$ be a finitely generated $RG$-module. Then $U/\mathrm{Rad}U$ is a finitely generated semisimple $RG$-module. Write $U/\mathrm{Rad}U\simeq S_1\oplus \cdots \oplus S_t$ for simple modules $S_i$. As ${\hat{P}}_{S_1}\oplus \cdots \oplus {\hat{P}}_{S_t}$ is a projective cover of $S_1,\cdots ,S_t$, there exists $\phi :{\hat{P}}_{S_1}\oplus \cdots \oplus {\hat{P}}_{S_t}\to U$. By Nakayama lemma, $\phi$ is an essential epimorphism.

iv) Let $P$ be a finitely generated projective $RG$-module. By iii), there exists a projective cover $\alpha :{\hat{P}}_{S_1}\oplus \cdots \oplus {\hat{P}}_{S_t}\to P$. Since $P$ is projective, there exists $\beta :P\to {\hat{P}}_{S_1}\oplus \cdots \oplus {\hat{P}}_{S_t}$ such that $\alpha \beta ={\mathrm{id}}_P$. Since $\alpha$ and ${\mathrm{id}}_P$ are an essential epimorphism, we know $\beta$ is also an essential epimorphism. Furthermore, note that $\beta$ is an injective map by $\alpha \beta ={\mathrm{id}}_P$, so $\beta$ is an isomorphism. In particular, if $P$ is indecomposable, then $P\simeq {\hat{P}}_S$.

v) It suffices to prove the statement holds for indecomposable projective $kG$-modules, that are isomorphic to $P_S$ for some simple $kG$-module $S$. Remark that ${\hat{P}}_S$ is the projective cover of $P_S$ and ${\hat{P}}_S$ is an $RG$-lattice. Now we proved the first statement.

To show uniqueness, suppose $L$ is any $RG$-lattice for which $L/\pi L\simeq P_S$. Note that $\pi L\subseteq \mathrm{Rad}(L)$ and $L/\mathrm{Rad}(L)\simeq S$. The projective cover ${\hat{P}}_S\to S$ factors as ${\hat{P}}_S\to L\to S$. So ${\hat{P}}_S$ is surjective and then ${\hat{P}}_S\simeq L$. \end{proof}

FG-module & full RG-lattice

Definition

Let $R$ be a PID, and let $F=\mathrm{Frac}(R)$. For a $G$-lattice $L$, $F{\otimes }_{R}L$ is an $FG$-module.

Conversely, if $U$ is an $FG$-module, a full $RG$-lattice $U_0$ in $U$ is an $RG$-lattice $U_0\subseteq U$ that has an $R$-basis is also an $F$-basis of $U$. In this situation, $U\simeq F{\otimes }_R{U}_0$.

Lemma

Let $R$ be a PID, and let $F=\mathrm{Frac}(R)$. Let $U$ be a finitely generated $F$-vector space. Any finitely generated $R$-module $U_0\subseteq U$ that contains an $F$-basis of $U$ is a full $R$-lattice in $U$.

\begin{proof} $U_0$ is a finitely generated and torsion-free $R$-module. Then $U_0$ is a free $R$-module and it has an $R$-basis $x_1,\cdots ,x_n$. Since $U_0$ contains an $F$-basis of $U$. Then $x_1,\cdots ,x_n$ span $U$ as $F$-space.

One can show $x_1,\cdots ,x_n$ are linearly independent, and then we finish the proof. \end{proof}

Corollary

Let $R$ be a PID, and let $F$ be a field of fractions. Let $U$ be a finitely generated $FG$-module. Then there exists an full $RG$-lattices of $U$.

\begin{proof} Let $u_1,\cdots ,u_n$ be an $F$-basis of $U$. Consider $U_0=\mathrm{span}\{g{u}_i:g\in G,i=1,\cdots ,n\}$, which is a finitely generated $R$-submodule of $U$ and it is $G$-invariant. Thus $U_0$ is a full $RG$-lattice of $U$ by ^vnn9oj. \end{proof}

Brauer-Nesbitt Theorem

Brauer-Nesbitt

Let $(F,R,k)$ be a $p$-modular system with $k=R/(\pi )$, $G$ a finite group, and $U$ a finite-dimensional $FG$-module. Let $L_1$ and $L_2$ be full $RG$-lattices in $U$, then $L_1/\pi L_1$, $L_2/\pi L_2$ have the same composition factors with the same multiplications as $kG$-modules.

\begin{proof} Observe that $L_1+L_2$ is also a full $RG$-lattices in $U$. So we can assume that $L_1\subseteq L_2$. (If it holds, then apply it to $L_1\subseteq L_1+L_2$ and $L_2\subseteq L_1+L_2$.)

As $R$-module, $L_1$ and $L_2$ are free of the same rank, and then $L_2/L_1$ is a torsion $R$-module. Thus, as an $R$-module, $L_2/L_1$ has a composition series. Hence $L_2/L_1$ has a composition series as an $RG$-module. So it suffices to assume that $L_1$ is maximal in $L_2$, i.e. $L_2/L_1$ is simple. Thus $\pi (L_2/L_1)=0$ and so $\pi L_2\subseteq L_1$. It deduces that

$$\pi L_1\subseteq \pi L_2\subseteq L_1\subseteq L_2.$$

We claim that to show $L_2/L_1\simeq \pi L_2/\pi L_1$. Define

$$L_2\to \pi L_2/\pi L_1,x\mapsto \pi x+\pi L_1,$$

which is surjective as has kernel $L_1$. Now we prove the claim, and it remains to show $\pi L_1$ and $\pi L_2$ has the same composition factors with the same multiplications. \end{proof}

DCE Triangle

decomposition matrix

Let $(F,R,k)$ be a splitting $p$-modular system for $G$. The decomposition matrix $D$ of $G$ is the matrix with

• rows indexed by the simple $FG$-modules,

• columns indexed by the simple $kG$-modules,

• entries are the number

  $$d_{TS}=\text{multiplicities of }S\text{ as a composition factor of }{T}_0/\pi T_0.$$

  where $S$ is a simple $kG$-module, $T$ is a simple $FG$-module, and $T_0$ is a full $RG$-lattices in $T$.

Definition

Let $A$ be a finite-dimensional $k$-algebra, where $k$ is any field.

Define $G_0(A)$ as the free abelian group with the isomorphism types of simple $A$-modules as a basis, which is called Grothendieck group.

Define $K_0(A)$ as the free abelian group with isomorphism types of indecomposable projective $A$-modules as a basis.

If $T$ is a simple $A$-module, then write $[T]$ for the corresponding basis element of $G_0(A)$. If $P$ is a indecomposable projective $A$-module, then write $[P]$ for the corresponding basis element of $K_0(A)$.

If $U$ is an $A$-module with composition factors $S_1,\cdots ,S_r$ with multiplications $n_1,\cdots ,n_r$, we write

$$[u]=n_1[S_1]+\cdots +n_r[S_r]\in G_0(A).$$

If $P$ is a projective $A$-module with $P=P_1^{n_1}\oplus \cdots \oplus P_r^{n_r}$ where $P_1,\cdots ,P_r$ are indecomposable projective $A$-module. Put

$$[P]=n_1[P_1]+\cdots +n_r[P_r]$$

Let $(F,R,k)$ with $k=R/(\pi )$ be a splitting $p$-modular system for a finite group $G$. Suppose the valuation of $F$ is complete. Notice that

• $\mathrm{rank}{G}_0(FG)=|Cl(G)|$

• $\mathrm{rank}{G}_0(kG)=\mathrm{rank}{K}_0(kG)=$ the number of $p$-regular conjugacy class

• dce triangle:

  

• If $P_S$ is a indecomposable projective $kG$-module, then $e([P_S])=[F\otimes {\hat{P}}_S]\in G_0(FG)$, where ${\hat{P}}_S$ is an $RG$-module that lifts $P_S$.

• Now $P_S$ is a projective cover of $S$. Define $c([P_S])=[P_S]\in G_0(kG)$. Remark that $[P_T]={\sum }_{\text{simple }S}{c}_{ST}[S]$ for each simple $kG$-module $T$, where $c_{ST}$ is the entries of Cartan matrix.

• If $V$ is a simple $FG$-module containing a full $RG$-lattice $V_0$. Put $d(V)=[V_0/\pi V_0]$. The matrix of $d$ is the transpose of decomposition matrix.

Proposition

$c=de$, i.e., the diagram commutes.

\begin{proof} Let $P_S$ be an indecomposable projective $kG$-module. Then

• $c([P_S])=[P_S]\in G_0(kG)$
• $e([P_S])=[F{\otimes }_R{\hat{P}}_S]$ for some ${\hat{P}}_S$ lifts $P_S$
• $de([P_S])=d([F{\otimes }_R{\hat{P}}_S])$

Taking ${\hat{P}}_S$ to be the full $RG$-lattices, its reduction (modulo $\pi$) is $P_S$. Then $de([P_S])=[P_S]\in G_0(kG)$. \end{proof}

Proposition

Let $(F,R,k)$ be a $p$-modular system. Let $U,V$ be $FG$-modules, containing full $RG$-lattices $U_0,V_0$, respectively.

• ${\mathrm{Hom}}_{RG}(U_0,V_0)$ is a full $RG$-lattices of ${\mathrm{Hom}}_{FG}(U,V)$

• Suppose further that $U_0$ is projective. Then

  $${\mathrm{Hom}}_{RG}(U_0,V_0)/\pi {\mathrm{Hom}}_{RG}(U_0,V_0)\simeq {\mathrm{Hom}}_{RG}(U_0,V_0/\pi V_0)\simeq {\mathrm{Hom}}_{kG}(U_0/\pi U_0,V_0/\pi V_0).$$

\begin{proof} i) Take $R$-basis $u_1,\cdots ,u_n$ and $v_1,\cdots ,v_m$ of $U_0$ and $V_0$, respectively. These are $F$-basis of $U,V$. Any $RG$-morphism $U_0\to V_0$ can be represented as a matrix over $R$. Regard it as a matrix over $F$, then it represents an $FG$-morphism $U\to V$.

Notice that

$${\mathrm{Hom}}_{RG}(U_0,V_0)\subseteq {\mathrm{Hom}}_R(U_0,V_0)\simeq R^{mn},$$

then by $R$ PID, we know ${\mathrm{Hom}}_{RG}(U_0,V_0)$ is an $RG$-lattice.

Let $\phi \in {\mathrm{Hom}}_{FG}(U,V)$, then $\phi (U_i)={\sum }_{i,j}{\lambda }_{ji}{v}_j$. Assume that ${\lambda }_{ij}=a_{ji}/b_{ji}$ for all $i,j$. Let $b={\prod }_{i,j}{b}_{ji}$. Then $b\phi (u_i)\in V_0$, and so $b\phi :U_0\to V_0$ is an $RG$-morphism. Thus ${\mathrm{Hom}}_{RG}(U_0,V_0)$ is a full $RG$-lattice in ${\mathrm{Hom}}_{FG}(U,V)$.

ii) Claim that ${\mathrm{Hom}}_{RG}(U_0,\pi V_0)\simeq \pi {\mathrm{Hom}}_{RG}(U_0,V_0)$ since $U_0$ is projective.

Consider ${\mathrm{Hom}}_{RG}(U_0,V_0)\to {\mathrm{Hom}}_{RG}(U_0,V_0/\pi V_0),\phi \mapsto \ell \phi$, where $\ell :V_0\to V_0/\pi V_0$. Its kernel is ${\mathrm{Hom}}_{RG}(U_0,pi{V}_0)$. It is surjective since $U_0$ is projective. Thus ${\mathrm{Hom}}_{RG}(U_0,V_0)/\pi {\mathrm{Hom}}_{RG}(U_0,V_0)\simeq {\mathrm{Hom}}_{RG}(U_0,V_0/\pi V_0)$.

For the second isomorphism, all morphisms $\alpha :U\to V_0/\pi V_0$ contain $\pi U_0$ in the kernel, so factor as $U_0\to U_0/\pi U_0\to V_0\to \pi V_0$. \end{proof}

Corollary

Suppose $U_0,V_0$ are full $RG$-lattices of $FG$-modules $U,V$, and $U_0$ is projective. Then ${\dim }_F{\mathrm{Hom}}_{FG}(U,V)={\dim }_k{\mathrm{Hom}}_{kG}(U_0/\pi U_0,V_0/\pi V_0)$.

Brauer Reciprocity

Let $(F,R,k)$ be a splitting $p$-modular system for $G$, where $F$ is complete.

• Let $S$ be a simple $kG$-module and $T$ a simple $FG$-module containing a full $RG$-lattice $T_0$. The multiplicity of $T$ in $F{\otimes }_R{\hat{P}}_S$ equals the multiplicity of $S$ as a composition factor in $T_0/\pi T_0$.
• The matrix of $e$ is $D$, where $D$ is the decomposition matrix.

\begin{proof} i) ${\dim }_F{\mathrm{Hom}}_{FG}(F{\otimes }_R{\hat{P}}_S,T)={\dim }_k{\mathrm{Hom}}_{kG}(P_S,T_0/\pi T_0)$.

• LHS equals ${\dim }_F{\mathrm{End}}_{FG}(T)\cdot (\text{multiplicity of }T\text{ in }F{\otimes }_R{\hat{P}}_S)$.
• RHS equals ${\dim }_k{\mathrm{End}}_{kG}(S)\cdot (\text{multiplicity of }S\text{ as the composition factor of }{T}_0/\pi T_0)$.

By splitting, ${\dim }_{{\mathrm{End}}_{FG}}(T)={\dim }_k{\mathrm{End}}_{kG}(S)=1$. \end{proof}

Corollary

Let $(F,R,k)$ be the splitting $p$-modular system for $G$, where $F$ is complete. Then the Cartan matrix $C=D^{T}D$, where $D$ is the decomposition matrix. In particular, $C$ is a symmetric matrix.