15 250603

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} See ^k0k1kv. \end{proof}

Definition

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

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} See ^5awwqe. \end{proof}

Definition

Let $R$ be a PID, and let $F=\mathrm{Frac}(R)$. Given $L$ as a subset of $F{\otimes }_{R}L$. In this situation, the $FG$-module $F{\otimes }_{R}L$ can be written in $R$.

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} See ^vnn9oj. \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} See ^qsdviu. \end{proof}

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} See ^7sv8b9. \end{proof}

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 index 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} See ^j43lpr. \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} See ^hfu3dq. \end{proof}

Corollary

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

Theorem

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} See ^5kl5ya. \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.