Modular Representations and Projective Covers - Examples

By [L1], the algebra $A:=k{C}_p$ is local. A nontrivial decomposition of the left regular module would give a nontrivial idempotent projection in ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{o}\mathrm{p}}$, but a local algebra has no nontrivial idempotents. Hence the regular module is indecomposable.

The augmentation ideal is maximal because its quotient is $k$, and it is the unique maximal left ideal because $A$ is local. Every simple quotient of the regular module has a maximal left ideal as its kernel, so it is the augmentation quotient $k$, with the trivial $C_p$-action. Thus the regular module has the asserted unique simple quotient.

In characteristic $p$, one has $g^p-1=(g-1{)}^p=x^p$, so the relation $g^p=1$ becomes $x^p=0$. Every element of $k{C}_p$ is a polynomial in $g$, hence in $x=g-1$, and the basis $1,g,\dots ,g^{p-1}$ becomes the basis $1,x,\dots ,x^{p-1}$. Therefore $k{C}_p\cong k[x]\mathrm{/}(x^p)$.

Under that identification, the augmentation map kills $x$ and sends $1$ to $1$, so its kernel is $(x)$. Since [L1] makes the ring local, $(x)$ is its radical. Therefore [F1] gives the displayed Loewy series by successive powers of $(x)$.

The regular module $kP$ is projective, and $\epsilon$ is surjective onto the trivial module. Its kernel is the augmentation ideal, which is the unique maximal ideal because [L2] makes $kP$ local. Hence the kernel is superfluous.

By step 1.1, $\epsilon$ is a projective cover of the trivial module. The uniqueness theorem [L3] says every projective cover of that target is isomorphic to this one over the target.

The extension $K\mathrm{/}{\mathbb{Q}}_3$ is totally ramified of degree $2$, with uniformizer $1-{\zeta }_3$, valuation ring $\mathcal{O}$, and residue field $k={\mathbb{F}}_3$. The field $K$ splits the subgroups of $S_3$: it contains the values needed for the cyclic subgroups, and the trivial, sign, and standard representations split $S_3$. If $V$ is a simple $k{S}_3$-module and $g$ generates the normal subgroup $C_3$, then $(g-1{)}^3=g^3-1=0$, so $V^{C_3}\mathrm{\ne }0$; normality and simplicity give $V^{C_3}=V$. Thus $V$ factors through $S_3\mathrm{/}{C}_3\cong C_2$ and is trivial or sign. The same calculation handles the subgroups, so $k$ also splits all of them. Hence $(K,\mathcal{O},k)$ is a splitting $3$-modular system for $S_3$.

The lattice $L$ is free of rank $2$ over $\mathcal{O}$, with basis $(1,0,-1),(0,1,-1)$, so [F1] and [L1] give the two-dimensional $k{S}_3$-module $\bar{L}=L\mathrm{/}\mathfrak{m}L$. The vector $v:=(1,1,1)\in k^3$ lies in $\bar{L}$ because it is the reduction of $(1,1,-2)\in L$, and it is fixed by every permutation in $S_3$.

The nonzero line $kv\subseteq \bar{L}$ is therefore an $S_3$-stable proper submodule of the two-dimensional module $\bar{L}$. Hence $\bar{L}$ is reducible.

The module $M$ is itself induced from the trivial $kH$-module, so it is relatively $H$-projective by definition.

Applying [L1] to the relatively $H$-projective module of step 1.1 produces an $H$-endomorphism $\alpha$ with ${\mathrm{Tr}}_H^G(\alpha )={\mathrm{id}}_M$. Thus Higman's criterion detects this permutation-induced module exactly as expected.

Extending scalars gives $V=K{\otimes }_{\mathcal{O}}L=\{(a,b,c)\in K^3:a+b+c=0\}$ with the same two basis vectors. If $W\subseteq V$ is a nonzero invariant subspace, choose $0\mathrm{\ne }v\in W$. Its coordinates cannot all agree, since $3a=0$ implies $a=0$ in characteristic zero. For unequal coordinates $v_i,v_j$, subtracting the image of $v$ under the transposition $(ij)$ gives $(v_i-v_j)(e_i-e_j)\in W$. Thus $e_i-e_j\in W$, and permuting coordinates gives all differences, which span $V$. Hence $W=V$ and $V$ is irreducible.

By [F1] and [L2], reducing $L$ modulo $\mathfrak{m}$ gives a two-dimensional $k{S}_3$-module $\bar{L}$. The vector $(1,1,1)$ is the reduction of $(1,1,-2)\in L$ and spans a nonzero proper invariant line, so $\bar{L}$ is reducible. The residue characteristic is $3$, which divides $\mathrm{\mid }{S}_3\mathrm{\mid }=6$, consistent with [L1].

Therefore an ordinary irreducible representation can have reducible reduction modulo $p$, refuting the statement.