Nonunique indecomposable projective decompositions over the localized group ring of PSL2(F11)

Problem

Does Krull–Schmidt uniqueness hold for finitely generated projective modules over $\mathbb Z_{(p)}[G]$ , where $G$ is finite and $p$ is prime?

\bigoplus_{i=1}^{r}P_i\cong\bigoplus_{j=1}^{s}Q_j \quad\stackrel{?}{\Longrightarrow}\quad r=s,\quad P_i\cong Q_{\sigma(i)}

for nonzero indecomposable projectives and some permutation $\sigma$ ?

Failure of uniqueness over a noncomplete local ring

Kourovka Problem 4.55 asks whether the Krull–Schmidt theorem holds for projective modules over $\mathbb Z_{(p)}[G]$ , for a finite group $G$ and prime $p$ (1). For a noncomplete local coefficient ring, one must distinguish uniqueness of indecomposable decompositions from the stronger assertion that indecomposable projectives have local endomorphism rings. Johnston and Rumynin discuss these distinctions and study the semiperfectness question (2). We exhibit failure of uniqueness itself, already among finitely generated projectives.

Throughout, all modules are left modules. Put

O=\mathbb Z_{(11)}=\{a/b\in\mathbb Q:11\nmid b\},\qquad G=\operatorname{SL}_2(\mathbb F_{11})/\{\pm1\},\qquad \Lambda=O[G].

An $O$ -lattice means a $\Lambda$ -module that is free of finite rank as an $O$ -module.

Theorem 1. There are nonzero finitely generated indecomposable projective $\Lambda$ -modules $A,B,C,D$ of $O$ -rank $44$ satisfying

\tag{2} A\oplus D\cong B\oplus C,\qquad A\not\cong B,\qquad A\not\cong C.

Consequently uniqueness of decomposition into indecomposable projectives fails for $\Lambda$ .

Three projectives obtained by induction

Write a bar for the image of a matrix in $G$ . Define

\tag{3} r=\overline{\begin{pmatrix}0&-1\\1&5\end{pmatrix}},\qquad s=\overline{\begin{pmatrix}2&2\\3&9\end{pmatrix}},\qquad H=\langle r,s\rangle.

Multiplication in $\operatorname{SL}_2(\mathbb F_{11})$ gives

|r|=6,\quad |s|=2,\quad srs^{-1}=r^{-1},\quad s\notin\langle r\rangle.

Hence $H$ is dihedral of order $12$ . Let $\varepsilon,\lambda:H\to O^\times$ be the linear characters

\varepsilon(r)=1,\quad\varepsilon(s)=-1,\qquad \lambda(r)=-1,\quad\lambda(s)=1.

Also put

\tag{6} a=\overline{\begin{pmatrix}0&-1\\1&0\end{pmatrix}},\qquad b=\overline{\begin{pmatrix}0&5\\2&1\end{pmatrix}},\qquad K=\langle a,b\rangle.

Here $|a|=2$ , $|b|=3$ , and $|ab|=5$ . The $(2,3,5)$ presentation of $A_5$ gives an epimorphism $A_5\to K$ . Since $A_5$ is simple and $a\neq1$ , this is an isomorphism. Thus $|K|=60$ .

Let $\Omega$ be the six Sylow $5$ -subgroups of $K$ , with its conjugation action, and let

W=\ker\left(O[\Omega]\longrightarrow O,\quad \sum_{\omega}c_\omega\omega\longmapsto\sum_\omega c_\omega\right).

The lattice $W$ has rank five. Since $|G|=660$ , each of

\mathop{\mathrm{Ind}}_H^G O_\varepsilon,\qquad \mathop{\mathrm{Ind}}_K^G W,\qquad \mathop{\mathrm{Ind}}_H^G O_\lambda

has rank $55$ .

Let $\mathrm{St}$ be the character of the augmentation representation of $G$ on $\mathbb P^1(\mathbb F_{11})$ . This action is doubly transitive, so $\mathrm{St}$ is irreducible of degree $11$ , and

\mathrm{St}(g)=|\operatorname{Fix}_{\mathbb P^1(\mathbb F_{11})}(g)|-1\in\mathbb Z.

The associated central idempotent is

\tag{10} e=\frac1{60}\sum_{g\in G}\mathrm{St}(g^{-1})g\in\Lambda.

Define the three lattices

\tag{11} \begin{aligned} B&=(1-e)\mathop{\mathrm{Ind}}_H^G O_\varepsilon,\\ C&=(1-e)\mathop{\mathrm{Ind}}_K^G W,\\ D&=(1-e)\mathop{\mathrm{Ind}}_H^G O_\lambda. \end{aligned}

Lemma 2. The modules $B,C,D$ are finitely generated projective $\Lambda$ -modules.

Proof. If a finite group $L$ has order invertible in $O$ , every $O$ -free $O[L]$ -module is projective over $O[L]$ . Indeed, an $O$ -linear section of an epimorphism onto it can be replaced by the $L$ -equivariant section

x\longmapsto\frac1{|L|}\sum_{l\in L}l^{-1}t(lx).

Apply this with $L=H,K$ . Induction carries projectives to projectives, since $\Lambda$ is free as a right subgroup-ring module. The central idempotent $1-e$ selects a direct summand of each induced projective. Finite generation is preserved in both operations. ◻

Descent of isomorphisms and split maps

Write $\widehat O=\mathbb Z_{11}$ , $\widehat\Lambda=\widehat O[G]$ , and $\widehat X=\widehat O\otimes_O X$ .

Lemma 3. Let $X,Y$ be $\Lambda$ -lattices. If $\widehat X\cong\widehat Y$ , then $X\cong Y$ . If $\widehat X$ is isomorphic to a direct summand of $\widehat Y$ , then $X$ is isomorphic to a direct summand of $Y$ .

Proof. Choosing finite $O$ -bases expresses equivariance of a linear map as a finite system of linear equations over $O$ . Flatness of $\widehat O$ therefore gives

\tag{13} \widehat O\otimes_O\mathop{\mathrm{Hom}}_\Lambda(X,Y) \cong\mathop{\mathrm{Hom}}_{\widehat\Lambda}(\widehat X,\widehat Y).

The module on the left is the completion of the finite $O$ -module $\mathop{\mathrm{Hom}}_\Lambda(X,Y)$ . Hence every completed homomorphism can be approximated modulo $11$ by a homomorphism over $O$ .

Approximate a completed isomorphism by $f:X\to Y$ . Its reduction modulo $11$ is invertible, so $\det(f)\in O^\times$ . Thus $f$ is an $O$ -linear isomorphism, and its inverse is automatically $\Lambda$ -linear.

For the second assertion, let $\widehat u:\widehat X\to\widehat Y$ and $\widehat v:\widehat Y\to\widehat X$ satisfy $\widehat v\widehat u=1$ . Approximate both modulo $11$ by $u,v$ . Then $vu\equiv1\pmod{11}$ , so $vu$ is invertible over $O$ . Replacing $u$ by $u(vu)^{-1}$ gives $vu=1$ . Consequently $Y\cong X\oplus\ker v$ . ◻

Since $\widehat\Lambda$ is a finite algebra over a complete discrete valuation ring, it is semiperfect. Its finitely generated projectives are finite direct sums of the projective covers of simple $\mathbb F_{11}[G]$ -modules, and the multiplicities are unique. Only this finite Krull–Schmidt theorem for the completed ring is used below.

The projective characters in characteristic eleven

We use the ordinary-character numbering of the Modular Atlas (3). The ordinary irreducible degrees of $G$ are

\chi_1(1),\ldots,\chi_8(1)=1,5,5,10,10,11,12,12, \qquad\chi_6=\mathrm{St}.

The simple $\mathbb F_{11}[G]$ -modules are the absolutely simple modules

S_d=\operatorname{Sym}^{d-1}(\mathbb F_{11}^2),\qquad d\in\{1,3,5,7,9,11\}.

The even symmetric powers descend from $\operatorname{SL}_2(\mathbb F_{11})$ to $G$ . Their Brauer characters are denoted by $\varphi_d$ . The subscript $d$ is the dimension. The principal-block decomposition matrix is

\tag{16} \begin{array}{c|ccccc} &\varphi_1&\varphi_3&\varphi_5&\varphi_7&\varphi_9\\\hline \chi_1&1&0&0&0&0\\ \chi_2&0&0&1&0&0\\ \chi_3&0&0&1&0&0\\ \chi_4&0&1&0&1&0\\ \chi_5&1&0&0&0&1\\ \chi_7&0&0&1&1&0\\ \chi_8&0&1&0&0&1 \end{array}

The remaining character $\chi_6$ forms a defect-zero block and reduces to $S_{11}$ . These are the two blocks of $G$ in the characteristic-eleven decomposition table (3).

Let $Q_d$ be the projective $\widehat\Lambda$ -module whose reduction is the projective cover of $S_d$ . Brauer reciprocity applied to (16) gives

\tag{17} \begin{aligned} \Phi_1&=\chi_1+\chi_5,& \Phi_3&=\chi_4+\chi_8,\\ \Phi_5&=\chi_2+\chi_3+\chi_7,& \Phi_7&=\chi_4+\chi_7,\\ \Phi_9&=\chi_5+\chi_8,& \Phi_{11}&=\chi_6. \end{aligned}

Here $\Phi_d$ is the ordinary character of $Q_d$ . One may compute these after extending to a finite splitting extension of $\mathbb Q_{11}$ . Absolute simplicity of the $S_d$ implies that their projective covers remain indecomposable under the corresponding extension of complete discrete valuation rings; the multiplicities in (17) are unchanged. The six displayed projective characters are linearly independent, as is also immediate by comparing coefficients of the $\chi_i$ . Thus an ordinary character determines the multiplicities of a completed projective module.

Put $u=(-1+\sqrt5)/2$ and $v=(-1-\sqrt5)/2$ . On the eleven-regular classes the values needed below are

\tag{18} \begin{array}{c|rrrrrr} &1&2A&3A&5A&5B&6A\\\hline \chi_1&1&1&1&1&1&1\\ \chi_2,\chi_3&5&1&-1&0&0&1\\ \chi_4&10&-2&1&0&0&1\\ \chi_5&10&2&1&0&0&-1\\ \chi_6&11&-1&-1&1&1&-1\\ \chi_7&12&0&0&v&u&0\\ \chi_8&12&0&0&u&v&0 \end{array}

These values also follow from (16) and symmetric-power traces. On a class of order five, choose a primitive fifth root $\zeta$ so that $\zeta+\zeta^{-1}=u$ . The trace of $S_d$ is

\sum_{j=-(d-1)/2}^{(d-1)/2}\zeta^j.

For $d=1,3,5,7,9,11$ this gives $1,1+u,0,v,-1,1$ . Substituting into (16) yields the $5A$ column. Replacing $\zeta$ by $\zeta^2$ gives $5B$ , and the same formula with the appropriate roots of unity gives the columns of orders two, three and six.

Characters and completions of the induced summands

Lemma 4. The modules in (11) have rank $44$ , and

\tag{20} \widehat B\cong Q_3\oplus Q_7,\qquad \widehat C\cong Q_5\oplus Q_9,\qquad \widehat D\cong Q_7\oplus Q_9.

Proof. In $H$ , the nonidentity rotations consist of one involution, two elements of order three and two of order six. Its six reflections are involutions. Frobenius reciprocity gives, for an ordinary irreducible character $\chi$ ,

\begin{align*} \langle\mathop{\mathrm{Ind}}_H^G\varepsilon,\chi\rangle &=\frac{\chi(1)-5\chi(2A)+2\chi(3A)+2\chi(6A)}{12},\\ \langle\mathop{\mathrm{Ind}}_H^G\lambda,\chi\rangle &=\frac{\chi(1)-\chi(2A)+2\chi(3A)-2\chi(6A)}{12}. \end{align*}

For the second formula, the values of $\lambda$ on the six reflections sum to zero. Substitution from (18) yields

\begin{align*} \mathop{\mathrm{Ind}}_H^G\varepsilon&=2\chi_4+\chi_6+\chi_7+\chi_8,\\ \mathop{\mathrm{Ind}}_H^G\lambda&=\chi_4+\chi_5+\chi_6+\chi_7+\chi_8. \end{align*}

The character of $W$ has values $5,1,-1,0$ on elements of orders $1,2,3,5$ in $K$ . Indeed, in the action on its six Sylow $5$ -subgroups an involution fixes two points, an element of order three fixes none, and an element of order five fixes one. Since $K$ has $15$ involutions and $20$ elements of order three,

\langle\mathop{\mathrm{Ind}}_K^G W,\chi\rangle =\frac{5\chi(1)+15\chi(2A)-20\chi(3A)}{60}.

Hence $\mathop{\mathrm{Ind}}_K^G W=\chi_2+\chi_3+\chi_5+\chi_6+\chi_7+\chi_8$ . Multiplication by $1-e$ removes the $\chi_6$ constituent. Therefore

\tag{24} \begin{aligned} \chi_B&=2\chi_4+\chi_7+\chi_8=\Phi_3+\Phi_7,\\ \chi_C&=\chi_2+\chi_3+\chi_5+\chi_7+\chi_8=\Phi_5+\Phi_9,\\ \chi_D&=\chi_4+\chi_5+\chi_7+\chi_8=\Phi_7+\Phi_9. \end{aligned}

Each character has degree $44$ . The modules are projective by Lemma 2. Their completed projective multiplicities are thus determined by (17), giving (20). ◻

The second decomposition and indecomposability

Put $P=B\oplus C$ . By Lemma 4,

\widehat P\cong Q_3\oplus Q_7\oplus Q_5\oplus Q_9,

so $\widehat D\cong Q_7\oplus Q_9$ is a direct summand of $\widehat P$ . Lemma 3 gives $\Lambda$ -maps $u:D\to P$ and $v:P\to D$ with $vu=1$ . Define $A=\ker v$ . Then

\tag{26} P\cong A\oplus D,\qquad \mathop{\mathrm{rank}}_O A=44, \qquad \widehat A\cong Q_3\oplus Q_5.

The module $A$ is finitely generated projective as a direct summand of $P$ . The last isomorphism follows by finite Krull–Schmidt cancellation over $\widehat\Lambda$ . Different choices of the splitting give isomorphic complements, because their completions are isomorphic and Lemma 3 applies.

Lemma 5. None of $Q_3,Q_5,Q_7,Q_9$ is the completion of an $O$ -lattice with a $G$ -action.

Proof. Equations (17) and (18) give

\Phi_3(5A)=u,\quad \Phi_5(5A)=v,\quad \Phi_7(5A)=v,\quad \Phi_9(5A)=u.

Both $u$ and $v$ are irrational over $\mathbb Q$ . But the trace of an element of $G$ on an $O$ -lattice belongs to $O\subseteq\mathbb Q$ , and completion leaves this trace unchanged. The stated descent is therefore impossible. ◻

Proof of Theorem 1. The completed decompositions are

\begin{array}{c|c} \text{module}&\text{completion}\\\hline A&Q_3\oplus Q_5\\ B&Q_3\oplus Q_7\\ C&Q_5\oplus Q_9\\ D&Q_7\oplus Q_9 \end{array}

Suppose one of these four modules split into two nonzero summands over $\Lambda$ . The summands are finite projective $O$ -lattices. Completion is faithfully flat, so both completed summands are nonzero. Finite Krull–Schmidt uniqueness forces each to be one of the two displayed projective covers. This contradicts Lemma 5. Hence all four modules are indecomposable. Their rank $44$ makes them nonzero.

The isomorphism in (2) is (26). Finally, at an involution of $G$ ,

\chi_A(2A)=\chi_B(2A)+\chi_C(2A)-\chi_D(2A)=0, \qquad\chi_B(2A)=-4,\qquad\chi_C(2A)=4.

Unequal traces exclude $A\cong B$ and $A\cong C$ . Thus neither possible matching of the two indecomposable decompositions is an isomorphism matching, proving failure of uniqueness. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 4.55. arXiv:1401.0300.

D. Johnston and D. Rumynin, On a question by Roggenkamp about group algebras, J. Algebra 687 (2026), 776–791. doi:10.1016/j.jalgebra.2025.09.014.

Modular Atlas, Decomposition matrices: $L_2(11)$ modulo $11$ . RWTH Aachen University.