A coefficient-ring obstruction to freeness of projective group-ring modules

Problem

Let $k$ be a commutative ring and $G$ a torsion-free almost polycyclic group. Must a finitely generated projective $kG$ -module containing two $kG$ -linearly independent elements be free?

P\text{ finitely generated projective},\quad (kG)^2\hookrightarrow P \quad\stackrel{?}{\Longrightarrow}\quad P\cong(kG)^n.

The coefficient ring is unrestricted. The counterexample below makes no assertion about the field-coefficient case.

The coefficient-ring obstruction

Problem 11.5 of the Kourovka Notebook (1), proposed by V. A. Artamonov, allows an arbitrary commutative coefficient ring. With this hypothesis, nonfreeness can already occur for the trivial group. The construction below uses the standard direct-summand criterion for projectivity (2); it is an elementary obstruction to the printed formulation.

Let

k=\mathbb F_2\times\mathbb F_2,\qquad G=1,\qquad R=kG.

Evaluation at the unique group element gives a ring isomorphism

\epsilon:R\longrightarrow k,\qquad \epsilon\left(\sum_{g\in G}a_g g\right)=a_1.

Write

I=\epsilon^{-1}(\mathbb F_2\times0),\qquad P=(R\oplus R)\oplus I.

All modules are left modules.

Theorem 1. The group $G$ is torsion-free and almost polycyclic. The $R$ -module $P$ is finitely generated and projective, contains two $R$ -linearly independent elements, and is not free.

Proof. The trivial group has a cyclic subnormal series of length zero and is a subgroup of itself of index one. It is therefore almost polycyclic; it is also torsion-free.

Under the identification $R\cong k$ , the map

\pi:R\longrightarrow I,\qquad (a,b)\longmapsto(a,0)

is $R$ -linear. If $\iota:I\hookrightarrow R$ is the inclusion, then $\pi\iota=1_I$ . Thus $I$ is a direct summand of the free module $R$ , and hence is projective. It follows that $P$ is projective. Moreover,

\tag{5} |R|=4,\qquad |I|=2,\qquad |P|=4\cdot4\cdot2=32,

so $P$ is finitely generated.

The two elements

u=((1,0),0),\qquad v=((0,1),0)

are $R$ -linearly independent: if $ru+sv=0$ , the first two coordinates give $r=s=0$ .

Suppose that $P$ is free. Since $P$ is finite, a basis is finite, say of cardinality $n$ . Consequently $|P|=|R|^n=4^n$ . This contradicts (5): for $n\leq2$ one has $4^n\leq16$ , and for $n\geq3$ one has $4^n\geq64$ . ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 11.5, arXiv:1401.0300v46.

The Stacks Project Authors, The Stacks Project, Lemma 10.78.2, Tag 00NX.