Throughout, all modules shall be right
$R$
-modules for some ring
$R$
. A module
$S$
is said to be stably free if
$S\oplus F\cong F^{\prime }$
for some free
$R$
-modules
$F,F^{\prime }$
. By a theorem of Gabel (see [Reference Gabel4]), any infinitely generated stably free module is necessarily free. As such, this paper is solely interested in finitely generated stably free modules. This finite generation hypothesis shall be hereafter applied without mention. We now say that
$S$
is stably free of rank
$m-n\geqslant 0$
if
$S\oplus R^{n}\cong R^{m}$
for some
$m,n\geqslant 0$
. Such a rank is well defined and positive whenever
$R$
is weakly finite in the sense of [Reference Cohn2]. In [Reference Cohn1], Cohn attributes the result that
$\mathbf{F}[G]$
is weakly finite for any field
$\mathbf{F}$
of characteristic zero to an unpublished result of Kaplansky. From this, it follows that integral group rings are necessarily weakly finite. When any stably free module over
$R$
is necessarily free,
$R$
is said to have the stably free cancellation (SFC) property.
Our main concern shall be an integral group ring
$\mathbf{Z}[G]$
, where
$G$
is obtained from a finite metacyclic group. Let
$p$
be prime and let
$\unicode[STIX]{x1D6FC}:C_{p}\rightarrow C_{p}$
be a generator of
$\operatorname{Aut}(C_{p})\cong C_{p-1}$
. As
$\operatorname{ord}(\unicode[STIX]{x1D6FC})=p-1$
, we choose a prime
$q$
such that
$q|p-1$
and set
$\unicode[STIX]{x1D703}=\unicode[STIX]{x1D6FC}^{(p-1)/q}$
. Evidently,
$\operatorname{ord}(\unicode[STIX]{x1D703})=q$
. Next, we identify
$\mathbf{Z}[C_{p}\rtimes C_{q}]$
with the cyclic algebraic construction
${\mathcal{C}}_{q}(\mathbf{Z}[C_{p}],\unicode[STIX]{x1D703})$
. As is well known,
${\mathcal{C}}_{q}(\mathbf{Z}[C_{p}],\unicode[STIX]{x1D703})$
is free of rank
$q$
over
$\mathbf{Z}[C_{p}]$
. For a basis, we take
$\{1,y,\ldots ,y^{q-1}\}$
in which
$y^{q}=1$
and
$y\cdot \unicode[STIX]{x1D706}=\unicode[STIX]{x1D703}(\unicode[STIX]{x1D706})\cdot y$
, where
$\unicode[STIX]{x1D706}\in \mathbf{Z}[C_{p}]$
. Now we set
$G=(C_{p}\rtimes C_{q})\times C_{\infty }^{n}$
and
$\unicode[STIX]{x1D6EC}=\mathbf{Z}[G]$
for the integral group ring of
$G$
. The main result of this paper is then expressed as follows.
Theorem A.
The group ring
$\unicode[STIX]{x1D6EC}$
has SFC.
1 Preliminaries
We start by briefly discussing the notation and terminology to be employed hereafter. By
$M_{k}(R)$
, we mean the
$k\times k$
matrices over
$R$
. This has the canonical
$R$
-basis
$\{\unicode[STIX]{x1D716}(i,j)\mid 1\leqslant i,j\leqslant k\}$
, where
$\unicode[STIX]{x1D716}(i,j)_{r,s}=\unicode[STIX]{x1D6FF}_{ir}\unicode[STIX]{x1D6FF}_{js}$
. There are two subgroups of interest, the first being
${\mathcal{T}}_{k}(R,\unicode[STIX]{x1D70B})$
, which is the quasi-triangular subring of
$M_{k}(R)$
given by
$$\begin{eqnarray}{\mathcal{T}}_{k}(R,\unicode[STIX]{x1D70B})=\{X=(x_{rs})_{1\leqslant r,s\leqslant k}\in M_{k}(R)\mid x_{rs}\in (\unicode[STIX]{x1D70B})~\text{if}~r>s\}.\end{eqnarray}$$
More importantly,
$\text{GL}_{k}(R)$
shall mean the group of invertible
$k\times k$
matrices over
$R$
.
When discussing this latter group, we let
$E_{k}(R)$
be the subgroup of
$\text{GL}_{k}(R)$
generated by matrices of the form
$E(i,j;\unicode[STIX]{x1D706})=I_{k}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D716}(i,j)$
, where
$i\neq j$
and
$\unicode[STIX]{x1D706}\in R$
. A ring
$R$
is said to be weakly Euclidean if, for all
$k\geqslant 2$
,
$\text{GL}_{k}(R)=U(R)\cdot E_{k}(R)$
, where
$U(R)$
denotes the units of
$R$
. In a natural way, we confuse
$U(R)$
with the subgroup
$D_{k}(R)=\{D(1,\unicode[STIX]{x1D6FF})\mid \unicode[STIX]{x1D6FF}\in U(R)\}$
of
$\text{GL}_{k}(R)$
, where
$D(1,\unicode[STIX]{x1D6FF})=I_{k}+(\unicode[STIX]{x1D6FF}-1)\unicode[STIX]{x1D716}(1,1)$
. Since
$D_{k}(R)$
normalizes
$E_{k}(R)$
, the weakly Euclidean property is equivalent to requiring that each
$X\in \text{GL}_{k}(R)$
has Smith normal form.
A useful generalization of this shall be of fundamental use in §2. A ring
$R$
is said to be weakly
$m$
-Euclidean if
$\text{GL}_{k}(R)=\text{GL}_{m}(R)E_{k}(R)$
for
$m\leqslant k$
. Here
$\text{GL}_{m}(R)$
embeds via stabilization in
$\text{GL}_{k}(R)$
. Clearly, weakly 1-Euclidean and weakly Euclidean are synonymous.
Finally, we consider the triple of rings
${\mathcal{A}}=(A_{+},A_{0},A_{-})$
and ring homomorphisms
$\unicode[STIX]{x1D711}_{+}:A_{+}\rightarrow A_{0}$
and
$\unicode[STIX]{x1D711}_{-}:A_{-}\rightarrow A_{0}$
. Such a triple shall be called a corner. For a group valued functor
${\mathcal{G}}:\mathit{rings}\rightarrow \mathit{groups}$
we write
$\overline{{\mathcal{G}}}({\mathcal{A}})=\operatorname{Im}{\mathcal{G}}(\unicode[STIX]{x1D711}_{+})\backslash {\mathcal{G}}(A_{0})/\text{Im}\,{\mathcal{G}}(\unicode[STIX]{x1D711}_{-})$
, or simply
$\overline{{\mathcal{G}}}({\mathcal{A}})={\mathcal{G}}(A_{+})\backslash {\mathcal{G}}(A_{0})/{\mathcal{G}}(A_{-})$
when
$\unicode[STIX]{x1D711}_{+},\unicode[STIX]{x1D711}_{-}$
are understood. The functors of main importance in this paper shall be
${\mathcal{G}}=\text{GL}_{k},E_{k}$
and also
$GE_{k,m}(R)=\text{GL}_{k}(R)E_{k+m}(R)$
. For
$m\geqslant 1$
, the stabilization operator
$$\begin{eqnarray}X\mapsto \left(\begin{array}{@{}cc@{}}X & 0\\ 0 & I_{m}\end{array}\right)\end{eqnarray}$$
induces mappings
$s_{k,m}:\overline{\text{GL}}_{k}({\mathcal{A}})\rightarrow \overline{\text{GL}}_{k+m}({\mathcal{A}})$
and
$s_{k,m}:\overline{E}_{k}({\mathcal{A}})\rightarrow \overline{E}_{k+m}({\mathcal{A}})$
. These satisfy
$s_{k,l}=s_{k+m,l-m}\circ s_{k,m}$
for
$1\leqslant m<l$
. The reader is directed to [Reference Johnson6] for further details.
Now consider the following commutative diagram of rings and ring homomorphisms:
Such a square is called a fibre square when
$\unicode[STIX]{x1D70B}_{+}\times \unicode[STIX]{x1D70B}_{-}$
maps
$A$
isomorphically onto the fibre product
$$\begin{eqnarray}A_{+}\times _{\unicode[STIX]{x1D70B}}A_{-}=\{(x_{+},x_{-})\in A_{+}\times A_{-}\mid \unicode[STIX]{x1D711}_{+}(x_{+})=\unicode[STIX]{x1D711}_{-}(x_{-})\}.\end{eqnarray}$$
The mappings
$\unicode[STIX]{x1D70B}_{+},\unicode[STIX]{x1D70B}_{-}$
are then the natural projections. It is straightforward to see that this is equivalent to the following sequence being exact.
$$\begin{eqnarray}0\rightarrow A\stackrel{\big(\begin{smallmatrix}\unicode[STIX]{x1D70B}_{+}\\ \unicode[STIX]{x1D70B}_{-}\end{smallmatrix}\big)}{\longrightarrow }A_{+}\oplus A_{-}\stackrel{(\begin{smallmatrix}\unicode[STIX]{x1D711}_{+},\unicode[STIX]{x1D711}_{-}\end{smallmatrix})}{\longrightarrow }A_{0}.\end{eqnarray}$$
Of interest to us is the fibre square decomposition of
$\unicode[STIX]{x1D6EC}$
. First, write
$\mathbf{Z}[C_{p}]=\mathbf{Z}[x]/(x^{p}-1)$
. As
$x^{p}-1=(x-1)(x^{p-1}+\cdots +x+1)$
, we decompose
$\mathbf{Z}[C_{p}]$
as
in which we have identified
$\mathbf{Z}[x]/(x^{p-1}+\cdots +x+1)$
with
$\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]$
,
$\unicode[STIX]{x1D701}_{p}=\exp (2\unicode[STIX]{x1D70B}\text{i}/p)$
. The map
$\unicode[STIX]{x1D708}$
is the canonical homomorphism in which
$\unicode[STIX]{x1D701}_{p}\mapsto 1$
. As the cyclic algebraic construction is an exact functor, we get
Since
$\unicode[STIX]{x1D703}$
acts trivially on
$\mathbf{Z}$
and
$\mathbf{F}_{p}$
, we identify
${\mathcal{C}}_{q}(\mathbf{Z},\unicode[STIX]{x1D703})=\mathbf{Z}[C_{q}]$
and
${\mathcal{C}}_{q}(\mathbf{F}_{p},\unicode[STIX]{x1D703})=\mathbf{F}_{p}[C_{q}]$
. Finally, by an application of Rosen’s theorem (see [Reference Rosen9] or [Reference Remez8]),
${\mathcal{C}}_{q}(\mathbf{Z}[\unicode[STIX]{x1D701}_{p}],\unicode[STIX]{x1D703})\cong {\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})$
, where
$A$
is the fixed ring of
$\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]$
under
$\unicode[STIX]{x1D703}$
and
$\unicode[STIX]{x1D70B}$
is the unique prime over
$p$
. If we write
$\unicode[STIX]{x1D707}_{p}=\unicode[STIX]{x1D701}_{p}+\unicode[STIX]{x1D701}_{p}^{-1}$
, then we think of
$A$
as
$A=A\cap \mathbf{Z}[\unicode[STIX]{x1D701}_{p}]=\mathbf{Z}[\unicode[STIX]{x1D707}_{p}]$
, and therefore as the ring of integers in
$\mathbf{Q}(\unicode[STIX]{x1D707}_{p})$
. Combining the above, we apply the exact functor
$-\bigotimes _{\mathbf{Z}}\mathbf{Z}[C_{\infty }^{n}]$
to yield the fibre square decomposition for
$\unicode[STIX]{x1D6EC}$
:
The remainder of this section will concern the application of fibre squares to the discussion of stably free modules. As such, the following shall be related to the general fibre square
$(\dagger )$
. Start by taking a module
$M$
over
$A$
. This gives rise to a triple
$(M\otimes _{\unicode[STIX]{x1D70B}_{+}}A_{+},M\otimes _{\unicode[STIX]{x1D70B}_{-}}A_{-};\natural )$
in which
$\natural$
is the canonical isomorphism
$$\begin{eqnarray}\natural :(M\otimes _{\unicode[STIX]{x1D70B}_{-}}A_{-})\otimes _{\unicode[STIX]{x1D711}_{-}}A_{0}\stackrel{\simeq }{\rightarrow }(M\otimes _{\unicode[STIX]{x1D70B}_{+}}A_{+})\otimes _{\unicode[STIX]{x1D711}_{+}}A_{0}.\end{eqnarray}$$
By abuse of notation, we denote this triple simply by
$M$
. Now we set
$M_{\unicode[STIX]{x1D70E}}=M\otimes _{\unicode[STIX]{x1D70B}_{\unicode[STIX]{x1D70E}}}A_{\unicode[STIX]{x1D70E}}$
for
$\unicode[STIX]{x1D70E}=+,-$
, and we say that
$(M_{+},M_{-})$
is the local type of
$M$
. Notably, when
$M$
gives rise to
$(A_{+}^{k},A_{-}^{k};\unicode[STIX]{x1D6FC})$
for some
$\unicode[STIX]{x1D6FC}\in \text{GL}_{k}(A_{0})$
, we say that
$M$
is locally free. More generally,
$M$
is said to be locally projective if the local type of
$M$
is
$(M_{+},M_{-})$
in which each
$M_{\unicode[STIX]{x1D70E}}$
is projective over
$A_{\unicode[STIX]{x1D70E}}$
. Related to this is global projectivity. If
$P$
is an
$A$
-module giving rise to
$(P_{+},P_{-};\unicode[STIX]{x1D6FC})$
, then
$P$
is said to be globally projective if there is a triple
$Q=(Q_{+},Q_{-};\unicode[STIX]{x1D6FD})$
such that
$P\oplus Q=(P_{+}\oplus Q_{+},P_{-}\oplus Q_{-};\unicode[STIX]{x1D6FC}\oplus \unicode[STIX]{x1D6FD})\cong (A_{+}^{k},A_{-}^{k};\text{Id}_{k})$
for some
$k\geqslant 1$
. Evidently, a globally projective module is locally projective, but the converse need not be true in general. However, if we suppose at least one of
$\unicode[STIX]{x1D711}_{+}$
or
$\unicode[STIX]{x1D711}_{-}$
is surjective, then these two concepts are equivalent (see [Reference Johnson6, Theorem 3.8]). In this case, we say that the fibre square satisfies Milnor’s condition. Additionally, we have the following result.
Proposition 1.1. If
$\dagger$
is a Milnor square, then
$\overline{E}_{k}(A)=E_{k}(A_{+})\backslash E_{k}(A_{0})/E_{k}(A_{-})$
consists of a single point for all
$k\geqslant 2$
.
Corollary 1.2. If
$\dagger$
is a Milnor square, then the mapping
$$\begin{eqnarray}s_{k,m}^{\prime }:\overline{\text{GL}}_{k}(A)\rightarrow \overline{GE}_{k,m}(A)\end{eqnarray}$$
is bijective for all
$m\geqslant 1$
.
Now we define
${\mathcal{L}}{\mathcal{F}}_{k}(A)$
to be the set of isomorphism classes of
$A$
-modules giving rise to a triple
$(A_{+}^{k},A_{-}^{k};\unicode[STIX]{x1D6FC})$
in which
$\unicode[STIX]{x1D6FC}:A_{-}^{k}\otimes A_{0}\rightarrow A_{+}^{k}\otimes A_{0}$
is an
$A_{0}$
isomorphism. By identifying
$A_{+}\otimes A_{0}=A_{0}=A_{-}\otimes A_{0}$
,
$\operatorname{Iso}(A_{-}^{k}\otimes A_{0},A_{+}^{k}\otimes A_{0})=\text{GL}_{k}(A_{0})$
. This gives rise to a bijective correspondence
$$\begin{eqnarray}\unicode[STIX]{x1D708}_{k}:\overline{\text{GL}}_{k}(A)=\text{GL}_{k}(A_{+})\backslash \text{GL}_{k}(A_{0})/\text{GL}_{k}(A_{-})\stackrel{\simeq }{\rightarrow }{\mathcal{L}}{\mathcal{F}}_{k}(A).\end{eqnarray}$$
Let
$\unicode[STIX]{x1D70E}_{k,m}:{\mathcal{L}}{\mathcal{F}}_{k}(A)\rightarrow {\mathcal{L}}{\mathcal{F}}_{k+m}(A)$
denote the stabilization operators induced from the correspondence
$P\mapsto P\oplus (A_{+}^{k},A_{-}^{k};I_{k})$
. We then have the commutative diagram
We write
$\ast$
for the class of
$A^{k+m}$
in
${\mathcal{L}}{\mathcal{F}}_{k+m}(A)$
and observe that
$Z\in {\mathcal{L}}{\mathcal{F}}_{k}(A)$
is stably free if and only if
$\unicode[STIX]{x1D70E}_{k,m}([Z])=\ast$
for some
$k\geqslant 0$
. Consequently, we define the sets
$$\begin{eqnarray}Z_{k}(A)=\{\unicode[STIX]{x1D701}\in \overline{\text{GL}}_{k}(A)\mid s_{k,m}(\unicode[STIX]{x1D701})=\ast ~\text{for some}~k\geqslant 1\},\end{eqnarray}$$
where
$Z(A)=\bigsqcup _{k\geqslant 1}Z_{k}(A)$
. Finally, by
${\mathcal{S}}{\mathcal{F}}_{k}(A)$
we mean the set of isomorphism classes of stably free modules of rank
$k$
over
$A$
. We now say that
$A$
is locally
$k$
-free when
${\mathcal{S}}{\mathcal{F}}_{k}(A_{+})={\mathcal{S}}{\mathcal{F}}_{k}(A_{-})=\{\ast \}$
and observe the bijection
$$\begin{eqnarray}\unicode[STIX]{x1D708}_{k}:Z_{k}(A)\stackrel{\simeq }{\rightarrow }{\mathcal{S}}{\mathcal{F}}_{k}(A)\cap {\mathcal{L}}{\mathcal{F}}_{k}(A).\end{eqnarray}$$
If
$A$
is locally
$k$
-free, then every stably free
$A$
-module of rank
$k$
is locally free over
$A$
. As such,
${\mathcal{S}}{\mathcal{F}}_{k}(A)={\mathcal{S}}{\mathcal{F}}_{k}(A)\cap {\mathcal{L}}{\mathcal{F}}_{k}(A)$
and the above bijection becomes
$$\begin{eqnarray}\unicode[STIX]{x1D708}_{k}:Z_{k}(A)\stackrel{\simeq }{\rightarrow }{\mathcal{S}}{\mathcal{F}}_{k}(A)~\text{is bijective when}~A~\text{is locally}~k\text{-free}.\end{eqnarray}$$
To understand when stably free modules of rank
$k$
, say, are free, we now have to consider the set
$Z_{k}(A)$
. In particular, if it is trivial, then we know that every stably free module of rank
$k$
is free. The most obvious way of ensuring that
$Z_{k}(A)$
is trivial is to show that each
$s_{k,m}$
is injective. To summarize, we have the following method to check stably free modules of a particular rank.
Proposition 1.3. Suppose that
$\dagger$
is a fibre square decomposition for a module
$A$
such that
$A$
is locally
$k$
-free. If each
$s_{k,m}:\overline{\text{GL}}_{k}(A)\rightarrow \overline{\text{GL}}_{k+m}(A)$
is injective, then
${\mathcal{S}}{\mathcal{F}}_{k}(A)=\{\ast \}$
.
2 Stably free modules of rank greater than one are free
Return once more to the fibre square decomposition
$\heartsuit$
for
$\unicode[STIX]{x1D6EC}$
, derived in §1. In particular, note that this square satisfies Milnor’s condition.
Proposition 2.1.
$\unicode[STIX]{x1D6EC}$
is locally free: that is,
$\unicode[STIX]{x1D6EC}$
is locally
$k$
-free for all
$k\geqslant 1$
.
Proof. As
$C_{q}\times C_{\infty }^{n}$
is commutative, we know that
$\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
has SFC, by [Reference Johnson7]. It remains to show that
${\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]$
has SFC. For this, we intend to use [Reference Johnson6, Corollary 3.48]:
Let
$\dagger$
be a Milnor square of locally free type. Suppose, further, that
$A_{0}$
is weakly Euclidean and that
$U(A_{+})\backslash U(A_{0})/U(A_{-})$
is a singleton. Then
$A$
has SFC.
Consider the following fibre square decomposition that clearly satisfies Milnor’s condition:
As already noted,
$A$
is the ring of integers in
$\mathbf{Q}(\unicode[STIX]{x1D707}_{p})$
and is therefore a Dedekind domain. It follows that
$A[C_{\infty }^{n}]$
has SFC and therefore so does
$M_{q}(A[C_{\infty }^{n}])$
, by [Reference Johnson6, (9.9) and (9.20)]. Now as
$\unicode[STIX]{x1D70B}$
is the unique prime in
$A$
over
$p$
and as
$A$
is a Dedekind domain, it follows that
$A/\unicode[STIX]{x1D70B}$
is a field. Consequently, we observe that
$M_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
is weakly Euclidean by a consequence of Suslin’s stability theorem (see [Reference Suslin10]).
For the bottom left-hand corner, consider the obvious surjection
$$\begin{eqnarray}\unicode[STIX]{x1D713}:{\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}]){\twoheadrightarrow}\underbrace{(A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}]\times \cdots \times (A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}]}_{q}\end{eqnarray}$$
given by
$\unicode[STIX]{x1D713}(X)=(X_{11},\ldots ,X_{qq})$
. In particular, observe that each
$(A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}]$
has SFC (again by [Reference Johnson6, (9.9)]) and hence so does any finite product of these. To deduce that
${\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
has SFC, we seek to apply a proposition of Bourbaki–Nakayama [Reference Johnson6, Proposition 9.16]:
Let
$\mathfrak{m}$
be a two-sided radical ideal in a ring
$R$
. If
$R/\mathfrak{m}$
has SFC, then so does
$R$
.
Now if
$X\in \operatorname{Ker}(\unicode[STIX]{x1D713})$
, then
$X$
is strictly upper triangular. Evidently, for any
$T\in {\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
,
$(TX)^{m}=0$
for some
$m$
. Thus
$I_{q}+TX$
is a unit and so
$X\in \operatorname{rad}({\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}]))$
. Hence
$\operatorname{Ker}(\unicode[STIX]{x1D713})$
is radical and one can apply Bourbaki–Nakayama’s proposition alongside Noether’s isomorphism theorem to show that
${\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
has SFC.
Finally, consider the units of
$\ddagger$
. As
$M_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
is weakly Euclidean, any unit
$X\in \text{GL}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
may be written as
$X=\unicode[STIX]{x1D6E5}E$
. Here
$\unicode[STIX]{x1D6E5}=D(1,\unicode[STIX]{x1D6FF})$
for some
$\unicode[STIX]{x1D6FF}\in U((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
and
$E\in E_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])\subset \text{GL}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
. Clearly,
$\unicode[STIX]{x1D6E5}\in {\mathcal{T}}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
and, since
$\natural$
is surjective, it follows that there exists an
$E^{\prime }\in E_{q}(A[C_{\infty }^{n}])$
such that
$\natural (E^{\prime })=E$
. We therefore conclude, by [Reference Johnson6, Corollary 3.48], that
${\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]$
has SFC.◻
Proposition 2.2. The rings
$\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
and
$\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]$
are weakly 2-Euclidean.
Proof. The proof hinges on the commutativity of
$\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
and
$\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]$
. Consequently, denote both by
$R$
. Now, as
$\text{GL}_{2}(R)$
normalizes
$E_{k}(R)$
, we have the subgroup
$$\begin{eqnarray}\text{GL}_{2}(R)\cdot E_{k}(R)\subset \text{GL}_{k}(R).\end{eqnarray}$$
Utilizing Suslin’s stability theorem once more,
$\text{GL}_{k}(R)=U(R)\,\cdot \,E_{k}(R)$
whenever
$k\geqslant 3$
. By writing
$U(R)=\text{GL}_{1}(R)$
, which embeds into
$\text{GL}_{2}(R)$
,
$$\begin{eqnarray}\text{GL}_{k}(R)\subset \text{GL}_{2}(R)\cdot E_{k}(R).\end{eqnarray}$$
Thus
$\text{GL}_{k}(R)=\text{GL}_{2}(R)E_{k}(R)$
, as required.◻
Proposition 2.3. The ring
${\mathcal{T}}={\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]$
is weakly 2-Euclidean.
Proof. First, consider the induced map
$\natural :M_{q}(A[C_{\infty }^{n}])\rightarrow M_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
in which it follows that
${\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]=\natural ^{-1}({\mathcal{T}}_{q}(A)[C_{\infty }^{n}])$
. Likewise, from the induced map on unit groups
$\natural :\text{GL}_{q}(A[C_{\infty }^{n}])\rightarrow \text{GL}_{q}((A/\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
, we see that
$$\begin{eqnarray}U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}])=\natural ^{-1}(U({\mathcal{T}}_{q}(A)[C_{\infty }^{n}])).\end{eqnarray}$$
Now [Reference Johnson6, Proposition 2.43] states that if
$\unicode[STIX]{x1D711}:X\rightarrow Y$
is a surjective ring homomorphism with the strong lifting property for units, then
$X$
is weakly Euclidean whenever
$Y$
is. This proof may be suitably modified to show that the same holds for weakly 2-Euclidean rings. We apply this latter result to the induced map
$\natural :{\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]\rightarrow {\mathcal{T}}_{q}(A[C_{\infty }^{n}])$
. First, note the map
$$\begin{eqnarray}{\mathcal{T}}_{q}(A[C_{\infty }^{n}])\rightarrow \underbrace{A[C_{\infty }^{n}]\times \cdots \times A[C_{\infty }^{n}]}_{q},\end{eqnarray}$$
where
$T\mapsto (T_{11},\ldots ,T_{qq})$
. Clearly, this map has the strong lifting property for units. Furthermore, since each
$A[C_{\infty }^{n}]$
is weakly 2-Euclidean, then so too is any finite product. It follows from the above remark that
${\mathcal{T}}_{q}(A[C_{\infty }^{n}])$
is weakly 2-Euclidean.
To conclude the proof, we show that
$\natural$
has the strong lifting property for units. Clearly, the induced map on units
$U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}])\rightarrow U({\mathcal{T}}_{q}(A)[C_{\infty }^{n}])$
is surjective, and so
$\natural$
has the lifting property for units. Now suppose that
$\natural (T)\in U({\mathcal{T}}_{q}(A)[C_{\infty }^{n}])$
for some
$T\in {\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}]$
. By applying
$\natural ^{-1}$
, one gets
$T=\natural ^{-1}(\natural (T))\in U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[C_{\infty }^{n}])$
and so
$T$
is necessarily a unit. Thus
$\natural$
has the strong lifting property for units, as required.◻
Proposition 2.4. Stably free modules of rank
${\geqslant}2$
over
$\unicode[STIX]{x1D6EC}$
are necessarily free.
Proof. For ease of notation, we write
${\mathcal{A}}=\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
and
${\mathcal{B}}=\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]$
. We also let
$GE_{k,m}(-)$
denote the subgroup
$GE_{k,m}(-)=\text{GL}_{k}(-)E_{k+m}(-)$
. Consider the commutative diagram
where
$\natural _{1},\natural _{2}$
are the canonical mappings and
$s_{k,m},s_{k,m}^{\prime }$
are the mappings induced by stabilization. By Corollary 1.2,
$s_{k,m}^{\prime }$
is bijective. Now, by Propositions 2.2 and 2.3:
-
∙
$GE_{k,m}({\mathcal{A}})=\text{GL}_{2}(A)E_{k+m}({\mathcal{A}})=\text{GL}_{k+m}({\mathcal{A}})$
; -
∙
$GE_{k,m}({\mathcal{B}})=\text{GL}_{2}({\mathcal{B}})E_{k+m}({\mathcal{B}})=\text{GL}_{k+m}({\mathcal{B}})$
; and -
∙
$GE_{k,m}({\mathcal{T}})=\text{GL}_{2}({\mathcal{T}})E_{k+m}({\mathcal{T}})=\text{GL}_{k+m}({\mathcal{T}})$
.
Therefore, it follows that
$\natural _{1},\natural _{2}$
are the identity mappings. In particular, this implies that
$s_{k,m}$
is injective. By Proposition 1.3,
$\unicode[STIX]{x1D6EC}$
has no non-trivial stably free module of rank
${\geqslant}2$
.◻
3 Stably free modules of rank one are free
Proposition 2.4 leaves us with the possibility of nontrivial stably free modules of rank one. In order to demonstrate that these are free, we need a result due to Swan and Jacobinski (see [Reference Curtis and Reiner3], vol. 2 (51.28), pp. 324). By a
$\mathbf{Z}[G]$
-lattice
$M$
, we mean a
$\mathbf{Z}[G]$
-module whose underlying abelian group is finitely generated and torsion free. Regard
$\mathbf{Z}[G]$
as embedded in
$\mathbf{Z}[G]\otimes \mathbf{R}$
, which is semisimple over
$\mathbf{R}$
. This has the Wedderburn decomposition
$\mathbf{R}[G]\cong \unicode[STIX]{x1D6EC}_{1}\oplus \cdots \oplus \unicode[STIX]{x1D6EC}_{r}$
. If
$V_{i}$
is a simple right ideal in
$\unicode[STIX]{x1D6EC}_{i}$
, then any simple module over
$\mathbf{R}[G]$
is isomorphic to
$V_{i}$
for some unique index
$i$
. By setting
$D_{i}=\operatorname{End}_{\mathbf{R}[G]}(V_{i})$
, we observe that
$D_{i}$
is a finite-dimensional division algebra over
$\mathbf{R}$
and hence is isomorphic to either
$\mathbf{C},\mathbf{R}$
or
$\mathbf{H}=((-1,-1)/\mathbf{R})$
.
For a
$\mathbf{Z}[G]$
-lattice
$M$
, note the following isotypic decomposition
$M\otimes \mathbf{R}\cong \bigoplus _{i=1}^{r}V_{i}^{(f_{i})}$
, in which we apply the convention that
$V^{(0)}=0$
. Observe that the ring
$\operatorname{End}_{\mathbf{R}[G]}(M\otimes \mathbf{R})$
is also a semisimple algebra and hence decomposes as a direct sum of two-sided ideals
$$\begin{eqnarray}\operatorname{End}_{\mathbf{R}[G]}(M\otimes \mathbf{R})\cong \bigoplus _{i=1}^{r}M_{f_{i}}(D_{i}).\end{eqnarray}$$
We say that
$M$
is an Eichler lattice when no simple factor of
$\operatorname{End}_{\mathbf{R}[G]}(M\otimes \mathbf{R})$
is isomorphic to
$\mathbf{H}$
. We now have the following special case of the Swan–Jacobinski theorem.
Theorem 3.1 (Swan–Jacobinski).
Suppose that
$M$
is an Eichler lattice over
$\unicode[STIX]{x1D6EC}$
with the property that
$M\cong M_{0}\oplus \unicode[STIX]{x1D6EC}$
for some module
$M_{0}$
. Then
$M$
has the cancellation property: that is, if
$N$
is a
$\unicode[STIX]{x1D6EC}$
-lattice such that
$rk_{\mathbf{Z}}(M)\leqslant rk_{\mathbf{Z}}(N)$
and
$N\oplus \unicode[STIX]{x1D6EC}^{a}\cong M\oplus \unicode[STIX]{x1D6EC}^{b}$
, then
$N\cong M\oplus \unicode[STIX]{x1D6EC}^{b-a}$
.
In particular,
$\unicode[STIX]{x1D6EC}$
is said to satisfy the Eichler condition when, considered as a
$\unicode[STIX]{x1D6EC}$
-module,
$\unicode[STIX]{x1D6EC}$
is an Eichler lattice. The reader is directed to [Reference Johnson5, Ch. 3] for more details.
Proposition 3.2. If
$\unicode[STIX]{x1D6EC}$
satisfies the Eichler condition, then every
$\unicode[STIX]{x1D6EC}$
-lattice
$M$
is an Eichler lattice.
With the above discussion in mind, observe that
$\mathbf{Z}[C_{p}\rtimes C_{q}]$
is a retract of
$\unicode[STIX]{x1D6EC}$
: that is, there are ring homomorphisms
$i:\mathbf{Z}[C_{p}\rtimes C_{q}]\rightarrow \unicode[STIX]{x1D6EC}$
and
$r:\unicode[STIX]{x1D6EC}\rightarrow \mathbf{Z}[C_{p}\rtimes C_{q}]$
such that
$r\circ i=1$
. As such, we have the collection of mappings
where
$\clubsuit ,\heartsuit$
are the fibre squares defined in §1 and
$i_{k},r_{k}$
are the maps induced from
$i,r$
, respectively. Since
$r\circ i=1$
, it follows that each
$r_{k}$
is surjective. Moreover, by Swan–Jacobinski,
${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )\cong {\mathcal{L}}{\mathcal{F}}_{k+1}(\clubsuit )$
for each
$k\geqslant 1$
, since
$\mathbf{Z}[C_{p}\rtimes C_{q}]$
satisfies Eichler’s condition. In particular,
$$\begin{eqnarray}|{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|=|U(\mathbf{Z}[C_{q}])\backslash U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B}))|=N\end{eqnarray}$$
for some finite
$N$
. Now compare this with the size of
${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$
. To do so, we say that a unit
$\unicode[STIX]{x1D706}\in U(R[G])$
of a group ring
$R[G]$
is trivial if
$\unicode[STIX]{x1D706}=ug$
for
$u\in U(R)$
and
$g\in G$
.
Proposition 3.3.
$U(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\cong U(\mathbf{Z}[C_{q}])\times C_{\infty }^{n}$
.
Proof. Return to the fibre square decomposition
$\diamondsuit$
of
$\mathbf{Z}[C_{q}]$
(simply alter
$p$
to
$q$
) and let
$u\in U(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])$
. Then
$\unicode[STIX]{x1D70B}_{+}(u)\in U(\mathbf{Z}[C_{\infty }^{n}])$
. Since
$\mathbf{Z}$
is an integral domain, we know that the units of
$\mathbf{Z}[C_{\infty }^{n}]$
are trivial, by [Reference Johnson6, Theorem 2.51]: that is,
$U(\mathbf{Z}[C_{\infty }^{n}])\cong U(\mathbf{Z})\times C_{\infty }^{n}$
. Thus
$$\begin{eqnarray}u=\unicode[STIX]{x1D706}_{g}g+\mathop{\sum }_{h\in C_{\infty }^{n}\backslash \{g\}}\unicode[STIX]{x1D706}_{h}h,\end{eqnarray}$$
where
$\unicode[STIX]{x1D706}_{g}(1)=\pm 1,g\in C_{\infty }^{n}$
and
$\unicode[STIX]{x1D706}_{h}(1)=0$
for all
$h\in C_{\infty }^{n}\backslash \{g\}$
. Thus
$(x-1)|\unicode[STIX]{x1D706}_{h}$
for all
$h\in C_{\infty }^{n}\backslash \{g\}$
.
Now
$\unicode[STIX]{x1D70B}_{-}(u)\in U((\mathbf{Z}[\unicode[STIX]{x1D701}_{q}])[C_{\infty }^{n}])$
, where again we only have trivial units since
$\mathbf{Z}[\unicode[STIX]{x1D701}_{q}]$
is an integral domain. Thus
$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{+}(u)=\unicode[STIX]{x1D706}_{g}(\unicode[STIX]{x1D701}_{q})+\mathop{\sum }_{h\in C_{\infty }^{n}\backslash \{g\}}\unicode[STIX]{x1D706}_{h}(\unicode[STIX]{x1D701}_{q})h=\unicode[STIX]{x1D706}_{h^{\prime }}h^{\prime }\end{eqnarray}$$
for some
$h^{\prime }\in C_{\infty }^{n}$
. Suppose that
$\unicode[STIX]{x1D706}_{g}(\unicode[STIX]{x1D701}_{q})=0$
. Then
$(1+x+\cdots +x^{q-1})|\unicode[STIX]{x1D706}_{g}(x)$
. Therefore
$\unicode[STIX]{x1D706}_{g}(x)=(1+x+\cdots +x^{q-1})\unicode[STIX]{x1D707}_{g}(x)$
for some
$\unicode[STIX]{x1D707}_{g}\in \mathbf{Z}[x]$
. Since
$\unicode[STIX]{x1D706}_{g}(1)=\pm 1$
,
$\unicode[STIX]{x1D707}_{g}(1)=\pm 1/q$
. Clearly, this is a contradiction and we conclude that
$\unicode[STIX]{x1D706}_{g}(\unicode[STIX]{x1D701}_{q})\neq 0$
. Consequently,
$(1+x+\cdots +x^{q-1})|\unicode[STIX]{x1D706}_{h}$
for all
$h\in C_{\infty }^{n}$
. Recall that
$(x-1)$
also divides
$\unicode[STIX]{x1D706}_{h}$
for all such
$h$
. Hence
$\unicode[STIX]{x1D706}_{h}=0$
for all
$h\in C_{\infty }^{n}\backslash \{g\}$
and
$u=\unicode[STIX]{x1D706}_{g}g$
, as required.◻
Corollary 3.4.
$U(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])\cong U(\mathbf{F}_{p}[C_{q}])\times C_{\infty }^{n}$
.
Proof. Suppose that
$u\in U(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])$
and let
$\unicode[STIX]{x1D6FC}:\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]{\hookrightarrow}\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
be the obvious injection. By the previous proposition,
$\unicode[STIX]{x1D6FC}(u)=\unicode[STIX]{x1D706}_{g}g$
for some
$\unicode[STIX]{x1D706}_{g}\in \mathbf{Z}[C_{q}]$
and
$g\in C_{\infty }^{n}$
. Moreover, there is a surjection
$\unicode[STIX]{x1D6FD}:\mathbf{Z}[C_{q}\times C_{\infty }^{n}]\rightarrow \mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]$
such that
$\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}=1$
. The result now follows.◻
In particular, observe that
$\mathbf{F}_{p}[C_{q}]$
is semisimple. Indeed, it is a product of
$q$
copies of the finite field
$\mathbf{F}_{p}$
. From this it may be shown that
$|{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|=1$
. However, for our purposes, we need only use the fact that
$|{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|$
is finite.
Proposition 3.5. For
$k\geqslant 3$
,
$|{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )|$
. In particular,
${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$
is finite.
Proof. Set
$k\geqslant 3$
and write
$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )=\text{GL}_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\backslash \text{GL}_{k}(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])/\text{GL}_{k}({\mathcal{T}}).\end{eqnarray}$$
By Suslin’s stability theorem, we write
$$\begin{eqnarray}\text{GL}_{k}(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])=U(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])E_{k}(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]).\end{eqnarray}$$
Since
$\mathbf{Z}[C_{q}\times C_{\infty }^{n}]$
is abelian, we write
$$\begin{eqnarray}\text{GL}_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])=U(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\text{SL}_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\end{eqnarray}$$
and
$$\begin{eqnarray}\text{SL}_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])=E_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}]),\end{eqnarray}$$
by Suslin also. As there is a surjection
$\mathbf{Z}[C_{q}\times C_{\infty }^{n}]\rightarrow \mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}]$
, we induce a surjection
$E_{k}(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\rightarrow E_{k}(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])$
. Thus
${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$
becomes
$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )=U(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\backslash U(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])/\text{GL}_{k}({\mathcal{T}}).\end{eqnarray}$$
Now it is clear that
$U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B}))\subset U({\mathcal{T}})\subset \text{GL}_{k}({\mathcal{T}})$
, where the last inclusion is a consequence of the standard stabilization operator. Therefore, by Proposition 3.3 and Corollary 3.4,
$$\begin{eqnarray}|{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )|\leqslant |U({\mathcal{Z}}[C_{q}])\backslash U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B}))|=|{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|.\end{eqnarray}$$
The result follows since
$|{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )|=|{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|$
, by Swan–Jacobinski.◻
Corollary 3.6. For any
$k\geqslant 3$
,
${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )\cong {\mathcal{L}}{\mathcal{F}}_{k+1}(\heartsuit )$
.
Proof. Recall that
$r_{k}:{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$
is a surjective mapping. By Proposition 3.5, we know that
$|{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )|$
for
$k\geqslant 3$
. Thus, for any
$k\geqslant 3$
,
$r_{k}$
is an isomorphism. The result now follows since
${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )\cong {\mathcal{L}}{\mathcal{F}}_{k+1}(\clubsuit )$
, by Swan–Jacobinski.◻
Corollary 3.7. Any stably free module of rank one over
$\mathbf{Z}[(C_{\!p}\rtimes C_{\!q})\times C_{\!\infty }^{n}]$
is necessarily free.
Proof. Recall that
${\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )=U(\mathbf{Z}[C_{q}\times C_{\infty }^{n}])\backslash U(\mathbf{F}_{p}[C_{q}\times C_{\infty }^{n}])/U({\mathcal{T}})$
and that
$U({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B}))\subset U({\mathcal{T}})$
. Applying this along with Proposition 3.3 and Corollary 3.4, we deduce that
$|{\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|$
. From the surjection
$r_{1}:{\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )$
, we therefore deduce that
$r_{1}$
is an isomorphism. In particular,
${\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )\cong {\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$
for any
$k\geqslant 3$
. It follows that
$s_{1,k}$
is therefore an isomorphism for any
$k\geqslant 2$
. However, we may also write
$s_{1,k}=s_{2,k-1}\circ s_{1,1}$
. Thus
$s_{1,1}$
is injective and we conclude that
$s_{1,k}$
is injective for any
$k\geqslant 1$
.◻
By combining the results of Proposition 2.4 and Corollary 3.7, we therefore obtain the desired conclusion.
Theorem A.
There are no non-trivial stably free modules over
$\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{n}]$
.
Acknowledgement
The author would like to thank Professor F. E. A. Johnson for the numerous stimulating conversations on the problem of stably free modules over group rings. His knowledge and guidance has been invaluable.
