STABLY FREE MODULES OVER $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ ARE FREE

In the paper [1] I claimed to show that there are no non-trivial stably free modules over integral group rings of the groups $\,(C_{p}\rtimes C_{q})\times C_{\infty }^{m}$ . Unfortunately there are a number of erroneous statements in [1] which vitiate the attempted proof. To explain where these occur, recall that in [1] two Milnor fibre squares $\,(\clubsuit )\,$ and $\,(\heartsuit )\,$ were introduced as follows:

Here $\unicode[STIX]{x1D6E4}=C_{\infty }^{m}$ , and ${\mathcal{T}}_{q}={\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})$ is the ring of quasi-triangular $q\times q$ matrices where $A=\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]^{C_{q}}$ is the subring of the cyclotomic integers $\mathbf{Z}[\unicode[STIX]{x1D701}_{p}]$ fixed under the Galois action of $C_{q}$ and $\unicode[STIX]{x1D70B}\in \text{Spec}(A)$ is the unique prime over $p$ .

The most obvious errors [1, Corollary 3.4] include a misdescription of the unit group $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])$ , and the possibility of non-trivial rank-one stably free modules over ${\mathcal{T}}_{2}[\unicode[STIX]{x1D6E4}]$ . A slightly less obvious but more significant error concerns the possibility of lifting units from $\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]$ to ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ . In consequence we must amend the original statement of [1] as follows.

Theorem A. Let $S$ be a stably free module of rank $n$ over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ where $m\geqslant 2$ . Then:

• ∙ if $q$ is an odd prime, $S$ is free provided $n\neq 2$ ; and

• ∙ if $q=2$ , $S$ is free provided $n\geqslant 3$ .

Nevertheless, when $m=1$ the original statement continues to hold:

Theorem B. Any stably free module over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }]$ is free.

Rather than try to patch up the proof in [1] piecemeal we give a more straightforward approach which isolates the real difficulty and avoids it where possible. We first establish four propositions.

Proposition 1. $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ is finite; in fact

$$\begin{eqnarray}|U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])|\leqslant |U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})|.\end{eqnarray}$$

Proof. As $q$ is a divisor of $p-1$ , we have

$$\begin{eqnarray}\mathbf{F}_{p}[C_{q}]\;\cong \;\underbrace{\mathbf{F}_{p}\times \cdots \times \mathbf{F}_{p}}_{q}.\end{eqnarray}$$

Consequently

(*) $$\begin{eqnarray}\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]\cong \underbrace{\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}]\times \cdots \times \mathbf{F}_{p}}_{q}[\unicode[STIX]{x1D6E4}].\end{eqnarray}$$

Observe that $U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ contains a copy of $\unicode[STIX]{x1D6E4}^{(q)}=\underbrace{\unicode[STIX]{x1D6E4}\times \cdots \times \unicode[STIX]{x1D6E4}}_{q}$ , namely the diagonal matrices

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FE}_{1},\ldots ,\unicode[STIX]{x1D6FE}_{q})=\left(\begin{smallmatrix}\unicode[STIX]{x1D6FE}_{1}\\ & \unicode[STIX]{x1D6FE}_{2} & \\ & & \ddots \\ & & & \unicode[STIX]{x1D6FE}_{q}\end{smallmatrix}\right),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FE}_{i}\in \unicode[STIX]{x1D6E4}$ . Combining this with the obvious inclusion $U({\mathcal{T}}_{q})\subset U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ gives an injection $U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}{\hookrightarrow}U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . Hence we now have a surjection

(**) $$\begin{eqnarray}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}{\twoheadrightarrow}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

The ring isomorphism (*) now gives an isomorphism of unit groups

$$\begin{eqnarray}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])\cong \underbrace{U(\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}])\times \cdots \times U(\mathbf{F}_{p}}_{q}[\unicode[STIX]{x1D6E4}]\,).\end{eqnarray}$$

Now $\unicode[STIX]{x1D6E4}=C_{\infty }^{m}$ is a t.u.p. group so $\mathbf{F}_{p}[\unicode[STIX]{x1D6E4}]$ has only trivial units (cf. [2, Appendix C]). Hence

$$\begin{eqnarray}\displaystyle U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]) & \cong & \displaystyle \underbrace{(U(\mathbf{F}_{p})\times \unicode[STIX]{x1D6E4})\times \cdots \times (U(\mathbf{F}_{p})\times \unicode[STIX]{x1D6E4})}_{q}\nonumber\\ \displaystyle & \cong & \displaystyle \underbrace{U(\mathbf{F}_{p})\times \cdots \times U(\mathbf{F}_{p})}_{q}\times \unicode[STIX]{x1D6E4}^{(q)}\nonumber\end{eqnarray}$$

so that, by (*), there are bijections

$$\begin{eqnarray}\displaystyle U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)} & \leftrightarrow & \displaystyle U(\mathbf{F}_{p}[C_{q}])\times \unicode[STIX]{x1D6E4}^{(q)}/U({\mathcal{T}}_{q})\times \unicode[STIX]{x1D6E4}^{(q)}\nonumber\\ \displaystyle & \leftrightarrow & \displaystyle U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}).\nonumber\end{eqnarray}$$

From (**) we obtain a surjection $U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q}){\twoheadrightarrow}U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U$ $({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . The stated result now follows as $U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})$ is finite.◻

Proposition 2. Let $p$ be an odd prime and $q$ be a divisor of $p-1$ . Then, for all $n\geqslant 3$ ,

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])=U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])\cdot E_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

Proof. Given rings $A,B$ such that $\text{GL}_{n}(A)=U(A)E_{n}(A)$ and $\text{GL}_{n}(B)=U(B)E_{n}(B)$ , we have $\text{GL}_{n}(A\times B)=U(A\times B)E_{n}(A\times B)$ . The result thus follows from (*) by induction on $q$ , the case $q=1$ being Suslin’s theorem [3], namely that

$$\begin{eqnarray}\text{GL}_{k}(\mathbf{F}[\unicode[STIX]{x1D6E4}])=U(\mathbf{F}[\unicode[STIX]{x1D6E4}])\cdot E_{k}(\mathbf{F}[\unicode[STIX]{x1D6E4}])\end{eqnarray}$$

for any field $\mathbf{F}$ and any integer $k\geqslant 3$ .◻

Proposition 3. $\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ is finite for $n\geqslant 3$ ; in fact

$$\begin{eqnarray}|\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])|\leqslant |U(\mathbf{F}_{p}[C_{q}])/U({\mathcal{T}}_{q})|.\end{eqnarray}$$

Proof. Evidently $U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])E_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])\subset \text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ and so there is a natural surjection $\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])E_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]){\twoheadrightarrow}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ . Also, the ring homomorphism $\natural :{\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]\rightarrow \mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]$ is surjective and so induces surjections $\natural _{\ast }:E_{k}({\mathcal{T}}_{q}(A,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])\rightarrow E_{k}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])$ for all $k\geqslant 2$ . By Proposition 2 we may write

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])=U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])E_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

We obtain a surjection $U(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/U({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]){\twoheadrightarrow}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{n}$ $({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}])$ and so the stated result now follows from Proposition 1.◻

Proposition 4. Let $p$ be an odd prime and $q$ be a divisor of $p-1$ . Then, for all $n\geqslant 1$ ,

$$\begin{eqnarray}\text{GL}_{n}(\mathbf{F}_{p}[C_{q}\times C_{\infty }])=U(\mathbf{F}_{p}[C_{q}\times C_{\infty }])\cdot E_{n}(\mathbf{F}_{p}[C_{q}\times C_{\infty }]).\end{eqnarray}$$

Proof. We follow the same line of argument as Proposition 2 with the exception that, in establishing the induction base, we do not use Suslin’s theorem. Instead we note that, as $\mathbf{F}_{p}[C_{\infty }]$ is a Euclidean domain, we may use the Smith normal form to show that $\text{GL}_{k}(\mathbf{F}_{p}[C_{\infty }])=U(\mathbf{F}_{p}[C_{\infty }])\cdot$ $E_{k}(\mathbf{F}_{p}[C_{\infty }])$ .◻

As in [1], we denote the set of isomorphism classes of locally free $\mathbf{Z}[C_{p}\rtimes C_{q}]$ -modules of rank $k$ by ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ . By Milnor’s classification, this corresponds to the two-sided quotient

$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )=\text{GL}_{k}(\mathbf{Z}[C_{q}])\backslash \text{GL}_{k}(\mathbf{F}_{p}[C_{q}])/\text{GL}_{k}({\mathcal{T}}_{q}).\end{eqnarray}$$

Likewise, the locally free $\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}]$ -modules of rank $k$ correspond to the quotient

$$\begin{eqnarray}{\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )=\text{GL}_{k}(\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}])\backslash \text{GL}_{k}(\mathbf{F}_{p}[C_{q}\times \unicode[STIX]{x1D6E4}])/\text{GL}_{k}({\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]).\end{eqnarray}$$

In particular, if neither $\mathbf{Z}[C_{q}]$ nor ${\mathcal{T}}_{q}$ admits non-trivial stably free modules of rank $k$ , then any stably free module of rank $k$ over $\mathbf{Z}[C_{p}\rtimes C_{q}]$ is locally free. Consequently, the set ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[C_{p}\rtimes C_{q}])$ of stably free modules of rank $k$ over $\mathbf{Z}[C_{p}\rtimes C_{q}]$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ . Similarly, ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ if neither $\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}]$ nor ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ admits non-trivial stably free modules of rank $k$ .

There are obvious mappings of fibre squares $i:(\clubsuit ){\hookrightarrow}(\heartsuit )$ and $r:(\heartsuit )\rightarrow (\clubsuit )$ such that $r\circ i=\text{Id}$ . Consequently, there is a commutative ladder of mappings

where $s_{k,1}$ and $\unicode[STIX]{x1D70E}_{k,1}$ are the obvious stabilization mappings. We note that the mappings $i_{k}$ are injective in view of the fact that $r\circ i=\text{Id}$ .

The argument is now divided into two cases: $q$ is odd, and $q=2$ . First, suppose $q$ is an odd prime dividing $p-1$ . As noted in [1], in $(\clubsuit )$ , the rings $\mathbf{Z}[C_{q}]$ and ${\mathcal{T}}_{q}$ both have property SFC. Consequently, ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[C_{p}\rtimes C_{q}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )$ for all $k\geqslant 1$ . Similarly, in the fibre square $(\heartsuit )$ , the rings $\mathbf{Z}[C_{q}\times \unicode[STIX]{x1D6E4}]$ and ${\mathcal{T}}_{q}[\unicode[STIX]{x1D6E4}]$ also have SFC and once again ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ for all $k\geqslant 1$ . The essence of the argument now consists of the following five statements.

1. (I) For all $n$ , ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ is finite and $s_{n,1}:{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n+1}(\clubsuit )$ is bijective.

2. (II) $i_{1}:{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )$ is bijective.

3. (III) $i_{n}:{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )$ is bijective for all $n\geqslant 3$ .

4. (IV) $\unicode[STIX]{x1D70E}_{n,1}:{\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{n+1}(\heartsuit )$ is injective provided $n\neq 2$ .

5. (V) If $m=1$ (that is, $\unicode[STIX]{x1D6E4}\;=\;C_{\infty }$ ) then $i_{2}:{\mathcal{L}}{\mathcal{F}}_{2}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{2}(\heartsuit )$ is bijective.

To prove (I) we note that, as $C_{p}\rtimes C_{q}$ is finite, the finiteness of ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ follows from the Jordan–Zassenhaus theorem, together with Milnor’s classification of projectives. Moreover, as $\mathbf{Z}[C_{p}\rtimes C_{q}]$ satisfies the Eichler condition, the Swan–Jacobinski theorem shows that each $s_{k,1}:{\mathcal{L}}{\mathcal{F}}_{k}(\clubsuit )\rightarrow {\mathcal{L}}{\mathcal{F}}_{k+1}(\clubsuit )$ is bijective. It follows from Proposition 1 that $|{\mathcal{L}}{\mathcal{F}}_{1}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )|$ . Thus (II) is true as $i_{1}$ is injective and ${\mathcal{L}}{\mathcal{F}}_{1}(\clubsuit )$ is finite. Likewise it follows from Proposition 3 that $|{\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )|\leqslant |{\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )|$ for $n\geqslant 3$ . Thus (III) is true as $i_{n}$ is injective and ${\mathcal{L}}{\mathcal{F}}_{n}(\clubsuit )$ is finite; (IV) now follows from (I), (II) and (III) by diagram chasing using the fact that $i_{2}$ is injective. Finally, (V) follows by the same argument as (III) on substituting Proposition 4 for Proposition 2.

To proceed with the proof of Theorem A, put $\unicode[STIX]{x1D70E}_{n,k}=\unicode[STIX]{x1D70E}_{n+k-1,1}\circ \ldots \circ \unicode[STIX]{x1D70E}_{n,1}$ whenever $k\geqslant 1$ . It follows from (IV) that $\unicode[STIX]{x1D70E}_{n,k}$ is injective provided $n\geqslant 3$ . A straightforward diagram chase using (I), (II) and (III) also shows that each $\unicode[STIX]{x1D70E}_{1,k}$ is injective. Now suppose that $S$ is a stably free module of rank $n\neq 2$ over $\unicode[STIX]{x1D6EC}=\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ and denote its class in ${\mathcal{L}}{\mathcal{F}}_{n}(\heartsuit )$ by $[S]$ . Then $S\oplus \unicode[STIX]{x1D6EC}^{k}\cong \unicode[STIX]{x1D6EC}^{n+k}$ for some $k\geqslant 1$ so that $\unicode[STIX]{x1D70E}_{n,k}[S]=\unicode[STIX]{x1D70E}_{n,k}[\unicode[STIX]{x1D6EC}^{n}]$ . As $\unicode[STIX]{x1D70E}_{n,k}$ is injective, $S\cong \unicode[STIX]{x1D6EC}^{n}$ . Consequently, when $n\neq 2$ there are no non-trivial stably free modules of rank $n$ over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }^{m}]$ , and this proves the first part of Theorem A.

In the case $q=2$ (i.e. dihedral groups) we cannot claim ${\mathcal{T}}_{2}[\unicode[STIX]{x1D6E4}]$ has property SFC. To see why, consider the square

As $(A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ is commutative, we have $\text{GL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])=U((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])\cdot \text{SL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . The unit group $U((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ lifts back to ${\mathcal{T}}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . However, it is not clear whether we can lift the elements of $\text{SL}_{2}((A/\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}])$ . Thus, it is conceivable that ${\mathcal{T}}_{2}(A,\,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ has non-trivial stably free modules of rank 1. Nevertheless, using Suslin’s theorem as before, it is clear that ${\mathcal{T}}_{2}(A,\,\unicode[STIX]{x1D70B})[\unicode[STIX]{x1D6E4}]$ admits no non-trivial stably free module of rank ${\geqslant}2$ . Consequently, we observe that ${\mathcal{S}}{\mathcal{F}}_{k}(\mathbf{Z}[(C_{p}\rtimes C_{q})\times \unicode[STIX]{x1D6E4}])$ is a subset of ${\mathcal{L}}{\mathcal{F}}_{k}(\heartsuit )$ for all $k\geqslant 2$ . We now proceed as above.

Finally, the proof of Theorem B follows exactly the same lines except that now, in the case where $m=1$ and $\unicode[STIX]{x1D6E4}=C_{\infty }$ , we see from (V) that $\unicode[STIX]{x1D70E}_{2,1}$ is also injective. Consequently, each $\unicode[STIX]{x1D70E}_{2,k}$ is injective. Thus, there are no non-trivial stably free modules of any rank over $\mathbf{Z}[(C_{p}\rtimes C_{q})\times C_{\infty }]$ .