# Branched covers and rational homology balls

Charles Livingston  Charles Livingston: Department of Mathematics, Indiana
University, Bloomington, IN 47405<EMAIL_ADDRESS>
###### Abstract.

The concordance group of knots in $S^{3}$ contains a subgroup isomorphic to
$({\mathbb{Z}}_{2})^{\infty}$, each element of which is represented by a knot
$K$ with the property that for every $n>0$, the $n$–fold cyclic cover of
$S^{3}$ branched over $K$ bounds a rational homology ball. This implies that
the kernel of the canonical homomorphism from the knot concordance group to
the infinite direct sum of rational homology cobordism groups (defined via
prime-power branched covers) contains an infinitely generated two-torsion
subgroup.

This work was supported by a grant from the National Science Foundation, NSF-
DMS-1505586.

##  1\. Introduction

There is a homomorphism

$\varphi\colon\mathcal{C}\to\prod_{q\in{\mathcal{Q}}}\Theta^{3}_{{\mathbb{Q}}},$

where $\mathcal{C}$ is the smooth concordance group of knots in $S^{3}$,
$\mathcal{Q}$ is the set of prime power integers, and
$\Theta^{3}_{{\mathbb{Q}}}$ is the rational homology cobordism group. For a
knot $K$ and $q\in\mathcal{Q}$, the $q$–component of $\varphi(K)$ is the class
of $M_{q}(K)$, the $q$–fold cyclic cover of $S^{3}$ branched over $K$.

In the paper [MR4357613], Aceto, Meier, A. Miller, M. Miller, Park, and
Stipsicz proved that $\ker\varphi$ contains a subgroup isomorphic to
$({\mathbb{Z}}_{2})^{5}$. Here we will prove that $\ker\varphi$ contains a
subgroup isomorphic to $({\mathbb{Z}}_{2})^{\infty}$. Our examples are of the
form $K\mathbin{\\#}-K^{r}$, where $-K$ denotes the concordance inverse of $K$
(the mirror image of $K$ with string orientation reversed), and $K^{r}$ is
formed from $K$ by reversing its string orientation. Such knots are easily
seen to be in the kernel of $\varphi$; the more difficult work is to find
nontrivial examples of order two.

The first known example of a nontrivial element in $\ker\varphi$ was
represented by the knot $K_{1}=8_{17}\mathbin{\\#}-8_{17}^{r}$, which is of
order two in $\mathcal{C}$. That $K_{1}$ is of order at most two is
elementary; that $K_{1}$ is nontrivial in $\mathcal{C}$ is one of the main
results of [MR1670424], proved using twisted Alexander polynomials.

The results of [2019arXiv190412014K] provide an infinitely generated free
subgroup of $\ker\varphi$. Conjecturely,
$\mathcal{C}\cong{\mathbb{Z}}^{\infty}\oplus({\mathbb{Z}}_{2})^{\infty}$; if
true, then
$\ker\varphi\cong{\mathbb{Z}}^{\infty}\oplus({\mathbb{Z}}_{2})^{\infty}$.

###  1.1. Main result.

Figure 1 illustrates a knot $P_{n}$ in a solid torus, where $J_{n}$ represents
the braid illustrated on the right in the case of $n=5$; $n$ will always be
odd. We let $K_{n}$ denote the satellite of $8_{17}$ built from $P_{n}$. In
standard notation, $K_{n}=P_{n}(8_{17})$. For future reference, we illustrate
the braid $J_{n}^{*}$ formed by rotating $J_{n}$ around the vertical axis.

$P_{n}$ at -10 165 $J_{n}$ at 170 106 $J_{n}$ at 460 95 $J_{n}^{*}$ at 740 95
[scale=.45]companion-oriented  [scale=.55]braid [scale=.55]braid2

Figure 1. The knot $P_{n}\subset S^{1}\times B^{2}$ , $J_{n}$, and
$J_{n}^{*}$.

$K$ at 115 80 $K$ at 370 80 [scale=.60]connectedsum

Figure 2.  $P_{5}(K)\mathbin{\\#}P_{5}(K)$.

######  Theorem 1.

Let $K_{n}=P_{n}(8_{17})$. For all odd $n$, the knot
$L_{n}=K_{n}\mathbin{\\#}-K_{n}^{r}$ satisfies $2L_{n}=0\in\mathcal{C}$ and
$L_{n}\in\ker\varphi$. There is an infinite set of prime integers
$\mathcal{P}$ for which $L_{\alpha}\neq L_{\beta}\in\mathcal{C}$ for all
$\alpha\neq\beta$ in $\mathcal{P}$. In particular, the set of knots
$\\{L_{n}\\}_{n\in\mathcal{P}}$ generates a subgroup of $\ker\varphi$ that is
isomorphic to $({\mathbb{Z}}_{2})^{\infty}$.

The rest of the paper presents a proof of this theorem. The first two claims
are easily dealt with in Sections 2 and 3. The more difficult step of the
proof calls on an analysis of twisted Alexander polynomials and their
relevance to knot slicing; a review of twisted polynomials is included in
Section 4. The proof of Theorem 1 is completed in Section 5, with the
exception of a number theoretic result that is described Appendix A.

Acknowledgements. Thanks to Darrell Haile for assisting me in proof of the
number theoretic result in Appendix A. Allison Miller provided valuable
feedback about an early draft of this paper.

##  2\. Proof: $2L_{n}=0\in\mathcal{C}$

Let $P_{n}^{*}\subset S^{1}\times B^{2}$ denote the knot formed using the
braid $J_{n}^{*}$ in Figure 2. For any knot $K$, let $P_{n}^{*}(K)$ denote the
satellite of $K$ built using $P_{n}^{*}$. It should be clear that $P_{n}$ and
$P_{n}^{*}$ are orientation preserving isotopic, and thus for all knots $K$,
$P_{n}(K)=P_{n}^{*}(K)$.

Figure 2 illustrates for an arbitrary knot $K$, the connected sum
$P_{n}(K)\mathbin{\\#}P_{n}^{*}(K)=P_{n}(K)\mathbin{\\#}P_{n}(K)$ in the case
of $n=5$. Performing $n-1$ band moves in the evident way yields the
$(0,n)$–cable of $K\mathbin{\\#}K$. Thus, if
$K\mathbin{\\#}K=0\in\mathcal{C}$, then the $n$ components of this link can be
capped off with parallel copies of the slice disk for $K\mathbin{\\#}K$,
implying that $P_{n}(K)\mathbin{\\#}P_{n}(K)=0\in\mathcal{C}$. In particular,
$2K_{n}=0\in\mathcal{C}$ and $2K_{n}^{r}=0\in\mathcal{C}$.

##  3\. Proof $L_{n}\in\ker\varphi$

We prove a stronger statement: For all odd $n$, and for all positive integers
$q$, $M_{q}(L_{n})$ is a rational homology sphere that represents
$0\in\Theta_{\mathbb{Q}}^{3}$.

The $q$–fold cyclic cover of $S^{3}$ branched over
$K_{n}\mathbin{\\#}-K_{n}^{r}$ is the same space as the $q$–fold cyclic cover
of $S^{3}$ branched over $K_{n}\mathbin{\\#}-K_{n}$. A slice disk for
$K_{n}\mathbin{\\#}-K_{n}$ is built from $(S^{3}\times I,K_{n}\times I)$ by
removing a copy of $B^{3}\times I$. Taking the $q$–fold branched cover shows
that the $q$–fold cyclic cover of $B^{4}$ branched over that slice disk is
diffeomorphic to $M_{q}(K_{n})^{*}\times I$, where $M_{q}(K_{n})^{*}$ denotes
a punctured copy of $M_{q}(K_{n})$. It remains to show that $M_{q}(K_{n})$ is
a rational homology 3–sphere.

A formula of Fox [MR95876] and Goeritz [MR1507011] states that the order of
the first homology of $M_{q}(K_{n})$ is given by the product of values
$\Delta_{K_{n}}(\omega_{q}^{i})$, where $\Delta_{K_{n}}(t)$ denotes the
Alexander polynomial, $\omega_{q}$ is a primitive $q$–root of unity, and $i$
runs from 1 to $q-1$.

A result of Seifert [MR0035436] shows that
$\Delta_{K_{n}}(t)=\Delta_{8_{17}}(t^{n})\Delta_{P_{n}(U)}$, where $U$ is the
unknot. We have that $P_{n}(U)=U$. The Alexander polynomial for $8_{17}$ is

$\Delta_{8_{17}}(t)=1-4t+8t^{2}-11t^{3}+8t^{4}-4t^{5}+t^{6}.$

A numeric computation confirms that this polynomial does not have roots on the
unit complex circle, and hence $\Delta_{8_{17}}(t^{n})$ has no roots on the
unit complex circle. From this is follows that
$\Delta_{K_{n}}(\omega_{q}^{i})\neq 0$ for all $i$; thus the order of the
homology of $M_{q}(K_{n})$ is finite.

##  4\. Review of twisted polynomials and $8_{17}$

In this section we review twisted Alexander polynomials and their application
in [MR1670420, MR1670424] showing that $8_{17}\mathbin{\\#}-8_{17}^{r}\neq
0\in\mathcal{C}$.

Let $(X,B)\to(S^{3},K)$ be the $q$–fold cyclic branched cover of a knot $K$
with $q$ a prime power. In particular, $X$ is rational homology sphere. There
is a canonical surjection $\epsilon\colon\thinspace
H_{1}(X-B)\to{\mathbb{Z}}$. Suppose that $\rho\colon\thinspace
H_{1}(X)\to{\mathbb{Z}}_{p}$ is homomorphism for some prime $p$. Then there is
an associated twisted polynomial
$\Delta_{K,\epsilon,\rho}(t)\in{\mathbb{Q}}(\omega_{p})[t]$. It is well-
defined, up to factors of the form $at^{k}$, where $a\neq
0\in{\mathbb{Q}}(\omega_{p})$. These polynomials are discriminants of Casson-
Gordon invariants, first defined in [MR900252].

In the case of $K=8_{17}$ and $q=3$, we have
$H_{1}(X)\cong{\mathbb{Z}}_{13}\oplus{\mathbb{Z}}_{13}$, and as a
${\mathbb{Z}}_{13}$–vector space this splits as a direct sum of a 3–eigenspace
and a 9–eigenspace under the order three action of the deck transformation.
Both eigenspaces are 1–dimensional. We denote this splitting as $E_{3}\oplus
E_{9}$. There are corresponding characters $\rho_{3}$ and $\rho_{9}$ of
$H_{1}(X)$ onto ${\mathbb{Z}}_{13}$; these are defined as the quotient maps
onto $H_{1}(X)/E_{3}$ and onto $H_{1}(X)/E_{9}$. We let $\rho_{0}$ denote the
trivial ${\mathbb{Z}}_{13}$–valued character.

The values of $\Delta_{8_{17},\epsilon,\rho_{i}}(t)$ are given in [MR1670424],
duplicated here in Appendix B. For $i=0$ it is polynomial in
${\mathbb{Q}}[t]$. For $i=3$ and $i=9$ it is in ${\mathbb{Q}}(\omega_{13})[t]$
and is not in ${\mathbb{Q}}[t]$. An essential observation is that for
$8_{17}^{r}$, the roles of $\rho_{3}$ and $\rho_{9}$ are reversed. All three
of the polynomials are irreducible in their respective polynomial rings, once
any factors of $(1-t)$ and $t$ are removed.

In [MR1670424] the proof that $8_{17}\mathbin{\\#}-8_{17}^{r}$ is not slice
comes down to the observation that no product of the form

$\sigma_{\delta}(\Delta_{8_{17},\epsilon,\rho_{3}}(t))\sigma_{\gamma}(\Delta_{8_{17},\epsilon,\rho_{i}}(t))\hskip
21.68121pt\text{or}\hskip
21.68121pt\sigma_{\delta}(\Delta_{8_{17},\epsilon,\rho_{9}}(t))\sigma_{\gamma}(\Delta_{8_{17},\epsilon,\rho_{j}}(t))$

is of the form $af(t)\overline{f(t^{-1})}(1-t)^{j}$ for some
$f(t)\in{\mathbb{Q}}(\omega_{13})[t]$. (That is, these products are not norms
in the polynomial ring ${\mathbb{Q}}(\omega_{13})[t,t^{-1}]$, modulo powers of
$(1-t)$ and $t$.) Here $i=0$ or $i=9$ and $j=0$ or $j=3$. The number $a$ is in
${\mathbb{Q}}(\omega)$ and the $\sigma_{\nu}$ are Galois automorphisms of
${\mathbb{Q}}(\omega_{p})$ (which acts by sending $\omega_{p}$ to
$\omega_{p}^{\nu}$).

Showing that the product of the polynomials does not factor in this way is
elementary once it is established that $\Delta_{8_{17},\epsilon,\rho_{3}}(t)$
and $\Delta_{8_{17},\epsilon,\rho_{9}}(t)$ are irreducible and not Galois
conjugate.

##  5\. Main Proof

Using the fact that $-P_{n}(8_{17})^{r}=P_{n}(8_{17})^{r}$, the knot
$L_{\alpha}\mathbin{\\#}L_{\beta}$ can be expanded as

$P_{\alpha}(8_{17})\mathbin{\\#}P_{\alpha}(8_{17})^{r}\mathbin{\\#}P_{\beta}(8_{17})\mathbin{\\#}P_{\beta}(8_{17})^{r}.$

We begin by analyzing the 3–fold cover of $S^{3}$ branched over
$P_{n}(8_{17})$ and assume that $3$ does not divide $n$. This cover is
$M_{3}(P_{n}(8_{17}))$ and we denote the branch set in the cover by
$\widetilde{B}$.

There is the obvious separating torus $T$ in $S^{3}\setminus P_{n}(8_{17})$.
Since 3 does not divide $n$, $T$ has a connected separating lift
$\widetilde{T}\subset M_{3}(P_{n}(8_{17}))$. One sees that $\widetilde{T}$
splits $M_{3}(P_{n}(8_{17}))$ into two components: $X$, the 3–fold cyclic
cover of $S^{3}\setminus 8_{17}$ and $Y$, the 3–fold cyclic branched cover of
$S^{1}\times B^{2}$, branched over $P_{n}$. A simple exercise shows that since
$P_{n}(U)$ is unknotted, $Y$ is the complement of some knot
$\widetilde{J_{n}}\subset S^{3}$.

A Mayer-Vietoris argument shows that
$H_{1}(M_{3}(P_{n}(8_{17})))\cong{\mathbb{Z}}_{13}\oplus{\mathbb{Z}}_{13}$ and
the two canonical representations $\rho_{3}$ and $\rho_{9}$ that are defined
on $X$ extend trivially on $Y$ and so to $M_{3}(P_{n}(8_{17}))$. We denote
these extension $\rho_{3}^{\prime}$ and $\rho_{9}^{\prime}$. Let
$\epsilon^{\prime}$ be the canonical surjective homomorphism
$\epsilon^{\prime}\colon\thinspace
H_{1}(M_{3}(P_{n}(8_{17})))\setminus\widetilde{B})\to{\mathbb{Z}}$. Restricted
to $X$ we have $\epsilon^{\prime}(x)=\epsilon(nx)$, where $\epsilon$ was the
canonical representation to ${\mathbb{Z}}$ defined for the cover of
$S^{3}\setminus 8_{17}$.

In [MR1670420, Theorem 3.7] there is a discussion of twisted Alexander
polynomials of satellite knots in $S^{3}$, working in the greater generality
of homomorphisms to the unitary group $U(m)$. (A map to ${\mathbb{Z}}_{p}$ can
be viewed as a representation to $U(1)$). The proof of that theorem, which
relies on the multiplicativity of Reidemeister torsion, applies in the current
setting, yielding the following lemma.

######  Lemma 2.

$\Delta_{P_{n}(8_{17}),\epsilon^{\prime},\rho_{3}^{\prime}}(t)=\Delta_{8_{17},\epsilon,\rho_{3}}(t^{n})\Delta_{\widetilde{J_{n}}}(t).$

Similar results hold for the knot $P_{n}(8_{17})^{r}$ and for the character
$\rho_{9}$.

As described in [MR1670424, MR1670420] Casson-Gordon theory implies that if
$L_{\alpha}\mathbin{\\#}L_{\beta}$ is slice, then for some 3–eigenvector or
for some 9–eigenvector the corresponding twisted Alexander polynomial is a
norm; that is, it factors as $at^{k}f(t)\overline{f(t^{-1})}$, modulo
multiples of $(1-t)$. If it is a 3–eigenvector, the relevant polynomial is of
the form

(1)
$\Delta(t)=\sigma_{a}(\Delta_{8_{17},\epsilon,\rho_{3}}(t^{\alpha}))^{x}\sigma_{b}(\Delta_{8_{17},\epsilon,\rho_{9}}(t^{\alpha}))^{y}\sigma_{c}(\Delta_{8_{17},\epsilon,\rho_{3}}(t^{\beta}))^{z}\sigma_{d}(\Delta_{8_{17},\epsilon,\rho_{9}}(t^{\beta}))^{w}\big{(}\Delta_{\widetilde{J_{\alpha}}}(t)\Delta_{\widetilde{J_{\beta}}}(t)\big{)}^{2},$

where one of $x,y,z,$ or $w$ is equal to 1, and each of the others are either
1 or 0.

The four ${\mathbb{Q}}(\omega_{13})[t]$–polynomials that appear here,

$\Delta_{8_{17},\epsilon,\rho_{3}}(t^{\alpha}),\hskip
7.22743pt\Delta_{8_{17},\epsilon,\rho_{9}}(t^{\alpha}),\hskip
7.22743pt\Delta_{8_{17},\epsilon,\rho_{3}}(t^{\beta}),\hskip
7.22743pt\text{and}\hskip
7.22743pt\Delta_{8_{17},\epsilon,\rho_{9}}(t^{\beta}),$

and all their Galois conjugates are easily seen to be distinct for any pair
$\alpha\neq\beta$. The following number theoretic result implies that there is
an infinite set of primes such that if $\alpha\in\mathcal{P}$ and
$\beta\in\mathcal{P}$, then no product as given in Equation 1 can be a norm in
${\mathbb{Q}}(\omega_{13})[t]$, proving that the connected sum
$L_{\alpha}\mathbin{\\#}L_{\beta}$ is not slice. We will present a proof in
Appendix A,

######  Lemma 3.

Let $f(t)\in{\mathbb{Z}}(\omega_{p})[t]$ be an irreducible monic polynomial.
If there exists $\zeta\in{\mathbb{C}}$ such that $f(\zeta)=0$ and
$\zeta^{n}\neq 1$ for all $n>0$, then the set of primes $p$ for which
$f(t^{p})$ is reducible is finite.

Proof of Theorem 1. The last factor in Equation 1 involving the
$\widetilde{J_{n}}$ is a norm, so it can be ignored in determining if the
product is a norm.

A numeric computation shows that the twisted polynomials
$\Delta_{8_{17},\epsilon,\rho_{i}}(t)$ for $i=3$ and $i=9$ do not have roots
on the unit circle, so Lemma 3 can be applied with
${\mathbb{F}}={\mathbb{Q}}(\omega_{13})$. Let $\mathcal{P}$ be the infinite
set of primes with the property that if $p\in\mathcal{P}$ then
$\Delta_{8_{17},\epsilon,\rho_{3}}(t^{p})$ and
$\Delta_{8_{17},\epsilon,\rho_{9}}(t^{p})$ are irreducible. Consider the case
of $x=1$ in Equation 1. Then, assuming that $\alpha\in\mathcal{P}$ and
$\beta\in\mathcal{P}$, the term
$\sigma_{a}(\Delta_{8_{17},\epsilon,\rho_{3}})(t^{\alpha})$ that appears in
Equation 1 is relatively prime to the remaining factors and all the factors
are irreducible, modulo powers of $t$ and $(1-t)$. Hence, the product cannot
be of the form $t^{k}(1-t)^{j}f(t)f(t^{-1})$ for any
$f(t)\in{\mathbb{Q}}(\omega_{13})[t]$. The cases of $y,z$, or $w=1$ are the
same.

##  Appendix A Factoring $f(t^{p})$

In this appendix we prove Lemma 3, stated in somewhat more generality as Lemma
4 below. We first summarize some background material. Further details can be
found in any graduate textbook on algebraic number theory.

  * •

${\mathbb{A}}\subset{\mathbb{C}}$ denotes the ring of algebraic integers. This
is the ring consisting of all roots of monic polynomials in ${\mathbb{Z}}[t]$.

  * •

For an extension field ${\mathbb{F}}/{\mathbb{Q}}$, the ring of algebraic
integers in ${\mathbb{F}}$ is defined by
$\mathcal{O}_{\mathbb{F}}={\mathbb{F}}\cap{\mathbb{A}}$.

  * •

The property of transitivity states that if
$f(t)\in\mathcal{O}_{{\mathbb{F}}}[t]$ is monic and $f(\zeta)=0$, then
$\zeta\in{\mathbb{A}}$.

  * •

$\mathcal{O}_{\mathbb{F}}^{\times}$ is defined to be the set of units in
$\mathcal{O}_{\mathbb{F}}$.

  * •

The norm of an element $x\in\mathcal{O}_{\mathbb{F}}$ is defined as
$N(x)=\prod x_{i}\in{\mathbb{Z}}$ where the $x_{i}$ are the complex Galois
conjugates of $x$. This map satisfies $N(xy)=N(x)N(y)$ for all
$x,y\in\mathcal{O}_{\mathbb{F}}$. An element $x\in\mathcal{O}_{{\mathbb{F}}}$
is in $\mathcal{O}_{\mathbb{F}}^{\times}$ if and only if $N(x)=\pm 1$.

  * •

The Dirichlet Unit Theorem states that for a finite extension
${\mathbb{F}}/{\mathbb{Q}}$, the abelian group
$\mathcal{O}_{\mathbb{F}}^{\times}$ is finitely generated, isomorphic to
$G\oplus{\mathbb{Z}}^{r+s-1}$, where $G$ is finite cyclic, $r$ is the number
of embeddings of ${\mathbb{F}}$ in ${\mathbb{R}}$, and $2s$ is the number of
non-real embeddings of ${\mathbb{F}}$ in ${\mathbb{C}}$.

######  Lemma 4.

Let ${\mathbb{F}}$ be a finite extension of ${\mathbb{Q}}$ and let
$f(t)\in\mathcal{O}_{{\mathbb{F}}}[t]$ be an irreducible monic polynomial. If
there exists $\zeta\in{\mathbb{C}}$ such that $f(\zeta)=0$ and $\zeta^{n}\neq
1$ for all $n>0$, then the set of primes $p$ for which $f(t^{p})$ is reducible
is finite.

###### Proof.

Step 1. If $f(\zeta)=0$, then $\zeta\in\mathcal{O}_{F(\zeta)}$.

This follows immediately from the assumption that $f(t)$ is monic.

Step 2. Suppose that $f(t)\in{\mathbb{F}}[t]$ is irreducible and $f(\zeta)=0$.
If for some prime $p$, $f(t^{p})$ is reducible over ${\mathbb{F}}$, then
$\zeta=\eta^{p}$ for some $\eta\in\mathcal{O}_{{\mathbb{F}}(\zeta)}$.

Let $\xi\in{\mathbb{C}}$ satisfy $\xi^{p}=\zeta$. Since $f(t)$ is irreducible
of degree $n$ and $f(t^{p})$ is reducible, we have the degrees of extensions
satisfying $[{\mathbb{F}}(\zeta)\colon\thinspace{\mathbb{F}}]=n$ and
$[{\mathbb{F}}(\xi)\colon\thinspace{\mathbb{F}}]<np$. It follows from the
multiplicity of degrees of extensions that
$[{\mathbb{F}}(\xi)\colon\thinspace{\mathbb{F}}(\zeta)]<p$.

The polynomial $t^{p}-\zeta\in{\mathbb{F}}(\zeta)[t]$ has $\xi$ as a root. For
all $i$, $\omega_{p}^{i}\xi$ is also a root, so $t^{p}-\zeta$ factors
completely in ${\mathbb{C}}[t]$ as

$t^{p}-\zeta=(t-\xi)(t-\omega_{p}\xi)\cdots(t-\omega_{p}^{p-1}\xi).$

By the degree calculation just given, $t^{p}-\zeta$ has an irreducible factor
$g(t)\in{\mathbb{F}}(\zeta)[t]$ of degree $l<p$. We can write
$g(t)=\prod(t-\omega_{p}^{i}\xi)$ where the product is over some proper subset
of $\\{0,\ldots,p-1\\}$. Multiplying this out, one finds that the constant
term is of the form $\omega_{p}^{j}\xi^{l}\in{\mathbb{F}}(\zeta)$ for some $j$
and $l<p$. Since $l$ and $p$ are relatively prime, there are integers $u$ and
$v$ such that $ul+vp=1$. Thus,
$(\omega_{p}^{j}\xi^{l})^{u}(\xi^{p})^{v}=\omega_{p}^{s}\xi$ for some $s$. In
particular, for some $s$, we have $\omega_{p}^{s}\xi\in{\mathbb{F}}(\zeta)$.
We let $\eta=\omega_{p}^{s}\xi$ and find that
$\eta^{p}=(\omega_{p}^{s})^{p}\xi^{p}=\zeta$. Finally, $\eta$ satisfies the
monic polynomial $f(t^{p})$ and thus is in
$\mathcal{O}_{{\mathbb{F}}(\zeta)}$.

Step 3. The set of primes $p$ such that $\zeta=\eta^{p}$ for some
$\eta\in\mathcal{O}_{{\mathbb{F}}(\zeta)}$ is finite.

If $\zeta=\eta^{p}$ then $N(\zeta)=N(\eta)^{p}$. If $N(\zeta)\neq\pm 1$, then
the set of $p$ for which $N(\zeta)=a^{p}$ for some integer $a$ is finite.

If $N(\zeta)=\pm 1$, then
$\zeta\in\mathcal{O}_{{\mathbb{F}}(\zeta)}^{\times}$. Hence $\zeta$ represents
a non-torsion element in a finitely generated abelian group, and thus it has a
finite number of roots. ∎

Comments. The argument just given is based on a summary of the proof for the
case ${\mathbb{F}}={\mathbb{Q}}$ presented in MathOverflow by Dimitrove
[vdimitrov]. Step (2) is a special case of the Vahlen-Capelli Theorem, proved
in the case of ${\mathbb{F}}={\mathbb{Q}}$ by Vahlen and for fields of
characteristic 0 by Capelli [capelli]. A proof for fields of finite
characteristic is given in the book by Rédei [MR0211820].

##  Appendix B Twisted polynomials of $8_{17}$

Here are the three needed polynomials. We write $\omega$ for $\omega_{13}$.

$\Delta_{8_{17},\epsilon,\rho_{0}}(t)=1-t-34t^{2}-101t^{3}-34t^{4}-t^{5}+t^{6}.$

$\Delta_{8_{17},\epsilon,\rho_{3}}(t)/(1-t)=$

$1\ +$

$t(2\omega+2\omega^{2}+2\omega^{3}+4\omega^{4}+2\omega^{5}+2\omega^{6}+\omega^{7}+\omega^{8}+2\omega^{9}+4\omega^{10}+\omega^{11}+4\omega^{12})\
+$

$t^{2}(-15\omega-10\omega^{2}-15\omega^{3}-15\omega^{4}-10\omega^{5}-10\omega^{6}-10\omega^{7}-10\omega^{8}-15\omega^{9}-15\omega^{10}-10\omega^{11}-15\omega^{12})\
+$

$t^{3}(4\omega+\omega^{2}+4\omega^{3}+2\omega^{4}+\omega^{5}+\omega^{6}+2\omega^{7}+2\omega^{8}+4\omega^{9}+2\omega^{10}+2\omega^{11}+2\omega^{12})\
+$

$t^{4}$

$\Delta_{8_{17},\epsilon,\rho_{9}}(t)/(1-t)=$

$1\ +$

$t(6\omega+5\omega^{2}+6\omega^{3}+6\omega^{4}+5\omega^{5}+5\omega^{6}+5\omega^{7}+5\omega^{8}+6\omega^{9}+6\omega^{10}+5\omega^{11}+6\omega^{12})\
+$

$t^{2}(-13\omega-12\omega^{2}-13\omega^{3}-13\omega^{4}-12\omega^{5}-12\omega^{6}-12\omega^{7}-12\omega^{8}-13\omega^{9}-13\omega^{10}-12\omega^{11}-13\omega^{12})\
+$

$t^{3}(6\omega+5\omega^{2}+6\omega^{3}+6\omega^{4}+5\omega^{5}+5\omega^{6}+5\omega^{7}+5\omega^{8}+6\omega^{9}+6\omega^{10}+5\omega^{11}+6\omega^{12})\
+$

$t^{4}$

## References