text:
\begin{document}

\maketitle

\begin{abstract}
Let $G$ be a finite abelian group. A number field $K$ is called a
Hilbert-Speiser field of type $G$ if every tame $G$-Galois extension $L/K$
has a normal integral basis, i.e.,
the ring of integers $\mathcal{O}_L$ is free as an $\mathcal{O}_KG$-module.
Let $C_{p}$ denote the cyclic group of prime order $p$.
We show that if $p \geq 7$ (or $p=5$ and extra conditions are met) and $K$ is totally
real with $K/\Q$ ramified at $p$, then $K$ is not Hilbert-Speiser of type $C_{p}$.
\end{abstract}

\section{Introduction}

Let $L/K$ be a finite Galois extension of number fields with Galois group $G$.
Then by a theorem of Noether it is well known that the ring of integers
$\mathcal{O}_{L}$ is a projective module over the group ring $\mathcal{O}_KG$
if and only if $L/K$ is tamely ramified. If $\mathcal{O}_{L}$ is in fact free (necessarily
of rank $1$) over $\mathcal{O}_KG$, then $L/K$ is said to have a
\emph{normal integral basis}.

A number field $K$ is called a \emph{Hilbert-Speiser field} if every finite abelian tamely
ramified extension $L/K$ has a normal integral basis. The celebrated Hilbert-Speiser
Theorem says that $\mathbb{Q}$ is such a field, and the main result of \cite{grrs} is that
$\mathbb{Q}$ is in fact the only such field.  By fixing a finite abelian group $G$ one can consider a finer problem: given a number field $K$, does every tame $G$-Galois extension $L/K$ have a normal integral basis? If so, $K$ is said to be a \emph{Hilbert-Speiser field of type $G$}. The simplest case to consider is when $G=C_{p}$, the cyclic group of prime order $p$.
This has been studied, for instance, in  \cite{carter-NIB-quad-cubic}, \cite{carter-erratum},
\cite{herreng-HS}, \cite{ichimura-kummer-prime-V}, \cite{ichimura-NIB-rayclassgroups},
\cite{Ichimura-HS-p}, \cite{Ichimura-imag-quad} and \cite{Ichimura-Sumida-Takahashi-HS-imag-quad}. We continue the investigation of this case by establishing the following result, the proof of which is based on a detailed analysis of locally free class groups and ramification indices.

\begin{theorem}\label{real-not-HS}
Let $K$ be a totally real number field and let $p\geq5$ be prime.
Suppose that $K/\Q$ is ramified at $p$. If $p=5$ and $[K(\zeta_{5}):K]=2$,
assume further that there exists a prime $\mathfrak{p}$ of $K$ above $p$ such that the
ramification index of $\mathfrak{p}$ in $K/\Q$ is at least $3$. Then $K$ is not
Hilbert-Speiser of type $C_{p}$.
\end{theorem}

\begin{remark}
Some extra conditions in the case $p=5$ and $[K(\zeta_{5}):K]=2$  are required
because, for example, as noted in \cite[Remark 1]{Ichimura-HS-p}, $K=\Q(\sqrt{5})$ is in fact Hilbert-Speiser of type $C_{5}$.
\end{remark}

Theorem \ref{real-not-HS} can be seen as an analogue of the following result of Herreng
(see \cite[\S 3]{herreng-HS}). The authors are grateful to Nigel P. Byott for pointing out that
the original hypothesis that $K/\Q$ is Galois can be weakened as below.

\begin{theorem}[Herreng]\label{imag-not-HS}
Let $K$ be a totally imaginary number field and let $p$ be an odd prime.
Suppose that every prime $\frakp$ of $K$ above $p$ is ramified. If
\begin{enumerate}
\item $p > [K:\Q]$, or
\item $p \geq 5$ and $\zeta_{p} \in K$, or
\item $p \geq 7$ and the ramification index in $K/\Q$ of every prime $\frakp$ of $K$ above $p$
is at least $3$,
\end{enumerate}
then $K$ is not Hilbert-Speiser of type $C_{p}$.
\end{theorem}

Combining Theorems \ref{real-not-HS} and \ref{imag-not-HS}(a) we immediately obtain the
following result, which in many (but not all) respects is a significant sharpening of
\cite[Theorems 1 and 2]{Ichimura-HS-p}.

\begin{theorem}
Let $K$ be a Hilbert-Speiser field of type $C_{p}$ for some odd prime $p$.
If either
\begin{enumerate}
\item $K$ is totally real and $p \geq 7$, or
\item $K$ is totally imaginary and $p > [K:\Q]$,
\end{enumerate}
then $K \cap \Q(\zeta_{p^{n}})=\Q$ for all $n \geq 1$.
\end{theorem}

\section{Realizable Classes}\label{realizable}

We briefly recall the work of McCulloh on realizable classes in the special case of
cyclic extensions of prime degree (see \cite{mcculloh-abelian} for further details).

Let $K$ be a number field and let $p$ be a prime.
Let $\Delta \simeq (\Z/p\Z)^{\times}$ be the group of automorphisms of $C_p$.
Then the locally free class group $\Cl(\mathcal{O}_KC_p)$ is a $\Delta$-module.
As $L/K$ varies over all tame $C_{p}$-Galois extensions of $K$,
the class $(\mathcal{O}_L)$ of $\mathcal{O}_L$ varies over a subset $R(\mathcal{O}_KC_p)$
of $\Cl(\mathcal{O}_KC_p)$. This subset is in fact a subgroup which can be described explicitly.

Let $\Cl(\mathcal{O}_K)$ denote the ideal class group of $K$ and let $\Cl'(\mathcal{O}_KC_p)$
be the kernel of the map $\Cl(\mathcal{O}_KC_p) \longrightarrow \Cl(\mathcal{O}_K)$ induced by augmentation.
Let $\mathcal{J}$ be the Stickelberger ideal in $\mathbb{Z}\Delta$ (the definition of  $\mathcal{J}$ will be given later).
The key result of relevance to the present paper is that $R(\mathcal{O}_KC_p)$ is the subgroup
$\Cl'(\mathcal{O}_KC_p)^{\mathcal J}$ of $\Cl(\mathcal{O}_KC_p)$ where
$\Cl'(\mathcal{O}_KC_p)^{\mathcal J}=\{c^{\alpha}: c\in \Cl'(\mathcal{O}_KC_p), \alpha \in {\mathcal J}\}$.

\section{The proof of Theorem \ref{real-not-HS}}

Let $K$ be a totally real number field and let $p \geq 5$ be prime.
Let $\mathfrak{p}$ be some prime of $K$ above $p$ and let $e$ denote the
ramification index of $\mathfrak{p}$ in $K/\Q$. We will assume that $p$ is
ramified in $K/\Q$ and so $\mathfrak{p}$ can be chosen such that $e \geq 2$.
Under these hypotheses we shall show that $K$ is not Hilbert-Speiser
of type $C_{p}$ (note that in the case $p=5$ and $[K(\zeta_{p}):K]=2$ we shall have to assume that $e \geq 3$).

The basic idea of the proof will be to construct certain $\mathcal{O}_{K}$-algebras
$\Gamma$ and $S$ such that $\Gamma \subseteq S$ with
${S/\Gamma}\simeq {\mathcal{O}_K}/{\mathfrak p}$ as $\mathcal{O}_{K}$-modules.
Together, $S$ and $\Gamma$ will be used to construct
a non-trivial subgroup of the realizable classes
$R(\mathcal{O}_KC_p)=\Cl'(\mathcal{O}_KC_p)^{\mathcal J}$ described in Section
\ref{realizable}, thereby giving the desired result.
At all primes ${\mathfrak q}\neq {\mathfrak p}$ of $K$
the completions $S_{\mathfrak q}$ and $\Gamma_{\mathfrak q}$ will be equal,
so the essential part of the argument will be local at ${\mathfrak p}$.

Let $\phi_p(z)$ be the $p$th cyclotomic polynomial. Then
$\Gamma:=\mathcal{O}_{K}[z]/(\phi_p(z))$ is an $\mathcal{O}_K$-algebra,
but is a domain if and only if $[K(\zeta_p):K]=p-1$. The group
$\Delta:=(\Z/p\Z)^{\times}$ acts on $\Gamma$ in the
following way: to each $\bar{a} \in \Delta$ we associate an automorphism
$\sigma_{a}$ of $\Gamma$ defined by $\sigma_a(z)=z^a$, where
the image of $z$ in $\Gamma$ is again written $z$.
Let $\omega :\Delta \rightarrow \Z_p^{\times}$ be the Teichm\"uller character, so that
$\omega(\sigma_a)=\tilde{a}$ where $\tilde{a}^{p-1}=1$, and $\tilde{a}\equiv a\pmod p$.

There exists an element $\lambda$ such that
${\mathbb Z}_p[\zeta_p]={\mathbb Z}_p[\lambda]$ with
$\lambda^{p-1}=-p$, $\lambda \equiv 1-\zeta_p \pmod{(1-\zeta_p)^2}$
(see, for example, \cite[Chapter 14, Lemma 3.1]{lang-cyclo-I-and-II}). Furthermore,
$\Delta$ acts on $\lambda^i{\mathbb Z}_p$ through the character $\omega^i$ with
$i\in\{0,\ldots, p-2\}$.
Note that $\Gamma_{\mathfrak p}=\mathcal{O}_{K_{\mathfrak p}} \otimes_{{\mathbb Z}_{p}} {\mathbb Z}_p[\zeta_p]$, for which
$\{ 1, \lambda, \lambda^{2}, \ldots \lambda^{p-2}\}$ is an
$\mathcal{O}_{K_{\mathfrak{p}}}$-basis .
Let $\pi$ denote a parameter of $\mathcal{O}_{K_{\mathfrak p}} $ and define
the element $x : = {\frac{1}{\pi}}\otimes{\lambda^{p-2}}$ in
$K_{\mathfrak p} \otimes_{\Q_{p}}\Q_{p}(\zeta_{p})=K_{\mathfrak p}\Gamma$ (we will abuse notation and write $x=\frac{\lambda^{p-2}}{\pi}$).

\begin{lemma}\label{elements-lemma}
We have $x^2, x^3, \lambda x, \pi x \in {\Gamma}_{\mathfrak p}$.
\end{lemma}

\begin{proof}
Since $e \geq 2$, we have $\frac{p}{\pi^{2}} \in \mathcal{O}_{K_{\mathfrak{p}}}$.
Hence
\[
x^2=\frac{\lambda^{2p-4}}{\pi^2}=\frac{-p\lambda^{p-3}}{\pi^2}
\quad \textrm{ and } \quad
x^3=\frac{(-p)^2\lambda^{p-4}}{\pi^3}
\]
are both in ${\Gamma}_{\mathfrak p}$ (we have used that $p \geq 5$ here).
Furthermore, it is clear that
\[
\lambda x = \frac{\lambda^{p-1}}{\pi} = \frac{-p}{\pi}
\quad
\textrm{and}
\quad
\pi x = \lambda^{p-2}
\]
are both in ${\Gamma}_{\mathfrak p}$.
\end{proof}

We shall now consider three cases, the first two of which overlap.

\subsection{The case $[K(\zeta_{p}):K]>2$}

A consequence of Lemma \ref{elements-lemma} is that the
$\mathcal{O}_{K_{\mathfrak{p}}}$-module
$T:= \Gamma_{\mathfrak p}+x\mathcal{O}_{K_{\mathfrak p}} $ is in fact an
$\mathcal{O}_{K_{\mathfrak p}}$-algebra. Furthermore, we have
\[
\pi T = \pi \Gamma_{\mathfrak p}+\lambda^{p-2}\mathcal{O}_{K_{\mathfrak p}}
\subseteq \Gamma_{\mathfrak p} \subseteq T
\]
since $\lambda^{p-2}$ is part of an $\mathcal{O}_{K_{\mathfrak p}}$-basis of $\Gamma_{\mathfrak p}$.
We now let $S$ be the
$\mathcal{O}_K$-order defined by
\begin{eqnarray*}
S_{\mathfrak q} & = & \Gamma_{\mathfrak q} \quad ({\mathfrak q}\neq {\mathfrak p}); \\
S_{\mathfrak p} & = & T.
\end{eqnarray*}

We find that $\Gamma \subseteq S$ and $\pi S \subseteq \Gamma$
(note that we have abused notation in the obvious way here). Furthermore, the ring
$\bar{S}:=S/\pi S$ is isomorphic to
$S_{\mathfrak{p}}/\pi S_{\mathfrak{p}}=T/\pi T$.
Let $\bar{\Gamma}$ be the image of $\Gamma$ under the canonical map
$S\rightarrow \bar{S}$. We have a Milnor square
\[
\begin{array}{ccc}
\Gamma & \hookrightarrow & S \\
\downarrow &  & \downarrow \\
\bar{\Gamma}& \hookrightarrow &\bar{S}
\end{array}
\]
where the horizontal arrows are the natural inclusions and the vertical arrows are the natural projections (note that this is a special case of a fiber product).
Note that we have $\delta(\lambda^{p-2}) \equiv \omega^{p-2}(\delta)\lambda^{p-2}$ modulo $p{\mathbb Z}_p[\zeta_p]$ for every $\delta\in \Delta$, so $\delta(x) \in x+\Gamma \subset S$. Hence $\Delta$ acts on $S$ and so acts on each of the rings in the Milnor square.
By \cite[p.242]{curtis-reiner-II} we have the following exact sequence
\[
\K_{1}(S) \times \K_{1}(\bar{\Gamma})
\longrightarrow
\K_{1}(\bar{S})
\longrightarrow
\Cl(\Gamma)
\longrightarrow
\Cl(S)
\longrightarrow 0.
\]
As all the rings above are commutative, this becomes
\[
S^{\times} \times \bar{\Gamma}^{\times}
\longrightarrow
\bar{S}^{\times}
\longrightarrow
\Cl(\Gamma)
\longrightarrow
\Cl(S)
\longrightarrow 0.
\]
Hence we have an embedding of $\Delta$-modules
\[
N:=\frac{\bar{S}^{\times}}{\bar{\Gamma}^{\times} \cdot {\rm im}(S^{\times})}
\hookrightarrow
\Cl(\Gamma),
\]
where ${\rm im}(S^{\times})$ is the image of $S^{\times}$ under the map
$S\rightarrow \bar{S}$.

For every $\Delta$-module $X$, let $X^-$ and $X^{\omega^{-1}}$ denote the
minus part and $\omega^{-1}$-part of $\mathbb{Z}_p \otimes_{\mathbb{Z}} X$, respectively.
Then $X^{\omega^{-1}}\subseteq X^-$. We will show that $N^{\omega^{-1}}$ contains a submodule $M$ of order $p$. Note that by the definition of $x$ and the action of $\Delta$,
we have $x \in S^{\omega^{-1}}$. We define $\bar{x} \in \bar{S}$ to be the image of $x \in T$ under the natural projection $T \rightarrow T/\pi T \simeq \bar{S}$ and note that
$\bar{x} \in \bar{S}^{\omega^{-1}}$.

Let $[{\rm exp}](z):=\sum_{i=0}^{p-1}\frac{1}{i!}z^i$ denote the truncated exponential series.
Whenever the ideal $(a,b)$ generated by $a$ and $b$ satisfies $(a, b)^p=0$, we have $[{\rm exp}](a+b)=[{\rm exp}](a)\cdot [{\rm exp}](b)$ (see the proof of
\cite[$p$-elementary group schemes---constructions and Raynaud's theory, Remark 1.1]{Hopf-poly-Raynaud}). Let $y:=[{\rm exp}](\bar{x})\in \bar{S}$.
Since $y^p=[{\rm exp}](p\bar{x})=[{\rm exp}](0)=1$, we have $y\in {\bar{S}}^{\times}$.
We note that $y\notin \bar{\Gamma}$ (the summand with $i=1$ is $\bar x$ and hence plainly outside $\bar{\Gamma}$, and all other summands are in $\bar{\Gamma}$ by Lemma \ref{elements-lemma}). Moreover, as [exp] is compatible with the $\Delta$-action, we have $y\in (\bar{S}^{\times})^{\omega^{-1}} = \bar{S}^{\times} \cap \bar{S}^{\omega^{-1}}$.

\begin{lemma}\label{omega-equality}
We have
$(\bar{\Gamma}^{\times}\cdot {\rm im}(S^{\times}))^{\omega^{-1}}
=({\bar{\Gamma}^{\times}})^{\omega^{-1}}$.
\end{lemma}

\begin{proof}
Let ${\mathfrak M}$ denote the maximal order in $KS=K\Gamma$. Then
${\mathfrak M}={\rm ind}_{\Delta_0}^{\Delta}\mathcal{O}_{K(\zeta_p)}$ with
$\Delta_0=\Gal(K(\zeta_p)/K)$.
We consider ${S^{\times}}^-\subseteq {{\mathfrak M}^{\times}}^-$; since $K$ is totally real, complex conjugation $j \in \Delta_{0}$ acts on each factor of $\mathfrak{M}$ separately, and we see that $\mathcal{O}_{K(\zeta_p)}^{\times -}$ is the multiplicative group of roots of unity $\langle \zeta_{{p}^{f}} \rangle$ for some $f \geq 1$ (see \cite[Theorem 4.12]{wash}).
Hence ${\mathfrak{M}^{\times}}^-={\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^{f}} \rangle$.

Suppose that $f=1$. Then $\Delta_0$ acts on $\zeta_p$ via $\omega \vert_{\Delta_0}$,
and from the Frobenius reciprocity theorem one obtains that
${\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_p \rangle$ has non-trivial
$\omega^{-1}$-part if and only if ${\omega^{-1}}\vert_{\Delta_0}=\omega \vert_{\Delta_0}$, that is, if and only if $\omega^2$ is trivial on $\Delta_0$. But this is not the case since
$[K(\zeta_{p}):K]=|\Delta_0|>2$ by hypothesis.
Now suppose $f>1$. Then considering short exact sequence
\[
1 \longrightarrow {\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^{f-1}} \rangle
\longrightarrow {\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^f} \rangle
\longrightarrow {\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p} \rangle
\longrightarrow 1,
\]
we see that the middle term has trivial $\omega^{-1}$-part if and only if
the same is true of both the outer terms. It now follows by induction on $f$ that
${\mathfrak{M}^{\times}}^-={\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^{f}} \rangle$
has trivial $\omega^{-1}$-part. Hence $(S^{\times})^{\omega^{-1}}$ is trivial,
and the lemma is proved.
\end{proof}

Let $\bar{y}$ denote the projection of $y$ to $N$. If $\bar{y}$ were trivial
in $N$, then $y$ would have to be in $(\bar{\Gamma}^{\times}\cdot {\rm im}(S^{\times}))^{\omega^{-1}}=({\bar{\Gamma}^{\times}})^{\omega^{-1}}$. However, we have already
noted that $y$ is not even in $\bar{\Gamma}$. Hence $M:= \langle \bar{y} \rangle$ is a
non-trivial $\Delta$-submodule of $N$ with $M^{\omega^{-1}}=M$.

\subsection{The case $e \geq 4$}

Let $x_{1}=x=\frac{\lambda^{p-2}}{\pi} = \frac{1}{\pi} \otimes \lambda^{p-2}$ be as above and define $x_{2}= \frac{\lambda^{p-2}}{\pi^{2}} = \frac{1}{\pi^{2}} \otimes \lambda^{p-2}$.

\begin{lemma}\label{elements-lemma2}
We have $x_{2}^2, x_{2}^3, \lambda x_{2}, \pi^2 x_{2}, x_{1}x_{2}, x_{1}^{2}x_{2},
x_{1}x_{2}^{2}  \in {\Gamma}_{\mathfrak p}$.
\end{lemma}

\begin{proof}
We use that $p \geq 5$ without further mention.
Since $e \geq 4$, we have $\frac{p}{\pi^{4}} \in \mathcal{O}_{K_\mathfrak{p}}$.
Hence
\[
x_{2}^2=\frac{\lambda^{2p-4}}{\pi^4}=\frac{-p\lambda^{p-3}}{\pi^4}
\quad \textrm{ and } \quad
x_{2}^3=\frac{p^2\lambda^{p-4}}{\pi^6}
\]
are both in ${\Gamma}_{\mathfrak p}$.
Furthermore, it is clear that
\[
\lambda x_{2} = \frac{\lambda^{p-1}}{\pi^{2}} = \frac{-p}{\pi^{2}}
\quad
\textrm{and}
\quad
\pi^{2} x_{2} = \lambda^{p-2}
\]
are both in ${\Gamma}_{\mathfrak p}$.
Finally,
\begin{align*}
x_{1}x_{2} = \frac{\lambda^{2p-4}}{\pi^{3}} &= \frac{(-p)\lambda^{p-3}}{\pi^{3}},
\quad
x_{1}^{2}x_{2} = \frac{\lambda^{3p-6}}{\pi^{4}} = \frac{(-p)^{2}\lambda^{p-4}}{\pi^{4}},
\\
\textrm{and } \quad x_{1}x_{2}^{2} &= \frac{\lambda^{3p-6}}{\pi^{5}} = \frac{(-p)^{2}\lambda^{p-4}}{\pi^{5}}
\end{align*}
are all in $\Gamma_{\mathfrak{p}}$.
\end{proof}

A consequence of Lemmas \ref{elements-lemma} and \ref{elements-lemma2} is that
the $\mathcal{O}_{K_\mathfrak{p}}$-module
$T:=\Gamma_{\mathfrak{p}}+x_{1}\mathcal{O}_{K_\mathfrak{p}}+x_{2}\mathcal{O}_{K_{\mathfrak{p}}}$ is in fact
an $\mathcal{O}_{K_{\mathfrak{p}}}$-algebra. Furthermore, we have
\[
\pi^{2}T = \pi^{2}\Gamma_{\mathfrak{p}} + \pi \lambda^{p-2}\mathcal{O}_{K_\mathfrak{p}}
+ \lambda^{p-2}\mathcal{O}_{K_\mathfrak{p}}
= \pi^{2}\Gamma_{\mathfrak{p}} + \lambda^{p-2}\mathcal{O}_{K_\mathfrak{p}}
\subseteq \Gamma_{\mathfrak{p}} \subseteq T
\]
since $\lambda^{p-2}$ is part of an $\mathcal{O}_{K_{\mathfrak{p}}}$-basis of $\Gamma_{\mathfrak{p}}$.
We now let $S$ be the
$\mathcal{O}_K$-order defined by
\begin{eqnarray*}
S_{\mathfrak q} & = & \Gamma_{\mathfrak q} \quad ({\mathfrak q}\neq {\mathfrak p}); \\
S_{\mathfrak p} & = & T.
\end{eqnarray*}
Then $\Gamma \subseteq S$ and $\pi^{2}S \subseteq \Gamma$.
The same argument as in the previous case gives
an embedding of $\Delta$-modules
\[
N:=\frac{\bar{S}^{\times}}{\bar{\Gamma}^{\times} \cdot {\rm im}(S^{\times})}
\hookrightarrow
\Cl(\Gamma).
\]
By definition of $x_{1}, x_{2}$ and the action of $\Delta$, we have
$x_{1}, x_{2} \in S^{\omega^{-1}}$. Let $y_{1} = [{\rm exp}](\bar{x}_{1})$,
$y_{2} = [{\rm exp}](\bar{x}_{2}) \in \bar{S}=S/\pi^2 S$. As in the previous case,
both $y_{1}, y_{2}$ are elements of order $p$ in $(\bar{S}^{\times})^{\omega^{-1}}$.

\begin{lemma}\label{cyclic}
$(S^{\times})^{\omega^{-1}}$ is cyclic.
\end{lemma}

\begin{proof}
Let ${\mathfrak M}$ denote the maximal order in $KS=K\Gamma$.
As $S \subseteq \mathfrak{M}$, we have
${(S^{\times}})^{\omega^{-1}} \subseteq {({\mathfrak M}^{\times}})^{\omega^{-1}}$,
and so it suffices to show that $({{\mathfrak M}^{\times}})^{\omega^{-1}}$ is a cyclic group.

By the same argument as for Lemma \ref{omega-equality}, we obtain ${\mathfrak{M}^{\times}}^-={\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^{f}} \rangle$.
Furthermore, as $\omega^{-1}$ is an odd character, we have
$({\mathfrak M}^{\times})^{\omega^{-1}} \subseteq {{\mathfrak M}^{\times}}^-$.
Now $\langle \zeta_{p^{f}} \rangle$ is trivially cyclic as a $\Z_{p}[\Delta_{0}]$-module;
hence
\[
{\mathfrak{M}^{\times}}^-={\rm ind}_{\Delta_0}^{\Delta}\langle \zeta_{p^{f}} \rangle =
\Z_{p}[\Delta] \otimes_{\Z_{p}[\Delta_{0}]}\langle \zeta_{p^{f}} \rangle
\]
is cyclic as a $\Z_{p}[\Delta]$-module. Thus $({\mathfrak{M}^{\times}})^{\omega^{-1}} =
\Z_{p}(\omega^{-1}) \otimes_{\Z_{p}[\Delta]} {\mathfrak{M}^{\times}}^-$
is cyclic as a $\Z_{p}(\omega^{-1})$-module, where $\Z_{p}(\omega^{-1})$ is the ring extension of $\Z_{p}$ obtained by adjointing the image of $\omega^{-1}$. However, $\omega^{-1}$ takes its values in $\Z_{p}^{\times}$, and so $\Z_{p}(\omega^{-1})=\Z_{p}$. Therefore
$({\mathfrak M}^{\times})^{\omega^{-1}}$ is cyclic as a $\Z_{p}$-module, and hence is cyclic
as a group.
\end{proof}

Let $\tilde{y}_{1}, \tilde{y}_{2} \in
(\bar{S}^{\times}/\bar{\Gamma}^{\times})^{\omega^{-1}}$ be the images
of $y_{1}, y_{2}$ under the natural projection. Since $y_{1}, y_{2} \notin \bar{\Gamma}$
(the summand with $i=1$ is outside $\bar{\Gamma}$ and all others are in $\bar{\Gamma}$ by Lemmas \ref{elements-lemma} and \ref{elements-lemma2}), $\tilde{y}_{1}, \tilde{y}_{2}$ are also each
of order $p$.
Suppose that
$\tilde{y}_{1}^{k_{1}}\tilde{y}_{2}^{k_{2}}$
is trivial for some $k_{1}, k_{2} \in \{ 1, 2, \ldots, p-1\}$.
This would mean that $[{\rm exp}](k_{1}\bar{x}_{1}+k_{2}\bar{x}_{2})$ is in $\bar \Gamma$.
By virtue of Lemma \ref{elements-lemma2}, we  have $[{\rm exp}](k_{1}\bar{x}_{1}+k_{2}\bar{x}_{2})
\equiv 1+k_{1}\bar{x}_{1}+k_{2}\bar{x}_{2}$ modulo $\bar{\Gamma}$. Therefore we would obtain
$k_{1}\bar{x}_{1}+k_{2}\bar{x}_{2} \in \bar{\Gamma}$ and so
$k_{1}x_{1}+k_{2}x_{2} \in \Gamma_{\mathfrak{p}}$, which is impossible.
Hence the subgroup
$\langle \tilde{y}_{1}, \tilde{y}_{2} \rangle \simeq \langle y_{1}, y_{2}
\rangle \simeq \Z/p\Z \times \Z/p\Z$ is non-cyclic.
Let $\bar{y}_{1}, \bar{y}_{2}$ be the projections of $\tilde{y}_{1}, \tilde{y}_{2}$ to $N$
and let $M := \langle \bar{y}_{1}, \bar{y}_{2} \rangle \subseteq N^{\omega^{-1}}$.
Note that $M$ is non-trivial because $({\rm im}(S^{\times}))^{\omega^{-1}}$ is cyclic by Lemma
\ref{cyclic}, but $\langle \tilde{y}_{1}, \tilde{y}_{2} \rangle$ is non-cyclic. Hence $M$ is a non-trivial
$\Delta$-submodule of $N$ with $M^{\omega^{-1}}=M$.

\subsection{The case $[K(\zeta_{p}):K]=2$ and  $e=2$ or $3$}

Note that the condition $[K(\zeta_{p}):K]=2$ implies that $\frac{p-1}{2}$ divides $e$.
Hence we are reduced to considering the cases $p=5$ and $p=7$
(since $p \geq 11$ forces $e \geq 5$). If $p=5$, then $e$ must be even
and so in fact $e=2$. However, this case is excluded by hypothesis. If $p=7$, then
we must have $e=3$. In this case, we let $x_{1}=\frac{\lambda^{5}}{\pi}$,
$x_{2}=\frac{\lambda^{5}}{\pi^{2}}$ and $x_{3}=\frac{\lambda^{4}}{\pi}$. It is straightforward
to check that the $\mathcal{O}_{K_{\mathfrak{p}}}$-module
$T:=\Gamma_{\mathfrak{p}}+x_{1}\mathcal{O}_{K_{\mathfrak{p}}}+x_{2}\mathcal{O}_{K_{\mathfrak{p}}}
+x_{3}\mathcal{O}_{K_{\mathfrak{p}}}$ is in fact an $\mathcal{O}_{K_{\mathfrak{p}}}$-algebra.
The result is then given by a slight variant of the proof of
the previous case (note that $x_{1}, x_{2} \in S^{\omega^{-1}}$ but $x_{3} \notin S^{\omega^{-1}}$).

\subsection{The proof of Theorem \ref{real-not-HS}}

In each of the above cases, we have shown that there exists a non-trivial $\Delta$-submodule
$M$ of $\Cl(\Gamma)$ such that $M^{\omega^{-1}}=M$.

\begin{proof}[Proof of Theorem \ref{real-not-HS}]

Recall that the Stickelberger ideal is defined to be ${\mathcal J}={\mathbb Z}\Delta\cap \theta\cdot {\mathbb Z}\Delta={\rm Ann}_{\Delta}(\langle \zeta_p \rangle ) \cdot \theta$ where $\theta$ is the Stickelberger element $\frac{1}{p}\sum_{j=1}^{p-2}j\sigma_j^{-1}$. Let ${\mathcal J}_p\subseteq {\mathbb Z}_p\Delta$ be the $p$-completion  of $\mathcal J$. Then $\omega^{-1}({\mathcal J}_p)=\omega^{-1}({\rm Ann}_{\Delta}(\langle \zeta_p \rangle))\cdot {\omega}^{-1}\theta$. The second factor of the last expression is the generalized Bernoulli number $B_{1, \omega}$. Since $p\geq 5$, the first factor  is ${\mathbb Z}_p$.
By \cite[Corollary 5.15]{wash} we have
$B_{1, \omega}\equiv \frac{B_2}{2}=\frac{1}{12} \pmod p$. Hence, $M^{\mathcal J}=M^{\omega^{-1}({\mathcal J}_p)}=M^{{\mathbb Z}_p}=M$ and therefore
$\Cl(\Gamma)^{\mathcal{J}} \neq 0$.

Let $\Sigma$ denote the sum of the elements of $C_p$.
Consider the following Milnor square
\[
\begin{array}{ccc}
\mathcal{O}_KC_p &\stackrel{\alpha}{ \longrightarrow} & \mathcal{O}_KC_p/\mathcal{O}_K\Sigma =: \Lambda \\
\downarrow {\scriptstyle \beta} &  & \downarrow {\scriptstyle \gamma}  \\
\mathcal{O}_K&\longrightarrow &\mathcal{O}_K/p\mathcal{O}_K,
\end{array}
\]
where the horizontal maps are the natural projections, $\beta$ is the augmentation
map, and $\gamma$ is the map induced by augmentation. The resulting map
\[
\Cl(\mathcal{O}_KC_p)\stackrel{(\alpha, \beta)}{\longrightarrow}\Cl(\Lambda)\times \Cl(\mathcal{O}_K)
\]
is surjective (see, for instance, \cite[Corollary 49.28]{curtis-reiner-II}).
It follows immediately that
\[
\Cl'(\mathcal{O}_KC_p) \longrightarrow \Cl(\Lambda)
\]
is surjective. However, $\Cl(\Lambda) \simeq \Cl(\Gamma)$ since
$\Lambda \simeq \Gamma$, and so $\Cl(\Lambda)^{\mathcal{J}} \neq 0$.
Therefore $\Cl'(\mathcal{O}_KC_p)^{\mathcal J}=R(\mathcal{O}_KC_p) \neq 0$,
and so $K$ is not a Hilbert-Speiser field of type $C_p$.
\end{proof}

\section{Acknowledgments}

The authors are grateful to the Deutscher Akademischer Austausch Dienst (German Academic Exchange Service) for a grant allowing the second named author to visit the first for the 2006-07 academic year, thus making this collaboration possible.

The authors are indebted to James E. Carter for suggesting the original problem,
to Nigel P. Byott for pointing out an error in an earlier version of this paper, and to
both the aforementioned and the referee for several helpful comments and suggestions.

\bibliography{not-p-HS-Bib}{}
\bibliographystyle{amsalpha}

 \end{document}


text:
\subsection{Cut Sets in the Decomposition Space}
The next two lemmas use essentially the same ideas that go into the
proofs of \fullref{theorem:Bestvina} in \cite{Mar95} and
\cite[Proposition 2.1]{Ota92}.

\begin{lemma}\label{lemma:disconnected}
 If for some choice of basis $\Wh(*)$ is disconnected, then $\decomp$ is disconnected.
\end{lemma}
\begin{proof}
  Let $*$ be the identity vertex in $\tree$. Let $\mathcal{B}$ be a free
  basis of $F$ such that $\Wh_\mathcal{B}(*)$ is not
  connected. Vertices of $\Wh(*)$ are in bijection with
  $\mathcal{B}\cup\bar{\mathcal{B}}$.
There is some partition of
  $\mathcal{B}\cup\bar{\mathcal{B}}$ into subsets $\mathcal{A}$ and
  $\mathcal{A}'$ so that no lines of $\lines$ connect
  $\mathcal{A}$ to $\mathcal{A}'$.
Let \[\hat{A}=\cup_{a\in \mathcal{A}}\shadow_\infty^*(a)\]
The sets $\hat{A}$ and $\hat{A}^c\subset\bdry \tree$ are both nonempty
clopens, sets that are both open and closed.
Since there are no lines of $\lines$ with one endpoint in
$\hat{A}$ and one in $\hat{A}^c$, their images in $\decomp$ are
disjoint nonempty clopens, so $\decomp$ is disconnected.
\end{proof}

\begin{lemma}\label{lemma:basicconnectivity}
Suppose there exists a free basis $\mathcal{B}$ of $F$ such that
$\Wh_\mathcal{B}(*)$ is connected without cut vertices.
Let
$\tree$ be the Cayley graph of $F$ corresponding to $\mathcal{B}$.
Pick any edge $e$ in $\tree$.
Let $*$ and $v$ be the endpoints of $e$.
Let $\hat{A}=\shadow_\infty^*(v)$. That is, $\hat{A}$ is the ``half''
of $\bdry \tree$ on the ``$v$-side'' of $e$.
The set $A=q(\hat{A})$ is connected in $\decomp$.
\end{lemma}

\begin{proof}
  \begin{figure}[h]
\labellist \small
\pinlabel $\hat{A}$ [l] at 169 59
\pinlabel $e$ [t] at 78 58
\pinlabel $*$ [tl] at 35 58
\pinlabel $v$ [tr] at 126 58
\endlabellist

    \centering
    \includegraphics[width=.4\textwidth]{Pictures/hyperbolictreeseparated}
    \caption{The boundary of the tree split into ``halves''.}
    \label{fig:halfboundary}
  \end{figure}

  Suppose  there are open sets $B$ and $C$ of $\decomp$ such that $A\subset B\cup C$ and $A\cap B\cap C=\emptyset$.
The set $\hat{A}$ is open in $\bdry \tree$, so $A'=\hat{A}\cap q^{-1}(B)$ and $A''=\hat{A}\cap q^{-1}(C)$ are open.
Assuming that $A'$ is nonempty, we will show that $A''$ must be empty, which implies $A$ is connected.

$\hat{A}$ is closed in $\bdry \tree$, so $A'$ and $A''$ are closed.
Compactness of $\bdry \tree$ implies $A'$ and $A''$ are compact clopens.
Since $A'$ is compact and open, there are finitely many vertices
$x_1,\dots ,x_a$ so that $A'=\cup_{i=1}^a\shadow_\infty^*(x_i)$

There is a similar finite collection $y_1,\dots , y_b$ that determines $A''$.

Consider the convex hull $\hull$ of $\{x_i\}_{i=1}^a\cup\{y_j\}_{j=1}^b\cup
\{v\}$; it is a finite tree.
Call the vertices of $\hull$ other than $\{x_i\}_{i=1}^a\cup\{y_j\}_{j=1}^b\cup
\{v\}$ the ``interior vertices''.
Since $A'\cup A''=A$, $\hull$ includes all
edges incident to its interior vertices.
Let $\X$ be the union of the set of interior vertices with $\{v\}$.

Construct the Whitehead graph
$\Wh(\X)$.
It has $a+b+1$ vertices corresponding to the $x_i$ and $y_j$ and the
edge $e$ of $\tree$.

The graph is connected without cut points, since it can be
constructed by splicing together finitely many copies of
$\Wh(*)$, which is connected without cut points. In
particular, the vertex $e$ is not a cut vertex.

Assume $A'$ is nonempty. If $x_1=v$ then $A'=A$, so $A''=\emptyset$,
and we are done. Otherwise, $x_1$ is a vertex of
$\Wh(\X)\setminus v$. An edge of $\Wh(\X)\setminus v$
incident to $x_1$ corresponds to a line $l\in\lines$ with one
endpoint in the shadow of $x_1$ and the other endpoint in the shadow
of $z$ for some $z\in \{x_i\}_{i=2}^a\cup\{y_j\}_{j=1}^b$.
In the decomposition space these two endpoints are identified, and we
already know that the image of the first endpoint is in $B$. This
means that $z$ must be in $\{x_2,\dots ,x_a\}$.
Since $\Wh(\X)\setminus v$ is connected we conclude that all the vertices of $\Wh(\X)\setminus v$ belong to $\{x_1,\dots ,x_n\}$, so $A''=\emptyset$. Thus, $A$ is connected in $\decomp$.
\end{proof}

Thus, if $\Wh(*)$ is connected without cut vertices, then for any edge $e$ in $\tree$ the boundaries of
the two connected components of $\tree\setminus e$ correspond to connected
sets in the decomposition space. Since $\Wh(*)$ is connected
there is also at least one line in $\lines$ crossing $e$. This
means that these two connected sets in the decomposition space have a
point in common.

\begin{corollary}\label{corollary:connected}
Suppose $\Wh(*)$ has no cut vertices. Then the
decomposition space is connected if and only if $\Wh(*)$ is connected.
\end{corollary}

\begin{proof}[Proof of \fullref{theorem:Bestvina}]
Conditions (1) and (2) are equivalent by definition.
The equivalence of (1) and (3) is
a consequence of Whitehead's
Algorithm.

(3)$\implies$(4) is \fullref{lemma:disconnected}.

\fullref{corollary:connected} implies the contrapositive of (4)$\implies$(5).

It is always possible to eliminate cut vertices with Whitehead
automorphisms, so (5)$\implies$(3).
\end{proof}

Here is another corollary of \fullref{lemma:basicconnectivity}:
\begin{corollary}\label{corollary:edgeset}
 Suppose $\Wh(*)$ is connected without cut vertices. Pick any
 edge $e$ in $\tree$. Let $\tilde{S}$ be the collection of (finitely many)
 lines of $\lines$ that cross $e$. Then $S=q_*(\tilde{S})$ is a cut set in $\decomp$.
\end{corollary}

We will call such a set $S$ coming from all the lines crossing an edge
an \emph{edge cut set}.

From now on, unless otherwise noted, we will assume that any Whitehead
graph $\Wh(*)$ is connected without cut vertices. Thus, the
decomposition space is connected.
Our goal is to identify finite minimal cut sets.

We have an easy sufficient condition to see that a set $\tilde{S}=\{l_1,\dots ,
l_k\}\subset\lines$ gives a cut set $S=q_*(\tilde{S})$ of $\decomp$.
\begin{proposition}\label{proposition:interioredges}
  Let $\tilde{S}=\{l_1,\dots, l_k\}$ be a finite collection of lines in $\lines$. Let
  $\X$ be any compact, connected set in $\tree$.
In $\Wh(\X)$, delete the interior of any edge corresponding to one of the lines  $l_i$.
 If the resulting graph $\Wh(\X)\setminus \tilde{S}$ is disconnected then
  $S=q_*(\tilde{S})$ is a cut set.
\end{proposition}

The proof of this proposition is similar to
\fullref{corollary:edgeset}, but it will also be a special case of the
next proposition. Before moving on, though, let us consider an example
that shows that this proposition does not give a necessary condition for
$S$ to be a cut set.

Consider the
pattern $\lines$ generated by the pair of words $b$ and
$ab\bar{a}\bar{b}$ in $F=\left<a,b\right>$. The Whitehead graph (with
loose ends) for this pattern is shown in
  \fullref{fig:a-abAB}.

  \begin{figure}[h!]
    \centering
    \includegraphics[angle=90,height=\figstandardheight]{Pictures/wha-abAB}
    \caption{$\Wh(*)\{a,ab\bar{a}\bar{b}\}$ (loose)}
    \label{fig:a-abAB}
  \end{figure}

This graph is connected without cut points, so the decomposition space
is connected.
We claim that the endpoints of any $b$--line give a cut point in the
decomposition space. For instance, the $b$--line through the identity
vertex has endpoints $b^\infty$, $b^{-\infty}$ in $\bdry \tree$. Let $A=q(b^\infty)=q(b^{-\infty})$.

Let $*$ be
the identity vertex.
Let \[\hat{B}=\cup_{m\in\mathbb{Z}}\shadow^*_\infty(b^ma),\]
that is, $\hat{B}$ consists of all the boundary points $\xi$ of $\tree$ such that the first occurrence of $a$ or $\bar{a}$ in the
geodesic from the identity to $\xi$ is an $a$.

Now $\hat{B}$ is open, and every line of $\lines$ with one
endpoint in $\hat{B}$ has both endpoints in $\hat{B}$.
Let $B=q(\hat{B})$; the preimage is $\hat{B}=q^{-1}(B)$, so $B$ is open in
$\decomp$.
Similarly, let $B'$ be the image in $\decomp$ of the boundary points of $\tree$ such that the first occurrence of $a$ or $\bar{a}$ in the
geodesic from the identity is an $\bar{a}$.

$\decomp=A\cup B\cup B'$, and $A=\closure{B}\setminus
B=\closure{B'}\setminus B'$, so $A$ is a cut point.

For any compact, connected $\X$, $\Wh(\X)$ looks like a
circle with a number of disjoint chords (see
\fullref{fig:whb2-a-abAB}).

  \begin{figure}[h!]
    \centering
    \includegraphics[angle=90]{Pictures/Whb2-a-abAB}
    \caption{$\Wh(N_1(*))$}
    \label{fig:whb2-a-abAB}
  \end{figure}
The edges of the circle
correspond to $ab\bar{a}\bar{b}$--lines, and the chords correspond to
$b$--lines. This graph has no cut points, so deleting the interior of
an edge does not disconnect it.

This example has shown that to decide if $\tilde{S}$ gives a cut set, it is
not enough to delete the interiors of edges in a Whitehead graph.

There will be several different notions of deleting parts of Whitehead
graphs, so let us standardize notation. Let
$\X\subset\Y\subset\closure{\tree}$, and let $l\in\lines$ and
$\tilde{S}\subset\lines$. Let $e$ be an edge of $\tree$ incident to
exactly one vertex of $\X$.

The edge $e$ corresponds to a vertex in $\Wh(\X)$. The graph
$\Wh(\X)\setminus e$
is obtained from $\Wh(\X)$ by deleting this vertex, but retaining the
incident edges as \emph{loose ends at} $e$.

If $v$ is a vertex of $\tree$ that is distance 1 from $\X$, then there
is a unique edge $e$ with one endpoint equal to $v$ and the other in
$\X$. Define $\Wh(\X)\setminus v=\Wh(\X)\setminus e$.

Similarly, $\Wh(\X)\setminus \Y$ is obtained from $\Wh(\X)$ by deleting each vertex
of $\Wh(\X)$ that corresponds to an edge in $\Y$.
Visualizing Whitehead graphs in the tree, $\Wh(\X) \setminus \Y$ is the
portion of $\Wh(\Y)$ that passes through the set $\X$.

$\Wh(\X)\setminus l$ is obtained from $\Wh(\X)$ by deleting the interior
of the edge corresponding to $l$, if such an edge exists.
Similarly, obtain $\Wh(\X)\setminus\tilde{S}$ by deleting the interiors
of any edges corresponding to a line in $\tilde{S}$.

$\Wh(\X)\setminus \ddot{l}$ is obtained from $\Wh(\X)$ by deleting the interior of the edge corresponding to $l$ as well as the two
vertices that are its endpoints, retaining loose ends at these vertices.

Consider the line pattern $\lines$ generated by $ab\bar{a}\bar{b}$ and
$b$. Let $l$ be the $b$--line through the identity vertex $*$. Let
$\X=*$ and let
$\Y=[\bar{b},b]$. Figures~\ref{fig:whbabAB}--\ref{fig:whbabAB-Y} illustrate
our different notions of deleting from $\Wh(\X)$.

\begin{figure}[ht]
\begin{minipage}[b]{0.45\textwidth}
\labellist \small
\pinlabel $a$ [tl] at 164 84
\pinlabel $\bar{a}$ [tr] at 4 84
\pinlabel $b$ [r] at 82 163
\pinlabel $\bar{b}$ [r] at 82 3
\endlabellist
  \centering
  \includegraphics[height=\figstandardheight]{Pictures/whbabAB}
 \caption{\newline$\Wh(*)$}
\label{fig:whbabAB}
\end{minipage}
\hfill
\begin{minipage}[b]{0.45\textwidth}
\labellist \small
\pinlabel $a$ [tl] at 164 84
\pinlabel $\bar{a}$ [tr] at 4 84
\pinlabel $b$ [r] at 82 163
\pinlabel $\bar{b}$ [r] at 82 3
\endlabellist
  \centering
  \includegraphics[height=\figstandardheight]{Pictures/whbabAB-l}
 \caption{\newline$\Wh(*)\setminus l$}
\end{minipage}
\\
\bigskip
\begin{minipage}[b]{0.45\textwidth}
\labellist \small
\pinlabel $a$ [tl] at 164 80
\pinlabel $\bar{a}$ [tr] at 3 80
\pinlabel $b$ [r] at 79 163
\pinlabel $\bar{b}$ [r] at 78 3
\endlabellist
  \centering
    \includegraphics[height=\figstandardheight]{Pictures/whbabAB-ll}
  \caption{\newline$\Wh(*)\setminus\ddot{l}$}
\end{minipage}
\hfill
\begin{minipage}[b]{0.45\textwidth}
\labellist \small
\pinlabel $a$ [tl] at 164 80
\pinlabel $\bar{a}$ [tr] at 3 80
\pinlabel $b$ [r] at 79 163
\pinlabel $\bar{b}$ [r] at 78 3
\endlabellist
  \centering
    \includegraphics[height=\figstandardheight]{Pictures/whbabAB-Y}
  \caption{\newline$\Wh(*)\setminus[\bar{b},b]$}
\label{fig:whbabAB-Y}
\end{minipage}
\end{figure}

\begin{lemma}[Hull Determines Connectivity]\label{lemma:connectedhull}
  Let $S$ be a nonempty, finite subset of $\decomp$ that is not just a single bad point. Let $\hull$ be the convex hull of $q^{-1}(S)$. There is a bijection between connected components of $\Wh(\hull)$ and connected components of $\decomp\setminus S$.
\end{lemma}
\begin{proof}
Components $\mathcal{A}_i$ of $\tree\setminus \hull$ are the kind of sets in \fullref{lemma:basicconnectivity}.
Therefore, $q(\bdry \mathcal{A}_i)$ is connected in $\decomp$.
The set $\bdry \mathcal{A}_i$ is open in $\bdry \tree$.
For a subcollection $\{\mathcal{A}_{i_j}\}$ corresponding to a
connected component of $\Wh(\hull)$, we have that $q(\bigcup_j \bdry \mathcal{A}_{i_j})$ is an open connected set in $\decomp\setminus S$, as in the proof of \fullref{lemma:basicconnectivity}.
The complement of this set in $\decomp\setminus S$ is either empty or
is a union of sets of a similar form, corresponding to other connected
components of $\hull$. Thus, $q(\bigcup_j \bdry A_{i_j})$ is closed, and is therefore a connected component of $\decomp\setminus S$.
\end{proof}

Pick any vertex $*\in \tree$. If $\xi\in \decomp$ is a bad point, the previous argument applies if we take $\hull$ to be the ray $[*,\xi]$.
If $\Wh(*)$ is connected without cut points then $\Wh(\hull)$ is connected. Therefore:

\begin{corollary}
 No bad point of $\decomp$ is a cut point.
\end{corollary}

In a sense, \fullref{lemma:connectedhull} achieves our goal of
relating the topology of the decomposition space to generalizations of
the Whitehead graph.
However, this generalized Whitehead graph is infinite.
In the next two sections we show that the same information can be
obtained from a finite portion of this Whitehead graph.

\subsection{Identifying Cut Points and Cut Pairs}
The previous corollary tells us that any cut point is a good point, so
its preimage in $\bdry \tree$ is a pair of points.
We have a similar situation if there is a cut pair consisting of two
bad points; the preimage of such a set in $\bdry\tree$ is a pair of
points.
In both cases, the convex hull is a line.

Suppose $g\in F\setminus\{1\}$ is cyclically reduced with
$\hull^+=g^\infty$ and $\hull^-=g^{-\infty}$.
Let $\hull$ be the convex hull of these two points.
Let $\X=[*,g)$ be the segment joining the identity vertex to the $g$
vertex in $\tree$.

We know, by \fullref{lemma:connectedhull}, that the connected
components of $\decomp\setminus q(\{\hull^-,\hull^+\})$ are in bijection
with components of $\Wh(\hull)$.
We can construct $\Wh(\hull)$ by splicing together $g$-translates of
$\Wh(\X)\setminus \hull$.

$\Wh(\X)\setminus\hull$ is $\Wh(\X)\setminus\{e,ge\}$ for some edge $e$
incident to $*$ and
$ge$ incident to $g=g*$, so  $\Wh(\X)\setminus\hull$ has a collection of loose ends at $e$
and at $ge$.
The action of $g$ identifies $\Wh(\X)\setminus\hull$ with
$\Wh(g\X)\setminus\hull$, which has loose ends at $ge$ and $g^2e$.
The line pattern determines for us a splicing map for splicing the loose ends of
$\Wh(\X)\setminus\hull$ at $ge$ to the loose ends of
$\Wh(g\X)\setminus\hull$ at $ge$.

It is an easy consequence of the hypothesis that $\Wh(*)$ is connected
without cut vertices that for any segment $[*,g^k)\subset\hull$, every
component of $\Wh([*,g^k))\setminus \hull$ contains a loose end at $e$
(and a loose end at $g^ke$).
Thus, the number of components of $\Wh(\hull)$ is bounded above by the
number of components of $\Wh(*)\setminus\hull$.
To bound the number of connected components of $\Wh(\hull)$ below we need to
know if distinct connected components of $\Wh(\X)\setminus
\hull$ become connected when we splice on more translates.

Let $P$ be a partition of the loose ends of
$\Wh(\X)\setminus\hull$ at $e$ that is at least as coarse as
connectivity in $\Wh(\X)\setminus\hull$, ie,  if two loose ends belong to the
same connected component of $\Wh(\X)\setminus\hull$ then they belong in
the same subset of the partition.

Let $|P|$ be the number of subsets in the partition;
$P$ is \emph{nontrivial} if $|P|>1$.

Since $P$ is at least as coarse as connectivity, every vertex and edge in $\Wh(\X)\setminus\hull$ is connected to loose
ends in exactly one subset of $P$.
Let $P'$ be the partition of the loose ends of
$\Wh(\X)\setminus\hull$ at $ge$ such that two loose ends are in the
same subset of the partition if and only if they are connected to
loose ends at $e$ in a common subset of the partition $P$.

The $g$ action determines a partition $gP$ of the loose ends
of $\Wh(g\X)\setminus\hull$ at $ge$ by pushing forward the partition $P$.

We say the partition $P$ is \emph{compatible with the
  splicing map} if there is a bijection between subsets of the
partitions of $P'$ and $gP$ and the splicing map splices edges in a
subset of $P'$ to edges in the corresponding subset of $gP$.

The trivial partition is always compatible with the splicing map, but
this gives us no information. Another obvious partition to consider
would be the partition that comes from connectivity in
$\Wh(\X)\setminus\hull$.
The is the partition in which two loose ends of
$\Wh(\X)\setminus\hull$ at $e$ belong to the same subset
of the partition if and only if they belong to the same connected
component of $\Wh(\X)\setminus\hull$.
Suppose this partition is compatible with the splicing map.
This would mean that two loose ends of $\Wh(\X)\setminus\hull$ at $ge$
in the same connected component of $\Wh(\X)\setminus\hull$ must splice
to two loose ends of $\Wh(g\X)\setminus\hull$ at $ge$ in the same
connected component of $\Wh(g\X)\setminus\hull$, so splicing
introduces no new connectivity.
In this case it follows that for all $k\geq 1$ the number of connected components of
$\Wh([*,g^k))\setminus\hull$ is equal to the number of connected
components of $\Wh(\X)\setminus\hull$.

However, this is not always the case.
Splicing may introduce new
connectivity.
Compatibility of the partition controls how much new connectivity is introduced.
If we have a partition compatible with the splicing map, then, after
splicing, the partition $P$ is at least as coarse as connectivity in
$\Wh([*,g^2))\setminus\hull$.
Moreover, $P$ will still be compatible with the splicing map at
$g^2e$, so we may continue by induction to show:

\begin{proposition}
Suppose $P$ is a partition that is compatible
with the splicing map.
Then for any
segment $\Y=[*,g^k)$ of $\hull$, the number of
connected components of $\Wh(\Y)\setminus\hull$ is greater than or
equal to $|P|$.
\end{proposition}

Given a compatible partition, there are two cases to consider.
If for some $\Y$ we have no free edges in
$\Wh(\Y)\setminus\hull$ then the number of connected components of
$\Wh(\hull)$ is greater than or equal to $|P|$. In
particular, if $|P|\geq 2$, then $q(\{\hull^+,\hull^-\})$ is a bad
cut pair in $\decomp$.

Any particular line $l\in\lines$ overlaps with $\hull$ for distance at
most
\[|g| \times (2+\text{maximum number of consecutive $g$'s in a
generating word for } \lines.)\]
This is infinite if and only if $g$ is a generator of the line pattern.
Thus, $g$ is a generator of the line pattern if and only if
in every $\Y$ there is a free edge in
$\Wh(\Y)\setminus\hull$, since in this case $\hull=\closure{l}$ where
$l\in\lines$ is the $g$--line through the identity.

In this case we could have chosen the partition $P$ so that one of the
subsets is the singleton consisting of the loose end of the free edge.
The partition $P'$ also has a subset that is a singleton, consisting of the
other loose end of the free edge.
Such a partition has a \emph{segregated free edge}.

We do not see the free edge in $\Wh(\hull)$, so in general we can only
conclude that $\Wh(\hull)$ has at least $|P|-1$ connected components.
If $|P|-1\geq 2$ then $q(\{\hull^+,\hull^-\})=q_*(l)$ is a cut point
in $\decomp$.

\begin{proposition}[$q(\{g^\infty,\,g^{-\infty}\})$ Cut Set Criterion]\label{proposition:gcriterion}
  Let $g\in F\setminus\{1\}$ be an element of the free group.
With notation as above, let $P$ be the finest partition that is compatible
  with the splicing map and at least as coarse as connectivity in $\Wh(\X)\setminus\hull$. Then:
  \begin{enumerate}
  \item If $P$ is trivial then $q(\{g^\infty,\,g^{-\infty}\})$ is not a cut
    set.
\item If $P$ is nontrivial and has no segregated free edge then
$q(\{g^\infty,\,g^{-\infty}\})$ is a bad cut pair.
\item If $P$ has a segregated free edge and $|P|=2$ then $q(\{g^\infty,\,g^{-\infty}\})$ is not a cut
    set.
\item If $P$ has a segregated free edge and $|P|>2$ then
 $q(\{g^\infty,\,g^{-\infty}\})$ is a cut point.
 \end{enumerate}
\end{proposition}
\begin{proof}
  If $P$ is trivial then $\Wh([*,g^2))\setminus\hull$ is connected, so $q(\{g^\infty,\,g^{-\infty}\})$ is not a cut
    set.
Similarly, if $|P|=2$ and there is a segregated free edge then
$\hull=\closure{l}$ for $l\in\lines$ and  $\Wh([*,g^2])\setminus l$ is
connected, so $q(\{g^\infty,\,g^{-\infty}\})$ is not a cut
    set.

In the other cases, $\Wh(\hull)$ has multiple components, so $q(\{g^\infty,\,g^{-\infty}\})$ is a cut
    set.
\end{proof}

The proposition tells us that given a $g$ we can decide if
$q(\{g^\infty,\,g^{-\infty}\})$ is a cut set. We call this a
\emph{periodic} cut set.
Next we show that if there are cut points or cut pairs then there are
periodic cut sets:

\begin{proposition}\label{proposition:cutpaircriterion}
If $\decomp$ has cut points or cut pairs then there is some $R$
depending on $\lines$ and some $g$ with $|g|\leq R$ such that
$q(\{g^\infty,\,g^{-\infty}\})$ is a cut set.
\end{proposition}

To identify cut points we just need to apply
\fullref{proposition:gcriterion} to the generators of $\lines$, so in
this case it
is sufficient to take $R$ to be  the length of
the longest generator of $\lines$.
The work of proving \fullref{proposition:cutpaircriterion} lies in
finding an $R$ that works for the cut pair case:

\begin{lemma}\label{lemma:periodicbadpair}
  If $q(\{\hull^+,\hull^-\})$ is a cut pair then there is some $R$
  depending on $\lines$ and some $g\in F\setminus\{1\}$ with $|g|\leq R$ such that
  $q(\{g^\infty,\,g^{-\infty}\})$ is a cut set.
\end{lemma}

Note that  $q(\{g^\infty,\,g^{-\infty}\})$ is either a cut point or a
bad cut pair.

\begin{proof}
 Let $\hull$ be the convex hull of $\{\hull^+,\hull^-\}$.
We may assume that $\hull$ contains the identity vertex $*$.

Use $\#$ to denote number of connected components.

Every connected component of $\Wh(*)\setminus\hull$ contains an edge,
so the number of components is at most the complexity of $\Wh(*)$.

For any segment $\X$ of $\hull$ we have:
\[2\leq \#\Wh(\hull)\leq\#\left(\Wh(\X)\setminus\hull\right)\leq\#\left(\Wh(*)\setminus\hull\right)\leq \text{complexity
  of }\Wh(*)\]

Number the vertices of $\hull$ consecutively with integers with $*=v_0$
and index increasing in the $\hull^+$ direction.
Number the edges of
$\hull$ so that $e_i$ is incident to $v_{i-1}$ and $v_i$.
We consider these edges oriented in the direction of increasing index.
An oriented edge of $\tree$ comes with a label that is a generator or
inverse of a generator of $F$.

The function $f(i)=\#(\Wh([\hull^{-\infty},v_i])\setminus\hull)$ is
nonincreasing and, for high enough $i$, stabilizes at $\#\Wh(\hull)$.
Since we started with a cut pair, for high enough $i$ there is no
free edge in $\Wh([\hull^{-\infty},v_i])\setminus\hull$.
After changing by an isometry and relabeling, if necessary, we may
assume that $i=0$ is ``high enough'' in the previous two statements.

Fix a numbering from 1 to $c= \#\Wh(\hull)$ on the components of $\Wh(\hull)$.
At each $v_i$ we get a partition $P_i$ into $c$ subsets of the loose ends of $\Wh(v_i)\setminus\hull$ at
$e_i$ by connectivity in $\Wh(\hull)$.
Similarly, we get a partition $P_i'$ of the  loose ends of $\Wh(v_i)\setminus\hull$ at
$e_{i+1}$.
These partitions are at
least as coarse as connectivity in $\Wh([v_i,v_j])\setminus\hull$
for any $j\geq i$.

By construction, the splicing map at
$e_{i+1}$ connecting loose ends of $\Wh(v_i)\setminus\hull$ at
$e_{i+1}$ to loose ends of $\Wh(v_{i+1})\setminus\hull$ at $e_{i+1}$ is
compatible with the partitions $P_i'$ and $P_{i+1}$.

For each edge pair $(e_i,e_{i+1})$ there is a corresponding label pair
$L_i$ that gives a nontrivial word of length two in $F$. There are
$2n(2n-1)$ such words.

Let $m$ be the number of partitions of (complexity of $\Wh(*)$)
things into $c$ nonempty subsets.
Consider the segment $[v_0,v_R]$, where $R=2n(2n-1)m$.
Some label pair appears at least $m$ times.
Let $\{i_j\}_{j=1}^m$ be a set of indices such that the $L_{i_j}$
are the same.

Let $g_{j,k}$ be the element of $F$ that takes $v_{i_j}$ to
$v_{i_k}$.

If we fix $P_{i_1}$ we get a map of the elements $g_{1,k}$ into the set
of all possible partitions by $g_{1,k}\mapsto g_{1,k}P_{i_1}$, so for some $1\leq j<k$ we have $g_{1,j}$
and $g_{1.k}$ mapping to the same partition.
Therefore, $g_{j,k}P_{i_j}=P_{i_k}$.

$g=g_{i,j}$ is then the desired element.
\end{proof}

\begin{remark}
  In the preceding proof we found an element $g$ such that the $g$--action preserved a partition.
We did not insist that the $g$--action also fixed the numbering of
components of $\Wh(\hull)$; these may be permuted.
The proof may easily be modified to fix the numbering, at the expense
of a larger bound on $|g|$.
\end{remark}

\begin{corollary}\label{proposition:pairimpliesbadpair}
  Existence of a cut pair implies existence of a cut point or bad cut pair.
\end{corollary}

\begin{corollary}\label{corollary:longsegments}
  With $R$ as in the previous proposition, for any pair of points
  $\{\hull^+,\hull^-\}\subset\bdry\tree$, if $\X$ is a segment of the
  convex hull $\hull$ of $\{\hull^+,\hull^-\}$ of length greater than
  $R$, and if there are no cut pairs in the decomposition space, then one of the
  following is true:
  \begin{enumerate}
\item $\Wh(\X)\setminus\hull$ is connected.
\item $\Wh(\X)\setminus\hull$ has two components, one of which is a free edge.
  \end{enumerate}
\end{corollary}

\begin{theorem}[Detecting Cut Pairs]\label{theorem:findcutpairs}
There is a finite algorithm for detecting cut pairs in the
decomposition space
\end{theorem}
\begin{proof}
Given a list of words generating a line
pattern, apply Whitehead automorphisms, if necessary, to eliminate cut vertices.
If the graph becomes disconnected, stop; the decomposition space is disconnected by \fullref{corollary:connected}.

If it is possible to disconnect the Whitehead graph by deleting the
interiors of two edges, stop; these two edges correspond to a cut
pair. In particular, this happens if the Whitehead graph has any
valence two vertices.

Use \fullref{proposition:gcriterion} to check if any of the generators
of the line pattern give a cut point in the decomposition space. If
so, stop.

Let $R$ be
the constant from \fullref{lemma:periodicbadpair}.
The idea now is to check segments of length $R$ to see if we can
find a disconnected Whitehead graph. There are a lot of these. We
streamline the process by only checking those long segments for which
every sub-segment gives a disconnected Whitehead graph.

Let $\X_0=\{*\}$.

We proceed by induction. Suppose $\X_i$ is defined.

Start with $\X_{i+1}=\X_i$.
Consider pairs of points $v$ and $v'$ such that
$d(v,\X_i)=d(v',\X_i)=1$, such that
$d(v,*)=d(v',*)=i+1$, and such that $*\in[v,v']$.
If $\Wh(\X_i\cap [v,v'])\setminus [v,v']$ is
not connected, add $v$ and $v'$ to $\X_{i+1}$.

Continue until stage $k=1+\lceil \frac{R}{2}\rceil$.
Apply \fullref{corollary:longsegments} and \fullref{proposition:cutpaircriterion}:
there are pairs $v$ and $v'$ in $\X_k\setminus \X_{k-1}$ with
$*\in[v,v']$ such that
$\Wh(\X_{k-1}\cap[v,v'])\setminus [v,v']$ has more than one component
that is not just a free edge if and only if there are cut pairs in the
decomposition space.

\end{proof}

\begin{corollary}
  If a Whitehead graph for a line pattern has the property
  that deleting any pair of vertices leaves at most one free edge and
  at most one other connected component, then the decomposition space
  has no cut pairs.
\end{corollary}

Unfortunately, this corollary does not apply if a Whitehead graph has
more than one edge between a pair of vertices. Indeed, consider the
pattern in $F_2=\left<a,b\right>$ generated by the word
$a^2ba^2\bar{b}^2$. The Whitehead graph in \fullref{fig:badcomplete} is reduced and contains the
complete graph on the four vertices, but $q(\{a^\infty,a^{-\infty}\})$
is a cut pair, as is evident from \fullref{fig:badcomplete2}.

\begin{figure}[ht]
\begin{minipage}[b]{0.37\textwidth}

  \centering
 \includegraphics[height=\figstandardheight]{Pictures/wha2ba2B2}
 \caption{$\Wh(*)\{a^2ba^2\bar{b}^2\}$}
\label{fig:badcomplete}
\end{minipage}
\hfill
\begin{minipage}[b]{0.62\textwidth}

  \centering
  \includegraphics[height=\figstandardheight]{Pictures/wha2ba2B2-2}
 \caption{ $\Wh([*,a])\{a^2ba^2\bar{b}^2\}\setminus [a^{-1},a^2]$}
\label{fig:badcomplete2}
\end{minipage}

\end{figure}

\subsection{Cut Sets When There are No Cut Pairs}
Let $S$ be a finite set in the decomposition space, and let $\hull$ be
the convex hull of $q^{-1}(S)$.
\fullref{lemma:connectedhull} tells us that $S$ is a cut set if and
only if $\Wh(\hull)$ is disconnected.
We will pass to a finite subset of $\hull$ whose Whitehead graph
contains the same connectivity information.

Define the \emph{core} $\core$ of $q^{-1}(S)$, to be the smallest closed, connected set such that $\hull\setminus \core$ is a collection of disjoint infinite geodesic rays $\ray_j\from [1,\infty]\to \closure{\tree}$.
We use $\ray_j(0)$ to denote the vertex of the core that is adjacent to
$\ray_j(1)$.

Let $\xi$ be a point in $\bdry\tree$.
If $q(\xi)$ is either a cut point or a bad point that is a member of a
cut pair, it is not hard to see that there is a geodesic ray $\ray$
with $\ray(\infty)=\xi$ such that
$\Wh(\ray([1,\infty))\setminus\ray(0)$ is not connected.

Conversely, if there exists a geodesic ray
$\ray\from [0,\infty]\to\closure{\tree}$ with $\ray(\infty)=\xi$ such that
$\Wh(\ray([1,\infty))\setminus\ray(0)$ is not connected, then, arguing
as in the proof of \fullref{lemma:periodicbadpair}, $q(\xi)$ is either
a cut point or is a bad point that belongs to a cut pair.

If there are no cut points or cut pairs, then
$\Wh(\ray([1,\infty))\setminus\ray(0)$ is connected for any ray
$\ray$.

\begin{proposition}\label{proposition:nobadpoints}
Suppose $\xi$ is a point in $\bdry{\tree}$ such that $q(\xi)$ is
a bad point that is not a member of  a cut pair. Then $q(\xi)$ is
not a member of any
minimal finite  cut set. In particular, if $\decomp$ has no cut pairs then no
bad point belongs to any minimal finite cut set.
\end{proposition}
\begin{proof}
The assumption that $q(\xi)$ is not a member of a cut pair implies that for any ray $\ray\from [0,\infty]\to \closure{\tree}$
with $\ray(\infty)=\xi$, the Whitehead graph $\Wh(\ray([1,\infty]))\setminus \ray(0)$ is
connected.

Let $S$ be a finite cut set in $\decomp$ with $q(\xi)\in S$.
Let $\hull$ and $\core$ be the hull and core of $q^{-1}(S)$,
respectively.
Consider the ray $\ray$ that is the
component of $\hull\setminus \core$ containing $\xi$.

Components of $\decomp\setminus S$ are in bijection with components of
$\Wh(\hull)$, which, in turn, are in bijection with components of
$\Wh(\hull\setminus \ray([1,\infty]))$,
since $\Wh(\ray([1,\infty]))\setminus\ray(0)$ is connected.
This is just the hull of
$q^{-1}(S\setminus\{q(\xi)\})$.

Thus, $S\setminus\{q(\xi)\}$ is still a cut set, so $S$ was not minimal.
\end{proof}

For a finite collection of lines $\tilde{S}=\{l_1,\dots, l_k\}\subset\lines$, the core is a finite tree.
The convex hull minus the core is a collection of $2k$ disjoint rays:
\[\{\ray_i^\e\from [1,\infty]\to \tree\mid\lim_{t\to
  \infty}\ray_i^\e(t)=l_i^\e,\,\e\in\{+,-\},\,i=1\dots k\}\]

\begin{lemma}\label{lemma:openedges}
Let $S$ be a finite set of good points of $\decomp$, none of which is
a cut point. Components of $\decomp\setminus S$ are in bijection with components of
$\Wh(\core)\setminus \tilde{S}$
\end{lemma}
\begin{proof}
Let $\tilde{S}=\{l_1,\dots, l_k\}$.

For each $i$ and $\e$, since $q_*(l_i)$ is not a cut point, $\Wh(\ray_i^\e([1,\infty]))\setminus \ray_i^\e(0)$ is connected.

$\Wh(\hull)$ is obtained from
$\Wh(\core)\setminus \{\ddot{l}_1,\dots, \ddot{l}_k\}$ by splicing on each
$\Wh(\ray_i^\e([1,\infty]))\setminus \ray_i^\e(0)$.

This means to each deleted vertex of $\Wh(\core)\setminus \{\ddot{l}_1,\dots, \ddot{l}_k\}$ we have spliced on a connected graph, so we might have just
as well not deleted those vertices.
\end{proof}

In fact, we can use the argument of \fullref{lemma:openedges} to
reduce the convex hull even further.
If $\core$ is not just a vertex, then it has some valence one
vertices, that we call \emph{leaves}. The edge connecting a leaf to
the rest of the core is called the \emph{stem}.

Suppose that for some leaf $v$ of $\core$ every line of the $\tilde{S}$
that goes through $v$ goes through the stem of $v$. From $\Wh(v)$,
delete the interiors of edges corresponding to the $l_i$ and the
vertex corresponding to the stem.
The resulting graph is connected, so connected components of
$\Wh(\core)\setminus\tilde{S}$ are in bijection with connected
components of $\Wh(\core\setminus\{v\})\setminus\tilde{S}$.

Thus, we may \emph{prune} some leaves off of the core without changing
the connectivity of the Whitehead graphs.

If $\tilde{S}$ is not an edge cut set then we may prune the core down to a
well defined nonempty tree $\pcore$, the \emph{pruned core}, such that every leaf
contains a line of $\tilde{S}$ that does not go through the stem.

If $\tilde{S}$ is an edge cut set then the core can be pruned down to a pair
of adjacent vertices, both of which look like prunable leaves.
In this case define $\pcore$ to be these two vertices.

\begin{proposition}\label{proposition:edgeset}
 An edge cut set that does not contain a cut point is minimal.
\end{proposition}
\begin{proof}
  Let $\tilde{S}=\{l_1,\dots, l_k\}$ be the set of lines of
  $\lines$ going through an edge $e$ of $\tree$, so that $S=q_*(\tilde{S})$ is
  an edge cut set.
Let $\pcore$ be the pruned core.
There are two connected components of $\Wh(\pcore)\setminus\tilde{S}$;
they lie on opposite sides of $e$.
By \fullref{lemma:openedges} these correspond to two connected
components of $\decomp\setminus S$.

Each of the $l_i$ has one endpoint in
each component, so if any $l_i$ is omitted from
the set the two components will have a point in common.
\end{proof}

\begin{corollary}\label{corollary:topologicalmorality}
  If $\decomp$ has no cut pairs, the good points and bad
  points are topologically distinguished.
\end{corollary}
\begin{proof}
  Bad points are the points that do not belong to any minimal finite
  cut set. Good points are the points that do.
\end{proof}

\begin{proposition}\label{proposition:twocomponents}
 Let $S$ be a minimal finite cut set, none of whose elements are members of a cut pair.
There are  exactly two connected components of $\decomp\setminus S$.
\end{proposition}
\begin{proof}
By \fullref{proposition:nobadpoints}, $S$ consists of good points.
Let $\tilde{S}=\{l_1,\dots, l_k\}=q_*^{-1}(S)$.
Components of $\decomp\setminus S$ are in bijection with components of
$\Wh(\core)\setminus \tilde{S}$.
This is a finite graph, so $\decomp\setminus S$ has only finitely many components.

Let $A_1,\dots, A_m$ be a list of the components of $\decomp\setminus S$.

If $q_*(l_i)$ is not a limit point of $A_j$ in $\decomp$ then $A_j$ is still a connected component in $\decomp\setminus (S\setminus q_*(l_i))$.
This contradicts minimality of $S$, so each point of $S$ is a limit point in $\decomp$ of every $A_j$.
 This implies that for each $i$ and $j$, at least one of the points $l_i^+$ and $l_i^-$ is a limit point of $q^{-1}(A_j)$.

Now $\hull\setminus \core$ is a collection of disjoint rays $\ray_i^\e$.
The graph $\Wh(\ray_i^\e([1,\infty]))\setminus \ray_i^\e(0)$ is connected, so  no $l_i^+$ or $l_i^-$ is a limit point of more than one $q^{-1}(A_j)$.

Thus, there are exactly two components $A_1$ and $A_2$ of $\decomp\setminus S$, and each line $l_i$ has one endpoint in $q^{-1}(A_1)$ and the other in $q^{-1}(A_2)$.
\end{proof}

\begin{corollary}\label{corollary:bothcomponents}
  Let $S$ be a minimal finite cut set that is not an edge cut set,
  none of whose elements are members of a cut pair.
For
  every vertex $v\in \pcore$, the portion of
  $\Wh(\pcore)\setminus\tilde{S}$ at $v$ contains an edge from each component of
    $\Wh(\pcore)\setminus\tilde{S}$.
\end{corollary}
\begin{proof}
If $v$ is a leaf such that the portion of
$\Wh(\pcore)\setminus\tilde{S}$ at $v$ belongs to a single component of
$\Wh(\pcore)\setminus\tilde{S}$ then $v$ should have been pruned
off.

If $v$ is not a leaf, $\pcore\setminus\{v\}$ has at least two components. If the Whitehead
graph over one of those components sees only one component of
$\Wh(\pcore)\setminus\tilde{S}$ then it would have been
possible to prune it off. Thus, every component of
$\pcore\setminus\{v\}$ must see two components of
$\Wh(\pcore)\setminus\tilde{S}$.
There are only two components of
$\Wh(\pcore)\setminus\tilde{S}$, so both must be able to
connect to all components of $\pcore\setminus\{v\}$. In particular,
they must connect through $v$.
\end{proof}

\endinput

config: arxiv