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\newcommand{\z}{ {\zeta} } 
\newcommand{\e}{ {\epsilon} }
\newcommand{\la}{\lambda}
\newcommand{\Qed}{ $\square$ } 
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\newcommand{\bb}{ \frak{b}}
\newcommand{\PP}{ \frak{P}}
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\newcommand{\Q}{{\Bbb Q}}
\newcommand{\ZZ}{{\Bbb Z}}
\newcommand{\Norm}{{\mbox{Norm}}}
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\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
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\newtheorem{prop}{Proposition}

\def\Proof{\par\noindent{\bf Proof\/}:\enspace}



\title{Non-existence of Siegel Zeros in Towers of Radical Extensions}
\author{Andrew M. Odlyzko}
\address{AT\&T Bell Laboratories, Murray Hill, NJ 07974}
\thanks{1991 Mathematics Subject Classification:
Primary 11R42,
Secondary 11M41, 11S20, 11S40}
\author{Chris M. Skinner}
\address{Current address of Skinner: Department of Mathematics, University of Michigan, Ann Arbor, MI 48109}
\thanks{
{\em Key words and phrases.}
Siegel zero, Dedekind zeta function,
radical extension, Artin's conjecture.
This is the final version of this paper}
\dedicatory{Dedicated to the memory of Emil Grosswald}
\begin{document}
\maketitle
\begin{abstract}
Given a number field $K$, it is shown that for a tower of radical
extensions of odd degree over $K$, a Siegel zero of any Dedekind
$\z$-function of a field in the tower
must be a zero of the $\z$-function of $K$.  In particular, towers of
radical extensions of odd degree over $\Q$ have no Siegel zeros.
This result is derived from
a more general theorem, covering wider classes of field extensions.
The basic results
are obtained by showing that a Siegel zero not coming from a lower field
would be a zero of high multiplicity of some normal extension of the
field under consideration, a multiplicity greater than allowed
by simple bounds.
\end{abstract}

\section{Introduction}

\ \ \ \ \ 
Let $K$ be an algebraic number field of degree $n_{K}$ and let $d_{K}$
be the absolute value of its discriminant.  Let $\z_{K}(s)$ denote the Dedekind $\z$-function
of $K$.  It is well known that there is an effectively computable constant $c$ such that
for any $K$, $\z_{K}(s)$ has at most one zero $\sigma+it$ in the range
$$ \sigma \geq 1-\frac{c}{\log d_K} ,  \:\:\:\:
   |t| \leq \frac{c}{\log d_K} \: ; $$
if such a zero exists, it must be real and simple.  Stark \cite{Stark} has shown that
one may take $c=\frac{1}{4}$.  Exceptional zeros of this kind are called
Siegel zeros.  It is conjectured that there are no Siegel zeros, but
this has not been proved in general.

The main motivation for studying Siegel zeros arises because
they determine the size of the residue $\kappa$ of $\z_{K}(s)$ at $s=1$,
and thus the product of the class number and the regulator of $K$.  Good
unconditional upper bounds for $\kappa$ can be derived easily, but
lower bounds are much harder to obtain.
It was first observed by Gronwall \cite{Gronwall} and
Hecke (see Landau \cite{Landau}) that for quadratic fields non-existence of Siegel
zeros yields good lower bounds for $\kappa$.  In general, it is known that if
$\z_{K}(s)$ has no Siegel zero, then $\kappa\geq c'(\log d_K)^{-1}$ for
an effectively computable constant $c'$ that depends only on $c$.
The Brauer-Siegel Theorem \cite{Brauer} yields a lower bound for $\kappa$ as $K$ runs
over certain sequences of normal extensions of $\Q$ by using the
argument of Siegel that few fields can have Siegel zeros.
However, the Brauer-Siegel result is ineffective.
  
Siegel zeros also play a role in determining the distribution of prime
ideals.  For example, the asymptotic size of the error in the prime
ideal theorem, which counts prime ideals with norm $\leq x$, is
determined by how close the zeros of $\z_K(s)$ come to the line
Re$(s)=1$, with the influence of any single zero being negligible for $x$
large.  However, if there is a Siegel zero, $x$ has to be extremely
large before the influence of that zero becomes small.
Therefore
in statements of the prime ideal theorem that give explicit dependence on
the parameters of the field, the influence of a possible Siegel zero is
usually included separately from that of other zeros.

The best general results to date about Siegel zeros and the concomitant bounds
for residues are due to Stark \cite{Stark}, who significantly generalized and
strengthened earlier results of Sunley \cite{Sun}, Goldstein \cite{Gold}, and Heilbronn \cite{Heilbronn}.  Other bounds on Siegel zeros of
certain number fields have been obtained recently by Hoffstein and
Jochnowitz \cite{Hoff}, \cite{Hoff2}.

In what follows, we redefine a Siegel zero by altering the constant $c$.
This is done to ensure that certain technical arguments go through.
This paper is concerned with showing that for an appropriate choice of
$c$, $\z_K(s)$ often has no Siegel zeros.  We
have not worried about getting the best possible constants, and it
may be possible to prove that the results below hold with $c_0 =\frac{1}{4}$ or
an even larger constant.

Recall that a radical extension of a number field $K$ is a field
$L=K[\alpha]$ such that $\alpha^n\in K$ for some integer $n$.  We prove
the following theorem about Siegel zeros of radical extensions.

\begin{theorem}
\label{the1}
There is an effectively computable constant $c_0 > 0$ with the
following property.
Let $F_0\subset F_1\subset \cdots \subset F_m$ be a tower of extensions
of number fields such that $[F_i:F_{i-1}]$ is odd and $F_i$ is a radical
extension of $F_{i-1}$ for $i=1,
\ldots ,m$.  If $\beta$
is a simple real zero of $\z_{F_{m}}(s)$ in the range
$$ 1-\frac{c_0}{\log d_{F_m}} \leq \beta < 1, $$
then $\beta$ is a simple real zero of $\z_{F_0}(s)$.
\end{theorem}

Since $\z_\Q(s)$ has no real zeros with $0<\beta<1$, we have the following
corollary.

\begin{corollary}
Let $\Q=F_0\subset F_1\subset \cdots \subset F_m$ be a tower of
extensions such that $[F_i:F_{i-1}]$ is odd and $F_i$ is a
radical        extension of $F_{i-1}$ for $i=1,
\ldots,m$.  Then $\z_{F_m}(s)$ has no zeros in the region
$$ \sigma \geq 1-\frac{c_0}{\log d_{F_m}} , \:\:\:\: 
   |t| \leq \frac{c_0}{\log d_{F_m}}. $$ 
\end{corollary}

For towers of radical extensions, Corollary 1 is much stronger than the corresponding result of Stark,
which has $n_{F_m}\log d_{F_m}$ in place of $\log d_{F_m}$
\cite[Lemma~8]{Stark}.
The results of Sunley, Goldstein, Heilbronn, and Stark are
strongest for normal extensions of number fields.
For radical extensions, their results give little
information. 

The restriction to odd degree extensions in Theorem~\ref{the1} is necessary to
exclude the possibility of a Siegel zero of $\z_{F_m}(s)$ coming from
$\z_E(s)$ where $E$ is a quadratic extension of some $F_i$.  In
the study of Siegel zeros, quadratic extensions are the hardest to
deal with.  As far as anyone knows, a quadratic number field $K$ can
have a Siegel zero within about $d_K^{-1/2}$ of 1.  A standard result
(cf. Lemma~11 of \cite{Stark}) shows that the hypothetical Siegel zero
is $< 1-c''d_K^{-1/2}$ for some $c''>0$, and the recent work of
Gross and Zagier \cite{Gross} allows one to increase the distance from 1
only
by a power of $\log d_K$.

Theorem~\ref{the1} is a consequence of the following more general result.
\begin{theorem}
\label{the2}
There is an effectively computable constant $c_0 > 0$ with the following property.
Let $F_0 \subset F \subset \cdots \subset F_m$ be a tower of
extensions of number fields.
Assume that for each $i$,
$1 \le i \le m$,
there is no number field $F'_i$ such that $F_{i-1} \subset F'_i \subset F_i$.
Let $F_i^\ast$ be the normal closure of $F_i$ over $F_{i-1}$,
and assume that the Galois group $G_i = \Gal (F_i^\ast / F_{i-1} )$ is solvable.
Let $A_i$ be the minimal normal subgroup of $G_i$ with $|A_i | > 1$, and let $H_i$ be the subgroup fixing $F_i$.
If for each $i$, $1 \le i \le m$,
either (i)~$G_i =A_i$ and $|G_i|$ is odd, or (ii)~$H_i$ is abelian with $|H_i| > 1$,
then any real zero $\beta$ of $\zeta_{F_m} (s)$ in the range
$$
1 - \frac{c_0}{\log d_{F_m}} \le \beta < 1$$
is a simple zero of $\z_{F_0} (s)$.
\end{theorem}

The derivation of Theorem \ref{the1} from Theorem \ref{the2}
will be sketched in Section~2.
At this point we note that Theorem~\ref{the2} is more general, as it shows, for example,
that if $L$ and $K$ are number fields with $L$ cubic over $K$, then any Siegel zero of $\zeta_L (s)$ is a
Siegel zero of $\zeta_K (s)$.

We note here that we define a Siegel zero with respect to the field, and
not to any of the $L$-functions it is a zero of.  Thus, for example, if
$\beta$ is a Siegel zero of the $\z$-function of the field $K=\Q[\sqrt{p}]$ for
a prime $p$, then it is also a zero of the $\z$-function of the field
$L=\Q[\e_p]$, where $\e_p$ is a primitive $p$th root of unity.  However,
$\beta$ is not a Siegel zero of $\z_L(s)$ since
$\beta<1-c'' p^{-1/2}$,
whereas $\log d_L$ is of order $p\log p$.  

Zeros of zeta functions of radical extensions are especially interesting
because of their connection with Artin's conjecture that any integer $a$
not equal to $-1,0,$ or a perfect square is a primitive root for
infinitely many primes (and even for a positive fraction of all primes).
Hooley \cite{Hooley}, \cite{Hooley2} proved this conjecture under the assumption of the Generalized
Riemann Hypothesis.  Unfortunately, our results do not help to obtain an
unconditional proof, because a much wider zero-free region appears to be
needed (cf. \cite{Hooley2}), and in any event, real zeros of $\z_K(s)$ for
$K=\Q[\sqrt[p]{a}]$ would only help in bounding the error terms that
arise in the proof.

The results of Sunley, Goldstein, Heilbronn, and Stark were proved
by showing that a Siegel zero of $\z_K(s)$ had to come from
$\z_F(s)$ for some subfield $F$ of $K$, usually a quadratic subfield.
Our proof relies on showing that for some normal extension $L$
of $K$, a Siegel zero of $\z_K(s)$
would imply the existence of many zeros of $\z_L(s)$ near $s=1$, more zeros than allowed by
known bounds.  This argument can be generalized to classes of
fields other than those we consider, but its applicability is limited by the
requirement that certain relations hold for the representations of the Galois group of
the field being considered. 
What is essential for our method of proof is that
in each stage of the tower of extensions,
the normal closure $M$ of an extension $L$ of $K$
is the
compositum of only a few of the fields conjugate to $L$,
so $d_M$
is not large, and that $\z_M (s) / \z_K (s)$ is the product of a high
power of $\z_L (s) / \z_K (s)$ and other well-behaved functions.
The group representation arguments we use are similar to those employed by
Uchida \cite{Uchida} and van~der Waall \cite{Waall,Waall1}
in their proofs that if $M$ is a normal extension of a number field $K$ with a solvable
Galois group, then the quotient $\zeta_L (s) / \zeta_K (s)$
is entire for any number field $L$ with $K \subseteq L \subseteq M$.
It would be
desirable
to prove a strengthening of Theorem~\ref{the2}, in which the only requirement
would be that
the groups $\Gal (F_i^\ast / F_{i-1} )$ have to be solvable.
Any such strengthening would have to allow for the possibility of a Siegel
zero coming from a quadratic subfield.
As we remark at the end of Section~3, there are
indications that our arguments cannot be extended too far.

We prove Theorems~\ref{the1} and \ref{the2} in Section~2.
We conclude in Section 3 with some observations about the residue
of $\z_{F_i}(s)$ at $s=1$ and the analyticity of the functions
$\z_{F_i} (s) / \z_{F_j} (s)$
$(i > j )$.

\section{Proof of Theorem~\protect\ref{the2} and derivation of Theorem~\protect\ref{the1} from Theorem~\protect\ref{the2}}

\ \ \ \ \ 
In this section, we prove some results that imply Theorem~\ref{the2}.
The first lemma presents
an estimate of the number of zeros of
$\z_K(s)$ in a region near 1
that is valid for all number fields $K$.

\begin{lemma}
\label{lem1}
Let $n(r)$ denote the number of zeros $\rho$ of $\z_K(s)$ with
$|1-\rho|\leq r$.  Then for all $r>0$,
\begin{equation}
\label{eq101}
 n(r) < 1+c_1r\log d_K 
\end{equation}
where $c_1$ is an effectively computable constant
that is independent of $K$.
\end{lemma}

\Proof For $r \leq (4\log d_K)^{-1}$ the previously mentioned
result of Stark implies the truth of (1).  Suppose that $r > (4\log
d_K)^{-1}$.  By Lemma 2.2 of \cite{LMO}, we know that
$$ n(r) < c_2(1+r\log d_K) $$
for some effectively computable constant $c_2$.  (We have used $n_K < c_3\log d_K.)$
Put $c_1=5c_2.$  Then
$$ c_2(\frac{1}{r\log d_K}+1) < 5c_2 = c_1. $$
Hence,
$$ n(r) < c_2(1+r\log d_K) \leq 1+c_1r\log d_K. \:\:\:$$
\begin{flushright} \Qed \end{flushright}

Theorem~\ref{the2} follows from
the following result.
\begin{theorem}
\label{the3}
There is an effectively computable constant $c_0 > 0$ with the following property.
Let $L \supset K$ be number fields with no intermediate field (i.e., no field $L'$
such that $L \supset L' \supset K$).
Let $M$ be the normal closure of $L$ over $K$, and assume that $G = \Gal (M/K )$ is solvable.
Let $A$ be the minimal normal subgroup of $G$ with
$|A| > 1$, and let $H$ be the subgroup of $G$ fixing $L$.
If either (i)~$G=A$ and $|G|$ is odd or (ii)~$H$ is abelian with $|H|>1$,
then any real zero of $\zeta_L (s)$ in the range
\begin{equation}
\label{eq102}
1 - \frac{c_0}{\log d_L} \le \beta < 1
\end{equation}
is a simple zero of $\zeta_K (s)$.
\end{theorem}

Theorem~\ref{the2} follows from repeated application of Theorem~\ref{the3} to each stage of the tower of extensions.

From now on we concentrate on proving Theorem~\ref{the3}.
We first note that if case (i) of Theorem~\ref{the3} holds,
then the conclusion of the theorem follows from Heilbronn's results \cite{Heilbronn},
which say that if $L/K$ is a normal extension of number fields,
and $K'$ is the compositum of quadratic subfields of $L/K$,
then any real simple zero of $\zeta_L (s)$ is a zero of $\zeta_{K'} (s)$.
Therefore from now on we assume that we are in case~(ii).

\begin{lemma}
\label{lem2}
Under the above conditions,
(i) $G= AH$, (ii)~$A \cap H = \{ 1 \}$,
(iii)~$A$ is elementary abelian, 
and (iv)~$H \cap (g Hg^{-1} ) = \{ 1 \}$ for any $g \in G \setminus H$~( so that
$H$ is a Frobenius complement in $G$).
\end{lemma}

\Proof (i)~Since $L/K$ has no proper subfields,
$H$ must be a maximal subgroup of $G$.
Therefore $AH = G$ or $H$.
If $AH = H$, then $A \subseteq H$.
Let $M' \subset M$ be the subfield fixed by $A$.
Then $L \subseteq M'$ and $M'$ is normal over $K$,
contrary to the choice of $M$.
This is a contradiction, so $AH =G$.

(ii)~$A \cap H$ is normal in $AH =G$, so $A \cap H = \{1\}$ as $A$ is minimal.

(iii)~See Satz 9.13 on p.~52 of \cite{Huppert}.

(iv)~Suppose this claim is not true, and that $h_1 = g h_2 g^{-1}$ for some $h_1$, $h_2 \in H$,
$g \in G$, $h_1 \neq 1$.
We can write $g = ah$, $a \in A$, $h \in H$.
We consider $a$ and $h_1$ to be fixed from now on.
Since $H$ is abelian,
$$h_1 = gh_2 g^{-1} = a h_2 a^{-1} ~.$$
Then
\begin{equation}
\label{Aeq3}
x = ah_2 a^{-1} h_2^{-1} = h_1 h_2^{-1} ~.
\end{equation}
Since $A$ is normal, $h_2 a^{-1} h_2^{-1} \in A$, so $x = a h_2 a^{-1} h_2^{-1} \in A$.
But by (\ref{Aeq3}), $x = h_1 h_2^{-1} \in H$, and therefore, by
part~(ii) of this lemma, $x=1$ and $h_1 = h_2$.

Let $H' = a Ha^{-1}$, and let $F$ be the subgroup of $G$ generated by
$\{ ha h^{-1} : h \in H \}$.
$F$ is a subgroup of $A$, since $A$ is normal, and is fixed under
conjugation by H. Since $A$ is abelian, this means that $F$ is a normal subgroup
of $G$, and therefore $F = A$ or $\{1\}$.
But $a \in F$, $a \neq\ 1$, so we must have $F=A$.

Since $F=A$, any element of $A$ can be written as a product of elements
of the form $h_3 a h_3^{-1}$,
$h_3 \in H$.
But $H$ is abelian, and $h_1$ commutes with $a$, so
$h_1$ commutes with every element of $A$.
Therefore the subgroup $H_1$ generated by $h_1$ is normal in $G$, and so the fixed field
of $H_1$ is a normal extension of $K$ that lies between $L$ and $M$.
Since $M$ is the minimal normal extension of $L$ over $K$, we must have $H_1 = \{1\}$, which means
that $h_1 = 1$, and this contradicts the basic assumption.
Thus we must have $H \cap (gHg^{-1} ) = \{1\}$ for all $g \in G \setminus H$.
\begin{flushright} \Qed \end{flushright}

\begin{lemma}
\label{lem3}
Under the conditions above,
\begin{equation}
\label{eq103}
d_M | d_L^{2f} ~,
\end{equation}
where $f = |H|$.
\end{lemma}

\Proof Let $L' \neq L$ be a conjugate of $L$ over $K$.
We will prove below that $M= L \cdot L'$.
Once this is established, we apply
Lemma~7 of \cite{Stark}
and obtain
$$d_M | d_L^f d_{L'}^f ~,$$
which yields (\ref{eq103}).

It remains to show that $M = L \cdot L'$.
The field $L'$ is fixed by a conjugate of $H$, say $gHg^{-1}$,
where $g \in G \setminus H$.
By Lemma~\ref{lem2}, $H \cap ( gHg^{-1} ) = \{1\}$.
If $L \cdot L'$ were a proper subgroup of $G$,
it would be fixed by a nontrivial subgroup of $G$, which
would also fix $L$ and $L'$.
This would contradict $H \cap (gHg^{-1} ) = \{1\}$.
\begin{flushright} \Qed \end{flushright}

We next investigate the characters of $G$.
If $\eta$ is a character of a subgroup of $G$, we let $\eta^\ast$ denote the character
of $G$ induced by $\eta$.

\begin{lemma}
\label{lem4}
Under the above conditions on the group $G$, the nonlinear irreducible characters of $G$ are of the form $\lambda^\ast$, where $\lambda$ denotes an
irreducible character of $A$, and
are of degree $f = |H|$.
\end{lemma}

\Proof See p.~199 of \cite{Isaacs}.
\begin{flushright} \Qed \end{flushright}

Let $\{ \chi_i \}$,
$\{ \phi_i \}$, and $\{ \la_i \}$ be the sets of irreducible characters of
$G$, $H$, and $A$, respectively.
Let $\chi_0$, $\phi_0$, and $\la_0$ be the identity characters of $G$,
$H$, and $A$.
We then have
\begin{equation}
\label{eq104}
\z_M (s) = \prod_i L(s, \chi_i )^{\chi_i (1)}
\end{equation}
and
\begin{equation}
\label{eq105}
\z_L (s) = L( s, \phi_o^\ast ) ~,
\end{equation}
where $L(s, \eta )$ is the Artin $L$-function of $\eta$.
Let
\begin{equation}
\label{eq106}
\phi_0^\ast = \sum_i s_i \chi_i ~,
\end{equation}
where the $s_i$ are integers, $s_i \ge 0$.
We will show that
\begin{itemize}
\item[(A)]
Every irreducible nonlinear character $\chi_i$ of $G$ appears in (\ref{eq106}) with
$s_i =1$.
\item[(B)]
The only linear character $\chi_i$ of $G$ with $s_i \neq 0$ is the identity
character $\chi_0$.
\end{itemize}


To prove (A),
note that if $a \in A$, $h,h_1 \in  H$, then since $H$ is abelian (as we are in case (ii)
of Theorem~\ref{the3}),
$$(a h_1 ) h ( ah_1)^{-1} =
aha^{-1} ~.
$$
If $aha^{-1} = a'$ for some $a' \in A$, then
$h = a^{-1} a' a \in A$, and so by Lemma~\ref{lem2},
$h = 1$.
Hence for $h \in H$,
\begin{eqnarray*}
\la_j^\ast
(h) & = & \frac{1}{|A|} \sum_{x \in G}
\la_j (xhx^{-1} )
~=~
\frac{1}{|A|} \sum_{a \in A} \sum_{h_1 \in H}
\la_j ((ah_1) h(ah_1)^{-1} ) \\
& = & \frac{|H|}{|A|}
\sum_{a \in A} \la_j (aha^{-1} )~=~
\left\{
\begin{array}{cl}
|H| & \mbox{if} ~~ h=1 ~, \\
0 & \mbox{otherwise} ~,
\end{array}
\right.
\end{eqnarray*}
since $G= AH$ by Lemma~\ref{lem2}.
Therefore
\begin{equation}
\label{eq107}
\la_j^\ast \bigl|_H = \sum_i \phi_i ~.
\end{equation}

Next, suppose $\chi_i$ is some nonlinear irreducible character of $G$.
Then, by Lemma~\ref{lem4}, $\chi_i = \la_j^\ast$ for some $j$, and
$$s_i =
( \phi_0^\ast , \chi_i ) = ( \phi_0^\ast , \la_j^\ast ) = ( \phi_0 , \la_j^\ast \bigl|_H ) = 1 ~,$$
and this proves (A).

To prove (B), note that since
$$|G| = \sum_i \chi_i (1)^2 ~,$$
and $|G| = |A| \cdot f$, if we let $k_1$ be the number of linear characters of $G$ and $k_2$ the number of
nonlinear characters, then by Lemma~\ref{lem4},
\begin{equation}
\label{eq108}
|A| f = k_1 + f^2 k_2 ~.
\end{equation}
Now in general $k_1 = |G| / |C|$, where $C$ is the commutator
subgroup.
In our case $C=A$,
so $k_1 = |G| / |A| =f$, and so (\ref{eq108}) yields
\begin{equation}
\label{eq109}
|A| = 1+ f k_2 ~.
\end{equation}
Since $A$ is normal, and $H \cap A = \{1\}$ by Lemma~\ref{lem2},
we find that for $a \in A$,
$$
\phi_0^\ast (a) = \frac{1}{|H|} \sum_{x \in G}
\phi_0 (x a x^{-1} ) =
\left\{
\begin{array}{cl}
|A| & \mbox{if}~~ a=1 ~, \\
0 & \mbox{otherwise} ~.
\end{array}
\right.
$$
Therefore
$$\phi_0^\ast \Bigl|_A = \sum_i \la_i ~.$$
Hence
\begin{equation}
\label{eq1010}
\phi_0^\ast (1) = \sum_i \la_i (1) = |A| ~,
\end{equation}
and therefore
$$\sum_i s_i \chi_i (1) = |A| ~.$$
Next,
$$1= ( \chi_0 \bigl|_H , \phi_0 ) = ( \chi_0 , \phi_0^\ast ) ~,$$
and so
by (\ref{eq109}) and (\ref{eq1010})
\begin{equation}
\label{eq1011}
\phi_0^\ast (1) = \chi_0 (1) + fk_2 +
\sum_{i \in X} s_i = |A| = 1+ f k_2 ~,
\end{equation}
where $X$ is the set of $i$ such that $\chi_i$ is a linear character
of $G$, $i \neq 0$.
Since $s_i \ge 0$ for all $i$,
(\ref{eq1011}) is possible only if $s_i =0$ for all $i \in X$,
which proves (B).

The claims (A) and (B), which were proved above, lead to the
following result about zeta functions.

\vspace{.1in}
\noindent
{\bf Proposition~1}
{\em
Under the conditions of case (ii) of Theorem~\ref{the3},}
\begin{eqnarray}
\label{eq1012}
\zeta_L (s) & = & \zeta_K (s) L(s) ~, \\
\zeta_M (s) & = & \zeta_K (s) L_1 (s) L(s)^f ~,
\label{eq1013}
\end{eqnarray}
{\em where $f= |H|$ and the functions
$L_1 (s)$ and $L(s)$ are entire.}

\vspace{.1in}
\Proof Let $Y$ be the set of nonlinear characters of $G$.
We have
$$\zeta_L (s) = L (s, \phi_0 ) = L ( s, \phi_0^\ast ) =
L(s, \chi_0 ) \prod_{\chi_i \in Y} L(s, \chi_i ) ~.$$
If we let
$$L(s) = \prod_{\chi_i \in Y} L(s, \chi_i ) ~,$$
we obtain (\ref{eq1012}).
By Lemma~\ref{lem4}, each $\chi_i$ is induced by a linear character $\la_j$ of $A$,
so each $L(s, \chi_i) = L(s, \lambda_j )$ is entire, and so is $L(s)$.

Next,
$$\z_M (s) =
\prod_{\chi_i} L(s, \chi_i )^{\chi_i (1)} =
\z_K (s) L_1 (s) L(s)^f ~,$$
where $L_1 (s)$ is the product of $L(s, \chi_i )$ over the
nonprincipal linear irreducible characters
$\chi_i$ of $G$.
Since all these $L$-functions are abelian, they are entire, and so is $L_1 (s)$.
\begin{flushright} \Qed \end{flushright}

Finally, we return to our original investigation of the
possible real zeros of $\z_{L}(s)$.
Suppose $\beta$ is a real zero of $\z_{L}(s)$ in the range
(\ref{eq102}).  If $\beta$ is not a zero of $\z_K(s)$, then it must be a zero of
$L(s)$, and so by Proposition~1,
$\beta$ is a zero of $\z_M (s)$
of multiplicity at least
$f$.
By Lemma~\ref{lem3},
$$\log d_M\leq 2f\log d_{L}.$$
Choose $c_0=(4c_1)^{-1}$.
Then for $r = c_0 / ( \log d_L )$,
Lemma~\ref{lem1}
applied to $\z_M(s)$
gives
$$ f\leq n(r) < 1+c_1r\log d_{M} \leq 1+\frac{f}{2} \leq f~, $$
which is a contradiction.  Hence $\beta$ must be a zero of $\z_K(s)$.
That it is a simple zero follows from $d_K \leq d_{L}$. This establishes Theorem~\ref{the3}.   \begin{flushright} \Qed \end{flushright}



We next show how Theorem~\ref{the1} can be derived from Theorem~\ref{the2}.
Since any radical extension can be obtained as a tower of radical extensions of prime degrees, it suffices to prove Theorem~\ref{the1} when all the
degrees $[F_i : F_{i-1} ]$ are odd primes.
When $[F_i : F_{i-1} ] = p$ is prime,
there are no intermediate fields $F'_i$,
$F_{i-1} \subset F'_i \subset F_i$.
Let $F_i = F_{i-1} [a^{1/p} ]$, $a \in F_{i-1}$.
Then it is easy to show
(and well-known, cf.~\cite{Tcheb,Vino}) that if $F_i^\ast$ is the normal closure of $F_i$ over $F_{i-1}$, then
$\Gal (F_i^\ast / F_{i-1} ) = A \cdot H$, where
$A$ is isomorphic to $\ZZ_p$ and $H$ is isomorphic to a subgroup of $\ZZ_p^\ast$.
Since $[F_i : F_{i-1} ]$ is odd, Theorem~\ref{the2} applies.
\section{Concluding Remarks}

\ \ \ \ \ 
We make some final remarks regarding
Theorem~\ref{the2} and some of its implications.
First, one immediately notices that Theorem~\ref{the2}
and Heilbronn's results \cite{Heilbronn} can be combined to get

\begin{theorem}
\label{the4}
Let $F_0\subset F_1\subset\cdots\subset F_m$ be a tower of extensions
such that $[F_i:F_{i-1}]$ is odd and $F_i/F_{i-1}$ is either normal or
of the form of Theorem~\ref{the2}.  Then any real zero of $\z_{F_m}(s)$ in the
range
$$1-\frac{c_0}{\log d_{F_m}}\leq \beta <1$$
is a simple real zero of $\z_{F_0}(s)$.
\end{theorem}

Also, there is the following application.

\begin{corollary}
Let $\Q=F_0\subset F_1\subset\cdots\subset F_m$ be a tower of fields as in
Theorem~\ref{the4}.
Let $\kappa_m$ denote the residue of
$\z_{F_m}(s)$ at $s=1$.  Then
$$ \kappa_m > \frac{c_4}{\log d_{F_m}} $$
where $c_4$ is an effectively computable constant depending only on $c_0$.
\end{corollary}

\Proof This follows from
Lemma~\ref{lem4} of \cite{Stark} and Theorem~\ref{the4}.
\begin{flushright} \Qed \end{flushright}

We also
make an observation about the analyticity of the ratios of
the $\z_{F_i}(s)$.  The result below follows already from the 
theorem of Uchida \cite{Uchida} and
van der Waall \cite{Waall,Waall1},
which states that if $M$ is a normal extension of a number field $K$
with a solvable Galois group,
then the quotient $\z_L (s) / \z_K (s)$
is entire for any number
field $L$ intermediate between $K$ and $M$.

\begin{theorem}
\label{the5}
Let $F_0\subset F_1\subset\cdots\subset F_m$ be a tower of fields as in
Theorem~\ref{the4} (or Theorem~\ref{the2}). Then, if $i>j$
$$\frac{\z_{F_i}(s)}{\z_{F_j}(s)}\:\:\:is\:\:\:entire. $$
\end{theorem}

\Proof It suffices to prove that
$\z_{F_i} (s) / \z_{F_{i-1}} (s)$ is
analytic when $F_i/F_{i-1}$ is either normal or an extension of the type
considered in Theorem~\ref{the3}, for then we may "build up" in much the same way
as Theorem~\ref{the3} was proved.  If $F_i/F_{i-1}$ is normal, then the
fundamental result of Aramata \cite{Aramata} tells us that
$\z_{F_i} (s) / \z_{F_{i-1}} (s)$ is analytic.
In the other cases, though, this follows from Proposition~1.
\begin{flushright} \Qed \end{flushright}

We also remark that our method sometimes yields stronger results than those given
by the statements of our theorems.
For example, if $L= Q [2^{1/p}]$, $p$ an odd prime, and $K=Q$, Theorem~\ref{the1} shows that
$\z_L (s)$ has no Siegel zeros.
If we let $M= L [ \exp ( 2 \pi i /p )]$ be the normal
closure of $L$, then $\z_M (s)$ has no Siegel zeros by
Heilbronn's result \cite{Heilbronn}, say, and easy estimates of discriminants.
This means that $\z_M (s)$ has no zeros within $(c_5 p^2 \log p )^{-1}$ of 1.
Since we showed that $\z_L (s)$ has no Siegel
zeros, it follows that $\z_M (s)$ has no zeros within $(c_6 p \log p )^{-1}$ of 1.

The proof of
Theorem~\ref{the3} relied on all the nonlinear irreducible characters of the group $G$
that satisfy the conditions of case (ii) having the same degree $f$.
I.~M. Isaacs has pointed out that if $f$ and $|G|/f$ are relatively prime,
then it follows from \cite{IP} that $G$ can be constructed as a semidirect
product of a cyclic group $H'$ acting regularly on an abelian group $A'$.
In this case $H'$ is necessarily a Frobenius complement in $G$.
Thus our argument cannot be readily extended to other classes of groups.
The basic approach can be used even if the degrees of the nonlinear
irreducible characters of $G$ differ, but we have not found any nice generalization.
\section*{Acknowledgements}
We
thank Eduardo Friedman, Marty Isaacs, Hendrik Lenstra, and the referee
for their comments on preliminary versions of this paper.
%{
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