1. Introduction
For any number field K and
$s \in {\mathbb C}$
with
$\Re(s) \gt1$
, the Dedekind zeta function of K is defined by
\begin{align*}\zeta_{\mathrm{K}}(s)=\sum_{0 \ne \mathcal{I} \subseteq {\mathcal{O}_{\mathrm{K}}}}\mathcal{N}_{\mathrm{K}}(\mathcal{I})^{-s},\end{align*}
where
$\mathcal{N}_{\mathrm{K}} (\mathcal{I})= [\mathcal{O}_{\mathrm{K}}\ :\ \mathcal{I}]$
denotes the absolute norm of K. The function
$\zeta_{\mathrm{K}}$
extends analytically to the whole complex plane except at
$s=1$
where it has a simple pole and satisfies the functional equation
\begin{align*}\xi_{\mathrm{K}}(s) = \xi_{\mathrm{K}}(1- s).\end{align*}
Here
\begin{align*}\xi_{\mathrm{K}} (s) = \left( \frac{\sqrt{d_{\mathrm{K}}} }{ 2^{r_2} \pi^{n/2}}\right)^{s}\Gamma\left(\frac{s}{2}\right)^{r_1} \Gamma(s)^{r_2} \zeta_{\mathrm{K}}(s),\end{align*}
$d_{\mathrm{K}}, r_{1}$
and
$r_{2}$
denote the absolute discriminant, the number of real places and the number of complex places of K respectively [
Reference Neukirch13
, p. 467]. Functional equation and non-vanishing of
$\zeta_{\mathrm{K}}$
on the line
$\Re(s) =1$
imply that all non-trivial zeros of
$\zeta_{\mathrm{K}}$
lie inside the critical strip
$0 \lt \Re(s) \lt 1$
. The extended Riemann hypothesis (ERH) predicts that all non-trivial zeros of Dedekind zeta functions lie on the line
$s=1/2$
.
In 1972, Armitage [
Reference Armitage1
], based on the work of Serre [
Reference Serre15
], proved that there exists a number field K such that
$\zeta_{\mathrm{K}}(1/2) = 0$
. Consequently, by the Aramata–Brauer theorem there are infinitely many such number fields. Since functional equation necessitates that the order of
$\zeta_{\mathrm{K}}$
at
$1/2$
is even, the extended Riemann Hypothesis in the least will ensue that order of any non-trivial real zero of
$\zeta_{\mathrm{K}}$
is necessarily even. In this paper, as special case of a general result, we will show that the Dedekind zeta function for a Galois number field of odd degree can not have non-trivial real zero of odd order.
Throughout the paper,
$\mathrm{K}, \mathrm{F}$
will denote number fields and
$d_{\mathrm{K}}, d_{\mathrm{F}}$
will denote absolute discriminants of K and F respectively. In 1973, Heilbronn [
Reference Heilbronn6
] proved the following result.
Theorem 1·1 (Hilbronn). Let
$\mathrm{K}/\mathrm{F}$
be a Galois extension and
$s_0$
be a real simple zero of
$\zeta_{\mathrm{K}}$
. Then there exist a number field
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
such that
$[\mathrm{M} : \mathrm{F}] \leq 2$
and
$\zeta_{\mathrm{M}}(s_0) = 0$
.
One year later, Stark [
Reference Stark17
] proved that if
$s_0 = \sigma + it $
is a zero of
$\zeta_{\mathrm{K}}$
in the region
\begin{equation}1- \frac{1}{4 \log d_{\mathrm{K}}} \leq \sigma \leq 1,\phantom{m} | t | \leq \frac{1}{4 \log d_{\mathrm{K}}},\end{equation}
then
$s_0$
is necessarily real and simple. The region (1·1) is now known as Stark’s region. Using the above observation, Stark proved the following theorem.
Theorem 1·2
(Stark). Let
$\mathrm{K}/\mathrm{F}$
be an extension of number fields of degree n and
$s_0$
be a real zero in the region
\begin{align*}1- \frac{1}{4\, n! \log d_{\mathrm{K}}} \leq s_0 \leq 1.\end{align*}
Then there exists a number field
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
such that
$[\mathrm{M} : \mathrm{F}] \leq 2$
and
$\zeta_{\mathrm{M}}(s_0) = 0$
.
We can extend these results for odd order real zeros of the Dedekind zeta function. More precisely, we prove the following theorem.
Theorem 1·3. Let
$\mathrm{K}/\mathrm{F}$
be a Galois extension of number fields with Galois group G and
$s_0$
be an odd order real zero of
$\zeta_{\mathrm{K}}$
. If the order of the commutator subgroup G’ of G is odd, then there exists a number field
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
with
$[\mathrm{M}\ :\ \mathrm{F}] \leq 2$
such that
$s_0$
is an odd order zero of
$\zeta_{\mathrm{M}}$
.
As immediate corollaries to Theorem 1·3, we get:
Corollary 1·4. Dedekind zeta function of a Galois number field of odd degree does not have any non-trivial real zero of odd order.
We refer to [
Reference Kandhil10
] for a discussion when
$s=1/2$
.
Corollary 1·5. Let
$\mathrm{K}/\mathrm{F}$
be an abelian extension of number fields and
$s_0$
be an odd order real zero of
$\zeta_{\mathrm{K}}$
. Then there exists a number field M with
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
such that
$[\mathrm{M}\, : \, \mathrm{F}] \leq 2$
and
$s_0$
is an odd order zero of
$\zeta_{\mathrm{M}}$
.
Remark 1·6. In the context of Heilbron’s theorem for odd degree Galois extensions
$\mathrm{K}/\mathrm{F}$
, a simple zero of
$\zeta_K$
, if it does descent to
$\zeta_{\mathrm{M}}$
, must necessarily be simple as
$\zeta_{\mathrm{K}}/\zeta_{\mathrm{M}}$
is entire. However, for a higher order odd zero it can descent to a zero of lower order. This can be seen by considering the splitting field K of the polynomial
$X^3 + p$
, where p is a rational prime, over
${\mathbb Q}$
. Clearly,
$\mathrm{K}/{\mathbb Q}$
is a
$S_3$
extension, where
$S_3$
is the symmetric group of three elements. The field K has a quadratic subfield
$\mathrm{M}= {\mathbb Q}(\omega)$
, where
$\omega^3 =1$
. Note that trivial zeros of
$\zeta_{\mathrm{K}}$
have order 3 whereas trivial zeros of
$\zeta_{\mathrm{M}}$
have order one. However, the parity of such a zero is necessarily preserved which is the crux of our theorem.
Remark 1·7. In case of even degree Galois extensions
$\mathrm{K}/\mathrm{F}$
, an even order zero
$\alpha$
of
$\zeta_K$
need not descent to
$\zeta_{\mathrm{M}}$
. The result of Armitage provides such instances with
$\alpha = 1/2$
. Furthermore, if such a zero does descent, its parity need not be preserved. Consider the Galois extension
$\mathrm{K} = {\mathbb Q}(\sqrt{2}, i)$
over
${\mathbb Q}$
. It has three quadratic subfields
$\mathrm{M}_1 = {\mathbb Q}(\sqrt{2}), \mathrm{M}_2 = {\mathbb Q}(i),\mathrm{M}_3 = {\mathbb Q}(\sqrt{-2})$
over
${\mathbb Q}$
. We know that any trivial zero of
$\zeta_K$
at odd negative integers has even order whereas the Dedekind zeta functions
$\zeta_{\mathrm{M}_1}, \zeta_{\mathbb Q}$
have no zeros and
$\zeta_{\mathrm{M}_2}, \zeta_{\mathrm{M}_3}$
have simple zeros at odd negative integers. This shows that
$-2n-1$
for
$n \geq 0$
is an even order zero of
$\zeta_{\mathrm{K}}$
but
$-2n -1$
is not a zero or is a simple zero of
$\zeta_{\mathrm{M}}$
, where
${\mathbb Q} \subseteq \mathrm{M} \subseteq \mathrm{K}$
with
$[\mathrm{M}\, : \, {\mathbb Q}] \leq 2$
.
In 1993, Odlyzko--Skinner [ Reference Odlyzko and Skinner14 ] extended the result of Stark for radical extensions. They proved the following theorem.
Theorem 1·8 (Odlyzko–Skinner). For an radical extension
$\mathrm{K}/\mathrm{F}$
of odd degree, a Stark zero of
$\zeta_{\mathrm{K}}$
is necessarily a Stark zero of
$\zeta_{\mathrm{F}}$
.
Before we proceed further, let us introduce some notation. For an extension of number fields
$\mathrm{K}/\mathrm{F}$
, we denote its Galois closure by
$\mathrm{K}^{cls}$
, the Galois group of
$\mathrm{K}^{cls}/\mathrm{F}$
by G and its commutator subgroup by G’. Let
\begin{align*}\mathcal{C}\ =\ \big\{ \mathrm{K}/\mathrm{F} \ :\ [\mathrm{K}^{cls}\, : \, \mathrm{K}] \text{ is odd},\ |{G}'| \text{ is odd} \big\}\end{align*}
and
\begin{align*}\mathcal{D}\ =\ \big\{ \mathrm{K}/\mathrm{F}\ :\ [\mathrm{K}\ :\ \mathrm{F}] \text{is odd}, |{G}'| \text{is odd}, {\mathrm{Gal}(\mathrm{K}^{cls}/\mathrm{K})} =H \leq N \ \text{with}\ N\ \text{abelian and}\ N\ \trianglelefteq G\ \big\}.\end{align*}
In this set-up, Murty [ Reference Murty11 ] proved the following theorem.
Theorem 1·9 (Murty). Let
$\mathrm{K}/\mathrm{F}$
be an extension of odd degree n with solvable Galois closure
$\mathrm{K}^{cls}$
. Suppose that
\begin{align*}e(n) \ =\ \text{ max}_{p^{\alpha}|| n} \alpha,\phantom{m}\gamma(n)\ =\ 12^{n-1} 3^{1/3}\phantom{m}\text{and}\phantom{m}\delta(n) \ =\ (e(n)+1)^2 \gamma(n).\end{align*}
If
$0 \lt c \lt 1/2$
and
$\zeta_{\mathrm{K}}$
has a zero
$s_0$
in
\begin{align*}1- \frac{c}{\delta(n) \log d_{\mathrm{K}}} \leq \sigma \leq 1,\phantom{m} | t | \leq \frac{c}{\delta(n) \log d_{\mathrm{K}}},\end{align*}
then
$\zeta_{\mathrm{F}}(s_0) =0$
.
In 2001, Murty [ Reference Murty12 ] removed the restriction of odd degree to show the following theorem.
Theorem 1·10 (Murty). Let
$\mathrm{K}/\mathrm{F}$
be an extension of degree n with solvable Galois closure
$\mathrm{K}^{cls}$
. Suppose that
$e(n), \gamma(n), \delta(n)$
and c are as in Theorem 1·9. If
$\zeta_{\mathrm{K}}$
has a zero
$s_0$
in
\begin{align*}1- \frac{c}{n^{e(n)}\delta(n) \log d_{\mathrm{K}}} \leq \sigma \leq 1,\phantom{m} | t | \leq \frac{c}{n^{e(n)}\delta(n) \log d_{\mathrm{K}}},\end{align*}
then there is a number field M with
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
such that
$[\mathrm{M}\, : \, \mathrm{F}] \leq 2$
and
$\zeta_{\mathrm{M}}(s_0) = 0$
.
For the set
$\mathcal{C}$
, we have the following theorem.
Theorem 1·11. Let
$\mathrm{K}/\mathrm{F} \in \mathcal{C}$
and
$s_0$
be an odd order real zero of
$\zeta_{\mathrm{K}}$
. Then there exists a number field M with
$\mathrm{F} \subseteq \mathrm{M} \subseteq \mathrm{K}$
such that
$[\mathrm{M}\, : \, \mathrm{F}] \leq 2$
and
$s_0$
is an odd order zero of
$\zeta_{\mathrm{M}}$
.
As a consequence of Theorem 9, we get the following corollary.
Corollary 1·12. There is an effectively computable constant
$c \gt 0$
such that for any
$\mathrm{K}/{\mathbb Q} \in \mathcal{C}$
of degree n, we have
\begin{align*}\rho_{\mathrm{K}} \ \gt\ c \ d_{\mathrm{K}}^{-1/n},\end{align*}
where
$d_{\mathrm{K}}$
is the absolute discriminant of K and
$\rho_{\mathrm{K}}$
is the residue of the Dedekind zeta function
$\zeta_{\mathrm{K}}$
at
$s=1$
.
This improves the bound derived by Stark [
Reference Stark17
, theorem 1) and Murty (see
Reference Murty12
, theorem 3·1] when
$\mathrm{K}/\mathrm{F} \in \mathcal{C}$
. For the set
$\mathcal{D}$
defined earlier, we have the following theorem.
Theorem 1·13. Let
$\mathrm{K}/\mathrm{F} \in \mathcal{D}$
. Then any odd order real zero of
$\zeta_{\mathrm{K}}$
is an odd order zero of
$\zeta_{\mathrm{F}}$
.
Let
$\mathcal{D}^t$
be a collection of extension of number fields
$\mathrm{K}/\mathrm{F}$
for which there exist towers of number fields
\begin{align*} \mathrm{F} \ =\ \mathrm{K}_0 \subseteq \mathrm{K}_1 \subseteq \mathrm{K}_2 \subseteq \cdots \subseteq \mathrm{K}_{\ell} \ =\ \mathrm{K} \end{align*}
such that for each
$0 \leq i \leq \ell -1$
, the extension
$\mathrm{K}_{i+1}/\mathrm{K}_i \in \mathcal{D}$
. Here we have the following theorem.
Theorem 1·14. Let
$\mathrm{K}/\mathrm{F} \in \mathcal{D}^t$
. Then any odd order real zero of
$\zeta_{\mathrm{K}}$
is an odd order zero of
$\zeta_{\mathrm{F}}$
.
Let
\begin{align*}\mathcal{D}_{{\mathbb Q}}^t\ =\ \left\{ \mathrm{K} \ :\ \mathrm{K} \text{ number field},\ \mathrm{K}/{\mathbb Q} \in \mathcal{D}^t \right\}.\end{align*}
Recall that the zeros of Dedekind zeta function lying in the Stark region (1) is simple and Riemann zeta function
$\zeta_{{\mathbb Q}}$
does not have any non-trivial real zero. Hence, as an immediate corollary to Theorem 1·14, we get
Corollary 1·15. If
$\mathrm{K} \in \mathcal{D}_{{\mathbb Q}}^t$
, then
$\zeta_{\mathrm{K}}$
does not have any non-trivial odd order real zero. In particular,
$\zeta_{\mathrm{K}}$
does not have any real zero in the Stark region (1).
Using Corollary 1·15 and applying [ Reference Stark17 , lemma 4], we get:
Corollary 1·16. Let
$\mathrm{K} \in \mathcal{D}_{{\mathbb Q}}^t$
and
$\rho_{\mathrm{K}}$
be the residue of
$\zeta_{\mathrm{K}}$
at
$s = 1$
. There is an effective and absolute constant
$c \gt 0$
such that
\begin{align*}\rho_{\mathrm{K}} \gt \frac{c}{\log d_{\mathrm{K}}}.\end{align*}
2. Preliminaries
From now onwards, G denotes a finite group, V a finite dimensional vector space over
${\mathbb C}$
and
$\rho\, : \, G \rightarrow \mathrm{GL}(V)$
a representation of G. The character
$\chi$
of a representation of
$\rho$
is defined by
\begin{align*}\chi(g) = \mathrm{Tr}(\rho(g)).\end{align*}
The dual representation
$\rho^*$
of a representation
$\rho$
of G is a group homomorphism G to
$\mathrm{GL}(V^*)$
satisfying
\begin{align*}\rho^{*}(g)(f) = f \circ \rho(g^{-1}), \phantom{m} \forall g\in G,\ f \in V^*,\end{align*}
where
$V^*$
is the dual of V. It is well known that
$\rho$
is irreducible if and only if
$\rho^*$
is irreducible ([
Reference Goodman and Wallach5
, p. 10]). If we denote by
$\chi^*$
, the character of
$\rho^*$
, then
$\chi^*(g) = \overline{\chi}(g) = \overline{\chi(g)}$
for all
$g \in G$
([
Reference Serre16
, chapter 2]). We denote the set of characters of G by
$\mathrm{Char}(G)$
and the set of all irreducible characters by
$\mathrm{Irr}(G)$
.
A class function on a group G is a function
$f\, : \, G \longrightarrow {\mathbb C}$
such that
$f(x^{-1}gx) = f(x)$
, for all
$g, x \in G$
. Clearly, characters are class functions. The set of all class functions of G forms a vector space over
${\mathbb C}$
, denoted by
$\mathrm{Cl}(G)$
. One can define an inner product on
$\mathrm{Cl}(G)$
as follows; for
$f_1, f_2 \in \mathrm{Cl}(G)$
, define
\begin{align*}\langle f_1, f_2 \rangle = \frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}.\end{align*}
The set of irreducible characters of G form an orthonormal basis for
$\mathrm{Cl}(G)$
with respect to this inner product. Note that if
$f \in \mathrm{Char}(G), \chi \in \mathrm{Irr}(G)$
, then
$\langle f, \chi \rangle$
is a non-negative integer. Following theorem of Clifford provides information about restriction of an irreducible character.
Theorem 2·1 (Clifford). Let G be a group, N be a normal subgroup of G and
$\chi \in \mathrm{Irr}(G)$
. Also let
$\phi$
be an irreducible constituent of
$\chi|_N$
. If
$\phi = \phi_1, \phi_2,....,\phi_t$
are the distinct conjugates of
$\phi$
in G, then
\begin{align*}\chi|_N = \ e \ \sum_{i=1}^t \phi_i,\end{align*}
where
$e = \ \lt\chi|_N, \phi \gt $
.
See [
Reference Isaacs8
, p. 79] for the details of the proof of Clifford’s theorem. Let H be a subgroup of G and
$\{g_1, g_2,..., g_r \}$
be a set of right coset representatives of H in G. For
$\psi \in \mathrm{Cl}(H)$
, one can define induction of
$\psi$
to G by
\begin{align*}\mathrm{Ind}_H^G \psi (g)\ =\ \sum_{j=1}^r \dot{\psi}(g_j^{-1} g g_j),\end{align*}
where
$\dot{\psi}(g) = \psi(g)$
if
$g \in H$
and
$\dot{\psi}(g) = 0$
if
$g \notin H$
.
Theorem 2·2 (Frobenius Reciprocity). Let G be a finite group and H be a subgroup of G. For class functions
$f \in \mathrm{Cl}(G), \psi \in \mathrm{Cl}(H)$
, the following equality holds
\begin{align*}\langle \psi, f\mid_H \rangle \ =\ \langle \mathrm{Ind}_H^G \psi, \ f \rangle.\end{align*}
See [ Reference Isaacs8 , chapter 5] for further details.
Throughout the paper, for number fields
$\mathrm{K}, \mathrm{F}$
,
$\mathcal{O}_{\mathrm{K}},\mathcal{O}_{\mathrm{F}}$
denote their ring of integers,
${\mathfrak p} , {\mathfrak P}$
primes ideals in
$\mathcal{O}_{\mathrm{K}}, \mathcal{O}_{\mathrm{F}}$
such that
${\mathfrak P}$
is lying over
${\mathfrak p}$
and
${\mathcal N}_{\mathrm{K}}$
denotes the absolute norm of K. When the extension
$\mathrm{K}/\mathrm{F}$
is Galois, we will denote the Galois group by G. The decomposition group and the inertia group of
${\mathfrak P}$
is denoted by
$D_{{\mathfrak P}}$
and
$I_{{\mathfrak P}}$
respectively. Let
$[\sigma_{{\mathfrak P}}]$
be the conjugacy class of Frobenius of
${\mathfrak P}$
in G. Then
$\sigma_{{\mathfrak P}}(\alpha) \equiv \alpha^{{\mathcal N}_{\mathrm{K}}({\mathfrak p})} ({\, \mathrm mod \, } {\mathfrak P})$
for any
$\alpha \in \mathcal{O}_{\mathrm{K}}$
.
For a Galois extension
$\mathrm{K}/\mathrm{F}$
with Galois group G, a representation
$\rho$
of G with character
$\chi$
, and
$s \in {\mathbb C}$
, with
$\Re(s) \gt 1$
, the Artin L-function attached to
$\chi$
is defined as
\begin{align*}L(s, \chi, \mathrm{K}/\mathrm{F}) \ =\ \prod_{{\mathfrak p} \subset \mathcal{O}_{\mathrm{F}} \atop {\mathfrak p} \neq 0} L_{{\mathfrak p}}(s, \chi, \mathrm{K}/\mathrm{F}),\end{align*}
where
\begin{align*}L_{{\mathfrak p}}(s, \chi, \mathrm{K}/\mathrm{F}) \ =\ \det( I - {\mathcal N}_{\mathrm{K}}({\mathfrak p})^{-s}\rho(\sigma_{{\mathfrak P}})|_{V^{I_{{\mathfrak P}}}})^{-1}.\end{align*}
Here
\begin{align*}V^{I_{{\mathfrak P}}} = \{v \in V \ :\ \rho(g)(v) = v, \ \ \forall g \in I_{{\mathfrak P}} \}.\end{align*}
It is well known that Artin L-functions satisfy the following properties :
-
(1) if
$1_G$
is the trivial character of G, then
\begin{align*}L(s, 1_G, \mathrm{K}/\mathrm{F}) = \zeta_{\mathrm{F}}(s);\end{align*}
-
(2) if
$\chi_1$
and
$\chi_2$
are characters of G, then
\begin{align*}L(s, \chi_1+\chi_2, \mathrm{K}/\mathrm{F}) = L(s, \chi_1, \mathrm{K}/\mathrm{F}) L(s, \chi_2, \mathrm{K}/\mathrm{F});\end{align*}
-
(3) if H is a subgroup of G and
$\psi \in \mathrm{Char}(H)$
, then where
\begin{align*}L(s, \mathrm{Ind}_H^G \psi, \mathrm{K}/\mathrm{F}) \ =\ L(s, \psi, \mathrm{K}/\mathrm{K}^H),\end{align*}
$\mathrm{Ind}_H^G \psi$
is the induced character of
$\psi$
from H to G and
$K^H$
is the fixed field of H;
-
(4) suppose that
$\mathrm{K}/\mathrm{F}$
and
$\mathrm{K}'/\mathrm{F}$
are Galois extensions with
$\mathrm{F} \subseteq \mathrm{K}' \subseteq \mathrm{K}$
. Also suppose that Galois groups of
$\mathrm{K}/\mathrm{F}$
and
$\mathrm{K}/\mathrm{K}'$
are G and H respectively. For any
$\chi' \in \mathrm{Char}(G/H)$
, define
$\chi \in \mathrm{Char}(G)$
by
$\chi(g) = \chi'(gH)$
for
$g \in G$
. Then
\begin{align*}L(s, \chi, \mathrm{K}/\mathrm{F}) \ =\ L(s, \chi', \mathrm{K}'/\mathrm{F}).\end{align*}
Using these properties, one can prove that
\begin{align*}\zeta_{\mathrm{K}}(s) \ =\ \prod_{\chi \in \mathrm{Irr}(G)} L(s, \chi, \mathrm{K}/\mathrm{F})^{\chi(1)}\end{align*}
(see [ Reference Neukirch13 ] for further details). It follows from Brauer induction theorem [ Reference Neukirch13 , p. 522] that Artin L-functions have meromorphic continuations to the whole complex plane. In this set-up, Stark [ Reference Stark17 ] proved the following theorem.
Theorem 2·3 (Stark). Let
$\mathrm{K}/\mathrm{F}$
be Galois with Galois group G and
$\chi \in \mathrm{Irr}(G)$
. Suppose
$s_0$
is a simple zero of
$\zeta_{\mathrm{K}}$
. Then
$L(s, \chi, \mathrm{K}/\mathrm{F})$
is analytic at
$s=s_0$
. Further, there is a field M, cyclic over F and contained in K, such that for any field N between F and K,
$\zeta_N(s_0)=0$
if and only if
$\mathrm{M}\subseteq N$
. If
$s_0$
is real, then M is either F itself or is a quadratic extension over F.
3 An intermediate Proposition and a Lemma
As before, let
$\mathrm{K}/\mathrm{F}$
be a Galois extension with Galois group G. Also let H be a subgroup of G,
$\mathrm{K}^H$
be the fixed field of H,
$\psi \in \mathrm{Char}(H)$
and
$s_0 \in {\mathbb C}$
. From now on, we shall denote the order of vanishing of
$L(s, \psi, \mathrm{K}/\mathrm{K}^H)$
at
$s = s_0$
by
$n(H, \psi, s_0)$
. For any
$\chi \in \mathrm{Char}(G)$
associated with a representation
$\rho$
, let
$\ker \chi$
be the kernel of
$\rho$
. Here we have the following proposition.
Proposition 3·1. Let
$\mathrm{K}/\mathrm{F}$
be a Galois extension with Galois group G. For
$s \in {\mathbb C}$
and
$\chi \in \mathrm{Char}(G)$
, we have
\begin{align*}L(s, \chi, \mathrm{K}/\mathrm{F})\ =\ \overline{L(\overline{s}, \overline{\chi}, \mathrm{K}/\mathrm{F})}.\end{align*}
Hence for any real number
$s_0$
, we have
\begin{align*}n(G, \chi, s_0) \ =\ n(G, \overline{\chi}, s_0).\end{align*}
Proof. By Brauer’s theorem, we know that
$\chi$
can be written as
\begin{align*}\chi = \sum_{i = 1}^t a_i \ \mathrm{Ind}_{H_i}^G \psi_i,\end{align*}
where for
$1 \le i \le t$
,
$a_i \in {\mathbb Z}$
,
$H_i$
are subgroups of G and
$\psi_i \in \mathrm{Char}(H_i)$
are of degree one. So
\begin{align*}\overline{\chi} =\sum_{i = 1}^t a_i \ \overline{\mathrm{Ind}_{H_i}^G \psi_i}.\end{align*}
Since
\begin{align*}\overline{\mathrm{Ind}_{H_i}^G \psi_i} = \mathrm{Ind}_{H_i}^G \overline{\psi_i},\end{align*}
we have
\begin{align*}\overline{\chi} ={\sum_{i = 1}^t a_i \ \mathrm{Ind}_{H_i}^G \overline{\psi_i}}.\end{align*}
Hence by applying the properties of Artin L-functions, for any
$s \in {\mathbb C}$
, we have
\begin{align*}L(s, \overline{\chi}, \mathrm{K}/\mathrm{F})\ =\ \prod_{i=1}^t L(s, \overline{\psi_i}, \mathrm{K}/\mathrm{K}^{H_i})^{a_i}.\end{align*}
This implies that
\begin{align*}\overline{L(\overline{s}, \overline{\chi}, \mathrm{K}/\mathrm{F})}\ =\ \prod_{i=1}^t \overline{L(\overline{s}, \overline{\psi_i}, \mathrm{K}/\mathrm{K}^{H_i})^{a_i}}.\end{align*}
So it is enough to show that for any degree one character
$\psi$
, we have
\begin{align*}\overline{L(\overline{s}, \overline{\psi}, \mathrm{K}/\mathrm{F})}\ =\ L(s, \psi , \mathrm{K}/\mathrm{F}).\end{align*}
This is true as for any
$s \in {\mathbb C}$
with
$\Re(s)\gt1$
and for any prime ideal
${\mathfrak p}$
of
$\mathcal{O}_{\mathrm{F}}$
, one has
\begin{equation}\overline{L_{{\mathfrak p}}(\overline{s}, \overline{\psi}, \mathrm{K}/\mathrm{F})}\ =\ \left(1 - \frac{\psi(\sigma_{{\mathfrak p}})}{N({\mathfrak p})^{s}} \right)^{-1}\ =\ L_{{\mathfrak p}}(s, \psi , \mathrm{K}/\mathrm{F}).\end{equation}
Since
$n(H_i, \psi_i, s_0) = n(H_i, \overline{\psi_i}, s_0)$
for
$1 \le i \le t$
, we see that
\begin{align*}n(G, \chi, s_0)\ =\ \sum_{i = 1}^t a_i \ n(H_i, \psi_i, s_0)\ =\ n(G, \overline{\chi}, s_0).\end{align*}
This completes the proof of the proposition.
Lemma 3·2. Let G be a finite group and N be a normal subgroup of G. Also let
$\chi \in \mathrm{Irr}(G)$
. If
\begin{align*}\langle \chi|_N, \ 1_N \rangle \ \neq\ 0,\end{align*}
then
$\chi|_N = \chi(1) \ 1_N$
, i.e. N is a subgroup of
\begin{align*}\ker \chi \ =\ \{ g \in G \ :\ \chi(g) = \chi(1) \}.\end{align*}
Proof. One can write
\begin{align*}\chi\mid_N \ =\ \sum_{\phi \in \mathrm{Irr}(N) } \langle \chi\mid_N, \phi\rangle \phi.\end{align*}
Since N is a normal subgroup of G, by Clifford’s theorem, all
$\phi \in \mathrm{Irr}(N)$
with
$\langle \chi\mid_N, \phi\rangle \neq 0$
are conjugate to each other. Hence from the given hypothesis they are conjugate to the trivial character and so each of them is the trivial character. This implies that
\begin{align*}\chi|_N = \chi(1) \ 1_N.\end{align*}
Therefore N is a subgroup of
$\ker \chi$
.
4 Proofs of the Theorems
4·1 Proof of Theorem 1·3
Let
$\mathrm{K}/\mathrm{F}$
be as in Theorem 1·3 and
$s_0$
be a real zero of
$\zeta_{\mathrm{K}}$
of odd order. Since
\begin{align*}\zeta_{\mathrm{K}}(s) = \prod_{\chi \in \mathrm{Irr}(G)} L(s, \chi, \mathrm{K}/\mathrm{F})^{\chi(1)},\end{align*}
we have
\begin{equation}\mathrm{ord}_{s = s_0} \zeta_{\mathrm{K}}(s) = \sum_{\chi \in \mathrm{Irr}(G)} \chi(1) \ n(G, \chi , s_0).\end{equation}
Let
$\mathrm{K}_{ab}$
be the fixed field of the commutator G
′ of G. We first assume that G
′ is identity, i.e.
$\mathrm{K}/\mathrm{F}$
is abelian. Since by Proposition 3·1, we have
$n(G, \chi, s_0) = n(G, \overline{\chi}, s_0)$
and by given hypothesis, left-hand side of (4·1) is odd, there exist a
$\chi' \in \mathrm{Irr}(G)$
such that
$\chi' = \overline{\chi'}$
and
$n(G, \chi', s_0) \ge 1$
odd. Let
\begin{align*}H\ =\ \{ g \in G \ :\ \chi'(g) = \chi'(1) \}\end{align*}
be the kernel of
$\chi'$
and M be the fixed field of H. Consider the diagram
Since
$\chi'$
is one dimensional and
$\chi' = \overline{\chi'}$
, i.e.
$\chi'^2 =1$
, we have
$[\mathrm{M}\, : \, \mathrm{F}] \leq 2$
. Let
$1_H$
be the trivial character of H and
$\mathrm{Ind}_H^G 1_H$
be the induced character of
$1_H$
from H to G. Applying orthogonality of characters and using Frobenius reciprocity theorem, we have
\begin{align*} \mathrm{Ind}_H^G 1_H \ =\ \sum_{\chi \in \mathrm{Irr}(G)} \langle \chi, \ \mathrm{Ind}_H^G \ 1_H \rangle \chi \ =\ \sum_{\chi \in \mathrm{Irr}(G)} \langle \chi\mid_H, \ 1_H \rangle \chi. \end{align*}
This implies that
\begin{align*}\zeta_{\mathrm{M}}(s)\ =\ L(s, 1_H, \mathrm{K}/ \mathrm{M})\ =\ L(s, \mathrm{Ind}_H^G 1_H, \mathrm{K}/\mathrm{F})\ =\ \prod_{\chi \in \mathrm{Irr}(G)} L(s, \chi, \mathrm{K}/\mathrm{F})^{\langle \chi\mid_H, 1_H \rangle}. \end{align*}
Thus
\begin{equation}n(H, 1_H, s_0) = \sum_{\chi \in \mathrm{Irr}(G)} n(G, \chi, s_0) \langle \chi|_H, 1_H \rangle.\end{equation}
Since
$L(s, \chi, \mathrm{K}/\mathrm{F})$
for any
$\chi \in \mathrm{Irr}(G)$
is holomorphic at
$s_0$
,
$n(G, \chi', s_0) \geq 1$
and
$\chi'|_H = 1_H$
, we have
$n(H, 1_H, s_0) \geq 1$
. Hence
$\zeta_{\mathrm{M}}(s_0) = 0$
. Note that if
$n(G, 1_{G}, s_0)$
is odd, then
$s_0$
is an odd order zero of
$\zeta_F$
and we are done. Hence suppose that
$n(G, 1_{G}, s_0)$
is even.
Now we claim that any irreducible character
$\chi \not\in \{1_G, \chi' \}$
, we have
$\langle \chi\mid_H, 1_H \rangle = 0$
. Suppose not, then Lemma 3·2, we know that
$\ker \chi =G$
i.e.
$\chi = 1_G$
or
$\ker \chi$
is equal to the group H of index 2, i.e.
$\chi = \chi'$
. Hence (4·2) can be rewritten as
\begin{align*}n(H, 1_{H}, s_0)\ =\ n(G, \chi', s_0)\ +\ n(G, 1_{G}, s_0).\end{align*}
Since
$n(G, \chi', s_0)$
is odd, we have
$n(H, 1_{H}, s_0) $
is odd. This completes the proof when G
′ is identity.
For the case, when
$G'=G$
, we use the fact only irreducible real character is the trivial character
$1_G$
since by given hypothesis order of G
′ is odd (see [
Reference Neukirch13
, p. 57, proposition 13]). Applying (4·1) and Proposition 3·1, we see that
$n(G, 1_G, s_0) \geq 1$
is odd i.e.
$s_0$
is an odd order zero of
$\zeta_{\mathrm{F}}$
.
From now on, assume that G ′ is a proper subgroup of G. Identity (4·1) can be rewritten as
\begin{eqnarray}\mathrm{ord}_{s=s_0}\zeta_{\mathrm{K}}(s)&=& \sum_{\substack{\chi \in \mathrm{Irr}(G)\\ \chi \neq \overline{\chi}}} n(G, \chi, s_0) \chi(1)\phantom{m}+ \ \ \sum_{\substack{\chi \in \mathrm{Irr}(G) \\ \chi = \overline{\chi}\\ n(G, \chi, s_0) \chi(1)\equiv 0 ({\, \mathrm mod \, }{2}) }} n(G, \chi, s_0) \chi(1) \nonumber \\&&\phantom{m}\ +\ \sum_{\substack{\chi \in \mathrm{Irr}(G) \\ \chi = \overline{\chi}\\ n(G, \chi, s_0) \chi(1) \equiv 1 ({\, \mathrm mod \, }{2}) }} n(G, \chi, s_0) \chi(1).\end{eqnarray}
Let
\begin{align*}\mathcal{S} \ =\ \{ \chi \in \mathrm{Irr}(G) \ :\ \chi = \overline{\chi},\ \ n(G, \chi, s_0) \chi(1) \equiv 1 ({\, \mathrm mod \, }{2}) \}.\end{align*}
Since the first two sums on the right-hand side of (4·3) are even and the order of
$\zeta_{\mathrm{K}}(s)$
at
$s_0$
is odd, we see that
$\mathcal{S}$
is non-empty. For any
$\psi \in \mathrm{Irr}(G')$
and
$\chi \in \mathrm{Irr}(G)$
for which
$\chi = \overline{\chi}$
, we have
\begin{equation}\langle \chi\mid_{G'}, \ \psi \rangle\ =\ \langle \chi\mid_{G^{\prime}}, \ \overline{\psi} \rangle.\end{equation}
We can write
\begin{equation}\chi\mid_{G^{\prime}}\ =\ \sum_{\psi \in \mathrm{Irr}(G'), \atop \psi \ne \overline{\psi}} \langle \chi\mid_{G^{\prime}}, \ \psi \rangle \psi\ \ +\ \sum_{\psi \in \mathrm{Irr}(G'), \atop \psi = \overline{\psi}} \langle \chi\mid_{G^{\prime}}, \ \psi \rangle \psi.\end{equation}
From (4·4), it follows that
\begin{align*}\sum_{\psi \in \mathrm{Irr}(G'), \atop \psi \ne \overline{\psi}}\langle \chi\mid_{G^{\prime}}, \psi \rangle \psi(1)\end{align*}
is even and as
$\chi\mid_{G^{\,\prime}}(1)$
is odd for any
$\chi \in \mathcal{S}$
, the second sum on the right hand side of (4·5) is non-empty. Hence there exists
$\psi \in \mathrm{Irr}(G')$
such that
$\psi = \overline{\psi}$
and
$\langle \chi|_{G'}, \psi \rangle \neq 0$
.
Since G
′ is odd, there does not exist any non-trivial
$\psi \in \mathrm{Irr}(G')$
such that
$\psi = \overline{\psi}$
(see Proposition 13 of [
Reference Collins3
]). Hence the second sum on the right hand side of (4·5) consists of only trivial character and
$\langle \chi|_{G'}, 1_{G'} \rangle \neq 0$
. Since G
′ is a normal subgroup of G and
$\langle \chi|_{G'}, 1_{G'} \rangle \neq 0$
, by Lemma 3·2, we have G
′ is a subgroup of
$\ker \chi$
. Hence
$\langle \chi|_{G'}, 1_{G'} \rangle = \chi(1)$
. Consider the diagram
As before, we see that
\begin{align*}\zeta_{\mathrm{K}_{ab}}(s)\ =\ \prod_{\chi \in \mathrm{Irr}(G)} L(s, \chi, \mathrm{K}/\mathrm{F})^{\langle \chi\mid_{G^{\prime}}, 1_{G'} \rangle}. \end{align*}
Hence
\begin{eqnarray}n(G', 1_{G'}, s_0)&=&\sum_{\substack{\chi \in \mathrm{Irr}(G) \\ \chi \neq \overline{\chi}}}n(G, \chi, s_0) \langle \chi|_{G'}, 1_{G'} \rangle\phantom{m}+\ \ \sum_{\substack{\chi \in \mathrm{Irr}(G) \\ \chi = \overline{\chi} \\ n(G, \chi, s_0) \chi(1) \equiv 0 ({\, \mathrm mod \, }{2}) }}n(G, \chi, s_0) \langle \chi|_{G'}, 1_{G'} \rangle \nonumber \\&&\phantom{m}+\ \ \sum_{\substack{\chi \in \mathrm{Irr}(G) \\ \chi = \overline{\chi} \\n(G, \chi, s_0) \chi(1) \equiv 1 ({\, \mathrm mod \, }{2}) }}n(G, \chi, s_0) \chi(1).\end{eqnarray}
The first two sums on the right-hand side of (4·6) are even and from (4·3) we see that the third sum
$\sum_{\chi \in \mathcal{S}} n(G, \chi, s_0) \chi(1)$
is odd, and hence
$n(G', 1_{G'}, s_0)$
is odd, i.e. the order of
$\zeta_{\mathrm{K}_{ab}}(s)$
at
$s_0$
is odd. Since
$\mathrm{K}_{ab}/\mathrm{F}$
is an abelian extension, applying the first part of the proof of the theorem, we get the result.
4·2 Proof of Corollary 1·4
Since the degree of the extension
$\mathrm{K}/{\mathbb Q}$
is odd and the Riemann zeta function
$\zeta_{\mathbb Q}$
does not have any non-trivial real zero, we are done.
4·3 Proof of Theorem 1·11
If
$\mathrm{K}/\mathrm{F}$
is Galois, then the theorem follows from Theorem 1·3. From now onwards, we assume that
$\mathrm{K}/\mathrm{F}$
is not Galois and
$\mathrm{K}^{cls}$
is the Galois closure of
$\mathrm{K}/\mathrm{F}$
with Galois group G. Let
$\mathrm{K}^{cls}_{ab}$
be the maximal abelian sub extension of
$\mathrm{K}^{cls}$
with Galois group G
′ and H be the Galois group of
$\mathrm{K}^{cls}/\mathrm{K}$
. By given hypothesis,
$|H|$
as well as
$|G'| $
are odd. Consider the diagram
As in the proof of Theorem 1·3, we can deduce
\begin{align*}n(H, 1_{H}, s_0)\ =\ \sum_{\chi \in \mathrm{Irr}(G)} n(G, \chi, s_0) \ \langle \chi|_{H}, 1_{H} \rangle.\end{align*}
Since order of
$\zeta_{\mathrm{K}}$
at
$s_0$
is odd, applying Proposition 3·1 and
\begin{align*}\langle \chi|_{H}, 1_{H} \rangle\ =\ \langle \overline{\chi}|_{H}, 1_{H} \rangle,\end{align*}
we see that there exists
$\chi' \in \mathrm{Irr}(G)$
with
$\chi' = \overline{\chi'}$
such that
\begin{equation}n(G, \chi', s_0)\langle \chi'|_{H}, 1_{H} \rangle\equiv 1 ({\, \mathrm{mod} \, }{2}).\end{equation}
Also
\begin{align*}\chi'(1)\ =\ \chi' \mid_{H}(1)\ =\ \sum_{\phi \in \mathrm{Irr}(H), \atop \phi \neq \overline{\phi}} \langle \chi'\mid_{H}, \ \phi \rangle \phi(1)\ +\ \sum_{\phi \in \mathrm{Irr}(H), \atop \phi = \overline{\phi}} \langle \chi'\mid_{H}, \ \phi \rangle \phi(1).\end{align*}
As order of H is odd, there does not exist any non-trivial
$\phi \in \mathrm{Irr}(H)$
such that
$\phi = \overline{\phi}$
. Thus the second sum consists of only trivial character. From (4·7), we have
$\langle \chi'|_{H}, 1_{H} \rangle$
is odd and hence
$\chi'(1)$
is odd. We can write
\begin{align*}\chi^{\prime}\mid_{G^{\prime}}\ =\ \sum_{\psi \in \mathrm{Irr}(G'), \atop \psi \ne \overline{\psi}} \langle \chi^{\prime}\mid_{G^{\prime}}, \psi \rangle \psi\ \ +\ \langle \chi^{\prime}\mid_{G^{\prime}}, 1_{G'} \rangle 1_{G'}\end{align*}
as the order of G
′ is odd. Since
$\chi'(1)$
is odd, we have
$\langle \chi^{\prime}\mid_{G^{\prime}}, 1_{G^{\prime}} \rangle \neq 0$
. Then by Lemma 3·2, we have
$G' \leq \ker \chi'$
. Applying properties of Artin L-function, we have
\begin{align*}\zeta_{\mathrm{K}^{cls}_{ab}}(s)=\prod_{\lambda \in \mathrm{Irr}(G/G')} L(s, \lambda, \ G/G')=\prod_{\chi \in \mathrm{Irr}(G) \atop \chi(1) = 1} L(s, \chi, \mathrm{K}/\mathrm{F})\end{align*}
(see [
Reference Isaacs8
, p. 25, corollary 2·23]). Hence for any
$ \chi \in \mathrm{Irr}(G)$
,
$\langle\chi \mid_{G^{\prime}}, 1_{G'}\rangle \neq 0$
if and only if
$\chi(1) = 1$
. In particular,
$\chi'(1) = 1$
.
We also know that
$\chi' = \overline{\chi'}$
and so index of
$\ker \chi'$
is at most two. Let
$H' = \ker\chi'$
and M be the fixed field of H’. Moreover, since
$\chi'(1) = 1$
, we have
$\langle\chi'\mid_H, 1_H \rangle = 1$
and hence H is a subgroup of H’. Therefore
$\mathrm{K}^{H'} \subseteq \mathrm{K}^H$
i.e.
$\mathrm{M} \subseteq \mathrm{K}$
. Also
$[\mathrm{M}\, : \, \mathrm{F}] \leq 2$
.
Now we are in the situation of the following diagram;
Since H
′ is of index 2 subgroup of G and
$\langle \chi'\mid_{H'}, 1_{H'} \rangle = 1$
, we have
\begin{align*}n(H', 1_{H'}, s_0)\ =\ n(G, \chi', s_0)\ +\ n(G, 1_{G}, s_0).\end{align*}
It follows from (4·7) that
$n(G, \chi', s_0) \ge 1$
and hence
$n(H', 1_{H'}, s_0) \geq 1$
, i.e.
$\zeta_{\mathrm{M}}(s_0) =0$
as
$n(G, 1_{G}, s_0) \geq 0$
. Finally, note that if
$n(G, 1_{G}, s_0)$
is odd, then
$s_0$
is an odd order zero of
$\zeta_F$
and we are done. Now suppose that
$n(G, 1_{G}, s_0)$
is even. From (4·7), we know that
$n(G, \chi', s_0)$
is odd and hence
$n(H', 1_{H'}, s_0) $
is odd i.e.
$s_0$
is an odd order zero of
$\zeta_{\mathrm{M}}$
. This completes the proof of Theorem 1·11.
4·4 Proof of Corollary 1·12
Let
$s_0$
be a stark zero of
$\zeta_{\mathrm{K}}$
. By Theorem 1·11, there exists a quadratic number field M such that
$\zeta_{\mathrm{M}}(s_0) = 0$
. By [
Reference Stark17
, lemma 11], there is a effectively computable constant
$c_1 \gt0 $
such that
$\zeta_{\mathrm{M}}(\sigma) \neq 0$
for
\begin{equation}\sigma \geq 1- (c_1 \sqrt{d_{\mathrm{M}}})^{-1}.\end{equation}
Since
$[\mathrm{K}\, : \, \mathrm{M}] = n/2$
, we have
$d_{\mathrm{K}}^{1/n} \geq \sqrt{d_{\mathrm{M}}}$
. Hence
$\zeta_{\mathrm{K}}(\sigma) \neq 0$
for
\begin{equation}\sigma \geq 1 - (c_1 \ {d_{\mathrm{K}}}^{1/n})^{-1}.\end{equation}
Indeed, if there is a zero of
$\zeta_{\mathrm{K}}$
in the region (4·9), it is a real and simple zero as region (4·9) lies inside the Stark region. Then applying Theorem 1·11 and the fact that (4·8) lies inside the Stark region, we see that this zero gives rise to a simple zero of
$\zeta_{\mathrm{M}}$
. Hence by [
Reference Stark17
, lemma 11], there is a effectively computable constant
$c_2\gt0$
, such that
\begin{align*}\rho_{\mathrm{K}} \gt (c_1 \ c_2 \ {d_{\mathrm{K}}}^{1/n})^{-1}.\end{align*}
This completes the proof of Corollary 1·12.
4·5 Proof of Theorem 1·13
By given hypothesis, H is contained in N, where N is an abelian normal subgroup of G. Since Index of H in G is odd, it follows that the index of N in G is odd. By [
Reference Serre16
, chapter 8, corollary to proposition 24], we know that
$\chi(1)$
divides the index of N in G for any
$\chi \in \mathrm{Irr}(G)$
as N is a abelian normal subgroup of G. Hence
$\chi(1)$
is odd for all
$\chi \in \mathrm{Irr}(G)$
.
Since order of
$\zeta_{\mathrm{K}}$
at
$s_0$
is odd, proceeding as in Theorem 1·3 and Theorem 1·11, there exists
$\chi' \in \mathrm{Irr}(G)$
with
$\chi' = \overline{\chi'}$
such that
\begin{equation}n(G, \chi', s_0)\langle \chi'|_{H}, 1_{H} \rangle\equiv 1 ({\, \mathrm{mod} \, }{2}).\end{equation}
Since
$\chi(1)$
is odd for any
$\chi \in \mathrm{Irr}(G)$
and by given hypothesis, G
′ is odd, arguing as in the proof of Theorem 1·11, we have
$\chi'(1) = 1$
. Hence
\begin{eqnarray*}\langle \chi'|_{H}, 1_{H} \rangle =1&\iff&H \leq \ker \chi' \\&\iff&[G\, : \, \ker \chi'] \mid [G\, : \, H].\end{eqnarray*}
As
$\chi'(1) = 1, \chi' = \overline{\chi'}$
and
$[G\, : \, H]$
is odd, we get that
$[G\, : \, \ker \chi'] = 1$
. This implies that
$\chi' = 1_G$
. It now follows from (4·10) that
$n(G, \chi', s_0)$
is odd, i.e. the order of
$\zeta_{\mathrm{F}}$
at
$s_0$
is odd.
4·6 Proof of Theorem 1·14
Let
$\mathrm{K}/\mathrm{F} \in \mathcal{D}^t$
, i.e.
\begin{align*} \mathrm{F} \ =\ \mathrm{K}_0 \subseteq \mathrm{K}_1 \subseteq \mathrm{K}_2 \cdots \subseteq \mathrm{K}_{\ell} \ =\ \mathrm{K} \end{align*}
be a tower of number fields such that
$\mathrm{K}_{i+1}/\mathrm{K}_i \in \mathcal{D}$
for
$0 \le i \le \ell-1$
. The theorem follows by applying Theorem 1·13 recursively to show that any odd order real zero of
$\zeta_{\mathrm{K}}$
is a odd order real zero of
$\zeta_{\mathrm{K}_{\ell -1}}$
and hence an odd order real zero of
$\zeta_{\mathrm{K}_{\ell - 2}}$
and so on.
Acknowledgements
We would like to thank Purusottam Rath for his valuable suggestions and to the referee for their suggestions, which helped improve the exposition of the paper.
