# Relative Ideal Classes
of Arbitrary Order

David L. Pincus  Department of Mathematics
University of Maryland
College Park, MD 20742, USA<EMAIL_ADDRESS>and  Lawrence C. Washington
Department of Mathematics
University of Maryland
College Park, MD 20742, USA<EMAIL_ADDRESS>
###### Abstract.

We adapt a known technique for searching for ideal classes of arbitrary order
and then apply it to three families of number fields. We show that a family of
cyclic sextic number fields has infinitely many fields in it that contain a
relative ideal class of order $r,$ where $r$ is a positive integer relatively
prime to the degree of the extension. We then show that the same holds true
for a family of cyclic quartic number fields. Though the technique is
traditionally applied to Galois extensions, we show how it may be adapted to
handle a family of non-Galois cubic number fields and prove that this family
contains infinitely many fields with an ideal class of arbitrary order
relatively prime to three.

###### Key words and phrases:

fractional ideal, class group, relative ideal class

###### 2020 Mathematics Subject Classification:

Primary 11R29; Secondary 11R16

##  1\. Introduction

Let $\mathfrak{F}:=\\{K_{n}\\}$ denote an indexed family of number fields and
let $r$ be a positive integer. Let $\mathcal{I}_{K_{n}}$ denote the group of
fractional ideals of $K_{n}\in\mathfrak{F}.$ Can we find a field
$K_{n}\in\mathfrak{F}$ that contains an ideal class of order $r?$ We may
answer in the affirmative if the following criteria are satisfied.

  1. (1)

We can find an index $n$ and a principal ideal
$w{\mathcal{O}}_{K_{n}}\in\mathcal{I}_{K_{n}}$ that factors as
$w{\mathcal{O}}_{K_{n}}=J^{r}$ for some fractional ideal
$J\in\mathcal{I}_{K_{n}}.$

  2. (2)

We can show that no smaller power of $J$ is principal.

This sort of approach has been used by many authors going back at least as far
as Nagell [7].

A well-known method for meeting Criterion (1) begins by looking for solutions
to the equation

(1.1) $Y^{r}=f_{n}(X),$

where $f_{n}(X)\in{\mathbb{Q}}[X]$ is the minimal polynomial of a primitive
element for $K_{n}/{\mathbb{Q}}.$ Indeed if $K_{n}/{\mathbb{Q}}$ is Galois
with primitive element $\rho,$ then a solution $(x,y)\in{\mathbb{Q}}^{2}$ to
Equation (1.1) gives

(1.2) $y^{r}=f_{n}(x)=\prod_{i=0}^{[K:{\mathbb{Q}}]-1}(x-\rho_{i}),$

where the $\rho_{i}$ run over the Galois conjugates of $\rho:=\rho_{0}.$ If
one can show that the exact power of a prime ideal that divides, say,
$(x-\rho_{i}){\mathcal{O}}_{K}$ is also the exact power of the same prime
ideal that divides $y^{r}{\mathcal{O}}_{K},$ then one may conclude that
$(x-\rho_{i}){\mathcal{O}}_{K_{n}}=J^{r}$ for some ideal
$J\in\mathcal{I}_{K_{n}}.$

A well-known method for meeting Criterion (2) is then to assume the contrary
and suppose that $(x-\rho_{i}){\mathcal{O}}_{K_{n}}$ is the $p$th power of a
principal ideal, where $p$ is a prime dividing $r.$ Element-wise this implies
that there is a $z\in K_{n}^{\times}$ and a unit
$u\in{\mathcal{O}}_{K_{n}}^{\times}$ such that

(1.3) $x-\rho_{i}=uz^{p}.$

By restricting our consideration to those solutions $(x,y)\in{\mathbb{Q}}^{2}$
that have the property that for each prime divisor $p$ of $r$ the prime
divisors of $y$ satisfy certain $p$th power residue conditions, we can ensure
that for each prime $p\mid r,$ Equation (1.3) fails to hold.

In this paper we extend the technique outlined above in the following three
ways.

A. If several of the (ideal) factors $(x-\rho_{i}){\mathcal{O}}_{K_{n}}$ on
the right hand side of Equation (1.2) factor as $r$th powers of ideals, then
so do products and quotients of these ideals. By taking appropriate products
and quotients of such factors, we may obtain a principal ideal that has
trivial norm down to each proper subfield of $K_{n}$ and which also factors as
an $r$th power, say as $J^{r}.$ The ideal class of $J$ then has trivial norm
down to each proper subfield of $K_{n}$ and is called a _relative ideal
class_. We use this technique in Section 2 (resp. Section 3) below to search
for relative ideal classes of order $r$ in a family of sextic (resp. quartic)
number fields. The qualifier _relative_ is important. Had we simply been
interested in searching for ideal classes of order $r,$ then it would have
sufficed to look for a principal fractional ideal belonging to any of the
proper subfields of $K_{n}$ that factors as an $r$th power. The fact that the
ideal classes we detect are relative ideal classes guarantees that these
classes are _new_ and not coming from ideal classes of any proper subfield.

B. If $K_{n}/{\mathbb{Q}}$ is non-Galois with primitive element
$\rho\in{\mathcal{O}}_{K_{n}}$ and $(x,y)\in{\mathbb{Z}}^{2}$ is such that
$y^{r}=f_{n}(x),$ then Equation (1.2) no longer holds in $K_{n}.$ It is still
true, however, that $y^{r}=f_{n}(x)=(x-\rho)q(x),$ where
$q(x)\in{\mathcal{O}}_{K_{n}}.$ In particular, if for each prime ideal
$\mathfrak{P}$ dividing $(x-\rho){\mathcal{O}}_{K_{n}},$ we have
$q(x)\not\equiv 0\pmod{\mathfrak{P}},$ then we still have that
$(x-\rho){\mathcal{O}}_{K_{n}}=J^{r}$ for some ideal
$J\in\mathcal{I}_{K_{n}}.$ We use this technique in Section 4 to search for
ideal classes of order $r$ in a family of non-Galois cubic number fields.

C. Congruence conditions are not always sufficient to guarantee that for each
prime $p$ dividing $r,$ $(x-\rho_{i})/u$ doesn’t have a $p$th root in $K_{n}.$
Likewise, congruence conditions don’t always suffice to guarantee that for
each prime $p\mid r,$ there is no $p$th root of $w/u$ in $K_{n},$ where $w$ is
a product of powers of the $(x-\rho_{i}).$ What the congruences can do,
however, is limit the possibilities for $u\in{\mathcal{O}}_{K_{n}}^{\times}.$
Eliminating the possibility that there is a prime $p\mid r$ such that $w/u$
has a $p$th root in $K_{n}$ can then be reduced to showing that the polynomial
$\varphi(X^{r})$ is irreducible in ${\mathbb{Q}}[X],$ where
$\varphi(X)\in{\mathbb{Q}}[X]$ is the minimal polynomial of $w/u.$ Since
$\varphi(X^{r})$ depends on both $r$ and the parameter $n$ indexing our
families of number fields, proving irreducibility is not easy. In Sections 2
and 3 we make use of Hilbert’s Irreducibility Theorem [4] to do so.

Using these techniques we are able to show that each of the families mentioned
above has infinitely many fields that contain an ideal class of order $r$. In
the case of the sextic and quartic families, these are _relative_ ideal
classes. In all cases $r$ is a positive integer relatively prime to the degree
of the extension, but otherwise arbitrary. Although the methods we employ may
be extended to handle the case when $(r,[K:{\mathbb{Q}}])>1,$ the result
clutters the presentation and obscures the techniques we are trying to
illustrate. Thus our presentation sticks to the case in which
$(r,[K:{\mathbb{Q}}])=1.$

##  2\. A family of sextic number fields.

Let $n\in{\mathbb{Z}}\smallsetminus\\{0,\pm 6,\pm 26\\}$ and let
$K_{n}={\mathbb{Q}}(\rho),$ where $\rho$ is a root of

(2.1)
$\begin{split}f_{n}(X):=&X^{6}-\left(\frac{n-6}{2}\right)X^{5}-5\left(\frac{n+6}{4}\right)X^{4}\\\
&-20X^{3}+5\left(\frac{n-6}{4}\right)X^{2}+\left(\frac{n+6}{2}\right)X+1.\end{split}$

$K_{n}/{\mathbb{Q}}$ is a real cyclic sextic extension and was studied
extensively by M.-N. Gras in [3]. Let
$G=\operatorname{Gal}(K_{n}/{\mathbb{Q}}).$ Then
$G=\langle\sigma\mid\sigma^{6}=1\rangle,$ where
$\sigma:\rho\mapsto(\rho-1)/(\rho+2).$ The images of $\rho$ under $G,$ denoted
$\rho_{i}:=\sigma^{i}(\rho),$ $(i=0,\ldots,5),$ are given by

(2.2) $\begin{gathered}\rho_{0}=\rho\\\
\rho_{3}=-(\rho+2)/(2\rho+1)\end{gathered}\qquad\begin{gathered}\rho_{1}=(\rho-1)/(\rho+2)\\\
\rho_{4}=-(\rho+1)/\rho\end{gathered}\qquad\begin{gathered}\rho_{2}=-1/(\rho+1)\\\
\rho_{5}=-(2\rho+1)/(\rho-1).\end{gathered}$

Let $\mathfrak{F}:=\\{K_{n}\mid n\neq 0,\pm 6,\pm 26\\}.$ Let
$\mathcal{I}_{K}$ denote the group of fractional ideals of $K\in\mathfrak{F}$
and for a fractional ideal $J\in\mathcal{I}_{K},$ let $[J]$ denote its ideal
class. Let $k_{2}={\mathbb{Q}}(\sqrt{n^{2}+108})$ denote the unique quadratic
subfield of $K$ and let $k_{3}={\mathbb{Q}}(\rho_{0}\rho_{3})$ denote the
unique cubic subfield of $K.$ Then $[J]$ is said to be a _relative ideal
class_ if it has trivial norm down to both $k_{2}$ and $k_{3}.$

Note that $\rho_{i},$ $(i=0,\ldots,5)$ is an algebraic integer if and only if
$n\equiv 2\pmod{4}.$ So it is also worthwhile to note that
$K_{n}={\mathbb{Q}}(\rho_{0}/\rho_{3})$ and that $\rho_{0}/\rho_{3}$ is a zero
of the polynomial

(2.3)
$\displaystyle\begin{split}\varphi_{n}(X)&:=(X-1)^{6}-(n^{2}+108)(X^{2}+X)^{2}\\\
&=X^{6}-6X^{5}-(n^{2}+93)X^{4}-(2n^{2}+236)X^{3}-(n^{2}+93)X^{2}-6X+1.\end{split}$

In particular, $\rho_{0}/\rho_{3}$ and its Galois conjugates are algebraic
integers for any $n,$ i.e., regardless of whether or not $n\equiv 2\pmod{4}.$
Since $N_{K\mid k_{2}}(\rho_{0}/\rho_{3})=1$ and $N_{K\mid
k_{3}}(\rho_{0}/\rho_{3})=1$ the Galois conjugates of $\rho_{0}/\rho_{3}$ are,
in fact, relative units of $K_{n}.$

Our primary result is

######  Theorem 2.1.

Let $r>0$ be an integer with $(6,r)=1.$ There are infinitely many fields
$K\in\mathfrak{F}$ containing a relative ideal class of order $r.$

It is clear that $K_{n}=K_{-n},$ since the zeros of $\varphi_{n}(X)$ serve as
primitive elements for $K_{n}/{\mathbb{Q}}$ and since
$\varphi_{n}(X)=\varphi_{-n}(X).$ As we make no additional assumptions
regarding the injectivity of the parametrization of $\mathfrak{F}$ given by
$n\mapsto K_{n},$ it would be insufficient for us to show that there are
infinitely many $n$ for which $K_{n}$ contains a relative ideal class of order
$r.$ Instead we put most of our effort into first proving

######  Proposition 2.2.

Let $r>0$ be an integer with $(6,r)=1$ and let $q\in{\mathbb{Z}}$ be a prime
such that

$\mathcal{P}(X):=(X^{r}+143-30\sqrt{-108})(X^{r}+143+30\sqrt{-108})\in{\mathbb{Z}}[X]$

splits into linear factors $\pmod{q}.$ Assume that $q\nmid 210r.$ There is a
field $K\in\mathfrak{F}$ that satisfies the following three conditions:

  1. (i)

There is an ideal $J\in\mathcal{I}_{K}$ whose ideal class, $[J],$ has order
$r.$

  2. (ii)

$N_{K|k_{i}}(J),$ $(i=2,3)$ is a principal ideal.

  3. (iii)

$q$ ramifies in $k_{2}.$

Once Proposition 2.2 is established we have the following.

###### Proof of Theorem 2.1.

There are infinitely many primes $q\nmid 210r$ for which $\mathcal{P}(X)$
splits into linear factors $\pmod{q}.$ Hence there are infinitely many primes
$q$ that ramify in the quadratic subfield of some $K\in\mathfrak{F}$
(depending on $q$) for which $\mathcal{I}_{K}$ contains a relative ideal class
of order $r.$ Since only finitely many primes can ramify in the quadratic
subfields of a finite subcollection of $\mathfrak{F},$ we conclude that the
collection of $K\in\mathfrak{F}$ for which $\mathcal{I}_{K}$ contains a
relative ideal class of order $r$ is infinite. ∎

We remark that we could have proved Theorem 2.1 using the argument given in
the proof of Theorem 4.1 below. That proof finds infinitely many fields
containing a class of order $r$ by finding cyclic subgroups of the class group
of orders that are increasingly higher powers of $r.$ Since the proof we’ve
given above doesn’t rely on such an argument, it reveals slightly more about
this family of fields.

To provide some insight into the motivation behind Proposition 2.2, we note
that when $\mathcal{P}(X)$ splits into linear factors $\pmod{q},$ there is a
root of the equation

(2.4) $X^{r}=30n-143=f_{n}(-3)$

in the field ${\mathbb{Z}}/q{\mathbb{Z}},$ where $n^{2}+108\equiv 0\pmod{q}.$

Let

$w:=\frac{(3+\rho_{1})(3+\rho_{2})}{(3+\rho_{4})(3+\rho_{5})}.$

######  Proposition 2.3.

Let $n\equiv 2\pmod{4}$ be an integer. If there is a $y\in{\mathbb{Z}}$ with
$7\nmid y$ and such that $y^{r}=30n-143=f_{n}(-3),$ then
$w{\mathcal{O}}_{K_{n}}=J^{r}$ for some ideal $J\in\mathcal{I}_{K_{n}}.$

For the proof of this proposition we require the following easy lemma:

######  Lemma 2.4.

If a prime $p$ is a common divisor of $30n-143$ and
$\operatorname{disc}(f_{n}),$ then $p=7.$

###### Proof.

The congruences $30n-143\equiv 0\pmod{p}$ and
$\operatorname{disc}(f_{n})=3^{6}(n^{2}+108)^{5}/2^{14}\equiv 0\pmod{p}$ hold
simultaneously only if $p=7.$ ∎

###### Proof of Proposition 2.3.

It suffices to prove that for each index $i,$ the ideal
$(3+\rho_{i}){\mathcal{O}}_{K}$ is an $r$th power. Since $n\equiv 2\pmod{4},$
each $\rho_{i}$ is an algebraic integer of $K.$ Since $7\nmid y,$ no prime
dividing $y$ also divides $\operatorname{disc}(f_{n}).$ If a prime $p\mid y$
and $\mathfrak{P}\mid p{\mathcal{O}}_{K}$ is a prime ideal then
$\prod_{i=0}^{5}(3+\rho_{i})=f_{n}(-3)\in\mathfrak{P}.$ Hence
$3+\rho_{i}\in\mathfrak{P}$ for some $0\leq i\leq 5.$ Each of the six Galois
conjugates of $3+\rho_{i}$ is contained in a Galois conjugate of
$\mathfrak{P},$ but since $p\nmid\operatorname{disc}(f_{n}),$ no two distinct
conjugates of $3+\rho_{i}$ can be contained in the same Galois conjugate of
$\mathfrak{P}.$ Hence the prime ideals $\mathfrak{P}^{\sigma^{i}},$
$(i=0,\ldots,5),$ are all distinct and each contains exactly one of the Galois
conjugates of $3+\rho_{i}.$ So for each prime $p\mid y,$ let
$\mathfrak{P}_{i}$ denote the unique prime common divisor of
$p{\mathcal{O}}_{K}$ and $(3+\rho_{i}){\mathcal{O}}_{K},$ $(i=0,\ldots,5).$ If
$p^{\nu(p)}$ is the exact power of $p$ dividing $y,$ then
$\mathfrak{P}_{i}^{r\nu(p)}$ is the exact power of $\mathfrak{P}_{i}$ dividing
$(3+\rho_{i}){\mathcal{O}}_{K}.$ Hence
$(3+\rho_{i}){\mathcal{O}}_{K}=\left(\prod_{p\mid
y}\mathfrak{P}_{i}^{\nu(p)}\right)^{r}.$ ∎

Let $\varepsilon$ denote the fundamental unit of $k_{2},$ let
$\mu_{0},\mu_{1}$ be a fundamental system of units for $k_{3}$ and let
$U:=\langle-1,\varepsilon,\mu_{0},\mu_{1},\rho_{0}/\rho_{3},\rho_{1}/\rho_{4}\rangle\subseteq{\mathcal{O}}_{K}^{\times}.$
Without loss of generality [1] assume that $\mu_{i}=\sigma^{i}(\mu_{0}),$
$(i=0,1,2).$ We will need to know that
$[{\mathcal{O}}_{K}^{\times}:U]<\infty.$ Since
$\operatorname{rk}{\mathcal{O}}_{K}^{\times}=5,$ it will suffice to prove the
following.

######  Lemma 2.5.

The subset
$S:=\\{\varepsilon,\mu_{0},\mu_{1},\rho_{0}/\rho_{3},\rho_{1}/\rho_{4}\\}\subset{\mathcal{O}}_{K}^{\times}$
is multiplicatively independent.

###### Proof.

Let $R$ denote the regulator of $S.$ Then $R$ is the absolute value of

$\displaystyle\begin{split}&\;\;\;\;\;\det\begin{pmatrix}\log|\mu_{0}|&\log|\mu_{1}|&\log|\rho_{0}/\rho_{3}|&\log|\rho_{1}/\rho_{4}|&\log|\epsilon|\\\
\log|\mu_{1}|&\log|\mu_{2}|&\log|\rho_{1}/\rho_{4}|&\log|\rho_{2}/\rho_{5}|&\log|\epsilon^{\sigma}|\\\
\log|\mu_{2}|&\log|\mu_{0}|&\log|\rho_{2}/\rho_{5}|&\log|\rho_{3}/\rho_{0}|&\log|\epsilon|\\\
\log|\mu_{0}|&\log|\mu_{1}|&\log|\rho_{3}/\rho_{0}|&\log|\rho_{4}/\rho_{1}|&\log|\epsilon^{\sigma}|\\\
\log|\mu_{1}|&\log|\mu_{2}|&\log|\rho_{4}/\rho_{1}|&\log|\rho_{5}/\rho_{2}|&\log|\epsilon|\\\
\end{pmatrix}\\\ \\\
&=\det\begin{pmatrix}\log|\mu_{0}|&\log|\mu_{1}|&\log|\rho_{0}/\rho_{3}|&\log|\rho_{1}/\rho_{4}|&\log|\epsilon|\\\
\log|\mu_{1}|&\log|\mu_{2}|&\log|\rho_{1}/\rho_{4}|&\log|\rho_{2}/\rho_{5}|&\log|\epsilon^{\sigma}|\\\
0&0&2\log|\rho_{1}/\rho_{4}|&2\log|\rho_{2}/\rho_{5}|&\log|\epsilon|\\\
0&0&2\log|\rho_{2}/\rho_{5}|&2\log|\rho_{3}/\rho_{0}|&\log|\epsilon^{\sigma}|\\\
0&0&0&0&3\log|\epsilon|\end{pmatrix}.\end{split}$

It is clear that

$\det\begin{pmatrix}\log|\mu_{0}|&\log|\mu_{1}|\\\
\log|\mu_{1}|&\log|\mu_{2}|\end{pmatrix}\neq 0,$

because up to sign this is the regulator of $\\{\mu_{0},\mu_{1}\\},$ a
multiplicatively independent subset of $k_{3}.$ It is equally clear that
$3\log|\varepsilon|\neq 0,$ because $\varepsilon>1.$

When $n\geq 76,$ three of the zeros of $\varphi_{n}(X)$ lie in the intervals
$(3-n,4-n),$ $(-1,-9/10),$ and $(1/(n+5),1/(n+4))$; the other three zeros of
$\varphi_{n}(X)$ are reciprocals of these zeros. Since $f_{n}(1)=-27$ and
since $f_{n}(x)\to\infty$ as $x\to\infty,$ we deduce that $f_{n}(X)$ has a
zero $>1.$ From the form of this zero’s Galois conjugates, it is clear that
$f_{n}(X)$ has another positive zero in the interval $(0,1),$ and that the
remaining zeros of $f_{n}(X)$ are all negative. If we take $\rho_{0}$ to be
the zero of $f_{n}(X)$ greater than $1,$ then we may identify
$\rho_{0}/\rho_{3}$ with the zero of $\varphi_{n}(X)$ in the interval
$(3-n,4-n).$ We may also then identify $\rho_{1}/\rho_{4}$ with the zero in
$(-1,-9/10)$ and $\rho_{2}/\rho_{5}$ with the zero in $(1/(n+5),1/(n+4)).$
Hence

$\displaystyle\log|\rho_{1}/\rho_{4}|\log|\rho_{3}/\rho_{0}|-\log^{2}|\rho_{2}/\rho_{5}|$
$\displaystyle<-\log(9/10)\log(n-3)-\log^{2}(n+4)$
$\displaystyle<-\log(n+4)(\log(9/10)+\log(n+4))$ $\displaystyle<0,$

and we conclude that $R\neq 0$ when $n\geq 76.$ A computer verifies that
$\log|\rho_{1}/\rho_{4}|\log|\rho_{3}/\rho_{0}|-\log^{2}|\rho_{2}/\rho_{5}|\neq
0$ when $0<n<76$ and $n\not\in\\{0,6,26\\}.$ Hence $R\neq 0$ and $S$ is a
multiplicatively independent subset of units of $K.$ ∎

Having shown that the index of $U$ in ${\mathcal{O}}_{K}^{\times}$ is finite,
we will use the following lemma to prevent certain primes from dividing this
index. We also note that in the sequel we employ the notation
$(a\,|\,\ell)_{p}=1$ (resp. $(a\,|\,\ell)_{p}\neq 1$) to mean that $a$ is a
$p$th power $\textmd{residue}\pmod{\ell}$ (resp. $a$ is a $p$th power
$\textmd{nonresidue}\pmod{\ell}.$) When the subscript $p=2,$ the notation is
the Legendre symbol.

######  Lemma 2.6.

Let $n\equiv 2\pmod{4}$ be an integer and let $p\neq 2,3$ be a prime. If there
are primes $\ell_{i}\mid 30n-143=f_{n}(-3),$ $(i=1,2),$ not equal to $7$ and
such that

  1. (1)

$(2\,|\,\ell_{1})_{p}=(3\,|\,\ell_{1})_{p}=1$ and $(5\,|\,\ell_{1})_{p}\neq 1$

  2. (2)

$(3\,|\,\ell_{2})_{p}=(5\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq
1,$

then $p\nmid[{\mathcal{O}}_{K}^{\times}:U].$

###### Proof.

If $p\mid[{\mathcal{O}}_{K}^{\times}:U],$ then there is a
$u\in{\mathcal{O}}_{K}^{\times}\smallsetminus U$ such that
$u^{p}=\pm\varepsilon^{a}\mu_{0}^{b}\mu_{1}^{c}(\rho_{0}/\rho_{3})^{d}(\rho_{1}/\rho_{4})^{e}.$
Without loss of generality, we may take
$u^{p}=\varepsilon^{a}\mu_{0}^{b}\mu_{1}^{c}(\rho_{0}/\rho_{3})^{d}(\rho_{1}/\rho_{4})^{e}$
with $0\leq a,b,c,d,e<p.$ Taking field norms, we find that $(N_{K\mid
k_{2}}(u))^{p}=\pm\varepsilon^{3a}.$ Hence $p\mid a$ and $a=0.$ Likewise the
fact that $(N_{K\mid k_{3}}(u))^{p}=\pm\mu_{0}^{2b}\mu_{1}^{2c}$ implies that
$p$ divides both $b$ and $c$ and hence that $b=c=0.$ Choose prime ideals
$\mathcal{L}_{i}\subset{\mathcal{O}}_{K},$ $(i=1,2),$ with
$\mathcal{L}_{i}\cap{\mathbb{Z}}=\ell_{i}{\mathbb{Z}}$ and such that
$\rho_{0}\equiv-3\pmod{\mathcal{L}_{i}}$. Then $\rho_{0}/\rho_{3}\equiv
15\pmod{\mathcal{L}_{i}},$ and
$\rho_{1}/\rho_{4}\equiv-6\pmod{\mathcal{L}_{i}},$ and we deduce that
$u^{p}\equiv(-1)^{e}2^{e}3^{d+e}5^{d}\pmod{\mathcal{L}_{i}}.$ Note that every
Galois conjugate ideal $\mathcal{L}_{i}$ contains a conjugate of $3+\rho_{0},$
but that no two such ideals can contain the same conjugate since this would
imply that $\ell_{i}$ was a common divisor of $30n-143$ and
$\operatorname{disc}(f_{n})$ and so would contradict Lemma 2.4. Hence
$\ell_{i}$ completely splits in $K$ and the residual degree of
$\mathcal{L}_{i}$ is equal to $1.$ Applying our power residue hypotheses we
conclude that $d=e=0.$ But then $u^{p}=1$ and so $u=1\in U$; contradiction.
Hence $p\nmid[{\mathcal{O}}_{K}^{\times}:U].$ ∎

######  Proposition 2.7.

Let $n\equiv 2\pmod{4}$ be an integer. If there is an integer $y$ not
divisible by $7$ satisfying $y^{r}=30n-143=f_{n}(-3),$ and if for each prime
$p\mid r$ the following conditions hold:

  1. (1)

there are primes $\ell_{i}\mid 30n-143$ $(i=1,2),$ such that

    1. (a)

$(2\,|\,\ell_{1})_{p}=(3\,|\,\ell_{1})_{p}=1$ and $(5\,|\,\ell_{1})_{p}\neq 1$

    2. (b)

$(3\,|\,\ell_{2})_{p}=(5\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq 1$

  2. (2)

$w\rho_{4}/\rho_{1}$ fails to have a $p$th root in $K_{n},$

then $w{\mathcal{O}}_{K}=J^{r}$ where $[J]$ has order $r.$

###### Proof.

From Proposition 2.3 we deduce that $w{\mathcal{O}}_{K}=J^{r}$ for some ideal
$J\in\mathcal{I}_{K}$ and so the order of $[J]$ is $\leq r.$ If it was
strictly $<r,$ then $w{\mathcal{O}}_{K}$ would be a prime power of a principal
ideal for some prime $p\mid r.$ Supposing this to be the case and writing
$w{\mathcal{O}}_{K}=(z{\mathcal{O}}_{K})^{p},$ for some $z\in K^{\times},$ we
have that $w=uz^{p}$ for some $u\in{\mathcal{O}}_{K}^{\times}.$ Since
$p\nmid[{\mathcal{O}}_{K}^{\times}:U],$ we may assume $u\in U.$ Since $p$ is
odd, we may ignore any factor of $-1$ and assume that

$u=\varepsilon^{a}\mu_{0}^{b}\mu_{1}^{c}(\rho_{0}/\rho_{3})^{d}(\rho_{1}/\rho_{4})^{e}.$

By absorbing any $p$th powers into $z^{p},$ we may further assume that
$a,b,c,d,$ and $e$ are integers in the interval $[0,p).$ Since $N_{K\mid
k_{2}}(w)=1,$ we deduce that $N_{K\mid k_{2}}(1/z)^{p}=N_{K\mid
k_{2}}(u)=\pm\varepsilon^{3a}.$ Recalling that $p\mid r$ and that $(r,6)=1,$
we find that $p\mid a$ and so conclude that $a=0.$ Since $N_{K\mid
k_{3}}(w)=1,$ we deduce that $N_{K\mid k_{3}}(1/z)^{p}=N_{K\mid
k_{3}}(u)=\pm\mu_{0}^{2b}\mu_{1}^{2c}.$ Hence $p\mid b$ and $p\mid c$ and so
$b=c=0.$ Thus $w=(\rho_{0}/\rho_{3})^{d}(\rho_{1}/\rho_{4})^{e}z^{p}.$ Choose
prime ideals $\mathcal{L}_{i}\subset{\mathcal{O}}_{K},$ $(i=1,2),$ with
$\mathcal{L}_{i}\cap{\mathbb{Z}}=\ell_{i}{\mathbb{Z}}$ and such that
$\rho_{0}\equiv-3\pmod{\mathcal{L}_{i}}$. Then $w\equiv
6\pmod{\mathcal{L}_{i}},$ $\rho_{3}/\rho_{0}\equiv
1/15\pmod{\mathcal{L}_{i}},$ and
$\rho_{4}/\rho_{1}\equiv-1/6\pmod{\mathcal{L}_{i}},$ and we deduce that
$z^{p}\equiv(-1)^{e}2^{1-e}3^{1-d-e}5^{-d}\pmod{\ell_{i}}.$ Applying our power
residue hypotheses we conclude that $d=0,$ $e=1$ and thereby deduce that
$w\rho_{4}/\rho_{1}$ has a $p$th root in $K.$ Since this contradicts our
hypotheses $w{\mathcal{O}}_{K}$ is not a prime power of a principal ideal for
any prime $p\mid r$ and $[J]$ has order $r.$ ∎

To apply Proposition 2.7, we will need condition(s) sufficient to guarantee
that $w\rho_{4}/\rho_{1}$ fails to have a $p$th root in $K_{n}$ for each prime
$p\mid r.$ The next proposition provides us with such a condition.

######  Proposition 2.8.

Let

$p_{n,r}(X):=\sum_{k=0}^{6}a_{k}X^{rk},$

where

$\displaystyle a_{0}=a_{6}=(30n-143)^{2}$ $\displaystyle
a_{1}=a_{5}=-6(30n-143)^{2}$ $\displaystyle
a_{2}=a_{4}=-104149n^{2}-128700n-12399357$
$\displaystyle\;\;\;\;=-\frac{104149}{900}(30n-143)^{2}-\frac{16823807}{450}(30n-143)-\frac{13841287201}{900}$
$\displaystyle a_{3}=-253298n^{2}+171600n-25821164$
$\displaystyle\;\;\;\;=-\frac{253298}{900}(30n-143)^{2}-\frac{16823807}{225}(30n-143)-\frac{13841287201}{450}$

If $p_{n,r}(X)$ is irreducible in ${\mathbb{Q}}[X],$ then for each $p\mid r,$
$w\rho_{4}/\rho_{1}$ fails to have a $p$th root in $K_{n}.$

###### Proof.

If there is a $z\in K_{n}^{\times}$ such that $z^{p}=w\rho_{4}/\rho_{1}$ with
$p$ a prime dividing $r,$ then $X^{r}-w\rho_{4}/\rho_{1}$ factors non-
trivially in $K_{n}[X].$ Let $m(X)\in{\mathbb{Q}}[X]$ denote the minimal
polynomial of $w\rho_{4}/\rho_{1}$ over ${\mathbb{Q}}.$ We note that $\deg
m(X)=6,$ since $w\rho_{4}/\rho_{1}$ is a primitive element for
$K_{n}/{\mathbb{Q}}.$ (Note that $w\rho_{4}/\rho_{1}$ has trivial norm down to
each proper subfield of $K_{n}.$ If $w\rho_{4}/\rho_{1}$ was contained in such
a proper subfield we would be forced to conclude that $w=\rho_{1}/\rho_{4}.$
Using Equations (2.2), we rewrite this as
$14\rho^{6}+42\rho^{5}+28\rho^{4}-14\rho^{3}-42\rho^{2}-28\rho=0.$ Since
$\rho\neq 0,$ we deduce that $\rho$ is a zero of a degree $5$ polynomial with
integer coefficients. Since $\rho$ is a primitive element for
$K_{n}/{\mathbb{Q}}$ we’ve reached a contradiction and so conclude that
$w\rho_{4}/\rho_{1}$ is not contained in any proper subfield of $K_{n}$ and so
is a primitive element for the extension.) Let $\zeta$ be a zero of
$X^{r}-w\rho_{1}/\rho_{4}.$ Since $X^{r}-w\rho_{1}/\rho_{4}$ is reducible in
$K_{n}[X]$ and since $\zeta^{r}=w\rho_{1}/\rho_{4},$ we deduce that
$[{\mathbb{Q}}(\zeta):{\mathbb{Q}}]=[{\mathbb{Q}}(\zeta):K_{n}][K_{n}:{\mathbb{Q}}]<6r.$
But since $m(\zeta^{r})=0,$ we deduce that $m(X^{r})$ factors non-trivially in
${\mathbb{Q}}[X]$ (because otherwise $[{\mathbb{Q}}(\zeta):{\mathbb{Q}}]=\deg
m(X^{r})=6r.$) Letting $d$ denote the least common multiple of the
denominators of the coefficients of $m(X^{r}),$ it is clear that $d(m(X^{r}))$
factors non-trivially in ${\mathbb{Q}}[X],$ since $m(X^{r})$ factors non-
trivially in ${\mathbb{Q}}[X]$ if and only if $d(m(X^{r}))$ does as well. A
straightforward calculation reveals that $d(m(X^{r}))$ is the polynomial,
$p_{n,r}(X)$ given in the statement of the proposition. Thus we have shown
that the existence of a prime $p\mid r$ such that
$w\rho_{4}/\rho_{1}\in(K_{n}^{\times})^{p}$ implies that $p_{n,r}(X)$ admits a
non-trivial factorization in ${\mathbb{Q}}[X].$ ∎

Hilbert’s Irreducibility Theorem [4] states that if $g(X,Y)$ is an irreducible
polynomial in ${\mathbb{Z}}[X,Y],$ then there are infinitely many
$m\in{\mathbb{Z}}$ such that $g(X,m)$ is irreducible in ${\mathbb{Z}}[X].$ In
particular, if $g(X,Y)$ is an irreducible polynomial in ${\mathbb{Z}}[X,Y]$
and if for each $m\in{\mathbb{Z}}$ there is an $n_{m}\in{\mathbb{Z}}$ such
that

$g(X,m)=p_{n_{m},r}(X),$

then Hilbert’s Irreducibility Theorem implies that there are infinitely many
$m\in{\mathbb{Z}}$ such that $p_{n_{m},r}(X)$ is irreducible in
${\mathbb{Q}}[X].$ We turn now to the construction of such a polynomial
$g(X,Y)\in{\mathbb{Z}}[X,Y].$

######  Proposition 2.9.

Let $r>0$ be an integer with $(r,6)=1$ and let $y_{0},c_{r}\in{\mathbb{Z}},$
where the latter is chosen such that $(c_{r}(-143))^{r}\equiv-143\pmod{30}.$
Let

$g(X,Y):=\sum_{k=0}^{6}g_{k}(Y)X^{rk},$

where

$\displaystyle g_{0}(Y)=g_{6}(Y)=(c_{r}(30(y_{0}+8q^{2}Y)-143))^{2r}$
$\displaystyle g_{1}(Y)=g_{5}(Y)=-6g_{0}(Y)$ $\displaystyle
g_{2}(Y)=g_{4}(Y)=-\frac{104149}{900}g_{0}(Y)-\frac{16823807}{450}\sqrt{g_{0}(Y)}-\frac{13841287201}{900}$
$\displaystyle
g_{3}(Y)=-\frac{253298}{900}g_{0}(Y)-\frac{16823807}{225}\sqrt{g_{0}(Y)}-\frac{13841287201}{450}.$

Then $g(X,Y)$ is irreducible in ${\mathbb{Q}}[X,Y]$ and for each
$m\in{\mathbb{Z}},$ $g(X,m)=p_{n_{m},r}(X),$ where $n_{m}$ is the unique
integer satisfying

$(c_{r}(30(y_{0}+8q^{2}m)-143))^{r}=30n_{m}-143.$

To establish the proposition, we first need the following

######  Lemma 2.10.

Let $t\in{\mathbb{Q}}$ be such that $c_{r}(30(y_{0}+8q^{2}t)-143)=1.$ Then
$g(X,t)$ is irreducible in ${\mathbb{Q}}[X].$

###### Proof.

Note that $g(X,t)=h(X^{r}),$ where

$h(X)=X^{6}-6X^{5}-\frac{385417749}{25}X^{4}-\frac{770836748}{25}X^{3}-\frac{385417749}{25}X^{2}-6X+1.$

Since $25\times h(X+1)$ is Eisenstein at $13,$ $h(X)$ is irreducible in
${\mathbb{Q}}[X].$ Let $v$ be a zero of $h(X).$ If $g(X,t)=h(X^{r})$ has a
non-trivial factorization in ${\mathbb{Q}}[X],$ then $X^{r}-v$ has a non-
trivial factorization in ${\mathbb{Q}}(v)[X].$ So it suffices to show that
$X^{r}-v$ is irreducible in ${\mathbb{Q}}(v)[X].$ By [5, Chapter VI, §9,
Theorem 9.1], it suffices to show that $v\not\in({\mathbb{Q}}(v))^{p}$ for
each prime $p\mid r.$ Towards this end, we note that
$5^{6}\;h\left(\frac{X}{5}\right)$ is a monic polynomial in ${\mathbb{Z}}[X]$
that has $5v$ as a zero. Hence $5v$ is an algebraic integer in
${\mathbb{Q}}(v).$ Since $v$ is not an algebraic integer in ${\mathbb{Q}}(v)$
it has some prime ideals in its denominator. But since $5v$ is an algebraic
integer in ${\mathbb{Q}}(v)$ and since $5$ splits completely in
${\mathbb{Q}}(v)$ (as is easily checked using PARI [8], for example), $v$ must
have the first power of some ideal in its denominator. Hence
$v\not\in({\mathbb{Q}}(v))^{p}$ for every prime $p\mid r.$ ∎

###### Proof of Proposition 2.9.

Suppose that $g(X,Y)=k(X,Y)j(X,Y)$ is a non-trivial factorization of $g(X,Y)$
in ${\mathbb{Q}}[X,Y].$ If $k$ and $j$ each have degree $\geq 1$ in $X,$ then
choosing $t\in{\mathbb{Q}}$ such that $c_{r}(30(y_{0}+8q^{2}t)-143)=1$ implies
that $g(X,t)=k(X,t)j(X,t)$ is a non-trivial factorization of $g(X,t)$ in
${\mathbb{Q}}[X]$; a contradiction to Lemma 2.10. We conclude that if $g(X,Y)$
factors non-trivially in ${\mathbb{Q}}[X,Y],$ then one of the factors must
have degree $0$ in $X;$ i.e., $g(X,Y)=k(Y)j(X,Y).$ Without loss of generality
we may assume that $k(Y)$ is irreducible in ${\mathbb{Q}}[X,Y].$ Hence $k(Y)$
divides each $g_{i}(Y)$ as well as $\sqrt{g_{0}(Y)}.$ We conclude that $k(Y)$
divides $900g_{2}(Y)+104149g_{0}(Y)+2(16823807)\sqrt{g_{0}(Y)}=-7^{12}.$ But
$k(Y)$ is a non-trivial factor of $g(X,Y)$ in ${\mathbb{Q}}[X,Y]$ and so must
have degree greater than $0$ in $Y.$ This contradiction reveals that $g(X,Y)$
is irreducible in ${\mathbb{Q}}[X,Y].$

Note that for each $m\in{\mathbb{Z}},$
$(c_{r}(30(y_{0}+8q^{2}m)-143))^{r}\equiv(-143c_{r})^{r}\equiv-143\pmod{30}.$
Hence for each $m\in{\mathbb{Z}}$ there is an $n_{m}\in{\mathbb{Z}}$ such that
$(c_{r}(30(y_{0}+8q^{2}m)-143))^{r}=30n_{m}-143$ and hence such that
$g(X,m)=p_{n_{m},r}(X).$ The uniqueness of $n_{m}$ is clear. ∎

We now prove

######  Proposition 2.11.

There is a $y\in{\mathbb{Z}}$ for which Proposition 2.7 holds and hence a
field $K_{n}\in\mathfrak{F}$ for which

  1. (i)

there is an ideal $J\in\mathcal{I}_{K_{n}}$ whose ideal class, $[J],$ has
order $r.$

  2. (ii)

$N_{K|k_{i}}(J),$ $(i=2,3)$ is a principal ideal.

Our proof makes use of the following Lemma

######  Lemma 2.12.

There is a constant $c_{r}$ with $c_{r}^{r}\equiv 7^{1-r}\pmod{30}$ and
$(210q,c_{r})=1$ and which is such that for each prime $p\mid r$ there are
primes $\ell_{i}\mid c_{r},$ $(i=1,2),$ for which

  1. (1)

$(2\,|\,\ell_{1})_{p}=(3\,|\,\ell_{1})_{p}=1$ and $(5\,|\,\ell_{1})_{p}\neq 1$

  2. (2)

$(3\,|\,\ell_{2})_{p}=(5\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq
1.$

###### Proof.

Let $L_{1}={\mathbb{Q}}(\zeta_{p},2^{1/p},3^{1/p})$ and
$L_{2}={\mathbb{Q}}(\zeta_{2p},5^{1/p}),$ where $\zeta_{n}$ denotes a
primitive $n$th root of unity. A prime $\ell$ totally splits in $L_{1}$ if and
only if $\ell\equiv 1\pmod{p}$ and $(2\,|\,\ell)_{p}=(3\,|\,\ell)_{p}=1.$
Likewise a prime $\ell$ totally splits in $L_{2}$ if and only if $\ell\equiv
1\pmod{2p}$ and $(5\,|\,\ell)_{p}=1.$ Since $L_{2}\not\subset L_{1},$ we may
apply Bauer’s Theorem to conclude that there are infinitely many primes $\ell$
that totally split in $L_{1}$ but not in $L_{2}.$ Hence there are infinitely
many primes $\ell$ such that $(2\,|\,\ell)_{p}=(3\,|\,\ell)_{p}=1$ and
$(5\,|\,\ell)_{p}\neq 1.$ Choose such a prime $\ell_{1}$ and use the
infinitude of such primes to ensure that $(\ell_{1},210q)=1.$ Similarly,
choose $\ell_{2}$ with $(\ell_{2},210q)=1,$ and such that
$(3\,|\,\ell_{2})_{p}=(5\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq
1.$ Repeat this procedure for each prime $p\mid r.$ Choose
$a,b\in{\mathbb{Z}}$ such that $ra+4b=1$ and choose $s\in{\mathbb{N}}$ with
$(s,7q)=1$ and such that $s\left(\prod_{p\mid
r}\ell_{1}(p)\ell_{2}(p)\right)\equiv 7^{(1-r)a}\pmod{30}.$ Setting
$c_{r}=s\left(\prod_{p\mid r}\ell_{1}(p)\ell_{2}(p)\right),$ we conclude that
$c_{r}^{r}=s^{r}\left(\prod_{p\mid r}\ell_{1}(p)\ell_{2}(p)\right)^{r}\equiv
7^{(1-r)ar}\equiv\left(7^{1-r}\right)^{ra+4b}=7^{1-r}\pmod{30}.$ ∎

###### Proof of Proposition 2.11.

Choose $c_{r}$ according to Lemma 2.12, let $y_{0}\in{\mathbb{Z}},$ and let
$y:=dc_{r}(30(y_{0}+8q^{2}m)-143),$ where $d$ satisfies the system of
congruences $d\equiv 5/(c_{r}(30y_{0}-143))\pmod{8}$ and $d\equiv 1\pmod{30}.$
Then $y^{r}\equiv(-143c_{r})^{r}\equiv(7c_{r})^{r}\equiv 7\pmod{30}.$ Hence
$y^{r}=30n-143$ for some $n\in{\mathbb{Z}}.$ Furthermore, since $r$ is odd and
since $y\equiv 5\pmod{8},$ the congruence $5\equiv y^{r}=30n-143\pmod{8}$
implies that $n\equiv 2\pmod{4}.$ According to Lemma 2.12, for each prime
$p\mid r$ there are primes dividing $c_{r},$ and hence dividing $30n-143,$
that satisfy condition (1) of Proposition 2.7. Since
$(dc_{r}(-143))^{r}\equiv-143\pmod{30},$ Proposition 2.9 and the discussion of
Hilbert’s Irreducibility Theorem preceding it reveal that $p_{n_{m},r}(X)$ is
irreducible in ${\mathbb{Q}}[X]$ for infinitely many $m\in{\mathbb{Z}}.$
Choosing such a pair $(m,n_{m}),$ Proposition 2.8 reveals that
$w\rho_{4}/\rho_{1}$ fails to have a $p$th root in $K_{n_{m}}$ for each prime
$p\mid r.$ We see that for such a pair $(m,n_{m})$ all of the hypotheses of
Proposition 2.7 are satisfied and we conclude that there is a field
$K:=K_{n_{m}}\in\mathfrak{F}$ for which $w{\mathcal{O}}_{K}=J^{r}$ where $[J]$
has order $r.$ Finally, since $J^{r}=w{\mathcal{O}}_{K}$ and since $N_{K\mid
k_{i}}(w)=1$ $(i=2,3),$ it is clear that $N_{K\mid
k_{i}}(J)={\mathcal{O}}_{k_{i}}$ $(i=2,3).$ ∎

At this point we have found a field $K\in\mathfrak{F}$ that satisfies
conditions $(i)$ and $(ii)$ of Proposition 2.2. All that remains to establish
this proposition is to show that $q$ ramifies in $k_{2}.$ This is accomplished
through a judicious choice of $y_{0}$ in the following proof.

###### Proof of Proposition 2.2.

Since $\mathcal{P}(X)$ splits into linear factors$\pmod{q},$ we deduce that
$(-108\,|\,q)_{2}=1.$ Choose $n_{0}\in{\mathbb{Z}}$ such that
$n_{0}^{2}+108\equiv 0\pmod{q}$ and $n_{0}^{2}+108\not\equiv 0\pmod{q^{2}}.$
Use the fact that $\mathcal{P}(X)$ splits into linear factors $\pmod{q}$ to
choose $b_{0}\in{\mathbb{Z}}$ such that $b_{0}^{r}\equiv 30n_{0}-143\pmod{q}.$
Note that $q\neq 7$ implies that $q\nmid b_{0},$ since otherwise $q$ would be
a prime common divisor of $30n_{0}-143$ and
$n_{0}^{2}+108\mid\operatorname{disc}(f_{n_{0}}(X)).$ Since $q\nmid b_{0}$ and
since $q\nmid r,$ we conclude that we may lift $b_{0}$ to an $r$th root of
$30n_{0}-143\pmod{q^{2}}.$ Choose $c_{r}$ according to Lemma 2.12 and note
that since $q\nmid 30c_{r},$ we may choose $y_{0}\in{\mathbb{Z}}$ such that
$b_{0}\equiv c_{r}(30y_{0}-143)\pmod{q^{2}}.$

Let $m\in{\mathbb{Z}}$ and let $y:=dc_{r}(30(y_{0}+8q^{2}m)-143),$ where $d$
satisfies the system of congruences $d\equiv 5/(c_{r}(30y_{0}-143))\pmod{8}$
and $d\equiv 1\pmod{30q^{2}}.$ Then $y^{r}\equiv b_{0}^{r}\equiv
30n_{0}-143\pmod{q^{2}}.$ Since $y^{r}\equiv-143\pmod{30},$ we have
$y^{r}=30n_{m}-143,$ for some $n_{m}\in{\mathbb{Z}}.$ Since $n_{m}\equiv
n_{0}\pmod{q^{2}},$ we deduce that $n_{m}^{2}+108\equiv 0\pmod{q}$ and
$n_{m}^{2}+108\not\equiv 0\pmod{q^{2}}.$ This implies that $q$ ramifies in
$k_{2}={\mathbb{Q}}(\sqrt{n_{m}^{2}+108})\subset K_{n_{m}},$ since $q$ divides
the square-free part of $\operatorname{disc}(X^{2}-(n_{m}^{2}+108))$ and so
divides $\operatorname{disc}({\mathcal{O}}_{k_{2}}/{\mathbb{Z}}).$

We conclude that for each $m\in{\mathbb{Z}},$ the prime $q$ ramifies in
$K_{n_{m}}.$ To prove that one of these fields satisfies conditions $(i)$ and
$(ii)$ of Proposition 2.2, one may copy the proof of Proposition 2.11 verbatim
beginning with the fourth sentence. ∎

##  3\. A family of quartic number fields.

In this section we apply the method introduced in the previous section to a
family of quartic number fields. Since the results and proofs of this section
are highly similar to those presented in the previous section, much of this
section will be abridged.

Let $n\in{\mathbb{Z}}\smallsetminus\\{0,\pm 3\\}$ and let
$K_{n}={\mathbb{Q}}(\rho),$ where $\rho$ is a root of

(3.1) $f_{n}(X):=X^{4}-nX^{3}-6X^{2}+nX+1.$

$K_{n}/{\mathbb{Q}}$ is a real cyclic quartic extension and was studied by
M.-N. Gras in [2]. Let $G:=\operatorname{Gal}(K_{n}/{\mathbb{Q}}).$ Then
$G=\langle\sigma\mid\sigma^{4}=1\rangle,$ where
$\sigma:\rho\mapsto(\rho-1)/(\rho+1).$ The images of $\rho$ under $G,$ denoted
$\rho_{i}:=\sigma^{i}(\rho),$ $(i=0,\ldots,3),$ are given by

(3.2) $\begin{gathered}\rho_{0}=\rho\\\
\rho_{2}=-1/\rho\end{gathered}\qquad\begin{gathered}\rho_{1}=(\rho-1)/(\rho+1)\\\
\rho_{3}=-(\rho+1)/(\rho-1).\end{gathered}$

Let $\mathfrak{F}:=\\{K_{n}\mid n\neq 0,\pm 3\\}.$ Let $\mathcal{I}_{K}$
denote the group of fractional ideals of $K\in\mathfrak{F}$ and for a
fractional ideal $J\in\mathcal{I}_{K},$ let $[J]$ denote its ideal class. Let
$k_{2}={\mathbb{Q}}(\sqrt{n^{2}+16})$ denote the unique quadratic subfield of
$K.$ Then $[J]$ is said to be a _relative ideal class_ if it has trivial norm
down to $k_{2}.$

Note that the $\rho_{i}$ $(i=0,\ldots,3)$ are algebraic integers and, in fact,
relative units of $K_{n}.$

Our primary result is

######  Theorem 3.1.

Let $r>0$ be an odd integer. There are infinitely many fields
$K\in\mathfrak{F}$ containing a relative ideal class of order $r.$

It is straightforward to see that $K_{n}=K_{-n}.$ However, the parametrization
of $\mathfrak{F}$ given by $n\mapsto K_{n}$ fails to be injective even when we
restrict to $n>0.$ For example Lazarus [6] notes that $K_{2}=K_{22}.$ Hence in
proving Theorem 3.1 it is insufficient to simply show that there are
infinitely many $n$ such that $K_{n}$ contains a relative ideal class of order
$r.$ Instead we put most of our effort into first proving

######  Proposition 3.2.

Let $r>0$ be an odd integer. Let $q\in{\mathbb{Z}}$ be a prime such that

$\mathcal{P}(X):=(X^{r}+7-6\sqrt{-16})(X^{r}+7+6\sqrt{-16})\in{\mathbb{Z}}[X]$

splits into $\textmd{linear factors}\pmod{q}.$ Assume that $q\nmid 5r.$ There
is a field $K\in\mathfrak{F}$ that satisfies the following three conditions:

  1. (i)

There is an ideal $J\in\mathcal{I}_{K}$ whose ideal class, $[J],$ has order
$r.$

  2. (ii)

$N_{K\mid k_{2}}(J)$ is a principal ideal.

  3. (iii)

$q$ ramifies in $k_{2}.$

Once Proposition 3.2 is established we have the following

###### Proof of Theorem 3.1.

Replace $q\nmid 210r$ with $q\nmid 5r$ in the proof of Theorem 2.1. ∎

Let

$w:=\frac{2+\rho_{1}}{2+\rho_{3}}$

######  Proposition 3.3.

If there is a $y\in{\mathbb{Z}}$ with $5\nmid y$ and such that
$y^{r}=6n-7=f_{n}(-2),$ then $w{\mathcal{O}}_{K_{n}}=J^{r}$ for some ideal
$J\in\mathcal{I}_{K_{n}}.$

For the proof of this proposition we require the following easy lemma:

######  Lemma 3.4.

If a prime $p$ is a common divisor of $6n-7$ and $\operatorname{disc}(f_{n}),$
then $p=5.$

###### Proof.

The congruences $6n-7\equiv 0\pmod{p}$ and
$\operatorname{disc}(f_{n})=4(n^{2}+16)^{3}\equiv 0\pmod{p}$ hold
simultaneously only if $p=5.$ ∎

###### Proof of Proposition 3.3.

In the proof of Proposition 2.3 above, replace $7\nmid y$ with $5\nmid y.$
Also restrict the indices $0\leq i\leq 3$ and replace $(3+\rho_{i})$ with
$(2+\rho_{i})$ and $f_{n}(-3)$ with $f_{n}(-2).$ Finally note that no
restriction on the congruence class of $n$ is required, because $\rho$ and its
conjugates are algebraic integers. ∎

Let $\varepsilon$ denote the fundamental unit of $k_{2}$ and let
$U:=\langle-1,\varepsilon,\rho_{0},\rho_{1}\rangle\leq{\mathcal{O}}_{K}^{\times}.$
We will need to know that $[{\mathcal{O}}_{K}^{\times}:U]<\infty.$ It suffices
to prove the following.

######  Lemma 3.5.

The subset
$S:=\\{\varepsilon,\rho_{0},\rho_{1}\\}\subset{\mathcal{O}}_{K}^{\times}$ is
multiplicatively independent.

###### Proof.

Let $R$ denote the regulator of $S.$ Since $f_{n}(1)=-4$ and since
$f_{n}(x)\to\infty$ as $x\to\infty,$ we know that $f_{x}(X)$ has a zero $>1.$
From the from of this zero Galois conjugates, we deduce that $f_{n}(X)$ has
another positive zero in the interval $(0,1)$ and that the remaining two zeros
of $f_{n}(X)$ are negative reciprocals of the two positive zeros. Taking
$\rho_{0}$ to be the zero of $f_{n}(X)$ greater than $1,$ we find that

$\displaystyle R$
$\displaystyle=\left|\det\begin{pmatrix}\log\rho_{0}&\log\rho_{1}&\log\varepsilon\\\
\log\rho_{1}&-\log\rho_{0}&\log|\varepsilon^{\sigma}|\\\
-\log\rho_{0}&-\log\rho_{1}&\log\varepsilon\end{pmatrix}\right|$
$\displaystyle=2\log\varepsilon\;(\log^{2}\rho_{0}+\log^{2}\rho_{1}).$

Since $\varepsilon>1$ and $\rho_{0}>1,$ we conclude that $R>0$ and that $S$ is
a multiplicatively independent subset of ${\mathcal{O}}_{K}^{\times}.$ ∎

We gently remind the reader that the notation (introduced in Section 1 above)
$(a\,|\,\ell)_{p}=1$ (resp. $(a\,|\,\ell)_{p}\neq 1$) means that $a$ is a
$p$th power $\textmd{residue}\pmod{\ell}$ (resp. $a$ is a $p$th power
$\textmd{nonresidue}\pmod{\ell}.$)

######  Lemma 3.6.

Let $p$ be an odd prime. If a prime $\ell\mid 6n-7=f_{n}(-2)$ with $\ell\neq
5$ is such that $(2\,|\,\ell)_{p}=1$ and $(3\,|\,\ell)_{p}\neq 1,$ then
$p\nmid[{\mathcal{O}}_{K}^{\times}:U].$

###### Proof.

If $p\mid[{\mathcal{O}}_{K}^{\times}:U],$ then there is a
$u\in{\mathcal{O}}_{K}^{\times}\smallsetminus U$ such that
$u^{p}=\pm\varepsilon^{a}\rho_{0}^{b}\rho_{1}^{c}.$ Without loss of generality
we may take $u^{p}=\varepsilon^{a}\rho_{0}^{b}\rho_{1}^{c}$ with $0\leq
a,b,c<p.$ Since $\ell$ divides $f_{n}(-2)=\prod_{i=0}^{3}(2+\rho_{i}),$ any
prime ideal $\mathcal{L}$ dividing $\ell{\mathcal{O}}_{K}$ must contain one of
the conjugates $2+\rho_{0}.$ Furthermore every conjugate $2+\rho_{0}$ must be
contained in some prime ideal over $\ell,$ since the Galois group acts
transitively on the set of conjugates of $2+\rho_{0}$ and since it takes prime
ideals of ${\mathcal{O}}_{K}$ to prime ideals of ${\mathcal{O}}_{K}.$ Since
$\ell\neq 5$ and since $\ell\mid 6n-7,$ we know that
$\ell\nmid\operatorname{disc}(f_{n}),$ and so we deduce that every conjugate
of $2+\rho_{0}$ is contained in a unique prime ideal over $\ell.$ Let
${\mathcal{L}}$ denote the unique prime ideal
${\mathcal{L}}\mid\ell{\mathcal{O}}_{K}$ such that
$-2\equiv\rho_{0}\pmod{{\mathcal{L}}}.$ Modulo ${\mathcal{L}}$ this forces
$\rho_{1}\equiv 3,$ $\rho_{2}\equiv 1/2,$ and $\rho_{3}\equiv-1/3.$ Hence

$(uu^{\sigma})^{p}=(\pm
1)^{a}(\rho_{0}\rho_{1})^{b}(\rho_{1}\rho_{2})^{c}\equiv(\pm
1)^{a}(-1)^{b}2^{b-c}3^{b+c}\pmod{{\mathcal{L}}}$

and

$(u^{\sigma}u^{\sigma^{2}})^{p}=(\pm
1)^{a}(\rho_{1}\rho_{2})^{b}(\rho_{2}\rho_{3})^{c}\equiv(\pm
1)^{a}(-1)^{c}2^{-(b+c)}3^{b-c}\pmod{{\mathcal{L}}}.$

Applying the power residue hypotheses to these two congruences we find that
$b\equiv c$ and $b\equiv-c\pmod{p}.$ Hence $b\equiv c\equiv 0\pmod{p}$ and so
$b=c=0.$ Therefore
$u^{p}=\varepsilon^{a}=(\varepsilon^{a})^{\sigma^{2}}=(u^{p})^{\sigma^{2}}.$
Hence $u=u^{\sigma^{2}}.$ But then $u\in{\mathcal{O}}_{K}^{\times}\cap
k_{2}\subset U.$ We’ve reached a contradiction, and so are forced to conclude
that $p\nmid[{\mathcal{O}}_{K}^{\times}:U].$ ∎

######  Proposition 3.7.

If there is an integer $y$ not divisible by $5$ satisfying
$y^{r}=6n-7=f_{n}(-2),$ and if for each prime $p\mid r$ the following
conditions hold:

  1. (1)

there are primes $\ell_{i}\mid$ $(i=1,2),$ such that

    1. (a)

$(2\,|\,\ell_{1})_{p}=1$ and $(3\,|\,\ell_{1})_{p}\neq 1$

    2. (b)

$(3\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq 1$

  2. (2)

$w/\rho_{1}$ fails to have a $p$th root in $K_{n},$

then $w{\mathcal{O}}_{K}=J^{r}$ where $[J]$ has order $r.$

###### Proof.

The proof is nearly identical to that used in the proof of Proposition 2.7
except that we assume the unit $u$ to have the form
$u=\varepsilon^{a}\rho_{0}^{b}\rho_{1}^{c}.$ Taking norms down to $k_{2}$ then
allows us to deduce that $w=\rho_{0}^{b}\rho_{1}^{c}z^{p}.$ Choosing the
unique prime ideals $\mathcal{L}_{i}\mid\ell_{i}{\mathcal{O}}_{K}$ containing
$\rho_{0}+2$ determines the classes of $\rho_{0},\rho_{1},$ and $w$
$\pmod{\mathcal{L}_{i}}$ and thereby allows us to deduce that
$z^{p}\equiv(-1)^{b}2^{b}3^{1-c}\pmod{\ell_{i}}.$ Applying the power residue
hypotheses then forces $b=0$ and $c=1$ and so brings us to the contradiction
$z^{p}=w/\rho_{1}.$ ∎

To apply Proposition 3.7, we will need condition(s) sufficient to guarantee
that $w/\rho_{1}\not\in(K_{n}^{\times})^{p}$ for each prime $p\mid r.$ The
next proposition provides us with such a condition.

######  Proposition 3.8.

Let

$p_{n,r}(X):=\sum_{k=0}^{4}a_{k}X^{rk},$

where

$\displaystyle a_{0}=a_{4}=-a_{2}/6=6n-7$ $\displaystyle a_{1}=-a_{3}=7n+96$

If $p_{n,r}(X)$ is irreducible in ${\mathbb{Q}}[X],$ then $w/\rho_{1}$ fails
to have a $p$th root in $K_{n}$ for every prime $p\mid r.$

###### Proof.

If there is a $z\in K_{n}^{\times}$ such that $z^{p}=w/\rho_{1}$ with $p$ a
prime dividing $r,$ then $X^{r}-w/\rho_{1}$ factors non-trivially in
$K_{n}[X].$ Let $m(X)\in{\mathbb{Q}}[X]$ denote the minimal polynomial of
$w/\rho_{1}$ over ${\mathbb{Q}}.$ We note that $\deg m(X)=4,$ since
$w/\rho_{1}$ is a primitive element for $K_{n}/{\mathbb{Q}}.$ (Note that
$N_{K_{n}\mid k_{2}}(w/\rho_{1})=-1$, so the real number $w/\rho_{1}$ can’t be
contained in a proper subfield of $K_{n}.$) Let $\zeta$ be a zero of
$X^{r}-w/\rho_{1}.$ Since $X^{r}-w/\rho_{1}$ is reducible in $K_{n}[X]$ and
since $\zeta^{r}=w/\rho_{1},$ we deduce that
$[{\mathbb{Q}}(\zeta):{\mathbb{Q}}]=[{\mathbb{Q}}(\zeta):K_{n}][K_{n}:{\mathbb{Q}}]<4r.$
But since $m(\zeta^{r})=0,$ we deduce that $m(X^{r})$ factors non-trivially in
${\mathbb{Q}}[X]$ (because otherwise $[{\mathbb{Q}}(\zeta):{\mathbb{Q}}]=\deg
m(X^{r})=4r.$) Since $p_{n,r}(X)=a_{0}m(X^{r}),$ we deduce that $p_{n,r}(X)$
factors non-trivially in ${\mathbb{Q}}[X].$ Thus we have shown that the
existence of a prime $p\mid r$ such that $w/\rho_{1}\in(K_{n}^{\times})^{p}$
implies that $p_{n,r}(X)$ admits a non-trivial factorization in
${\mathbb{Q}}[X].$ ∎

We now turn to the construction of a polynomial $g(X,Y)$ that is irreducible
in ${\mathbb{Z}}[X,Y]$ and which has the property that for each
$m\in{\mathbb{Z}}$ there is an $n_{m}\in{\mathbb{Z}}$ such that
$g(X,m)=p_{n_{m},r}(X).$ Applying Hilbert’s Irreducibility Theorem [4], as was
done in Section 1, will then allow us to conclude that there are infinitely
many $m\in{\mathbb{Z}}$ such that $p_{n_{m},r}(X)$ is irreducible in
${\mathbb{Q}}[X].$

######  Proposition 3.9.

Let $r>0$ be an odd integer and let $y_{0},c_{r}\in{\mathbb{Z}},$ where the
latter is chosen such that $c_{r}\equiv 1\pmod{6}.$ Let

$g(X,Y):=\sum_{k=0}^{3}g_{k}(Y)X^{rk},$

where

$\displaystyle g_{0}(Y)=g_{4}(Y)=-g_{2}(Y)/6=(c_{r}(30(y_{0}+q^{2}Y)-1))^{r}$
$\displaystyle g_{1}(Y)=-g_{3}(Y)=7(g_{0}(Y)+7)/6+96.$

Then $g(X,Y)$ is irreducible in ${\mathbb{Q}}[X,Y]$ and for each
$m\in{\mathbb{Z}},$ $g(X,m)=p_{n_{m},r}(X),$ where $n_{m}$ is the unique
integer satisfying $(c_{r}(30(y_{0}+q^{2}m)-1))^{r}=6n_{m}-7.$

To establish the proposition, we first need the following

######  Lemma 3.10.

Let $t\in{\mathbb{Q}}$ be such that $c_{r}(30(y_{0}+q^{2}t)-1)=-1.$ Then
$g(X,t)$ is irreducible in ${\mathbb{Q}}[X].$

###### Proof.

Note that $g(X,t)=-f_{-103}(X^{r}).$ For this proof, let $f(X):=f_{-103}(X)$
and $K:=K_{-103}=K_{103}.$ If $g(x,t)=0,$ then
$\prod_{i=0}^{3}(x^{r}-\rho_{i})=f(x^{r})=0,$ and so $x$ is zero of
$X^{r}-\rho_{i}\in K[X]$ for some index $i.$ Let $\rho:=\rho_{i}$ and note
that it will suffice to show that $X^{r}-\rho$ is irreducible in $K[X].$ By
[5, Chapter VI, §9, Theorem 9.1] it is sufficient to show that
$\rho\not\in(K^{\times})^{p}$ for each prime $p\mid r.$ Using PARI [8] we find
that a fundamental system of units of $K$ is $\\{u_{i}\\}_{i=1}^{3},$ where

$\displaystyle u_{1}$
$\displaystyle=(1/125)\rho^{3}+(101/125)\rho^{2}-(208/125)\rho+63/125$
$\displaystyle u_{2}$
$\displaystyle=(1/250)\rho^{3}+(53/125)\rho^{2}+(156/125)\rho+83/250$
$\displaystyle u_{3}$
$\displaystyle=(1/250)\rho^{3}+(48/125)\rho^{2}-(364/125)\rho-207/250.$

In terms of these units, $\rho=u_{1}^{2}u_{2}^{3}.$ If $z\in K^{\times}$ is
such that $z^{p}=\rho$ for some prime $p,$ then
$z\in{\mathcal{O}}_{K}^{\times}$; say $z=\prod_{i=1}^{3}u_{i}^{a_{i}}.$ So
$\left(\prod_{i=1}^{3}u_{i}^{a_{i}}\right)^{p}=u_{1}^{2}u_{2}^{3}.$ Since
there is no prime $p$ dividing both $2$ and $3,$ we conclude that
$\rho\not\in(K^{\times})^{p}$ for each prime $p\mid r.$ ∎

###### Proof of Proposition 3.9.

Suppose that $g(X,Y)=k(X,Y)j(X,Y)$ is a non-trivial factorization of $g(X,Y)$
in ${\mathbb{Q}}[X,Y]$ and then by choosing $t\in{\mathbb{Q}}$ such that
$c_{r}(30(y_{0}+q^{2}t)-1)=-1$ proceed as in the proof of Lemma 2.9 to show
that one of the factors must have degree $0$ in $X$; say $g(X,Y)=K(Y)j(X,Y).$
Without loss of generality assume that $k(Y)$ is irreducible in
${\mathbb{Q}}[X,Y].$ Hence $k(Y)$ divides each $g_{i}(Y)$ and we conclude that
$k(Y)$ divides $6g_{2}(Y)-7g_{1}(Y)=625.$ But $k(Y)$ is a non-trivial factor
of $g(X,Y)$ in ${\mathbb{Q}}[X,Y]$ and so must have degree greater than $0$ in
$Y.$ This contradiction reveals that $g(X,Y)$ is irreducible in
${\mathbb{Q}}[X,Y].$ For each $m\in{\mathbb{Z}},$
$(c_{r}(30(y_{0}+q^{2}m)-1))^{r}\equiv(-c_{r})^{r}\equiv-1\pmod{6}.$ Hence for
each $m\in{\mathbb{Z}}$ there is an $n_{m}\in{\mathbb{Z}}$ such that
$(c_{r}(30(y_{0}+q^{2}m)-1))^{r}=6n_{m}-7$ and hence such that
$g(X,m)=p_{n_{m},r}(X).$ The uniqueness of $n_{m}$ is clear. ∎

We now prove the following.

######  Proposition 3.11.

There is a $y\in{\mathbb{Z}}$ for which Proposition 3.7 holds and hence a
field $K_{n}\in\mathfrak{F}$ for which

  1. (i)

there is an ideal $J\in\mathcal{I}_{K_{n}}$ whose ideal class, $[J],$ has
order $r.$

  2. (ii)

$N_{K|k_{2}}(J)$ is a principal ideal.

Our proof makes use of the following Lemma.

######  Lemma 3.12.

There is a constant $c_{r}\equiv 1\pmod{6}$ with $(c_{r},5q)=1$ and which is
such that for each prime $p\mid r$ there are primes $\ell_{i}\mid c_{r},$
$(i=1,2),$ for which

  1. (1)

$(2\,|\,\ell_{1})_{p}=1$ and $(3\,|\,\ell_{1})_{p}\neq 1$

  2. (2)

$(3\,|\,\ell_{2})_{p}=1$ and $(2\,|\,\ell_{2})_{p}\neq 1.$

###### Proof.

Let $L_{1}={\mathbb{Q}}(\zeta_{p},2^{1/p})$ and
$L_{2}={\mathbb{Q}}(\zeta_{2p},3^{1/p}),$ and then for each prime $p$ dividing
$r$ use Bauer’s Theorem as in the proof of Lemma 2.12 to choose primes
$\ell_{1}(p)$ (resp. $\ell_{2}(p)$) satisfying condition (1) (resp. condition
(2)) of the lemma with each $\ell_{i}\nmid 30q.$ Then set
$c_{r}=s\left(\prod_{p\mid r}\ell_{1}(p)\ell_{2}(p)\right),$ where $s$ is
chosen to ensure that $c_{r}\equiv 1\pmod{6}.$ ∎

###### Proof of Proposition 3.11.

Choose $c_{r}$ according to Lemma 3.12, let $y_{0}\in{\mathbb{Z}},$ and let
$y:=c_{r}(30(y_{0}+q^{2}m)-1).$ Then
$y^{r}\equiv(-c_{r})^{r}\equiv-1\pmod{6}.$ Hence $y^{r}=6n-7$ for some
$n\in{\mathbb{Z}}.$ According to Lemma 3.12, for each prime $p\mid r$ there
are primes dividing $c_{r},$ and hence dividing $6n-7,$ that satisfy condition
(1) of Proposition 3.7. Since $c_{r}\equiv 1\pmod{6},$ Proposition 3.9 and the
discussion of Hilbert’s Irreducibility Theorem preceding it reveal that
$p_{n_{m},r}(X)$ is irreducible in ${\mathbb{Q}}[X]$ for infinitely many
$m\in{\mathbb{Z}}.$ Choosing such a pair $(m,n_{m}),$ Proposition 3.8 reveals
that $w/\rho_{1}$ fails to have a $p$th root in $K_{n_{m}}$ for each prime
$p\mid r.$ We see that for such a pair $(m,n_{m})$ all of the hypotheses of
Proposition 3.7 are satisfied and we conclude that there is a field
$K:=K_{n_{m}}\in\mathfrak{F}$ for which $w{\mathcal{O}}_{K}=J^{r}$ where $[J]$
has order $r.$ Finally, since $J^{r}=w{\mathcal{O}}_{K}$ and since $N_{K\mid
k_{2}}(w)=1,$ it is clear that $N_{K\mid k_{2}}(J)={\mathcal{O}}_{k_{2}}.$ ∎

At this point we have found a field $K\in\mathfrak{F}$ that satisfies
conditions $(i)$ and $(ii)$ of Proposition 3.2. All that remains to establish
this proposition is to show that $q$ ramifies in $k_{2}.$ This is accomplished
through a judicious choice of $y_{0}$ in the following proof.

###### Proof of Proposition 3.2.

Since $\mathcal{P}(X)$ splits into linear factors$\pmod{q},$ we deduce that
$(-16\,|\,q)_{2}=1.$ Choose $n_{0}\in{\mathbb{Z}}$ such that
$n_{0}^{2}+16\equiv 0\pmod{q}$ and $n_{0}^{2}+16\not\equiv 0\pmod{q^{2}}.$ Use
the fact that $\mathcal{P}(X)$ splits into linear factors $\pmod{q}$ to choose
$b_{0}\in{\mathbb{Z}}$ such that $b_{0}^{r}\equiv 6n_{0}-7\pmod{q}.$ Note that
$q\neq 5$ implies that $q\nmid b_{0},$ since otherwise $q$ would be a prime
common divisor of $6n_{0}-7$ and
$n_{0}^{2}+16\mid\operatorname{disc}(f_{n_{0}}(X)).$ Since $q\nmid b_{0}$ and
since $q\nmid r,$ we conclude that we may lift $b_{0}$ to an $r$th root of
$6n_{0}-7\pmod{q^{2}}.$ Choose $c_{r}$ according to Lemma 3.12 and note that
since $q\nmid 30c_{r},$ we may choose $y_{0}\in{\mathbb{Z}}$ such that
$b_{0}\equiv c_{r}(30y_{0}-1)\pmod{q^{2}}.$

Let $m\in{\mathbb{Z}}$ and let $y:=c_{r}(30(y_{0}+q^{2}m)-1).$ Then
$y^{r}\equiv b_{0}^{r}\equiv 6n_{0}-7\pmod{q^{2}}.$ Since
$y^{r}\equiv-1\pmod{6},$ we have $y^{r}=6n_{m}-7,$ for some
$n_{m}\in{\mathbb{Z}}.$ Since $n_{m}\equiv n_{0}\pmod{q^{2}},$ we deduce that
$n_{m}^{2}+108\equiv 0\pmod{q}$ and $n_{m}^{2}+108\not\equiv 0\pmod{q^{2}}.$
This implies that $q$ ramifies in
$k_{2}={\mathbb{Q}}(\sqrt{n_{m}^{2}+16})\subset K_{n_{m}},$ since $q$ divides
the square-free part of $\operatorname{disc}(X^{2}-(n_{m}^{2}+16))$ and so
divides $\operatorname{disc}({\mathcal{O}}_{k_{2}}/{\mathbb{Z}}).$

We conclude that for each $m\in{\mathbb{Z}},$ the prime $q$ ramifies in
$K_{n_{m}}.$ To prove that one of these fields satisfies conditions $(i)$ and
$(ii)$ of Proposition 3.2, one may copy the proof of Proposition 3.11 verbatim
beginning with the fourth sentence. ∎

##  4\. A family of non-Galois cubic number fields.

Let $K_{n}={\mathbb{Q}}(\rho),$ where $\rho$ is a root of

(4.1) $f_{n}(X):=X^{3}+nX^{2}+nX-1.$

Since $f_{n}(\rho)=0$ if and only if $f_{-n}(1/\rho)=0,$ we have that
$K_{n}=K_{-n}$ and so we restrict our attention to $n\geq 0.$ When $n>0,$ we
find that $f_{n}(\pm 1)\neq 0$ and so conclude that $f_{n}(X)$ is irreducible
in ${\mathbb{Q}}[X]$ (since $\pm 1$ are the only possible zeros of $f_{n}(X)$
in ${\mathbb{Q}}$). Since this clearly doesn’t hold when $n=0,$ we restrict
our attention to the cubic extensions $K_{n}/{\mathbb{Q}}$ corresponding to
$n>0.$ When $n\geq 5,$ $f_{n}(X)$ changes sign in the intervals $(1-n,2-n),$
$(-1-4/n,-1-1/n),$ and $(1/(n+1),1/n).$ In particular, $f_{n}(X)$ has real
zeros, two of which are $<-1,$ while the third is $>0.$ When $|n|<5,$
$f_{n}(X)$ has some complex zeros and so we restrict our attention to the real
cubic extensions $K_{n}/{\mathbb{Q}}$ corresponding to $n\geq 5.$ Note that
$\operatorname{disc}(f_{n}(X))=n^{4}-18n^{2}-27=(n^{2}-9)^{2}-108.$ Since
$108$ is a difference of two squares in only two ways and since neither $12$
or $28$ is of the form $n^{2}-9,$ we may conclude that
$\operatorname{disc}(f_{n}(X))$ is not a square in ${\mathbb{Q}},$ and hence
that the Galois group of $f_{n}(X)$ is $S_{3}.$ Thus $K_{n}/{\mathbb{Q}}$ is a
non-Galois real cubic extension. We place one last restriction on the
parameter $n$; we restrict $n$ to have odd parity. This ensures the existence
of the explicit unit $\mu\in{\mathcal{O}}_{K}^{\times}$, distinct from $\rho$,
and introduced in Lemma 4.6 below.

Let $\mathfrak{F}:=\\{K_{n}\mid n\textmd{ is an odd integer }\geq 5\\}.$ Let
$\mathcal{I}_{K}$ denote the group of fractional ideals of $K\in\mathfrak{F}$
and for a fractional ideal $J\in\mathcal{I}_{K},$ let $[J]$ denote its ideal
class. Let $h_{K}$ denote the class number of $K.$ Our primary result is

######  Theorem 4.1.

Let $r\not\equiv 0\pmod{3}$ be a positive integer. There are infinitely many
fields $K\in\mathfrak{F}$ containing an ideal class of order $r.$

We do not know whether or not the parametrization of $\mathfrak{F}$ given by
$n\mapsto K_{n}$ is injective and so in proving Theorem 4.1 we do not attempt
to show that there are infinitely many $n$ such that $K_{n}$ contains an ideal
class of order $r.$ Instead we put most of our effort into first proving

######  Proposition 4.2.

Let $r\not\equiv 0\pmod{3}$ be an integer $>1.$ Then there is a field
$K\in\mathfrak{F}$ and an ideal $J\in\mathcal{I}_{K}$ whose ideal class,
$[J],$ has order $r.$

Once Proposition 4.2 is established we have the following

###### Proof of Theorem 4.1.

Since every class group contains the trivial class, the Theorem will hold for
$r=1$ if it holds for some $r>1,$ so assume $r>1.$ The proposition tells us
that the set $S$ consisting of all $K\in\mathfrak{F}$ containing an ideal
class of order $r$ is nonempty. If $S$ was a finite set, then there would be a
natural number $\alpha\geq 1$ and a field $K\in S$ with class number $h_{K}$
satisfying $r^{\alpha}\mid h_{K}$ and $r^{\alpha+1}\nmid h_{K^{\prime}}$ for
each $K^{\prime}\in S.$ By the proposition there is a field $L\in\mathfrak{F}$
and an ideal $J$ of $L$ whose class has order $r^{\alpha+1}.$ Hence $L$ also
contains a class of order $r,$ but since $L\not\in S$ we’ve reached a
contradiction. We conclude that $S$ is infinite. ∎

To establish Proposition 4.2, we begin by establishing sufficient conditions
for the ideal $(2-\rho){\mathcal{O}}_{K}\in\mathcal{I}_{K}$ to factor as an
$r$th power of an ideal $J\in\mathcal{I}_{K}.$ This is accomplished in
Proposition 4.5 and $[J],$ when $J$ exists, serves as our candidate for an
ideal class of order $r.$ Note that in the previous two sections we
accomplished this task with the help of the Galois group, $G,$ of our
extensions. For if $y^{r}=f_{n}(m)=\prod_{i=0}^{d}(m-\rho_{i}),$ where
$d=\deg(f),$ then $G$ acts transitively on factors $(m-\rho_{i})$ and _takes
prime ideals of $K$ to prime ideals of $K.$_ Since our cubic extension
$K/{\mathbb{Q}}$ is non-Galois, this technique is no longer available to us.
Indeed, if
$\sigma\in\operatorname{Gal}(L/{\mathbb{Q}})\smallsetminus\operatorname{Gal}(L/K),$
where $L$ is the Galois closure of $K/{\mathbb{Q}},$ then $\sigma|_{K}$
carries a prime ideal of $K$ to a prime ideal of an isomorphic but _distinct_
cubic extension $K^{\prime}/{\mathbb{Q}}.$ At its core, however, this
technique requires us to show that a prime ideal $\mathfrak{P}$ dividing
$(m-\rho_{0}){\mathcal{O}}_{K}$ doesn’t also divide
$\prod_{i=1}^{d}(m-\rho_{i}){\mathcal{O}}_{K},$ because this ensures that if
$\nu(\mathfrak{P})$ is the exact power of $\mathfrak{P}$ dividing
$(m-\rho_{0}),$ then it is also the exact power of this ideal dividing
$y^{r}=f_{n}(m)$ and hence that $r\mid\nu(\mathfrak{P}).$ This sufficiently
motivates the following pair of lemmas building up to Proposition 4.5 and in
regard to these we add that

(4.2) $6n+7=f_{n}(2)=\prod_{i=0}^{2}(2-\rho_{i}).$

######  Lemma 4.3.

Let $\rho_{i}$ $(i=0,1,2)$ denote the roots of $f_{n}(X)$ with
$\rho:=\rho_{0}.$ Then

$(2-\rho_{1})(2-\rho_{2})=\rho^{2}+(2+n)\rho+(4+3n)\in{\mathcal{O}}_{K_{n}}.$

###### Proof.

Since $f_{n}(X)=X^{3}+nX^{2}+nX-1,$ we have that $\sum_{i=0}^{2}\rho_{i}=-n$
and $\prod_{i=0}^{2}\rho_{i}=1.$ ∎

######  Lemma 4.4.

If a prime $p\mid(6n+7)$ and $\mathfrak{P}\mid p{\mathcal{O}}_{K}$ is a prime
common divisor of $(2-\rho){\mathcal{O}}_{K}$ and
$(2-\rho_{1})(2-\rho_{2}){\mathcal{O}}_{K},$ then $p=37.$

###### Proof.

Applying Lemma 4.3, we find that
$5n+12=(2-\rho_{1})(2-\rho_{2})+(2-\rho)(\rho+4+n)\in\mathfrak{P}.$ We thus
have that $5n\equiv-12\pmod{p}$ and $6n\equiv-7\pmod{p}.$ These hold
simultaneously only if $p=37.$ ∎

######  Proposition 4.5.

If $37\nmid(6n+7)$ and $6n+7=y^{r}$ for some $y\in{\mathbb{Z}},$ then there is
an ideal $J\in\mathcal{I}_{K}$ such that $(2-\rho){\mathcal{O}}_{K}=J^{r}.$

###### Proof.

Let $\mathfrak{P}$ be a prime ideal of ${\mathcal{O}}_{K}$ with
$\mathfrak{P}\cap{\mathbb{Z}}:=p{\mathbb{Z}}$ and with
$\mathfrak{P}\mid(2-\rho){\mathcal{O}}_{K}.$ Then
$6n+7=f_{n}(2)=\prod_{i=0}^{2}(2-\rho_{i})\in\mathfrak{P}.$ Hence $p\mid 6n+7$
and since $37\nmid 6n+7,$ we find that
$\mathfrak{P}\nmid(2-\rho_{1})(2-\rho_{2}){\mathcal{O}}_{K}.$ Let
$\mathfrak{P}^{\nu_{\mathfrak{P}}}$ denote the exact power of $\mathfrak{P}$
dividing $(2-\rho){\mathcal{O}}_{K}.$ Then it must also be the exact power of
$\mathfrak{P}$ dividing
$y^{r}{\mathcal{O}}_{K}=(6n+7){\mathcal{O}}_{K}=\prod_{i=0}^{2}(2-\rho_{i}){\mathcal{O}}_{K}.$
Hence $r\mid\nu_{\mathfrak{P}}.$ We now set

$J:=\prod_{\begin{subarray}{c}\textmd{prime ideals }\\\
\mathfrak{P}\mid(2-\rho){\mathcal{O}}_{K}\end{subarray}}\mathfrak{P}^{\nu_{\mathfrak{P}}/r},$

and note that $J^{r}=(2-\rho){\mathcal{O}}_{K}.$ ∎

In Proposition 4.12 we add sufficient conditions to those of Proposition 4.5
to ensure that $[J]$ has order $r.$ We turn first to a sequence of lemmas that
we use in our proof of Proposition 4.12. We gently remind the reader that $n$
is odd.

######  Lemma 4.6.

The algebraic integer

$\mu:=\left(\frac{n+3}{2}\right)(1+\rho)+\rho^{2}$

is a unit in ${\mathcal{O}}_{K}$ with $N_{K\mid{\mathbb{Q}}}(\mu)=1.$

###### Proof.

Let $a,b,c\in{\mathbb{Z}}.$ A straightforward calculation shows that
$N_{K\mid{\mathbb{Q}}}(a+b\rho+c\rho^{2})=(a^{2}c+ac^{2}-abc)n^{2}-(2a^{2}c+a^{2}b-2ac^{2}-ab^{2}-bc^{2}+b^{2}c)n+(a^{3}+b^{3}+c^{3}-3abc).$
Substituting $a=b=(n+3)/2$ and $c=1$ into this expression yields
$N_{K\mid{\mathbb{Q}}}(\mu)=1.$ Since $\mu\in{\mathcal{O}}_{K},$ the lemma is
established. ∎

######  Lemma 4.7.

The unit $\mu\in{\mathcal{O}}_{K}^{\times}$ is totally positive, while the
unit $\rho\in{\mathcal{O}}_{K}^{\times}$ is neither totally positive nor
totally negative.

###### Proof.

The unit $\rho$ is a zero of $f_{n}(X).$ As mentioned in the introduction, two
of the zeros of $f_{n}(X)$ are $<-1,$ while the third is $>0.$ In particular,
$\rho$ is neither totally positive nor totally negative. Since
$\rho_{i}^{3}+n\rho_{i}^{2}+n\rho_{i}-1=0,$ $(i=0,1,2),$ we see that
$n=(1-\rho_{i}^{3})/(\rho_{i}^{2}+\rho_{i}).$ Substituting this expression for
$n$ into $\mu_{i}=\left(\frac{n+3}{2}\right)(1+\rho_{i})+\rho_{i}^{2}$ yields

(4.3) $\mu_{i}=(\rho_{i}+1)^{3}/(2\rho_{i})$

The location of the $\rho_{i}$ now confirms that each $\mu_{i}$ is positive. ∎

Let $U:=\langle-1,\mu,\rho\rangle\leq{\mathcal{O}}_{K}^{\times}.$ We will need
to know that $[{\mathcal{O}}_{K}^{\times}:U]<\infty.$ It will suffice to prove
the following.

######  Lemma 4.8.

The subset $S:=\\{\mu,\rho\\}\subset{\mathcal{O}}_{K}^{\times}$ is
multiplicatively independent.

###### Proof.

When $n\geq 7,$ the zeros of $f_{n}(X)$ satisfy $\rho_{0}\in(1-n,2-n),$
$\rho_{1}\in(-1-3/n,-1-1/n)$ and $\rho_{2}\in(1/(n+1),1/n).$ The regulator,
$R,$ of $S$ is

$R=\bigg{|}\log|\mu_{0}|\log|\rho_{1}|-\log|\rho_{0}|\log|\mu_{1}|\bigg{|}.$

For $n\geq 7,$ we have that $1+1/n<|\rho_{1}|<1+3/n$ and
$1/n<|\rho_{1}+1|<3/n.$ Using the relation given in Equation (4.3), we
conclude that

$\displaystyle 1$ $\displaystyle>\frac{27}{2n^{2}(n+1)}=(3/n)^{3}/(2(1+1/n))$
$\displaystyle>|\mu_{1}|=|\rho_{1}+1|^{3}/(2|\rho_{1}|)$
$\displaystyle>(1/n)^{3}/(2(1+3/n))=\frac{1}{2n^{2}(n+3)}.$

Since $1-n<\rho_{0}<2-n,$ we have that $n-2<|\rho_{0}|<n-1$ and
$n-3<|\rho_{0}+1|<n-2.$ Using Equation (4.3) again, we conclude that

$(n-2)^{2}/2=(n-2)^{3}/(2(n-2))>|\mu_{0}|=|\rho_{0}+1|^{3}/(2|\rho_{0}|)>(n-3)^{3}/(2(n-1))>1.$

Hence

$\displaystyle R$
$\displaystyle=\log|\mu_{0}|\log|\rho_{1}|-\log|\rho_{0}|\log|\mu_{1}|$
$\displaystyle\geq\log\left(\frac{(n-3)^{3}}{2(n-1)}\right)\log(1+1/n)-\log\left(n-2\right)\log\left(\frac{27}{2n^{2}(n+1)}\right)$
$\displaystyle\geq\log(n-2)\log\left(\frac{2n^{2}(n+1)}{27}\right)$
$\displaystyle>0,$

and we conclude that $S$ is a multiplicatively independent subset of $K_{n}$
for $n\geq 7.$ It may be verified by hand that $R\neq 0$ when $n=5,$ so that
$\\{\rho,\mu\\}$ is a multiplicatively independent subset of $K_{n}$ for all
$n$ under consideration. ∎

######  Lemma 4.9.

Let $\mathfrak{P}$ be a prime ideal of ${\mathcal{O}}_{K}$ with
$\mathfrak{P}\cap{\mathbb{Z}}=p{\mathbb{Z}}.$ If $\rho\equiv
2\pmod{\mathfrak{P}}$ and $p\nmid\operatorname{disc}(f_{n}),$ then
$[{\mathcal{O}}_{K}/\mathfrak{P}:{\mathbb{Z}}/p{\mathbb{Z}}]=1.$

###### Proof.

Noting that $f_{n}(X)=(X-\rho)(X^{2}+(\rho+n)X+\rho^{2}+n\rho+n)$ and making
use of the fact that $\rho\equiv 2\pmod{\mathfrak{P}}$ along with the
monomorphism
$\frac{{\mathbb{Z}}}{p{\mathbb{Z}}}[X]\hookrightarrow\frac{{\mathcal{O}}_{K}}{\mathfrak{P}}[X],$
we see that $f_{n}(X)$ factors as $(X-2)(X^{2}+(n+2)X+4+3n)$ in
$({\mathbb{Z}}/p{\mathbb{Z}})[X].$ Applying Dedekind’s Factorization Theorem,
we conclude that
$[{\mathcal{O}}_{K}/\mathfrak{P}:{\mathbb{Z}}/p{\mathbb{Z}}]=1.$ ∎

######  Lemma 4.10.

If a prime $p$ is a common divisor of $6n+7$ and $\operatorname{disc}(f_{n}),$
then $p=37$ or $p=47.$

###### Proof.

The congruences $n^{4}-18n^{2}-27=\operatorname{disc}(f_{n})\equiv 0\pmod{p}$
and $6n+7\equiv 0\pmod{p}$ hold simultaneously only if $p=37$ or $p=47.$ ∎

######  Lemma 4.11.

Let $y\in{\mathbb{Z}}$ not divisible by either $37$ or $47$ and such that
$y^{r}=6n+7,$ and let $p$ be a prime dividing $r.$ If there are primes
$\ell_{i}\mid(6n+7),$ $(i=1,2),$ and such that

  1. (i)

$(2\,|\,\ell_{1})_{p}=1$ and $(3\,|\,\ell_{1})_{p}\neq 1,$

  2. (ii)

$(2\,|\,\ell_{2})_{p}\neq 1$ and $(3\,|\,\ell_{2})_{p}\neq 1,$

then $p\nmid[{\mathcal{O}}_{K}^{\times}:U].$

###### Proof.

Choose prime ideals $\mathcal{L}_{i}\subset{\mathcal{O}}_{K},$ $(i=1,2),$ with
$\mathcal{L}_{i}\cap{\mathbb{Z}}=\ell_{i}{\mathbb{Z}}$ and such that
$\rho\equiv 2\pmod{\mathcal{L}_{i}}.$ Then by equation 4.3 $\mu\equiv
27/4\pmod{\mathcal{L}_{i}}.$ If $p\mid[{\mathcal{O}}_{K}^{\times}:U],$ then
there is a unit $u\in{\mathcal{O}}_{K}^{\times}\smallsetminus U$ and
$a,b\in{\mathbb{Z}}$ with $0\leq a,b<p$ and such that
$u^{p}=\pm\rho^{a}\mu^{b}.$ We remark that Lemma 4.10 guarantees that
$\ell_{i}\nmid\operatorname{disc}(f_{n})$ and so Lemma 4.9 then guarantees
that
$[{\mathcal{O}}_{K}/\mathcal{L}_{i}:{\mathbb{Z}}/\ell_{i}{\mathbb{Z}}]=1.$

If $p\neq 2,$ then by making use of the fact that
$[{\mathcal{O}}_{K}/\mathcal{L}_{i}:{\mathbb{Z}}/\ell_{i}{\mathbb{Z}}]=1$
along with the congruences for $\rho$ and $\mu,$ we find that
$(2^{a-2b}3^{3b}\,|\,\ell_{i})_{p}=1.$ We remind the reader that $p\neq 3,$
since $p\mid r$ and since $r\not\equiv 0\pmod{3}.$ Hence upon applying
condition (i) we conclude that $p\mid b$ and hence that $b=0.$ Setting $b=0$
we conclude that $(2^{a}\,|\,\ell_{i})_{p}=1.$ Applying condition (ii) we find
that $p\mid a$ and so $a=0.$ Hence $u^{p}=\pm 1$ and we reach the
contradiction that $u=\pm 1\in U.$

Hence $p=2$ and $u^{2}=\pm\rho^{a}\mu^{b}.$ Since $\pm u^{2}\mu^{-b}$ is
either totally positive or totally negative and since $\rho$ is neither
totally positive nor totally negative, $a=0.$ Furthermore, since both $\mu$
and $u^{2}$ are totally positive, we are forced to conclude that
$u^{2}=\mu^{b}.$ If $b=0,$ then $u=\pm 1\in U;$ contradiction. Therefore $b=1$
and $u^{2}=\mu.$ Since $\mu\equiv 27/4\pmod{\mathcal{L}_{2}},$ we conclude
that $(3\,|\,\ell_{2})_{2}=1$ and so contradict condition (ii). Having
exhausted the possibilities, we are forced to conclude that
$p\nmid[{\mathcal{O}}_{K}^{\times}:U].$ ∎

######  Proposition 4.12.

Let $y\in{\mathbb{Z}}$ with $(37\times 47,y)=1$ and such that $y^{r}=6n+7.$ If
for each prime $p\mid r$ there are primes $\ell_{i}\mid(6n+7)$ $(i=1,2,3)$
satisfying

  1. (i)

$(2\,|\,\ell_{1})_{p}=(37\,|\,\ell_{1})_{p}=1,$ and $(3\,|\,\ell_{1})_{p}\neq
1,$

  2. (ii)

$(37\,|\,\ell_{2})_{p}=1,$ $(2\,|\,\ell_{2})_{p}\neq 1,$ and
$(3\,|\,\ell_{2})_{p}\neq 1,$

  3. (iii)

$(37\,|\,\ell_{3})_{p}\neq 1,$

then $(2-\rho)=J^{r}$ where $[J]$ has order $r.$

###### Proof.

Since the hypotheses of Proposition 4.5 hold, we have that $(2-\rho)=J^{r}$
for some ideal $J\in\mathcal{I}_{K}$ and so the order of $[J]$ is $\leq r.$ If
it is strictly $<r,$ then $(2-\rho)$ is a prime power of a principal ideal for
some prime $p\mid r.$ Supposing this to be the case and writing
$(2-\rho)=(\beta)^{p},$ for some $\beta\in{\mathcal{O}}_{K},$ we have that
$2-\rho=u\beta^{p}$ for some $u\in{\mathcal{O}}_{K}^{\times}.$ Since the
hypotheses of Lemma 4.11 hold we have that
$p\nmid[{\mathcal{O}}_{K}^{\times}:U].$ Writing
$1=gp+h[{\mathcal{O}}_{K}^{\times}:U]$ for some $g,h\in{\mathbb{Z}},$ we find
that

$2-\rho=u\beta^{p}=u^{gp+h[{\mathcal{O}}_{K}^{\times}:U]}\beta^{p}=(\pm\rho^{a^{\prime}}\mu^{b^{\prime}})^{h}(u^{g}\beta)^{p}=\pm\rho^{a}\mu^{b}(\beta^{\prime})^{p}.$

For notational convenience, replace $\beta^{\prime}$ with $\beta$ and note
that the same relation holds in the other two isomorphic cubic number fields
to conclude that $2-\rho_{i}=\pm\rho_{i}^{a}\mu_{i}^{b}\beta_{i}^{p}$
$(i=0,1,2).$

Choose prime ideals $\mathcal{L}_{i}\subset{\mathcal{O}}_{K},$ $(i=1,2,3),$
with $\mathcal{L}_{i}\cap{\mathbb{Z}}=\ell_{i}{\mathbb{Z}}$ and such that
$\rho\equiv 2\pmod{\mathcal{L}_{i}}.$ Then $\mu\equiv
27/4\pmod{\mathcal{L}_{i}}.$ On the one hand,
$(2-\rho_{1})(2-\rho_{2})=(\rho_{1}\rho_{2})^{a}(\mu_{1}\mu_{2})^{b}(\beta_{1}\beta_{2})^{p}=\rho^{-a}\mu^{-b}(\beta_{1}\beta_{2})^{p}\equiv
2^{2b-a}3^{-3b}(\beta_{1}\beta_{2})^{p}\pmod{\mathcal{L}_{i}}.$ On the other
hand, $(2-\rho_{1})(2-\rho_{2})=\rho^{2}+(2+n)\rho+(4+3n)\equiv
5n+12\pmod{\mathcal{L}_{i}}\equiv 37/6\pmod{\mathcal{L}_{i}}.$ Hence $37\equiv
2^{2b-a+1}3^{1-3b}(\beta_{1}\beta_{2})^{p}\pmod{\mathcal{L}_{i}}.$ Lemma 4.10
guarantees that $\ell_{i}\nmid\operatorname{disc}(f_{n})$ and so Lemma 4.9
then guarantees that
$[{\mathcal{O}}_{K}/\mathcal{L}_{i}:{\mathbb{Z}}/\ell_{i}{\mathbb{Z}}]=1.$
Thus $(2^{a-2b-1}3^{3b-1}37\,|\,\ell_{i})_{p}=1.$ Applying condition (i), we
conclude that $3b-1\equiv 0\pmod{p}.$ Using this result along with condition
(ii), we find that $a-2b-1\equiv 0\pmod{p}.$ Since $p$ divides $3b-1$ and
$a-2b-1,$ we deduce that $(37\,|\,\ell_{3})_{p}=1$ and so contradict condition
(iii). We conclude that $(2-\rho)$ is not a prime power of a principal ideal
for any prime $p\mid r$ and hence that $[J]$ has order $r.$ ∎

Having established sufficient conditions for a field $K\in\mathfrak{F}$ to
posses an ideal class of arbitrary nontrivial order $r\not\equiv 0\pmod{3},$
we now prove Proposition 4.2 by showing that for each such $r$ there is a
field $K\in\mathfrak{F}$ satisfying these conditions.

###### Proof of Proposition 4.2.

Let $p$ be a prime dividing $r.$ Let
$L_{1}={\mathbb{Q}}(\zeta_{p},2^{1/p},37^{1/p})$ and
$L_{2}={\mathbb{Q}}(\zeta_{2p},3^{1/p}),$ where $\zeta_{n}$ denotes a
primitive $n$th root of unity. A prime $\ell$ totally splits in $L_{1}$ if and
only if $\ell\equiv 1\pmod{p}$ and $(2\,|\,\ell)_{p}=(37\,|\,\ell)_{p}=1.$
Similarly a prime $\ell$ totally splits in $L_{2}$ if and only if $\ell\equiv
1\pmod{2p}$ and $(3\,|\,\ell)_{p}=1.$ Since $L_{2}\not\subset L_{1},$ we apply
Bauer’s Theorem to conclude that there are infinitely many primes $\ell$ that
totally split in $L_{1}$ but not in $L_{2}.$ Hence there are infinitely many
primes $\ell$ for which $(2\,|\,\ell)_{p}=(37\,|\,\ell)_{p}=1,$ and
$(3\,|\,\ell)_{p}\neq 1.$ In other words, there are infinitely many primes
$\ell$ satisfying condition (i) of Proposition 4.12. Analogous arguments show
that there are infinitely many primes satisfying condition (ii) and infinitely
many primes satisfying condition (iii) of Proposition 4.12.

For each prime $p\mid r,$ choose a triple of primes
$\\{\ell_{i,p}\\}_{i=1}^{3}$ satisfying conditions (i)-(iii) of Proposition
4.12 making sure never to choose $2,3,37,$ or $47.$ Let $s\in{\mathbb{N}}$
with $(s,37\times 47)=1$ and such that
$s\left(\prod\limits_{\begin{subarray}{c}\textmd{primes}\\\ p\mid
r\end{subarray}}\prod_{i=1}^{3}\ell_{i,p}\right)\equiv 1\pmod{6}.$ Set
$y:=s\left(\prod\limits_{\begin{subarray}{c}\textmd{primes}\\\ p\mid
r\end{subarray}}\prod_{i=1}^{3}\ell_{i,p}\right).$ If $y^{r}\equiv 3\pmod{4},$
then replace $y$ with $y^{2}$ so that $y^{r}\equiv 1\pmod{4}.$ Since
$y^{r}\equiv 7\pmod{6},$ there is an $n\in{\mathbb{N}}$ such that
$y^{r}=6n+7.$ Reducing this equation $\pmod{4}$ reveals that $n$ is odd.
Furthermore, since $y\geq
s\left(\prod\limits_{\begin{subarray}{c}\textmd{primes}\\\ p\mid
r\end{subarray}}\prod_{i=1}^{3}\ell_{i,p}\right)$ and since $s\in{\mathbb{N}}$
and $\ell_{i}\neq 2,3$ we deduce that $y\geq 125$ and hence that $n\geq 5.$ We
see that the hypotheses of Proposition 4.12 are satisfied and hence that there
is an ideal $J\in\mathcal{I}_{K_{n}}$ whose class $[J]$ has order $r.$ ∎

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