# Heuristics for anti-cyclotomic $\mathbb{Z}_{p}$-extensions

Debanjana Kundu  Fields Institute
University of Toronto
Toronto ON, M5T 3J1, Canada<EMAIL_ADDRESS>and  Lawrence C.
Washington  Department of Mathematics
University of Maryland, College Park, MD 20742 USA<EMAIL_ADDRESS>
###### Abstract.

This paper studies Iwasawa invariants in anti-cyclotomic towers. We do this by
proposing two heuristics supported by computations. First we propose the
_Intersection Heuristics_ : these model ‘how often’ the $p$-Hilbert class
field of an imaginary quadratic field intersects the anti-cyclotomic tower and
to what extent. Second we propose the _Invariants Heuristics_ : these predict
that the Iwasawa invariants $\lambda$ and $\mu$ usually vanish for imaginary
quadratic fields where $p$ is non-split.

###### 2020 Mathematics Subject Classification:

Primary 11R23

##  1\. Introduction

Let $K$ be a number field and let $p$ be a prime, which we shall always take
to be odd. Let $K_{\infty}/K$ be a $\mathbb{Z}_{p}$-extension, which means
$\operatorname{Gal}(K_{\infty}/K)\simeq\mathbb{Z}_{p}$, the additive group of
$p$-adic integers. Let $K_{n}$ be the subfield of degree $p^{n}$ over $K$, let
$h_{n}$ be its class number, and let $p^{e_{n}}$ be the exact power of $p$
dividing $h_{n}$. A well-known result of K. Iwasawa says there exist integers
$\lambda,\mu\geq 0$ and $\nu$, independent of $n$, such that

$e_{n}=\lambda n+\mu p^{n}+\nu,$

for all $n$ sufficiently large.

When $K$ is an imaginary quadratic field, there are two exactly
$\mathbb{Z}_{p}$-extensions such that $K_{\infty}/\mathbb{Q}$ is a Galois
extension. One is the _cyclotomic_ $\mathbb{Z}_{p}$-extension
$K_{{\mathrm{cyc}}}/K$: this is contained in the field obtained by adjoining
all $p$-power roots of unity to $K$ and is such that
$\operatorname{Gal}(K_{{\mathrm{cyc}}}/{\mathbb{Q}})$ is abelian. The other is
the _anti-cyclotomic_ $\mathbb{Z}_{p}$-extension $K_{{\mathrm{ac}}}/K$ and is
such that $\operatorname{Gal}(K_{{\mathrm{ac}}}/{\mathbb{Q}})$ is dihedral,
with $\operatorname{Gal}(K/{\mathbb{Q}})$ acting on
$\operatorname{Gal}(K_{{\mathrm{ac}}}/K)$ by inversion.

The $\mu$-invariant for $K_{{\mathrm{cyc}}}$ is always 0 (see [FW79]), and the
distribution of $\lambda$-invariants when $p$ is fixed and $K$ varies has been
studied by several authors, see for example [San91, DFKS91, Oza01, Fuj13,
Mur19]. In [EJV11], J. Ellenberg–S. Jain–A. Venkatesh studied this question in
the special case when $K$ varies over all imaginary quadratic fields and $p$
is non-split in $K$; they proposed a heuristic based on random $p$-adic
matrices. These heuristics have recently been extended to abelian extensions
by D. Delbourgo–H. Knospe in [DK22]. The initial goal of our study was to see
if similar predictions could be made for $K_{{\mathrm{ac}}}$. However, our
calculations suggest that $K_{{\mathrm{ac}}}/K$ where $p$ is non-split in $K$
acts more like the cyclotomic $\mathbb{Z}_{p}$-extensions of totally real
number fields, where Greenberg’s Conjecture (see [Gre76]) predicts that
$\mu=\lambda=0$.

Let $L(K)$ be the $p$-Hilbert class field of $K$, that is, the maximal
unramified abelian $p$-extension of $K$. Then $\operatorname{Gal}(L(K)/K)$ is
isomorphic to $A_{0}$, the Sylow $p$-subgroup of the ideal class group of $K$.
A phenomenon that plays a prominent role when studying anti-cyclotomic
$\mathbb{Z}_{p}$-extensions is that $L(K)\cap K_{{\mathrm{ac}}}$ can be larger
than $K$. The question is how large can this intersection be? Based on
computational evidence, we propose the following:

######  Conjecture (Intersection Heuristics).

Fix a finite abelian $p$-group $G$. Let $K$ vary over imaginary quadratic
fields such that the $p$-part of its ideal class group is isomorphic to $G$.
The probability that $L(K)\cap K_{{\mathrm{ac}}}=K_{n}$ is given by

$\frac{\\#(\text{elements of $G$ of order exactly }p^{n})}{\\#G}.$

This number is also

$\frac{\\#(\text{homomorphisms
}G\longrightarrow{{\mathbb{Q}}/{\mathbb{Z}}}\text{ with image of size
}p^{n})}{\\#G}.$

In other words, the $p$-class group $G$ randomly maps to
$\operatorname{Gal}(K_{n}/K)$ for large $n$ and the frequency of possible
images is predicted to model the distribution of possible intersections.

The Cohen-Lenstra heuristics (introduced in [CL90]) predict that the $p$-part
of the ideal class group of $K$ is most often cyclic, and our heuristic says
that for cyclic $G$, it is most common that $L(K)$ is contained in
$K_{{\mathrm{ac}}}$. A result of S. Fujii (see Theorem 2.2) says that
$\mu=\lambda=0$ in this case. Moreover, our calculations suggest that $\mu=0$
is always true, and that $\lambda=0$ when $A_{0}$ is cyclic and $L(K)\cap
K_{{\mathrm{ac}}}=K$. Putting these together yields the following:

######  Conjecture (Invariants Heuristics).

Among the imaginary quadratic fields $K$ in which $p$ does not split, we
always have $\mu=0$, and the proportion for which $\lambda=0$ is at least

$\left(1-\frac{1}{p}\right)+\frac{1}{p-1}\prod_{j\geq
1}\left(1-p^{-j}\right).$

For $p=3$, this predicts that $\lambda=0$ at least $94\%$ of the time, and for
$p=5$ at least $99\%$ of the time. These are heuristic lower bounds, but we
find it hard to imagine that there is such a small number of actual
exceptions. We show that there is (theoretical and/or experimental) evidence
that $\lambda=\mu=0$ in the two extreme cases, i.e., when the $p$-part of the
class group of $K$ is contained in the anti-cyclotomic tower or completely
disjoint from it. The cases where we could not calculate the Iwasawa
invariants from our experiments are the ones that ‘interpolate’ between these
two extreme cases - it seems reasonable to expect these to act similarly.
Therefore we ask the following:

######  Question.

Is $\lambda=\mu=0$ always true for the anti-cyclotomic
$\mathbb{Z}_{p}$-extension of an imaginary quadratic field $K$ in which $p$
does not split?

We restrict ourselves to the case that $p$ does _not_ split in $K$ because in
the complementary case it is known that $\lambda>0$. This follows from a
result of [Oza01] (see for example [Fuj13, p. 283]) and can be explained
briefly as follows: looking at the inertia groups of the primes above $p$ for
the $\mathbb{Z}_{p}^{2}$-extension $\widetilde{K}/K$, one sees that the
$\mathbb{Z}_{p}$-extension $\widetilde{K}/K_{{\mathrm{ac}}}$ is unramified.
Therefore, the unramified Iwasawa module has positive $\mathbb{Z}_{p}$ rank,
so $\lambda>0$.

Another situation where there is possibly an analogue of Greenberg’s
conjecture was considered in [FK02]. Suppose $p$ splits in $K$ as
$\mathfrak{p}\overline{\mathfrak{p}}$. Let $K_{\infty}/K$ be the
$\mathbb{Z}_{p}$-extension in which only $\mathfrak{p}$ ramifies and assume it
is totally ramified. It is known that $\mu=0$ (see [Gil87, Sch87]), and the
computations in [FK02] suggest that also $\lambda=0$.

For cyclotomic $\mathbb{Z}_{p}$-extensions of imaginary quadratic fields, R.
Gold was able to evaluate the $\lambda$-invariant from the class numbers of
the first few levels, see [Gol74]. We have followed a similar strategy in our
situation but it is more challenging. In the cyclotomic case, Gold used the
fact that the inverse limit of the $p$-class groups has no finite submodules.
This is no longer the case for anti-cyclotomic $\mathbb{Z}_{p}$-extensions, as
can be seen from the several cases where $\lambda=\mu=0$ but $\nu>0$.
Consequently, our methods cannot recognize situations where $\lambda>0$ or
$\mu>0$, if they exist.

In addition to proposing the above heuristics, we prove the following result
(see Theorem 6.1) which asserts that the Sylow $p$-subgroup of the ideal class
group of the $n$-th layer of the anti-cyclotomic extension, denoted by
$A_{n}$, is non-cyclic if $A_{0}$ is non-trivial and disjoint from
$K_{{\mathrm{ac}}}$.

######  Theorem.

Let $K$ be an imaginary quadratic field and let $p$ be an odd prime that does
not split in $K$. Suppose that $A_{0}$ is non-trivial. If the $p$-Hilbert
class field of $K$ is disjoint from $K_{{\mathrm{ac}}}$, then $A_{n}$ is non-
cyclic for all $n\geq 1$.

_Organization:_ Including this Introduction, the article has seven sections.
In Section 2, we record facts from classical Iwasawa theory and remind the
reader of the Cohen–Lenstra heuristics. In Section 3, we consider the case
when the $p$-Hilbert class field of $K$ is disjoint from the anti-cyclotomic
tower. We present data and prove sufficient conditions for the Iwasawa
invariants to be trivial. In Section 4, we study the case when the $p$-Hilbert
class field partially intersects the anti-cyclotomic tower. The computational
methods for obtaining the tables in the above sections are explained in
Section 5. In Section 6, we prove that in the case that the $p$-Hilbert class
field of $K$ is disjoint from the anti-cyclotomic tower, the $p$-Hilbert class
field of each subsequent layer $K_{n}$ must be non-cyclic. Finally, in Section
7 we present the two heuristics and explain the rationale behind them. The
LABEL:H:_intersection model ‘how often’ the $p$-Hilbert class field intersects
the anti-cyclotomic tower and ‘how much’ is the intersection. The LABEL:H:_AC
aim to predict ‘how often’ the anti-cyclotomic Iwasawa invariants are trivial.

##  2\. Preliminaries

###  2.1. Anti-cyclotomic Iwasawa theory

Let $K$ be an imaginary quadratic field and $p$ be a fixed prime. There exists
a Galois extension $\widetilde{K}/K$ containing all the
$\mathbb{Z}_{p}$-extensions of $K$ such that

$\operatorname{Gal}\left(\widetilde{K}/K\right)\simeq\mathbb{Z}_{p}^{2}.$

This $2$-dimensional Galois extension contains uncountably many quotients
isomorphic to $\mathbb{Z}_{p}$, and each quotient corresponds to some
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$. There are two
$\mathbb{Z}_{p}$-extensions that are special because they are the only ones
that are Galois over ${\mathbb{Q}}$. One is the often-studied _cyclotomic_
$\mathbb{Z}_{p}$-extension which is abelian over ${\mathbb{Q}}$; the other is
the _anti-cyclotomic_ $\mathbb{Z}_{p}$-extension, denoted by
$K_{\mathrm{ac}}$, which a pro-dihedral group (and hence non-abelian).

######  Lemma 2.1.

Keep the same notations as introduced above. Then,

$L(K)\cap\widetilde{K}\subseteq K_{{\mathrm{ac}}}.$

###### Proof.

For a proof of this assertion, we refer the reader to [Fuj13, Lemma 2.2]. ∎

In view of the above lemma, the following situations may arise:

  1. (A)

$L(K)=K$.

  2. (B)

$L(K)\neq K$.

    1. (a)

$L(K)\cap K_{{\mathrm{ac}}}=L(K)$, i.e., the $p$-Hilbert class field of $K$ is
contained in $K_{{\mathrm{ac}}}$.

    2. (b)

$L(K)\cap K_{{\mathrm{ac}}}=K$, i.e., the $p$-Hilbert class field of $K$ is
disjoint from $K_{{\mathrm{ac}}}$.

    3. (c)

$K\subsetneq L(K)\cap K_{{\mathrm{ac}}}\subsetneq L(K)$, i.e., the $p$-Hilbert
class field of $K$ partially intersects $K_{{\mathrm{ac}}}$.

The following theorem shows that the Iwasawa invariants in case (a) are well
understood.

######  Theorem 2.2.

With notation as introduced above, if $L(K)\subsetneq K_{{\mathrm{ac}}}$ then

$\mu(K_{{\mathrm{ac}}}/K)=\lambda(K_{{\mathrm{ac}}}/K)=0.$

###### Proof.

This assertion is proved in [Fuj13, Remarks on p. 286] using Chevalley’s
formula. ∎

###  2.2. Cohen–Lenstra Heuristics

The Cohen–Lenstra heuristics are based on the observation that structures
often occur in nature with frequency inversely proportional to their number of
automorphisms. The idea of the Cohen–Lenstra heuristics is that the $p$-part
of the ideal class group of $K$ (with $p\neq 2$) is a finite abelian $p$-group
that is distributed in this way.

######  Cohen–Lenstra Heuristics.

Let $D^{-}(x)$ denote the set of imaginary quadratic fields $K$ such that
$\mathinner{\\!\left\lvert\operatorname{Disc}K\right\rvert}\leq x$ and let
$G(n)$ denote the set of all finite groups of order $n$. For any ‘reasonable’
function $f$ on finite abelian groups,

$\lim_{x\rightarrow{\infty}}\frac{\sum_{K\in
D^{-}(x)}\mathinner{\\!\left\lvert
f\left(\operatorname{Cl}(K)\right)\right\rvert}}{\\#\left\\{K\in
D^{-}(x)\right\\}}=\lim_{n\rightarrow{\infty}}\frac{\sum_{i=1}^{n}\sum_{G\in
G(i)}\frac{f(G)}{\\#\operatorname{Aut}(G)}}{\sum_{i=1}^{n}\sum_{G\in
G(i)}\frac{1}{\\#\operatorname{Aut}(G)}}.$

The main examples where these heuristics are applied are when $f$ depends on
the Sylow $p$-subgroup of $G$.

##  3\. Data and initial calculations: $p$-Hilbert class field is disjoint
from $K_{{\mathrm{ac}}}$

In this section, we consider the case where the $p$-Hilbert class field of $K$
is disjoint from $K_{{\mathrm{ac}}}$. For computational reasons, we restrict
to $p=3$ (when $p=5$, the extension $K_{2}/{\mathbb{Q}}$ has degree 50, and it
is hard to compute class numbers of fields of such a large degree). The
methods of computation are described in Section 5.

###  3.1. Data for the case that $A_{0}$ is cyclic

Calculations presented in the data below were performed using PARI/GP. In
these tables, we record the $p$-part of the ideal class group in the first few
layers of the anti-cyclotomic $\mathbb{Z}_{p}$-extension of
$K=\mathbb{Q}(\sqrt{-d})$, where we take $-d$ to be the discriminant of the
quadratic field. Here, $A_{n}$ is the Sylow $p$-subgroup of the ideal class
group of $K_{n}$. To keep the tables less cumbersome, we write $n$ in place of
${\mathbb{Z}}/n{\mathbb{Z}}$. Since the $p$-Hilbert class field of $K$ is
disjoint from $K_{{\mathrm{ac}}}$, it follows that the unique prime above $p$
is totally ramified in the anti-cyclotomic $\mathbb{Z}_{p}$-extension. First,
in Table LABEL:T:_inert we consider the case when $3$ is inert in the
imaginary quadratic field. In Table LABEL:T:_ramify we present the data when
$3$ ramifies in the imaginary quadratic field.

Table 1. $3$ is inert $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$
---|---|---|---|---|---|---|---
331 | 3 | $3\times 3\times 3$ | $3\times 3\times 3\times 3$ | 2491 | 3 | $3\times 3$ | $3\times 3$
643 | 3 | $3\times 3$ | $3\times 3$ | 2740 | 3 | $3\times 3$ | $3\times 3$
835 | 3 | $3\times 3$ | $3\times 3$ | 2791 | 3 | $3\times 3$ | $3\times 3$
1048 | 3 | $27\times 9$ | $27\times 27$ | 2824 | 3 | $3\times 3$ | $3\times 3$
1192 | 3 | $3\times 3$ | $3\times 3$ | 2923 | 3 | $3\times 3$ | $3\times 3$
1327 | 3 | $3\times 3$ | $3\times 3$ | 2344 | 9 | $9\times 3$ | $9\times 9$
1588 | 3 | $3\times 3$ | $3\times 3$ | 3643 | 9 | $9\times 3$ | $9\times 9$
1843 | 3 | $27\times 9$ | $81\times 27$ | 4363 | 9 | $9\times 3$ | $9\times 9$
1951 | 3 | $3\times 3$ | $3\times 3$ | 4819 | 9 | $9\times 3$ | $9\times 9$
2227 | 3 | $3\times 3\times 3$ | $3\times 3\times 3\times 3$ | 5464 | 9 | $9\times 3$ | $9\times 9$
2488 | 3 | $3\times 3$ | $3\times 3$ | 6763 | 9 | $9\times 3$ | $9\times 9$
Table 2. $3$ is ramified $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$
---|---|---|---|---|---|---|---
759 | 3 | $3\times 3$ | $3\times 3$ | 3048 | 3 | $9\times 9$ | $9\times 9$
771 | 3 | $3\times 3$ | $3\times 3$ | 3387 | 3 | $3\times 3$ | $3\times 3$
804 | 3 | $3\times 3$ | $3\times 3$ | 3459 | 3 | $9\times 9$ | $9\times 9$
1191 | 3 | $3\times 3$ | $3\times 3$ | 3540 | 3 | $3\times 3$ | $3\times 3$
1236 | 3 | $3\times 3$ | $3\times 3$ | 3687 | 3 | $3\times 3$ | $3\times 3$
1272 | 3 | $3\times 3$ | $3\times 3$ | 3783 | 3 | $3\times 3$ | $3\times 3$
1419 | 3 | $9\times 3$ | $27\times 9$ | 3912 | 3 | $3\times 3$ | $3\times 3$
1515 | 3 | $3\times 3$ | $3\times 3$ | 3999 | 3 | $3\times 3$ | $3\times 3$
1668 | 3 | $3\times 3$ | $3\times 3$ | 4308 | 3 | $3\times 3$ | $3\times 3$
2235 | 3 | $3\times 3$ | $3\times 3$ | 4827 | 3 | $3\times 3$ | $3\times 3$
2283 | 3 | $3\times 3$ | $3\times 3$ | 8331 | 9 | $9\times 3$ | $9\times 9$
2355 | 3 | $9\times 9$ | $9\times 9$ | 15243 | 9 | $9\times 3$ | $9\times 9$
2856 | 3 | $9\times 9$ | $9\times 9$ | 24207 | 9 | $9\times 3$ | $9\times 9$

######  Theorem 3.1.

In all examples in Tables LABEL:T:_inert and LABEL:T:_ramify, we have
$\mu(K_{\mathrm{ac}}/K)=0$. For all cases except possibly $d=1843$ and
$d=1419$, we have $\lambda(K_{\mathrm{ac}}/K)=0$.

###### Proof.

The proofs that $\lambda(K_{\mathrm{ac}}/K)=0$ for these examples will be
given in Sections 3.2 and 3.3. When $K_{\mathrm{ac}}/K$ is totally ramified,
[San91, Lemma, p. 672] asserts that

(1) $\\#A_{n}\geq\\#A_{0}\cdot p^{\mu(p^{n}-1)+\text{min}(p^{n}-1,\lambda)}$

for each $n\geq 0$. In all of the examples, we observe that

$\\#A_{2}<\\#A_{0}\times 3^{8}.$

Therefore, $\mu(K_{\mathrm{ac}}/K)=0$. ∎

###  3.2. Conditions for triviality of Iwasawa invariants

In this section, we provide sufficient conditions to prove that the anti-
cyclotomic Iwasawa $\mu$ and $\lambda$-invariants are $0$. Throughout this
section we assume that $K$ is an imaginary quadratic field, there exists
exactly one ramified prime in the $\mathbb{Z}_{p}$-extension $K_{\infty}/K$,
and it is totally ramified. In particular, the results we prove hold when the
$p$-Hilbert class field of $K$ is disjoint from $K_{{\mathrm{ac}}}$.

#### 3.2.1.

We begin by recording some results that we use several times for proving our
statements. These results are true for _general_ $\mathbb{Z}_{p}$-extensions,
not just the anti-cyclotomic one, and they apply to arbitrary number fields
$K$.

######  Lemma 3.2.

Suppose that there exists exactly one ramified prime in the
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$; further suppose that it is totally
ramified. Set
$G_{n}=\operatorname{Gal}\left(K_{n+1}/K_{n}\right)=\langle\tau_{n}\rangle$.
Then

$\\#A_{n+1}^{G_{n}}=\\#A_{n}.$

###### Proof.

This follows from Chevalley’s formula; see [Ger77, Lemma 4.2]. ∎

######  Theorem 3.3 ([Ger77, Theorem 1.3(c)]).

Suppose that there exists exactly one ramified prime in the
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$; further suppose that it is totally
ramified. If every ideal class in $A_{0}$ becomes trivial in some $A_{n}$,
then

$\lambda(K_{\infty}/K)=\mu(K_{\infty}/K)=0.$

The norm map from $A_{n+1}$ to $A_{n}$ is surjective since we are assuming
throughout that $K_{n+1}/K_{n}$ is totally ramified. The composition of this
with the natural map from $A_{n}$ to $A_{n+1}$ yields

(2) $\begin{split}A_{n+1}&\longrightarrow A_{n+1}\\\
[J]&\mapsto\left(1+\tau_{n}+\ldots+\tau_{n}^{p-1}\right)[J].\end{split}$

######  Proposition 3.4.

Suppose that there exists exactly one ramified prime in the
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$; further suppose that it is totally
ramified. Suppose that $A_{n}\simeq A_{n+1}$ for some $n\geq 0$. Then
$\lambda(K_{\infty}/K)=\mu(K_{\infty}/K)=0$.

###### Proof.

Write $L=\bigcup_{n\geq}L(K_{n})$ and
$X=\operatorname{Gal}\left(L/K_{\infty}\right)$. Since there is a unique prime
above $p$ in $K$ and this prime is totally ramified we know that (see [Was97,
Proposition 13.22])

$A_{n}=X/\left(\left(1+T\right)^{p^{n}}-1\right)X.$

Since we are assuming that $A_{n}=A_{n+1}$, therefore

$X/\left(\left(1+T\right)^{p^{n}}-1\right)X=X/\left(\left(1+T\right)^{p^{n+1}}-1\right)X.$

Nakayama’s Lemma now implies that
$\left(\left(1+T\right)^{p^{n}}-1\right)X=0$; therefore, $X$ must be finite. ∎

######  Remark 3.5.

This proposition allows us to take care of, for example, the cases $d=2856$
and $3048$ in Table LABEL:T:_ramify.

######  Proposition 3.6.

Suppose that there exists exactly one ramified prime in the
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$; further suppose that it is totally
ramified. Suppose that for some $n$ both $A_{n}$ and $A_{n+1}$ are elementary
$p$-groups satisfying

$\operatorname{rank}_{p}\left(A_{n+1}\right)\leq\operatorname{rank}_{p}\left(A_{n}\right)+p-2.$

Then $\lambda(K_{\infty}/K)=\mu(K_{\infty}/K)=0$.

###### Proof.

Write $\tau_{n}$ for a generator of $\operatorname{Gal}(K_{n+1}/K_{n})=G_{n}$.
By Lemma 3.2, $\\#A_{n+1}^{G_{n}}=\\#A_{n}$.

Consider $A_{n+1}$ as a vector space over ${\mathbb{F}}_{p}$ and $\tau_{n}$ as
a linear transformation such that $\tau_{n}^{p}=1$. Note that $1$ is the only
eigenvalue of $\tau_{n}$.

The assumption on ranks implies that the $1$-eigenspace of $\tau_{n}$ has
codimension $\leq p-2$. If we now look at the Jordan form of $\tau_{n}$, then
no Jordan block has size larger than $p-1$. Write

$\tau_{n}=I+N\text{ satisfying }N^{p-1}=0.$

Hence,

(3)
$\begin{split}1+\tau_{n}+\ldots+\tau_{n}^{p-1}&=I+(I+N)+(I+N)^{2}+\ldots+(I+N)^{p-1}\\\
&=pI+\sum_{k=1}^{p-1}\sum_{j=1}^{p-2}{{k}\choose{j}}N^{j}.\end{split}$

Observe that for a fixed $j\leq p-2$,

$\sum_{k=1}^{p-1}{k\choose j}=\sum_{k=j}^{p-1}{k\choose
j}={p\choose{j+1}}\equiv 0\pmod{p}.$

Putting this all together we rewrite (3) as follows:

$1+\tau_{n}+\ldots+\tau_{n}^{p-1}\equiv 0\pmod{p}.$

In particular, the map given in (2) is a multiple of $p$. By assumption,
$A_{n+1}$ is an elementary $p$-group; hence it is annihilated under this map.
But, the norm map is surjective; hence we conclude that $A_{n}\rightarrow
A_{n+1}$ is the $0$-map. By Theorem 3.3 the result follows. ∎

######  Proposition 3.7.

Suppose that there exists exactly one ramified prime in the
$\mathbb{Z}_{p}$-extension $K_{\infty}/K$; further suppose that it is totally
ramified. Suppose that

$A_{0}\simeq{\mathbb{Z}}/p^{2}{\mathbb{Z}},\quad
A_{1}={\mathbb{Z}}/p^{2}{\mathbb{Z}}\times{\mathbb{Z}}/p{\mathbb{Z}},\quad
A_{2}={\mathbb{Z}}/p^{2}{\mathbb{Z}}\times{\mathbb{Z}}/p^{2}{\mathbb{Z}}.$

Then,

$\lambda(K_{\infty}/K)=\mu(K_{\infty}/K)=0.$

###### Proof.

The elements of $A_{1}$ can be written as column vectors $\begin{pmatrix}x\\\
y\end{pmatrix}$ with $x\in{\mathbb{Z}}/p^{2}{\mathbb{Z}}$ and
$y\in{\mathbb{Z}}/p{\mathbb{Z}}$. Let $\tau_{1}$ be a generator for
$\operatorname{Gal}(K_{1}/K)=G$. The action of $\tau_{1}$ on $A_{1}$ is via
the matrix $M=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ such that

$\tau_{1}\begin{pmatrix}1\\\ 0\end{pmatrix}=\begin{pmatrix}a\\\
c\end{pmatrix}\text{ and }\tau_{1}\begin{pmatrix}0\\\
1\end{pmatrix}=\begin{pmatrix}b\\\ d\end{pmatrix}.$

Since $\begin{pmatrix}0\\\ 1\end{pmatrix}$ is an element of order $p$, it
forces that $b\equiv 0\pmod{p}$.

By Lemma 3.2, we know that

$A_{1}^{G}\simeq{\mathbb{Z}}/p{\mathbb{Z}}\times{\mathbb{Z}}/p{\mathbb{Z}}\text{
or }A_{1}^{G}\simeq{\mathbb{Z}}/p^{2}{\mathbb{Z}}.$

We claim that in both cases,
$1+\tau_{1}+\tau_{1}^{2}+\cdots+\tau_{1}^{p-2}=p$.

_Case 1 –
$A_{1}^{G}\simeq{\mathbb{Z}}/p{\mathbb{Z}}\times{\mathbb{Z}}/p{\mathbb{Z}}$_:
We may choose the fixed elements of $A_{1}$ to be the column vectors
$\begin{pmatrix}p\\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\\ 1\end{pmatrix}$
and this forces

$M=\begin{pmatrix}a&0\\\ c&1\end{pmatrix}\text{ such that }a\equiv 1\pmod{p}.$

Note that for $0\leq j\leq p-1$,

$M^{j}=\begin{pmatrix}a^{j}&0\\\
\sum_{i=0}^{j-1}a^{i}c&1\end{pmatrix}=\begin{pmatrix}a^{j}&0\\\
jc&1\end{pmatrix}$

since $a\equiv 1\pmod{p}$. Therefore, we calculate that

$I+M+\ldots+M^{p-1}=pI.$

_Case 2 – $A_{1}^{G}\simeq{\mathbb{Z}}/p^{2}{\mathbb{Z}}$_: There is a fixed
column vector of order $p^{2}$ and we may choose

$M=\begin{pmatrix}1&b\\\ 0&d\end{pmatrix}.$

Since $M^{p}=I$, this forces $d=1$. Note that for $0\leq j\leq p-1$,

$M^{j}=\begin{pmatrix}1&jb\\\ 0&1\end{pmatrix}.$

Therefore, we obtain that

$I+M+\ldots+M^{p-1}=pI.$

Continuing with the proof, let $[J_{0}]$ be an ideal class in $A_{0}$ and
$[J_{1}]$ be an ideal class in $A_{1}$ satisfying

$\operatorname{\mathsf{N}}_{K_{1}/K}([J_{1}])=[J_{0}].$

Here we have used the fact that $\operatorname{\mathsf{N}}_{K_{1}/K}$ is
surjective since the extension is totally ramified. On the other hand, the
image of $\operatorname{\mathsf{N}}_{K_{1}/K}[J_{1}]=[J_{0}]$ in $A_{1}$ is

$\left(I+M+\ldots+M^{p-1}\right)[J_{1}]=p[J_{1}].$

Suppose that $\langle\tau_{2}\rangle=\operatorname{Gal}(K_{2}/K_{1})=G_{2}$
and
$A_{2}\simeq{\mathbb{Z}}/p^{2}{\mathbb{Z}}\times{\mathbb{Z}}/p^{2}{\mathbb{Z}}$.
The action of $\tau_{2}$ on $A_{2}$ can be understood via an invertible
$2\times 2$ matrix $M_{2}$ with coefficients in
${\mathbb{Z}}/p^{2}{\mathbb{Z}}$. Lemma 3.2 implies that there are fixed
column vectors of order $p^{2}$ and order $p$, which we choose to be
$\begin{pmatrix}1\\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\\ p\end{pmatrix}$,
respectively. Therefore, we choose

$M_{2}=\begin{pmatrix}1&b\\\ 0&d\end{pmatrix}\text{ with }b\equiv
0\pmod{p}\text{ and }d\equiv 1\pmod{p}.$

Note that once again for $0\leq j\leq p-1$,

$M_{2}^{j}=\begin{pmatrix}1&\sum_{i=0}^{j-1}bd^{i}\\\
0&d^{j}\end{pmatrix}=\begin{pmatrix}1&jb\\\ 0&d^{j}\end{pmatrix}.$

Therefore, we obtain that

$I+M_{2}+\ldots+M_{2}^{p-1}=pI.$

Working with the class $[J_{1}]$ from before, we know there exists a class
$[J_{2}]$ in $A_{2}$ such that

$[J_{1}]=\operatorname{\mathsf{N}}_{K_{2}/K}[J_{2}].$

As before, $[J_{1}]\mapsto p[J_{2}]$ in $K_{2}$. Therefore,

$[J_{0}]\mapsto p[J_{1}]\mapsto p^{2}[J_{2}]=0.$

The result follows from Theorem 3.3. ∎

######  Remark 3.8.

  1. (i)

Theorem 3.1 and Propositions 3.6 and 3.7 show that
$\lambda(K_{\mathrm{ac}}/K)=\mu(K_{\mathrm{ac}}/K)=0$ for all imaginary
quadratic fields considered in Tables LABEL:T:_inert and LABEL:T:_ramify
except when $d=1048$, $1419$, or $1843$.

  2. (ii)

The above propositions do not apply to the case $d=1048$. However, the prime
above $7$ generates $A_{0}$ and this prime becomes principal in $A_{2}$. By
Theorem 3.3, we know that

$\lambda(K_{\mathrm{ac}}/K)=\mu(K_{\mathrm{ac}}/K)=0.$

  3. (iii)

Now, consider the case $d=1843$. The prime above $11$ generates $A_{0}$ but it
does not become principal in $A_{2}$. So, we can not draw such a conclusion.
The result of Sands mentioned in the proof of Theorem 3.1 implies that
$\lambda\leq 6$. It appears possible in this case that
$\lambda(K_{{\mathrm{ac}}}/K)=2$. But, it is perhaps unlikely that this is an
exceptional case where $\lambda(K_{{\mathrm{ac}}}/K)>0$; we have not yet been
able to compute the image of $A_{0}$ in $A_{3}$.

Similar remarks also hold for the case $d=1419$.

###  3.3. Data for the case that $A_{0}$ is non-cyclic

Here are a few cases where $A_{0}$ is non-cyclic and the $3$-Hilbert class
field is disjoint from $K_{{\mathrm{ac}}}$.

Table 3. $A_{0}$ is non-cyclic $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$
---|---|---|---
50983 | $3\times 3$ | $9\times 3\times 3\times 3$ | $9\times 9\times 3\times 3$
63079 | $3\times 3$ | $9\times 3\times 3\times 3$ | $9\times 9\times 3\times 3$
141412 | $3\times 3$ | $9\times 3\times 3\times 3$ | $27\times 9\times 3\times 3$

The results proven in Section 3.2 can not be applied to these examples. Here,
we do not prove general results but only comment on these specific examples.

  1. (a)

In each of these cases, $\mu(K_{\mathrm{ac}}/K)=0$ by the same argument as in
Theorem 3.1.

  2. (b)

When $d=50983$, the $10^{\text{th}}$ powers of the primes above $13$ and $23$
generate $A_{0}$. They both become principal in $A_{2}$. Theorem 3.3 implies
that

$\mu(K_{\mathrm{ac}}/K)=\lambda(K_{\mathrm{ac}}/K)=0.$

  3. (c)

When $d=63079$, the $17^{\text{th}}$ powers of the primes above $5$ and $41$
generate $A_{0}$. They both become principal in $A_{2}$. Once again, Theorem
3.3 implies that

$\mu(K_{\mathrm{ac}}/K)=\lambda(K_{\mathrm{ac}}/K)=0.$

  4. (d)

When $d=141412$, the $8^{\text{th}}$ powers of the primes above $7$ and $43$
generate $A_{0}$. The $8^{\text{th}}$-power of the prime above $43$ becomes
principal in $A_{2}$ but the $8^{\text{th}}$-power of the prime above $7$ has
order $3$ in $A_{2}$. Therefore, we do not draw any conclusions regarding the
$\lambda$-invariant from this. The result of Sands mentioned in the proof of
Theorem 3.1 implies that $\lambda(K_{{\mathrm{ac}}}/K)\leq 5$.

##  4\. Data and Calculations: $p$-Hilbert class field partially intersects
$K_{{\mathrm{ac}}}$

In this section, we first present the table where the $p$-Hilbert class field
partially intersects $K_{\mathrm{ac}}$.

Table 4. $p$-Hilbert class field of $K$ intersects with $K_{{\mathrm{ac}}}$ partially $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $d$ | $A_{0}$ | $A_{1}$ | $A_{2}$
---|---|---|---|---|---|---|---
367 | 9 | 3 | $9\times 3$ | 6883 | 9 | 3 | $9\times 3$
1087 | 9 | 3 | $9\times 3$ | 7908 | 9 | 3 | $9\times 3$
1291 | 9 | 3 | $9\times 3$ | 13092 | 9 | 3 | $9\times 3$
3547 | 9 | 3 | $3\times 3\times 3\times 3$ | 36276 | $3\times 3$ | $3\times 3\times 3$ | $9\times 9\times 9\times 3$
4012 | 9 | 3 | $9\times 3$ | 49128 | $3\times 3$ | $9\times 3$ | $27\times 9$
6607 | 9 | 3 | $9\times 3$ | 49812 | $3\times 3$ | $9\times 3$ | $27\times 9$
6871 | 9 | 3 | $9\times 3$ | 58920 | $3\times 3$ | $9\times 3$ | $27\times 9$

In this next proposition we provide a proof of an observation one is likely to
make immediately from the above table. As before, write $L(K_{n})$ to denote
the $p$-Hilbert class field of $K_{n}$.

######  Proposition 4.1.

Suppose that $A_{0}$ is cyclic and $L(K)\supseteq K_{n}$ for some $n\geq 0$.
Then $L(K_{n})=L(K)$.

###### Proof.

Note that $L(K)\subseteq L(K_{n})$. Suppose that $L(K)\neq L(K_{n})$, then

$1\neq\operatorname{Gal}\left(L(K_{n})/L(K)\right){\unlhd}\operatorname{Gal}\left(L(K_{n})/K\right).$

Since these are $p$-groups there exists an index $p$ subgroup $N$ of
$\operatorname{Gal}(L(K_{n})/L(K))$ that is normal in
$\operatorname{Gal}\left(L(K_{n})/K\right)$, see [DF04, Section 6.1, Theorem
1(3)]. Consider the exact sequence

$1\longrightarrow\operatorname{Gal}\left(L(K_{n})/L(K)\right)/N\longrightarrow\operatorname{Gal}\left(L(K_{n})/K\right)/N\longrightarrow
A_{0}\longrightarrow 1.$

By hypothesis, $A_{0}$ is cyclic and it acts on
$\operatorname{Gal}\left(L(K_{n})/L(K)\right)/N$, by lifting and then
conjugation. However, $\operatorname{Gal}\left(L(K_{n})/L(K)\right)/N$ has
order $p$ which means that there are no automorphisms of order $p$. The action
of $A_{0}$ must therefore be trivial. Hence,
$\operatorname{Gal}\left(L(K_{n})/K\right)/N$ is abelian. Write $F$ for the
number field fixed by $N$. We have the following tower of number fields:

$K\subseteq K_{n}\subseteq L(K)\subsetneq F\subseteq L(K_{n}).$

This means that $F/K$ is abelian and unramified but this contradicts the
assumption that $L(K)$ is the $p$-Hilbert class field. ∎

######  Remark 4.2.

  1. (i)

This proposition does not hold when $A_{0}$ is non-cyclic, as we can see from
the data.

  2. (ii)

It seems to be difficult to make any accurate predictions about the anti-
cyclotomic Iwasawa invariants from this data. For $d\neq 36276$, the result of
Sands mentioned in the proof of Theorem 3.1, applied to $K_{2}/K_{1}$, says
that $\mu(K_{{\mathrm{ac}}}/K)\leq 1$, and if $\mu(K_{{\mathrm{ac}}}/K)=1$
then $\lambda(K_{{\mathrm{ac}}}/K)=0$. For $d=36276$, we find that
$\mu(K_{{\mathrm{ac}}}/K)\leq 2$, and if $\mu(K_{{\mathrm{ac}}}/K)=2$ then
$\lambda(K_{{\mathrm{ac}}}/K)=0$.

##  5\. Computational Methods I

We now explain how the computations were performed to obtain data for the
$p$-part of the ideal class groups of the first few layers of the anti-
cyclotomic tower in the earlier sections. Our techniques are a modification of
those of [BHW19].

### 5.1.

Let $z$ be an element in the upper half plane and define

$\eta(z)=q^{1/24}\prod_{n\geq 1}(1-q^{n})\text{ where }q=e^{2\pi iz}.$

Define the _Weber functions_

$f(z)=e^{-\pi i/24}\frac{\eta\left(\frac{z+1}{2}\right)}{\eta(z)}\text{ and
}f_{1}(z)=\frac{\eta\left(\frac{z}{2}\right)}{\eta(z)}.$

The following result of R. Schertz [Sch02, Theorem 1] plays a crucial role in
our computations.

######  Schertz’s Theorem.

Let $\alpha$ be an element in the upper half plane satisfying the primitive
equation

$AX^{2}+BX+C=0\text{ with }2\nmid A\text{ and }B\equiv 0\pmod{32}$

with discriminant $D=B^{2}-4AC=-4m=-t^{2}d$. Then, each of the following
numbers generates the ring class of conductor $t$ over the imaginary quadratic
field of discriminant $-d$ in the case stated:

  * •

$\left(\left(\frac{2}{A}\right)\frac{1}{\sqrt{2}}f(\alpha)^{2}\right)^{3}$ if
$m\equiv 1\pmod{8}$.

  * •

$f(\alpha)^{3}$ if $m\equiv 3\pmod{8}$.

  * •

$\left(\frac{1}{2}f(\alpha)^{4}\right)^{3}$ if $m\equiv 5\pmod{8}$.

  * •

$\left(\left(\frac{2}{A}\right)\frac{1}{\sqrt{2}}f(\alpha)\right)^{3}$ if
$m\equiv 7\pmod{8}$.

  * •

$\left(\left(\frac{2}{A}\right)\frac{1}{\sqrt{2}}f_{1}(\alpha)^{2}\right)^{3}$
if $m\equiv 2,6\pmod{8}$.

  * •

$\left(\left(\frac{2}{A}\right)\frac{1}{2\sqrt{2}}f_{1}(\alpha)^{2}\right)^{3}$
if $m\equiv 4\pmod{8}$.

Here, $\left(\frac{2}{A}\right)$ denotes the Legendre symbol.

Let $\alpha_{1},\alpha_{2},\ldots,\alpha_{h_{t}}$ be the roots corresponding
to the set of all classes of conductor $t$ (satisfying the conditions that the
coefficient $A_{i}$ is odd and $32\mid B_{i}$), and let $g$ be one of the
above expressions. Then $\\{g(\alpha_{i})\\}_{i}$ gives a complete set of
conjugates of $g(\alpha_{1})$ over the imaginary quadratic field of
discriminant $-d$.

Start with a primitive quadratic form

$ax^{2}+bxy+cy^{2}\text{ with }b^{2}-4ac=-4m.$

If $2\mid a$, we perform the change of variables $y\mapsto(x+y)$ and obtain

$(a+b+c)x^{2}+(b+2c)xy+cy^{2}.$

Note that $2\mid b$ and the original quadratic form is primitive. Thus, $c$
and $a+b+c$ are both odd. Next, suppose we have the quadratic form

$a_{1}x^{2}+b_{1}xy+c_{1}y^{2}\text{ with }2\nmid a_{1}.$

Choose $m$ satisfying

$b_{1}+2a_{1}m\equiv 0\pmod{32}.$

Perform the change of variable $x\mapsto x+my$. This gives the quadratic form

$a_{1}x^{2}+(b_{1}+2a_{1}m)xy+(c_{1}+mb_{1}+m^{2}a_{1})y^{2}.$

Therefore, each class of quadratic forms is represented as a primitive
quadratic form $Ax^{2}+Bxy+Cy^{2}$ with $2\nmid A$ and $32\mid B$.

Let $\alpha$ be the root in the upper half plane of $Ax^{2}+Bx+C=0$. Compute
the value of the appropriate function $g(\alpha)$ from LABEL:Thm:_Schertz.
This number generates the ring class field over $K={\mathbb{Q}}(\sqrt{-d})$.
The minimal polynomial of $g(\alpha)$ over ${\mathbb{Q}}$ has roots
$g(\alpha_{1}),\ldots,g(\alpha_{h_{t}})$.

For our study in the case $p=3$, we need polynomials that generate a layer of
the anti-cyclotomic $\mathbb{Z}_{3}$-extension of $K$. Such a layer will be
contained in a ring class field of conductor $9^{k}$ for some $k$. To use
LABEL:Thm:_Schertz, multiply $d$ by $4$ (if required) and by a power $9^{k}$
to get $-4m$. For $k$ sufficiently large, compute the group of reduced
primitive quadratic forms of discriminant $-4m$, call it $G$. Choose an
integer $M$ and let $H$ be the subgroup of $G$ consisting of elements of $G$
whose orders divide $M$. As we explain below, an appropriate choice of $M$
makes $G/H$ cyclic of order a power of $3$. The integer $M$ in our
calculations was essentially the class number of $K$ times a power of $2$ and
a power of $3$.

Define the polynomial

$\prod_{aH\in G/H}\left(x-\sum_{h\in H}g\left(\alpha_{ah}\right)\right);$

here $g$ is the function in LABEL:Thm:_Schertz and $\alpha_{ah}$ is the
element of the upper half plane corresponding to the class $ah$. Observe that
$\sum_{h\in H}g\left(\alpha_{ah}\right)$ is the trace to the fixed field of
$H$; the roots of the polynomial must lie in this field, which is cyclic over
$K$ and of degree a power of $p$. In all the cases that were considered, these
roots generate the field and the polynomial has rational integral
coefficients. We explain in detail below how, from the structure of the ring
class group, we know that a cyclic $p$-power extension obtained in the manner
given above is a layer of the anti-cyclotomic $\mathbb{Z}_{3}$-extension.

The calculations were done in floating point (often to 10000- or 20000- digit
accuracy for the larger discriminants) and then rounded to get a polynomial
with integer coefficients (the many-digit accuracy was needed in order to make
the rounding well-defined; GP-Pari keeps track of this). The polynomial
generating the compositum of $K$ with the field generated by a root of the
polynomial yields a polynomial generating the layer of the
$\mathbb{Z}_{3}$-extension over$K$. As a check, the discriminant of the number
field was computed and it always had the correct prime factors to the correct
powers. As the power $9^{k}$ increases, we get higher layers of the
$\mathbb{Z}_{3}$-extension.

Here are the details of why we get the anticyclotomic $\mathbb{Z}_{3}$
extension by our method. We assume 3 is not split in $K/\mathbb{Q}$ and the
class field of $K$ intersects the anticyclotomic $\mathbb{Z}_{3}$-extension
$K_{\infty}$ in $K$. Moreover, we assume $K\neq\mathbb{Q}(\sqrt{-3})$ and
$K\neq{\mathbb{Q}}(\sqrt{-1})$.

Let $c$ be a positive integer and let $K_{c}$ be the ring class field over $K$
of conductor $c$. Let $G_{c}=\text{Gal}(K_{c}/K)$, let $C$ be the ideal class
group of $K$, and let $\mathcal{O}$ be the ring of integers of $K$. Let
$R_{c}=(\mathcal{O}/c\mathcal{O})^{\times}/(\mathbb{Z}/c\mathbb{Z})^{\times}$.
There is an exact sequence

$1\longrightarrow R_{c}/{\mathcal{O}}^{\times}\longrightarrow
G_{c}\longrightarrow C\to 1.$

By the Chinese Remainder Theorem, if $c=c_{1}c_{2}$ with
$\gcd(c_{1},c_{2})=1$, then

$R_{c_{1}c_{2}}=R_{c_{1}}\times R_{c_{2}}.$

Note that the $3$-part of $R_{3^{k}}/{\mathcal{O}}^{\times}$ is cyclic of
3-power order or cyclic of 3-power order times $\mathbb{Z}/3\mathbb{Z}$. From
the exact sequence, we see that

$\mathinner{\\!\left\lvert G_{c}\right\rvert}=\mathinner{\\!\left\lvert
C\right\rvert}\cdot\mathinner{\\!\left\lvert
R_{c}/{\mathcal{O}}^{\times}\right\rvert}$

for $c=2$ and $c=2\cdot 3^{k}$. It follows that

$\mathinner{\\!\left\lvert G_{2\cdot
3^{k}}\right\rvert}=\mathinner{\\!\left\lvert
G_{2}\right\rvert}\cdot\mathinner{\\!\left\lvert R_{3^{k}}\right\rvert}.$

The prime of $K$ above 3 ramifies in $K_{3^{k}}/K$ to degree equal to
$\mathinner{\\!\left\lvert R_{3^{k}}/{\mathcal{O}}^{\times}\right\rvert}$;
therefore, it ramifies at least to this degree in $K_{2\cdot 3^{k}}/K$. Let
$I_{k}$ be the inertia group in $G_{2\cdot 3^{k}}$ for a prime above 3. Since
3 doesn’t ramify in $K_{2}/K$, we see that

$\mathinner{\\!\left\lvert G_{2}\right\rvert}\cdot\mathinner{\\!\left\lvert
R_{3^{k}}\right\rvert}=\mathinner{\\!\left\lvert G_{2\cdot
3^{k}}\right\rvert}\geq\mathinner{\\!\left\lvert
G_{2}\right\rvert}\cdot\mathinner{\\!\left\lvert
I_{k}\right\rvert}\geq\mathinner{\\!\left\lvert
G_{2}\right\rvert}\cdot\mathinner{\\!\left\lvert
R_{3^{k}}/{\mathcal{O}}^{\times}\right\rvert}=\mathinner{\\!\left\lvert
G_{2}\right\rvert}\cdot\mathinner{\\!\left\lvert R_{3^{k}}\right\rvert}/2,$

and therefore $|I_{k}|=|R_{3^{k}}|$ or $|I_{k}|=|R_{3^{k}}|/2$.

Moreover,

$G_{2}\times I_{k}\text{ and }G_{2}\times R_{3^{k}}/{\mathcal{O}}^{\times}$

are each either isomorphic to $G_{2\cdot 3^{k}}$ or to a subgroup of index 2
in it since these two groups embed in $G_{2\cdot 3^{k}}$ and the orders are
equal up to a factor of 2. Since the $3$-part of
$R_{3^{k}}/{\mathcal{O}}^{\times}$ is either cyclic of 3-power order or
$\mathbb{Z}/3\mathbb{Z}$ times a cyclic group of 3-power order, we see that

$G_{2\cdot 3^{k}}\simeq A\times B_{k},$

where $A$ is independent of $k$ and $B_{k}$ is cyclic of 3-power order. This
decomposition is not unique, but we can arrange for $B_{k+1}\to B_{k}$ under
the restriction map of $G_{2\cdot 3^{k+1}}\to G_{2\cdot 3^{k}}$ and
$I_{k+1}\to I_{k}$ for each $k$.

Choose $M$ that annihilates $G_{2\cdot 3}$ and let $H_{k}=\\{g\in G_{2\cdot
3^{k}}\,|\,g^{M}=1\\}$. Then $A\subseteq H_{k}$ and

$G_{2\cdot 3^{k}}/H_{k}\simeq(A\times B_{k})/H_{k}\simeq B_{k}/(B_{k}\cap
H_{k}),$

where this last group is cyclic of 3-power order. Under the map $G_{2\cdot
3^{k+1}}\to G_{2\cdot 3^{k}}$, we have

$B_{k+1}/(B_{k+1}\cap H_{k+1})\to B_{k}/(B_{k}\cap H_{k})$

for each $k$. We see that the fixed fields of $B_{k}/(B_{k}\cap H)$, as $k$
varies, form the layers of a $\mathbb{Z}_{3}$-extension of $K$. By the Galois
action of $\operatorname{Gal}(K/\mathbb{Q})$, it must be the anticyclotomic
$\mathbb{Z}_{3}$-extension of $K$.

### 5.2.

We now give a concrete example. Let $d=1048$, $p=3$, and
$K=\mathbb{Q}(\sqrt{-1048})$ which has class number $6$. Note that
$d/4=262\equiv 6\pmod{8}$ so
$g(\alpha)=\left(\left(\frac{2}{A}\right)\frac{1}{\sqrt{2}}f_{1}(\alpha)^{2}\right)^{3}$.

Let $G$ be the group of reduced primitive quadratic forms of discriminant
$-1048\times 9^{k}$ for $k=3,4$ and $H$ be the subgroup of forms of order
dividing $24$. When $k=3$, we have $\\#G=216$ and $\\#H=72$. Therefore, we get
a polynomial of degree $3$. The command polredbest in GP-Pari reduces this
polynomial to one with smaller coefficients that generates the same field,
namely

$x^{3}+42x-40.$

This polynomial generates the first layer $K_{1}/K$. The polynomial that
generates $K_{1}/\mathbb{Q}$ is given by

$x^{6}-90x^{4}-56x^{3}+4383x^{2}+4092x+1046.$

When $k=4$, $\\#G=648$ and $\\#H=72$. The degree $9$ polynomial we obtain is

$x^{9}-99x^{7}-414x^{6}-2025x^{5}+13086x^{4}+396855x^{3}+830250x^{2}+1026756x-2776392.$

Finally, the degree $18$ polynomial that generates $K_{2}/\mathbb{Q}$ is given
by

$x^{18}-2970x^{14}-13080x^{12}+7331265x^{10}+259656840x^{8}+3345463512x^{6}\\\
+18497139840x^{4}+129166770960x^{2}+1140995344768.$

##  6\. Non-cyclicity of $A_{n}$ when $A_{0}$ is disjoint from
$K_{{\mathrm{ac}}}$

Recall that in Section 3 we considered the case that the $p$-Hilbert class
field of $K$ is non-trivial and completely disjoint from the anti-cyclotomic
tower. The data in Tables LABEL:T:_inert and LABEL:T:_ramify seem to indicate
that $A_{1}$ and $A_{2}$ are non-cyclic. In this section, we prove this
observation.

######  Theorem 6.1.

Let $K$ be an imaginary quadratic field and let $p$ be an odd prime that does
not split in $K$. Suppose that $A_{0}$ is non-trivial. If the $p$-Hilbert
class field of $K$ is disjoint from $K_{{\mathrm{ac}}}$, then $A_{n}$ is non-
cyclic for all $n\geq 1$.

The proof will occupy the remainder of this section. Let $A_{n}$ denote the
Sylow $p$-subgroup of the ideal class group of $K_{n}$. Since $K_{n}/K_{0}$ is
totally ramified, the norm map

$A_{n}\longrightarrow A_{0}$

is surjective. Therefore, if $A_{0}$ is non-cyclic, then $A_{n}$ must be non-
cyclic as well. It suffices to prove the theorem in the case that $A_{0}$ is
cyclic of order $p^{u}$ where $u\geq 1$. Once again, the same argument implies
that if we show $A_{1}$ is non-cyclic then $A_{n}$ must be non-cyclic for all
$n\geq 1$.

######  Remark 6.2.

In the case that $p=3$, we are assuming that
$A_{0}\simeq{\mathbb{Z}}/3^{u}{\mathbb{Z}}$. It follows that we need not worry
about the special case when $p=3$ and $K={\mathbb{Q}}(\sqrt{-3})$, where
$A_{0}=1$. Therefore, throughout our discussion, we have the additional
property that $\mathcal{O}_{K}^{\times}$ has order prime to $p$.

### 6.1.

First, we prove two lemmas that hold for a general $\mathbb{Z}_{p}$-extension
of an imaginary quadratic field $K$ in which $p$ is not split and such that
$K_{1}/K$ is ramified. Write
$G\mathrel{\mathop{\ordinarycolon}}=\operatorname{Gal}\left(K_{1}/K\right)\simeq\langle\tau\rangle$.
Let $\operatorname{\mathsf{N}}_{K_{1}/K}$ denote the norm for $K_{1}/K$. By
Chevalley’s formula (see also Lemma 3.2)

(4) $\\#A_{1}^{G}=\frac{\\#A_{0}\times
p}{p[\mathcal{O}_{K}^{\times}\mathrel{\mathop{\ordinarycolon}}\mathcal{O}_{K}^{\times}\cap\operatorname{\mathsf{N}}_{K_{1}/K}(K_{1}^{\times})]}=\\#A_{0}.$

######  Lemma 6.3.

With the assumptions introduced above, $\\#A_{1}\neq\\#A_{0}$.

###### Proof.

Suppose that $\\#A_{1}=\\#A_{0}$. It follows from (4) that $A_{1}=A_{1}^{G}$.

Denote by $I$ an ideal that represents a class in $A_{1}$. Since it is fixed
by $\tau$, we have

$I^{\tau}=I\alpha\text{ for some }\alpha\in K_{1}.$

Taking the norm to $K$ on both sides of the equation, we obtain that

$(1)=\left(\operatorname{\mathsf{N}}_{K_{1}/K}(\alpha)\right).$

Without loss of generality, we may assume that $N(\alpha)=1$. By Hilbert’s
Theorem 90 (for cyclic groups), we know that

$\alpha=\beta^{1-\tau}\text{ for some }\beta\in K_{1}.$

It follows that

$(I\beta)^{\tau}=I\beta.$

Consider the following field diagram. We write $L(K_{i})$ to denote the
$p$-Hilbert class field of $K_{i}$.

$\mathbb{Q}$$K$$K_{1}$$L(K)$$L(K)\cdot
K_{1}$$A_{0}$${\mathbb{Z}}/p{\mathbb{Z}}$

By hypothesis there is a unique prime $\mathfrak{p}\mid p$ in $K$ that
ramifies in $K_{1}$. Denote by $\mathfrak{P}$ this unique prime above $p$ in
$K_{1}$. Recall that a prime splits completely in the $p$-Hilbert class field
if and only if its ideal class has order prime to $p$. For a non-split prime
$\mathfrak{p}\mid p$, we know that either $\mathfrak{p}$ or $\mathfrak{p}^{2}$
is principal in $K$. Therefore, the ideal class has order $1$ or $2$. It
follows that $\mathfrak{p}$ splits completely in $L(K)/K$. Note that
$\mathfrak{P}\mid p$ splits completely in $L(K)\cdot K_{1}/K_{1}$. We have
assumed that $\\#A_{1}=\\#A_{0}$, so we conclude that $L(K)\cdot
K_{1}=L(K_{1})$ (since $K_{1}/K_{0}$ is a ramified extension). Since we have
shown that $\mathfrak{P}\mid p$ splits completely in $L(K_{1})$, it has order
prime to $p$ in the class group of $K_{1}$.

Again consider the ideal $I$. From the above discussion, we conclude that
there exists an integer $h^{\prime}$ that is not divisible by $p$ such that
the class of $I^{h^{\prime}}$ is represented by an ideal from $A_{0}$. We
choose $h^{\prime}$ to be the order of the prime-to-$p$ part of the class
number of $K_{1}$. Therefore, the natural map $A_{0}\rightarrow A_{1}$ is
surjective. But, the norm map $A_{1}\rightarrow A_{0}$ is also surjective
because $K_{1}/K$ is a ramified extension. This means that the composition

$A_{0}\longrightarrow
A_{1}\xrightarrow{\operatorname{\mathsf{N}}_{K_{1}/K}}A_{0},$

which is the $p^{\text{th}}$-power map must be surjective. But, this is a
contradiction. ∎

######  Lemma 6.4.

With the notation and setting as above, $A_{1}$ is not cyclic of order $p^{k}$
with $k\geq u+2$.

###### Proof.

_Claim:_ The extension $L(K_{1})/K$ is non-abelian.
_Justification:_ Suppose that $L(K_{1})/K$ is abelian. Let $\mathfrak{p}$ be
the unique prime above $p$ in $K$. The inertia group of $\mathfrak{p}$ has
order $p$; so, the fixed field is an abelian unramified extension of $K$ of
degree $p^{k}$. But this contradicts the assumption that $\\#A_{0}=p^{u}$.

Recall the notation
$G\mathrel{\mathop{\ordinarycolon}}=\operatorname{Gal}\left(K_{1}/K\right)=\langle\tau\rangle$.
As a consequence of the above claim, we see that $\tau$ gives a non-trivial
automorphism of $A_{1}$ of order $p$.

Now suppose that $A_{1}\simeq{\mathbb{Z}}/p^{k}{\mathbb{Z}}$. Note that
$\operatorname{Aut}(A_{1})$ has a unique subgroup of order p, which is
generated by multiplication by $1+p^{k-1}$. The fixed group of this subgroup
has order $p^{k-1}$ but we have observed before that $\\#A_{1}^{G}=p^{u}$.
This implies that $k=u+1$. ∎

### 6.2.

In the remainder of the discussion we will crucially require that we are
studying the anti-cyclotomic $\mathbb{Z}_{p}$-extension of $K$. Denote by
$\mathcal{G}$ the Galois group
$\operatorname{Gal}\left(L(K_{1})/K_{0}\right)$. Suppose that
$A_{1}\simeq{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$. We have the following exact
sequence

$0\longrightarrow{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}\longrightarrow\mathcal{G}\longrightarrow\langle\tau\rangle\longrightarrow
0.$

Since $\tau$ acts on ${\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$ as an automorphism of
order $p$, we can identify

$\langle\tau\rangle\simeq 1+p^{u}{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}.$

An element $x\in 1+p^{u}{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$ acts on
${\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$ by multiplication by $x$. The group
$\mathcal{G}$ has the following concrete description

$\mathcal{G}\simeq\left\\{\begin{pmatrix}a&b\\\ 0&1\end{pmatrix}\mid a\in
1+p^{u}{\mathbb{Z}}/p^{u+1}{\mathbb{Z}},\
b\in{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}\right\\}.$

The quotient group
$G=\operatorname{Gal}\left(K_{1}/K\right)=\langle\tau\rangle$ is represented
by matrices of the form $\begin{pmatrix}c&0\\\ 0&1\end{pmatrix}$. Note that

$\begin{pmatrix}c&0\\\ 0&1\end{pmatrix}\begin{pmatrix}a&b\\\
0&1\end{pmatrix}\begin{pmatrix}c&0\\\
0&1\end{pmatrix}^{-1}=\begin{pmatrix}c&0\\\
0&1\end{pmatrix}\begin{pmatrix}a&b\\\ 0&1\end{pmatrix}\begin{pmatrix}1/c&0\\\
0&1\end{pmatrix}=\begin{pmatrix}a&bc\\\ 0&1\end{pmatrix}.$

######  Lemma 6.5.

With the notation introduced above,

$\begin{pmatrix}a&b\\\ 0&1\end{pmatrix}^{p}=\begin{pmatrix}1&pb\\\
0&1\end{pmatrix}$

###### Proof.

First, we rewrite

$\begin{pmatrix}a&b\\\ 0&1\end{pmatrix}=\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\ 0&1\end{pmatrix}.$

Therefore

$\displaystyle\begin{pmatrix}a&b\\\ 0&1\end{pmatrix}^{p}$
$\displaystyle=\begin{pmatrix}1&b\\\ 0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}\times\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}\times\ldots\times\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\ 0&1\end{pmatrix}$
$\displaystyle=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{-1}\times\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{2}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\ 0&1\end{pmatrix}^{-2}\times\ldots$
$\displaystyle\times\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{p-1}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{-(p-1)}\times\begin{pmatrix}a&0\\\ 0&1\end{pmatrix}^{p}$
$\displaystyle=\prod_{j=0}^{p-1}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{j}\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\begin{pmatrix}a&0\\\
0&1\end{pmatrix}^{-j}\times\begin{pmatrix}a&0\\\ 0&1\end{pmatrix}^{p}$
$\displaystyle=\prod_{j=0}^{p-1}\begin{pmatrix}1&a^{j}b\\\
0&1\end{pmatrix}\times\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$
$\displaystyle=\begin{pmatrix}1&\sum_{j=0}^{p-1}a^{j}b\\\ 0&1\end{pmatrix}.$

Recall that $a=1+kp^{u}$ for some integer $k$. Therefore,

$\sum_{j=0}^{p-1}a^{j}\equiv
1+\left(1+kp^{u}\right)+\left(1+2kp^{u}\right)+\ldots+\left(1+(p-1)kp^{u}\right)\equiv
p\pmod{p^{u+1}}.$

This completes the proof of the lemma. ∎

######  Lemma 6.6.

There is no automorphism $\phi$ of $\mathcal{G}$ such that for each $a\in
1+p^{u}{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$, there exists
$b_{a}\in{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$ satisfying

$\phi\left(\begin{pmatrix}a&0\\\
0&1\end{pmatrix}\right)=\begin{pmatrix}1/a&b_{a}\\\ 0&1\end{pmatrix}.$

###### Proof.

We prove this by contradiction; suppose such an automorphism $\phi$ exists.

Consider the matrix $P=\begin{pmatrix}1+p^{u}&0\\\ 0&1\end{pmatrix}$. Then,
there exists $b$ such that

$\phi\left(P\right)=\begin{pmatrix}1-p^{u}&b\\\ 0&1\end{pmatrix}.$

Note that $P$ has order $p$. Therefore, using Lemma 6.5 we see that

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}=\begin{pmatrix}1-p^{u}&b\\\
0&1\end{pmatrix}^{p}=\begin{pmatrix}1&pb\\\ 0&1\end{pmatrix}.$

Consequently, $b\equiv 0\pmod{p^{u}}$.

Now consider the matrix $M=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$ of order
$p^{u+1}$. Applying the automorphism $\phi$, we obtain

$\phi\left(M\right)=\begin{pmatrix}d&e\\\ 0&1\end{pmatrix}$

which must also have order $p^{u+1}$. Therefore,

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\neq\begin{pmatrix}d&e\\\
0&1\end{pmatrix}^{p^{u}}=\begin{pmatrix}1&p^{u}e\\\ 0&1\end{pmatrix}.$

Consequently, $e\not\equiv 0\pmod{p}$.

Finally, consider

(5) $\phi\left(PMP^{-1}\right)=\phi\left(\begin{pmatrix}1&1+p^{u}\\\
0&1\end{pmatrix}\right)=\phi\left(M^{p^{u}+1}\right).$

We know that

$\displaystyle\phi\left(PMP^{-1}\right)=\phi(P)\phi(M)\phi(P^{-1})$
$\displaystyle=\begin{pmatrix}1-p^{u}&b\\\
0&1\end{pmatrix}\begin{pmatrix}d&e\\\
0&1\end{pmatrix}\begin{pmatrix}1-p^{u}&b\\\ 0&1\end{pmatrix}^{-1}$
$\displaystyle=\begin{pmatrix}d&e(1-p^{u})\\\ 0&1\end{pmatrix}.$

For the last equality, we have used the fact that $b\equiv 0\pmod{p^{u}}$ and
$d\equiv 1\pmod{p^{u}}$.

On the other hand

$\displaystyle\phi(M^{p^{u}+1})=\phi(M)^{p^{u}}\phi(M)$
$\displaystyle=\begin{pmatrix}d&e\\\
0&1\end{pmatrix}^{p^{u}}\begin{pmatrix}d&e\\\ 0&1\end{pmatrix}$
$\displaystyle=\begin{pmatrix}1&p^{u}e\\\
0&1\end{pmatrix}\begin{pmatrix}d&e\\\ 0&1\end{pmatrix}\quad\text{ by
Lemma~{}\ref{lemma to be used next}}$
$\displaystyle=\begin{pmatrix}d&e(1+p^{u})\\\ 0&1\end{pmatrix}.$

By (5), it follows that

$\begin{pmatrix}d&e(1-p^{u})\\\
0&1\end{pmatrix}=\begin{pmatrix}d&e(1+p^{u})\\\ 0&1\end{pmatrix}.$

Equivalently,

$e(1-p^{u})\equiv e(1+p^{u})\pmod{p^{u+1}}.$

This contradicts the fact that $e\not\equiv 0\pmod{p}$. This completes the
proof of the lemma. ∎

We now prove Theorem 6.1.

###### Proof.

We know from the calculations done in Section 6.1 that $A_{1}$ is not cyclic
of order $p^{u}$ or order $p^{k}$ where $k\geq{u+2}$. We are only left to
eliminate the case that $A_{1}$ is cyclic of order $p^{u+1}$.

The concrete description of $A_{1}$ is as follows:

$A_{1}=\operatorname{Gal}\left(L(K_{1})/K_{1}\right)\simeq\left\\{\begin{pmatrix}1&b\\\
0&1\end{pmatrix}\mid b\in{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}\right\\}.$

Complex conjugation gives an element $\phi\in\operatorname{Aut}(\mathcal{G})$
that maps $A_{1}$ to itself. Since $K_{1}$ is contained in
$K_{{\mathrm{ac}}}$, it means that $\phi$ acts by inversion on
$\langle\tau\rangle$; i.e.,

$\phi\left(\begin{pmatrix}a&0\\\
0&1\end{pmatrix}\right)\equiv\begin{pmatrix}1/a&0\\\
0&1\end{pmatrix}\pmod{A_{1}}.$

But, such a $\phi$ does not exist by Lemma 6.6. This contradicts the
assumption that $A_{1}\simeq{\mathbb{Z}}/p^{u+1}{\mathbb{Z}}$. This completes
the proof of the theorem. ∎

##  7\. Heuristics

In this section, we aim to understand, as $K$ varies over all imaginary
quadratic fields, how often $L(K)$ is disjoint from $K_{{\mathrm{ac}}}$,
partially intersecting $K_{{\mathrm{ac}}}$, and completely contained inside
$K_{{\mathrm{ac}}}$.

### 7.1.

Fix an odd prime $p$. Let $K$ be an imaginary quadratic field and
$K_{{\mathrm{ac}}}$ denote the anti-cyclotomic $\mathbb{Z}_{p}$-extension of
$K$. Let $K_{n}$ denote the $n$-th layer of the $\mathbb{Z}_{p}$-extension,
and $L(K)$ be the $p$-Hilbert class field of $K$. Write $A_{n}$ to denote the
$p$-part of the ideal class group of $K_{n}$ for all $n\geq 0$.

######  Intersection Heuristics.

Fix a finite abelian $p$-group $G$. Let $K$ vary over imaginary quadratic
fields such that the $p$-part of its ideal class group is isomorphic to $G$.
The probability that $L(K)\cap K_{{\mathrm{ac}}}=K_{n}$ is given by

$\frac{\\#(\text{elements of $G$ of order exactly }p^{n})}{\\#G}.$

This number is also

$\frac{\\#(\text{homomorphisms
}G\longrightarrow{{\mathbb{Q}}/{\mathbb{Z}}}\text{ with image of size
}p^{n})}{\\#G}.$

Note that the numerator of (Intersection Heuristics) can also be interpreted
as

$\\#(\text{surjective homomorphisms
}G\longrightarrow{{\mathbb{Z}}/p^{n}{\mathbb{Z}}}).$

In particular, we expect that as $K$ varies over imaginary quadratic fields
with $A_{0}\simeq G$, the probability that class field is disjoint from
$K_{{\mathrm{ac}}}$ is $\frac{1}{\\#G}$.

We further remark that our calculations suggest that the LABEL:H:_intersection
is independent of the splitting type. For example, the heuristics remain the
same even if we restrict ourselves to vary over imaginary quadratic fields
such that $p$ splits in $K$ and the $p$-part of the ideal class group is
isomorphic to $G$. This agrees with the computational evidence that the Cohen-
Lenstra predictions are the same for each splitting type.

#### 7.1.1.

As $K$ varies over all imaginary quadratic fields, the probability that the
class field is disjoint is given by

(6) $\frac{\sum_{G}(\text{frequency of }G)(1/\\#G)}{\sum_{G}(\text{frequency
of }G)}.$

Set $\operatorname{Aut}(G)$ to denote the group of all automorphisms of $G$.
Define

$w(p^{k})=\sum_{\\#G=p^{k}}\frac{1}{\\#\operatorname{Aut}(G)}.$

Here, the sum is over finite abelian $p$-groups of order $p^{k}$.

It follows from the Cohen–Lenstra heuristics (see for example [Woo16,
Proposition 5.7]) that the denominator of (6) is given by

$\sum_{G}(\text{frequency of
}G)=\sum_{\alpha=0}^{\infty}w(p^{\alpha})=\prod_{j\geq
1}\left(1-p^{-j}\right)^{-1}.$

On the other hand, the numerator of (6) is given by

$\sum_{G}(\text{frequency of
}G)(1/\\#G)=\sum_{\alpha=0}^{\infty}w(p^{\alpha})p^{-\alpha}=\prod_{j\geq
2}\left(1-p^{-j}\right)^{-1}.$

Therefore, we obtain that as $K$ varies over all imaginary quadratic number
fields, the (expected) probability that the ideal class group of $K$ is
_completely disjoint_ from $K_{{\mathrm{ac}}}$ is given by

(7) $\prod_{j\geq 2}\left(1-p^{-j}\right)^{-1}\times\prod_{j\geq
1}\left(1-p^{-j}\right)=1-\frac{1}{p}.$

######  Remark 7.1.

Note that this includes those imaginary quadratic fields for which $p\nmid
h_{K}$ and these fields are a major contribution to the total. As per the
Cohen–Lenstra heuristics, varying over all imaginary quadratic fields the
probability that $p\nmid h_{k}$ is given by

$\prod_{j\geq 1}\left(1-p^{-j}\right).$

#### 7.1.2.

Let us now calculate, as $K$ varies over all imaginary quadratic fields, the
probability that the $p$-Hilbert class field is contained in
$K_{{\mathrm{ac}}}$. For such a $K$, we must have $A_{0}$ cyclic.

If $G$ is a cyclic $p$-group of order $p^{k}$ with $k\geq 1$, its automorphism
group has order $\phi(p^{k})$. Also,

$\\#(\text{surjective homomorphisms
}G\rightarrow{\mathbb{Z}}/p^{k}{\mathbb{Z}})=\phi(p^{k}).$

As we vary over all imaginary quadratic fields $K$, the probability that the
$p$-class field is (non-trivial) and _completely contained_ in
$K_{{\mathrm{ac}}}$ is given by

(8) $\frac{\displaystyle{\sum_{k\geq
1,G\simeq{\mathbb{Z}}/p^{k}{\mathbb{Z}}}}\left(\text{frequency of
}G\right)\cdot\frac{\\#\left(\text{surjective homomorphisms
}G\rightarrow{\mathbb{Z}}/p^{k}{\mathbb{Z}}\right)}{\\#G}}{\sum_{G}(\text{frequency
of }G)}\\\ =\left(\sum_{k\geq
1}\frac{1}{\phi(p^{k})}\cdot\frac{\phi(p^{k})}{p^{k}}\right)\prod_{j\geq
1}\left(1-p^{-j}\right)=\frac{1}{p-1}\prod_{j\geq 1}\left(1-p^{-j}\right).$

### 7.2.

In this section we predict how often
$\mu(K_{{\mathrm{ac}}}/K)=\lambda(K_{{\mathrm{ac}}}/K)=0$ as $K$ varies over
all imaginary quadratic fields such that $p$ does not split.

The computational data presented in Section 3 and the results proven in
Section 3.2 suggest the following: in addition to the case, where the anti-
cyclotomic Iwasawa invariants are provably $0$, we should expect that
$\mu(K_{{\mathrm{ac}}}/K)=\lambda(K_{{\mathrm{ac}}}/K)=0$ when $L(K)\cap
K_{{\mathrm{ac}}}=K$. The probability that the $p$-Hilbert class field is
disjoint from the anti-cyclotomic tower was calculated in (7). On the other
hand, the probability that we are in the situation considered in Theorem 2.2
is calculated in (8). Therefore, we may propose the following:

######  Invariants Heuristics.

Among the imaginary quadratic fields $K$ in which $p$ does not split,

$\displaystyle\operatorname{Prob}\left(\mu(K_{{\mathrm{ac}}}/K)=\lambda(K_{{\mathrm{ac}}}/K)=0\right)$
$\displaystyle\geq\left(1-\frac{1}{p}\right)+\frac{1}{p-1}\prod_{j\geq
1}\left(1-p^{-j}\right)$ $\displaystyle=1-p^{-3}-p^{-4}-p^{-5}+p^{-7}+\cdots.$

The probability is obtained by combining the case $p\nmid h_{K}$ and the cases
where $K\subsetneq L(K)$ is either disjoint or contained in
$K_{{\mathrm{ac}}}$.

###  7.3. Computational Method and Data

In this section, we provide tables which show how much of the $p$-Hilbert
class field of $K$ intersects with $K_{\mathrm{ac}}$, when $p=3,5$. This
provides computational evidence in support of our heuristics, at least in the
case when $p=3,5$. Note that in each of the tables, $k$ starts with a large
value because Cohen–Lenstra heuristics are known to converge very slowly. We
explain how the computations were done using PARI/GP.

The starting point is a special case of the following result of D. Brink
[Bri07, Theorem 2].

######  Brink’s Theorem.

Let $p$ be an odd prime, and $K$ be an imaginary quadratic field of
discriminant $-d$ and class number $h=p^{\theta}u$ (such that $\gcd(p,u)=1$).
Let $\ell\neq p$ be a prime number that splits as
$\mathfrak{l}\overline{\mathfrak{l}}$ in $K$, and define $\nu$ as follows:

$[L(K)\cap K_{{\mathrm{ac}}}\mathrel{\mathop{\ordinarycolon}}K]=p^{\nu}.$

Further, write

$\ell^{h}=\begin{cases}a^{2}+db^{2}&\text{if }d\not\equiv 3\pmod{4}\\\
a^{2}+ab+\left((d+1)/4\right)b^{2}&\text{if }d\equiv 3\pmod{4}\end{cases}$

with relatively prime $a,b\in{\mathbb{Z}}$. Set

$\omega=\begin{cases}\sqrt{-d}&\text{if }d\not\equiv 3\pmod{4}\\\
\left(1+\sqrt{-d}\right)/2&\text{if }d\equiv 3\pmod{4}\end{cases}.$

The following conclusions hold for $n\geq 0$.

  1. (i)

_$p$ splits in $K$:_ Write $(a+b\omega)^{p-1}=a^{*}+b^{*}\omega$. Then

$\mathfrak{l}\text{ splits completely in }K_{n}\Longleftrightarrow b^{*}\equiv
0\pmod{p^{n+1+\theta-\nu}}.$

  2. (ii)

_$p$ is inert in $K$:_ Write $(a+b\omega)^{p+1}=a^{*}+b^{*}\omega$. Then

$\mathfrak{l}\text{ splits completely in }K_{n}\Longleftrightarrow b^{*}\equiv
0\pmod{p^{n+1+\theta-\nu}}.$

  3. (iii)

_$p$ ramifies in $K$:_ If the conditions of the exceptional case (case (iv)
below) don’t hold, then,

$\mathfrak{l}\text{ splits completely in }K_{n}\Longleftrightarrow b\equiv
0\pmod{p^{n+\theta-\nu}}.$

  4. (iv)

_$p=3$ and $d\equiv 3\pmod{9}$:_ Write $(a+b\omega)^{3}=a^{*}+b^{*}\omega$.
Then

$\mathfrak{l}\text{ splits completely in }K_{n}\Longleftrightarrow b^{*}\equiv
0\pmod{p^{n+2+\theta-\nu}}.$

######  Corollary 7.2.

Keep the notation introduced above and let $v_{p}$ denote the $p$-adic
valuation. Then,

$\nu=\begin{cases}1+\theta-\min\\{v_{p}(b^{*})\mid\ell\neq p\text{ splits in
}K\\}&\text{if }p\text{ splits in }K.\\\
1+\theta-\min\\{v_{p}(b^{*})\mid\ell\neq p\text{ splits in }K\\}&\text{if
}p\text{ is inert in }K.\\\ \theta-\min\\{v_{p}(b)\mid\ell\neq p\text{ splits
in }K\\}&\text{if }p\text{ ramifies in }K\text{ but not the exceptional
case}.\\\ 2+\theta-\min\\{v_{p}(b^{*})\mid\ell\neq p\text{ splits in
}K\\}&\text{if }p=3\text{ and }d\equiv 3\pmod{9}.\end{cases}$

The corollary follows from the case $n=0$ of the theorem since the theorem
then gives a characterization of the primes $\ell$ that split in $K$. The
degree $p^{\nu}$ of the intersection was determined using this corollary. The
fundamental discriminants $-d$ were considered in the ranges described below
each table. We found the minimum over only primes $\ell<400$ that split in
$K$. Hence, it is possible that the counts in our tables are slightly
incorrect, but not enough to affect the overall agreement with our heuristic
model (Intersection Heuristics). The computations of the class numbers were
done in PARI/GP, and therefore depend on the correctness of the Generalized
Riemann Hypothesis. We assume that this also does not affect the overall
agreement with the heuristic model.

#### How to read the tables that follow

The results are arranged according to the isomorphism class of $A_{0}$. For
example, in Table LABEL:T:_3k-1_100_to_101 in the second row, there are 31483
fields in the range mentioned such that
$A_{0}\simeq{\mathbb{Z}}/9{\mathbb{Z}}$. The fraction where $[L(K)\cap
K_{{\mathrm{ac}}}\mathrel{\mathop{\ordinarycolon}}K]=9$ is $0.678017$, the
fraction where $[L(K)\cap
K_{{\mathrm{ac}}}\mathrel{\mathop{\ordinarycolon}}K]=3$ is $0.213734$, and the
fraction where $L(K)$ is disjoint from $K_{{\mathrm{ac}}}$ is $0.108249$. The
row labelled ‘Expected’ gives the fractions predicted by (Intersection
Heuristics).

Table 5. $d=3k-1,\quad 100,000,000\leq k\leq 101,000,000$ $A_{0}$ | Number | $3^{4}$ | $3^{3}$ | $3^{2}$ | 3 | $1$
---|---|---|---|---|---|---
3 | 94133 | 0 | 0 | 0 | 0.674429 | 0.325571
Expected |  | 0 | 0 | 0 | 0.666667 | 0.333333
9 | 31483 | 0 | 0 | 0.678017 | 0.213734 | 0.108249
Expected |  | 0 | 0 | 0.666667 | 0.222222 | 0.111111
27 | 10262 | 0 | 0.678718 | 0.214481 | 0.072988 | 0.033814
Expected |  | 0 | 0.666667 | 0.222222 | 0.074074 | 0.037037
81 | 3551 | 0.669107 | 0.227542 | 0.069839 | 0.021121 | 0.012391
Expected |  | 0.666667 | 0.222222 | 0.074074 | 0.024691 | 0.012346
Table 6. $d=3k+1,\quad 100,000,000\leq k\leq 101,000,000$ $A_{0}$ | Number | $3^{4}$ | $3^{3}$ | $3^{2}$ | $3$ | $1$
---|---|---|---|---|---|---
$3$ | 93747 | 0 | 0 | 0 | 0.675328 | 0.324672
Expected |  | 0 | 0 | 0 | 0.666667 | 0.333333
$9$ | 31072 | 0 | 0 | 0.677620 | 0.214663 | 0.107718
Expected |  | 0 | 0 | 0.666667 | 0.222222 | 0.111111
$27$ | 10477 | 0 | 0.677770 | 0.216283 | 0.069486 | 0.036461
Expected |  | 0 | 0.666667 | 0.222222 | 0.074074 | 0.037037
81 | 3495 | 0.678970 | 0.214306 | 0.072103 | 0.023748 | 0.010873
Expected |  | 0.666667 | 0.222222 | 0.074074 | 0.024691 | 0.012346

Table 7. $d=9k+3,\quad 1,000,000\leq k\leq 1,100,000$ $A_{0}$ | Number | $3^{4}$ | $3^{3}$ | $3^{2}$ | $3$ | $1$
---|---|---|---|---|---|---
$3$ | 9298 | 0 | 0 | 0 | 0.687567 | 0.312433
Expected |  | 0 | 0 | 0 | 0.666667 | 0.333333
$9$ | 2961 | 0 | 0 | 0.676798 | 0.209389 | 0.113813
Expected |  | 0 | 0 | 0.666667 | 0.222222 | 0.111111
$27$ | 1103 | 0 | 0.675431 | 0.212149 | 0.074343 | 0.038078
Expected |  | 0 | 0.666667 | 0.222222 | 0.074074 | 0.037037
81 | 350 | 0.717143 | 0.182857 | 0.071429 | 0.020000 | 0.008571
Expected |  | 0.666667 | 0.222222 | 0.074074 | 0.024691 | 0.012346
Table 8. $d=9k+6,\quad 1,000,000\leq k\leq 1,100,000$ $A_{0}$ | Number | $3^{4}$ | $3^{3}$ | $3^{2}$ | $3$ | $1$
---|---|---|---|---|---|---
$3$ | 9279 | 0 | 0 | 0 | 0.692532 | 0.307468
Expected |  | 0 | 0 | 0 | 0.666667 | 0.333333
$9$ | 3029 | 0 | 0 | 0.689006 | 0.206339 | 0.104655
Expected |  | 0 | 0 | 0.666667 | 0.222222 | 0.111111
$27$ | 1065 | 0 | 0.712676 | 0.186854 | 0.066667 | 0.033803
Expected |  | 0 | 0.666667 | 0.222222 | 0.074074 | 0.037037
81 | 321 | 0.716511 | 0.190031 | 0.059190 | 0.018692 | 0.015576
Expected |  | 0.666667 | 0.222222 | 0.074074 | 0.024691 | 0.012346
Table 9. $d=3k-1,\quad 100,000,000\leq k\leq 150,000,000$ $A_{0}$ | Number | $3^{4}$ | $3^{3}$ | $3^{2}$ | $3$ | $1$
---|---|---|---|---|---|---
$9\times 9$ | 420 | 0 | 0 | 0.900000 | 0.078571 | 0.021429
Expected |  | 0 | 0 | 0.888889 | 0.098765 | 0.012346
$27\times 9$ | 215 | 0 | 0.688372 | 0.297674 | 0.013953 | 0.000000
Expected |  | 0 | 0.666667 | 0.296296 | 0.032922 | 0.004115
$81\times 9$ | 72 | 0.555556 | 0.333333 | 0.111111 | 0.000000 | 0.000000
Expected |  | 0.666667 | 0.222222 | 0.098765 | 0.010974 | 0.001372
Table 10. $d=3k+1,\quad 10,000,000\leq k\leq 100,000,000$ $A_{0}$ | Number | $3^{2}$ | 3 | $1$
---|---|---|---|---
$3\times 3\times 3$ | 597 | 0 | 0.988275 | 0.011725
Expected |  | 0 | 0.962963 | 0.037037
$9\times 3\times 3$ | 287 | 0.728223 | 0.268293 | 0.003484
Expected |  | 0.666667 | 0.320988 | 0.012346
Table 11. $d=5k+2,\quad 1,000,000\leq k\leq 2,000,000$ $A_{0}$ | Number | $5^{4}$ | $5^{3}$ | $5^{2}$ | $5$ | $1$
---|---|---|---|---|---|---
$5$ | 39002 | 0 | 0 | 0 | 0.802779 | 0.197221
Expected |  | 0 | 0 | 0 | 0.800000 | 0.200000
$25$ | 7639 | 0 | 0 | 0.807566 | 0.154078 | 0.038356
Expected |  | 0 | 0 | 0.800000 | 0.160000 | 0.040000
$125$ | 1383 | 0 | 0.806218 | 0.151844 | 0.033984 | 0.007954
Expected |  | 0 | 0.800000 | 0.160000 | 0.032000 | 0.008000
625 | 203 | 0.753695 | 0.201970 | 0.044335 | 0.000000 | 0.000000
Expected |  | 0.800000 | 0.160000 | 0.032000 | 0.006400 | 0.001600
Table 12. $d=5k,\quad 10,000,000\leq k\leq 20,000,000$ $A_{0}$ | Number | $5^{3}$ | $5^{2}$ | $5$ | $1$
---|---|---|---|---|---
$5\times 5$ | 2122 | 0 | 0 | 0.967484 | 0.032516
Expected |  | 0 | 0 | 0.960000 | 0.040000
$25\times 5$ | 493 | 0 | 0.833671 | 0.164300 | 0.002028
Expected |  | 0 | 0.800000 | 0.192000 | 0.008000
$125\times 5$ | 48 | 0.812500 | 0.145833 | 0.041667 | 0.000000
Expected |  | 0.800000 | 0.160000 | 0.038400 | 0.001600

## Acknowledgements

DK was supported by a PIMS postdoctoral fellowship at the time of writing the
paper. We thank Reinier Bröker for helpful discussions. We are grateful to the
anonymous referees for their timely reading of the manuscript and for their
valuable comments.

## References

  * [BHW19] Reinier Bröker, David Hubbard, and Lawrence C. Washington, _Explicit computations in Iwasawa theory_ , Proceedings of the Thirteenth Algorithmic Number Theory Symposium, Open Book Ser., vol. 2, Math. Sci. Publ., Berkeley, CA, 2019, pp. 137–153.
  * [Bri07] David Brink, _Prime decomposition in the anti-cyclotomic extension_ , Math. Comp. 76 (2007), no. 260, 2127–2138.
  * [CL90] Henri Cohen and Hendrik Lenstra, Jr., _Heuristics on class groups of number fields_ , Number theory, Noordwijkerhout 1983: Proceedings of the Journées arithmétiques held at Noordwijkerhout, The Netherlands, July 11-15, 1983, Springer, 1990.
  * [DF04] David S Dummit and Richard Foote, _Abstract algebra_ , John Wiley&Sons, 2004\.
  * [DFKS91] David Dummit, David Ford, Hershy Kisilevsky, and Jonathan Sands, _Computation of Iwasawa lambda invariants for imaginary quadratic fields_ , J. Number Theory 37 (1991), no. 1, 100–121.
  * [DK22] Daniel Delbourgo and Heiko Knospe, _On Iwasawa $\lambda$-invariants for abelian number fields and random matrix heuristics_, 2022, arXiv preprint 2207.06287.
  * [EJV11] Jordan Ellenberg, Sonal Jain, and Akshay Venkatesh, _Modeling $\lambda$-invariants by $p$-adic random matrices_, Comm. Pure Appl. Math. 64 (2011), no. 9, 1243–1262.
  * [FK02] Takashi Fukuda and Keiichi Komatsu, _Non-cyclotomic $\mathbb{Z}_{p}$-extensions of imaginary quadratic fields_, Exp. Math. 11 (2002), no. 4, 469–475.
  * [Fuj13] Satoshi Fujii, _On a bound of $\lambda$ and the vanishing of $\mu$ of $\mathbb{Z}_{p}$-extensions of an imaginary quadratic field_, J. Math. Soc. Japan 65 (2013), no. 1, 277–298.
  * [FW79] Bruce Ferrero and Lawrence Washington, _The Iwasawa invariant $\mu_{p}$ vanishes for abelian number fields_, Ann. Math. (1979), 377–395.
  * [Ger77] Frank Gerth, _Structure of $l$-class groups of certain number fields and $\mathbb{Z}_{l}$-extensions_, Mathematika 24 (1977), no. 1, 16–33.
  * [Gil87] Roland Gillard, _Transformation de Mellin–Leopoldt des fonctions elliptiques_ , J. Number Theory 25 (1987), no. 3, 379–393.
  * [Gol74] Robert Gold, _Examples of Iwasawa invariants_ , Acta Arithmetica 26 (1974), no. 1, 21–32.
  * [Gre76] Ralph Greenberg, _On the Iwasawa invariants of totally real number fields_ , Amer. J. Math. 98 (1976), no. 1, 263–284.
  * [Mur19] Kazuaki Murakami, _On an upper bound of $\lambda$-invariants of $\mathbb{Z}_{p}$-extensions over an imaginary quadratic field_, J. Math. Soc. Japan 71 (2019), no. 3, 1005–1026.
  * [Oza01] Manabu Ozaki, _Iwasawa invariants of $\mathbb{Z}_{p}$-extensions over an imaginary quadratic field_, Class field theory–its centenary and prospect, Advanced Studies in Pure Mathematics 30 (2001), 387–399.
  * [San91] Jonathan Sands, _On small Iwasawa invariants and imaginary quadratic fields_ , Proc. Amer. Math. Soc. 112 (1991), no. 3, 671–684.
  * [Sch87] Leila Schneps, _On the $\mu$-invariant of $p$-adic ${L}$-functions attached to elliptic curves with complex multiplication_, J. Number Theory 25 (1987), no. 1, 20–33.
  * [Sch02] Reinhard Schertz, _Weber’s class invariants revisited_ , J. Théor. Nombres Bordeaux 14 (2002), no. 1, 325–343.
  * [Was97] Lawrence Washington, _Introduction to cyclotomic fields_ , Graduate texts in Mathematics, vol. 83, Springer, 1997.
  * [Woo16] Melanie Matchett Wood, _Asymptotics for number fields and class groups_ , Directions in Number Theory (Ellen Eischen, Ling Long, Rachel Pries, and Katherine Stange, eds.), Springer International Publishing, 2016, pp. 291–339.