Minimal relative units of the cyclotomic $\mathbb{Z}_{2}$ -extension

Tomokazu Kashio
Hyuga Yoshizaki Tokyo University of Science, [email protected]Tokyo University of Science, [email protected]

Abstract

Let $\mathbb{B}_{n}:=\mathbb{Q}(\cos(\pi/2^{n+1}))$ . For the relative norm map $\mathrm{N}_{n/n-1}\,\colon\mathcal{O}_{\mathbb{B}_{n}}^{\times}\rightarrow\mathcal{O}_{\mathbb{B}_{n-1}}^{\times}$ on the units group, we define $RE_{n}:=\mathrm{N}_{n/n-1}^{-1}(\{\pm 1\})$ , $RE_{n}^{+}:=\mathrm{N}_{n/n-1}^{-1}(\{1\})$ . Komatsu conjectured that $\mathrm{Tr}\,\epsilon^{2}\geq 2^{n}(2^{n+1}-1)$ for $\epsilon\in RE_{n}-\{\pm 1\}$ . Morisawa and Okazaki showed that it holds for $\epsilon\in RE_{n}-RE_{n}^{+}$ . In this paper we study the case $\epsilon\in RE_{n}^{+}$ . We conjecture that $\min\{\mathrm{Tr}\,\epsilon^{2}\mid\epsilon\in RE_{n}^{+}-\{\pm 1\}\}=2^{n}(1+8c_{n})$ , where $c_{1}:=2$ and $c_{n}:=2\cdot\mathrm{round}(2^{n}/5)$ ( $n\geq 2$ ). We show that this holds for $n\leq 6$ and that a “half” of this: $\min\{\mathrm{Tr}\,\epsilon^{2}\mid\epsilon\in RE_{n}^{+}-\{\pm 1\}\}\leq 2^{n}(1+8c_{n})$ holds for even $n$ . We also observe a relation to the class number problem.

1 Introduction

Let $\mathbb{B}_{n}:=\mathbb{Q}(\cos(\pi/2^{n+1}))$ , which is the $n$ th layer of the cyclotomic $\mathbb{Z}_{2}$ -extension over $\mathbb{Q}$ . We put $RE_{n}^{+}:=\operatorname{Ker}\mathrm{N}_{n/n-1}\,$ , $RE_{n}^{-}:=\mathrm{N}_{n/n-1}^{-1}(\{-1\})$ , $RE_{n}:=\mathrm{N}_{n/n-1}^{-1}(\{\pm 1\})=RE_{n}^{+}\coprod RE_{n}^{-}$ , where $\mathrm{N}_{n/n-1}\,\colon\mathcal{O}_{\mathbb{B}_{n}}^{\times}\rightarrow\mathcal{O}_{\mathbb{B}_{n-1}}^{\times}$ denotes the relative norm map on the unit group. Then Komatsu, in personal communication with Morisawa and Okazaki, stated the following conjecture.

Conjecture 1.1 ([MO3, Conjecture 1.1]).

We have for $\epsilon\in RE_{n}-\{\pm 1\}$

\displaystyle\mathrm{Tr}\,\epsilon^{2}\geq 2^{n}(2^{n+1}-1).

(1)

Morisawa and Okazaki showed that

Theorem 1.2 ([MO3, Theorem 6.4]).

Ineq. (1) holds for $\epsilon\in RE_{n}^{-}$ .

Namely the unsolved problem is Ineq. (1) for $\epsilon\in RE_{n}^{+}-\{\pm 1\}$ . We provide the best possible refinement in this case as follows.

Conjecture (2.2).

Let $c_{1}=2$ , $c_{n}=2\cdot\mathrm{round}\,(2^{n}/5)$ $(n\geq 2)$ where $\mathrm{round}\,(x)$ denotes the nearest integer to $x$ . Then we have

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}=2^{n}(1+8c_{n}).

(2)

The first few terms of $c_{n}$ are $c_{1}=2$ , $c_{2}=2$ , $c_{3}=4$ , $c_{4}=6$ , $c_{5}=12,\dots$ . We also present some partial results.

Theorem (2.5).

For $n=1,3,5$ or for even $n$ , there exists $u_{n}\in RE_{n}^{+}-\{\pm 1\}$ satisfying

\displaystyle\mathrm{Tr}\,u_{n}^{2}=2^{n}(1+8c_{n}).

Hence a “half” of Eq. 2: $\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}\leq 2^{n}(1+8c_{n})$ holds for such $n$ .

Theorem (2.7).

Eq. 2 holds for $n\leq 6$ .

The proof of 2.7 relies on the fact that the class number $h_{n}$ of $\mathbb{B}_{n}$ is $1$ . On the other hand, in 2.6, we also provide a proof for $n\leq 3$ without using any information of $h_{n}$ .

We also see a relation between our Conjecture and the class numbers in §3, 4. Weber’s class number problem asks whether $h_{n}=1$ for all $n$ and some partial results follows by studying the unit group. For example, by using 1.2 concerning $RE_{n}^{-}$ , Fukuda and Komatsu [FK3, Theorem 1.3] showed that

l\nmid h_{n}

for all

n

and for all primes

l

with

l\not\equiv\pm 1\bmod 32

(3)

We may observe a “similar” phenomena also for $RE_{n}^{+}$ . Morisawa and Okazaki [MO3, Proposition 6.6] showed that

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}\geq 2^{n}\cdot 17\quad(n\geq 2).

(4)

The second author [Yo, Remark in §5.1] showed that

Ineq. (4) implies $h_{2}/h_{1}=1$ .

We generalize these results as follows.

Theorem (2.3).

We have

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}\geq 2^{n}\cdot 33\quad(n\geq 3).

(5)

Theorem (3.2).

Ineq. (5) implies $h_{3}/h_{2}=1$ .

In §4 we introduce some numerical results:

(i)

When $n=4,5$ , Eq. 2 implies the $l$ -indivisibility of $h_{n}/h_{n-1}$ for several primes $l$ (§4.1).
(ii)

When $n=7$ , Eq. 2 implies the $l$ -indivisibility of $h_{7}/h_{7-1}$ for the first $1000$ primes $l$ satisfying $l>10^{9}$ , $l\equiv 65\bmod 128$ (§4.2).

The known results for the $l$ -indivisibility is as follows.

\displaystyle\text{if $n<7$ or $l\not\equiv\pm 1\bmod 64$ or $l\leq 10^{9}$, then a prime $l$ does not divide $h_{n}$}.

(6)

We note that the primes $l$ in the case (ii) are out of this range.

2 Minimal relative units

Let $\mathbb{B}_{n}$ be the $n$ th layer of the cyclotomic $\mathbb{Z}_{2}$ -extension over $\mathbb{Q}$ . More explicitly we have

\displaystyle\mathbb{B}_{n}=\mathbb{Q}(X_{n}),\quad X_{n}:=2\cos\left(\frac{2\pi}{2^{n+2}}\right).

In this paper, we fix a generator $\sigma$ of $G_{n}:=\mathrm{Gal}(\mathbb{B}_{n}/\mathbb{Q})\cong\mathbb{Z}/2^{n}\mathbb{Z}$ by

\displaystyle\sigma\colon 2\cos\left(\frac{2\pi}{2^{n+2}}\right)\mapsto 2\cos\left(\frac{3\cdot 2\pi}{2^{n+2}}\right).

Definition 2.1.

Let $E_{n}$ be the unit group of $\mathbb{B}_{n}$ . We consider the following subgroups:

	$\displaystyle RE_{n}^{+}$	$\displaystyle:=\{\epsilon\in E_{n}\mid\mathrm{N}_{n/n-1}\,\epsilon=1\},$
	$\displaystyle RE_{n}$	$\displaystyle:=\{\epsilon\in E_{n}\mid\mathrm{N}_{n/n-1}\,\epsilon=\pm 1\},$
	$\displaystyle A_{n}$	$\displaystyle:=\langle\pm 1,\varepsilon_{n}\rangle_{\mathbb{Z}[G_{n}]}=\left\{\pm\prod_{i=0}^{2^{n-1}-1}\sigma^{i}(\varepsilon_{n})^{m_{i}}\mid m_{i}\in\mathbb{Z}\right\}\quad\text{for}\quad\varepsilon_{n}:=\frac{X_{n}+1}{X_{n}-1}.$

Here $\mathrm{N}_{n/n-1}\,\colon\mathbb{B}_{n}\rightarrow\mathbb{B}_{n-1}$ denotes the relative norm map.

We have $A_{n}\subset RE_{n}^{+}$ since $\mathrm{N}_{n/n-1}\,\sigma^{i}(\varepsilon_{n})=\sigma^{i}(\frac{X_{n}+1}{X_{n}-1}\cdot\frac{-X_{n}+1}{-X_{n}-1})=1$ . We embed $\mathbb{B}_{n}$ into $\mathbb{R}^{2^{n}}$ as usual:

\displaystyle\mathbb{B}_{n}\rightarrow\mathbb{R}^{2^{n}},\quad x\mapsto(\sigma^{i}(x))_{0\leq i\leq 2^{n}-1}.

In particular, $\sqrt{\mathrm{Tr}\,x^{2}}$ is equal to the length of $x$ in $\mathbb{R}^{2^{n}}$ . The ring of integers $\mathcal{O}_{\mathbb{B}_{n}}=\mathbb{Z}[X_{n}]$ has an orthogonal basis $\{b_{i}\mid 0\leq i\leq 2^{n}-1\}$ :

\displaystyle b_{i}:=\begin{cases}1&(i=0)\\ 2\cos\left(\frac{i*2\pi}{2^{n+2}}\right)&(1\leq i\leq 2^{n}-1)\end{cases},\quad\mathrm{Tr}\,(b_{i}b_{j})=\begin{cases}0&(i\neq j)\\ 2^{n}&(i=j=0)\\ 2^{n+1}&(i=j>0)\end{cases}.

(7)

In this paper, we repeatedly use the following relations:

\displaystyle b_{0}b_{i}=b_{i},\quad b_{i}b_{j}=b_{i+j}+b_{i-j},\quad b_{i}^{2}=2+b_{2i}\qquad(1\leq i,j\leq 2^{n}-1,\ i\neq j),

where we regard that

\displaystyle b_{2^{n}}=0,\quad b_{-k}:=b_{k},\quad b_{2^{n}+k}:=-b_{2^{n}-k}\quad(1\leq k\leq 2^{n}-1).

The following conjecture and the partial results below are the main results in this paper.

Conjecture 2.2.

We define $c_{n}$ for $n\in\mathbb{N}$ by

	$\displaystyle c_{1}$	$\displaystyle:=2,$
	$\displaystyle c_{n}$	$\displaystyle:=2\cdot\mathrm{round}\,(2^{n}/5)=\begin{cases}2(2^{n}-1)/5&(n\equiv 0\mod 4)\\ 2(2^{n}-2)/5&(n\equiv 1\mod 4)\\ 2(2^{n}+1)/5&(n\equiv 2\mod 4)\\ 2(2^{n}+2)/5&(n\equiv 3\mod 4)\\ \end{cases}\qquad(n\geq 2).$

Here $\mathrm{round}\,(x)$ denotes the nearest integer to $x$ . Then we have

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}=2^{n}(1+8c_{n}).

For example, $c_{1}=2$ , $c_{2}=2$ , $c_{3}=4$ , $c_{4}=6$ , $c_{5}=12$ , $c_{6}=26$ , $c_{7}=52$ , $c_{8}=102$ , $c_{9}=204$ , $c_{10}=410$ . Hereinafter in this section, we present partial results (Theorems 2.3, 2.5 and 2.6) for 2.2. First, we generalize Morisawa-Okazaki’s Ineq. (4) a little.

Theorem 2.3.

We have for $n\geq 3$

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}\geq 2^{n}\cdot 33.

Proof.

Let $\epsilon\in RE_{n}^{+}$ , $\neq\pm 1$ . Write

\displaystyle\epsilon=\sum_{i=0}^{2^{n}-1}a_{i}b_{i}\quad(a_{i}\in\mathbb{Z}).

We have by $\epsilon\in RE_{n}^{+}$

\displaystyle\mathrm{N}_{n/n-1}\,\epsilon=\biggl{(}\sum_{2\mid i}a_{i}b_{i}\biggr{)}^{2}-\biggl{(}\sum_{2\nmid i}a_{i}b_{i}\biggr{)}^{2}=1.

(8)

[MO3, Lemma 6.2] states that

\displaystyle\text{$a_{0}$ is odd, $a_{i}$ ($i\neq 0$) are even}.

(9)

We claim that it suffices to show that

(a)

at least four $a_{i}$ ’s are not equal to $0$ for non-zero $i$ , or,
(b)

at least two $a_{i}$ ’s are not equal to $0$ for odd $i$ .

First we note that

\displaystyle\mathrm{Tr}\,\epsilon^{2}=2^{n}(a_{0}^{2}+2a_{1}^{2}+\cdots+2a_{2^{n}-1}^{2}).

by (7). The statement (a) implies the assertion since we have

\displaystyle 2^{n}(a_{0}^{2}+2a_{1}^{2}+\cdots+2a_{2^{n}-1}^{2})\geq 2^{n}(1+2\cdot 4\cdot 2^{2})=2^{n}\cdot 33

by (9). Now assume (b). By taking the trace of (8), we have

\displaystyle 2^{n}a_{0}^{2}+2^{n+1}\sum_{2\mid i\neq 0}a_{i}^{2}-2^{n+1}\sum_{2\nmid i}a_{i}^{2}=2^{n}.

It follows that

\displaystyle 2^{n}(a_{0}^{2}+2a_{1}^{2}+\cdots+2a_{2^{n}-1}^{2})=2^{n}+2^{n+2}\sum_{2\nmid i}a_{i}^{2}.

This is greater than or equal to $2^{n}+2^{n+2}\cdot 2\cdot 4=2^{n}\cdot 33$ by (9) and (b) as desired.

Recall that $\pm 1\neq\epsilon\in RE_{n}^{+}$ . In particular $\epsilon\in\mathbb{B}_{n}-\mathbb{B}_{n-1}$ , so at least one $a_{i}$ is not equal to $0$ for odd $i$ . We may assume $i=1$ by considering the Galois action. If there exists at least one more odd $i$ satisfying $a_{i}\neq 0$ , then (b) holds. Assume that $a_{i}=0$ for odd $i\neq 1$ . Then (8) becomes

\displaystyle\biggl{(}\sum_{2\mid i}a_{i}b_{i}\biggr{)}^{2}=(1+2a_{1}^{2})+a_{1}^{2}b_{2}.

By (9), we have $a_{0}\neq 0$ . There exists at least one more even $i_{1}$ satisfying $a_{i_{1}}\neq 0$ , because otherwise it follows that $a_{1}^{2}b_{2}=a_{0}^{2}-1-2a_{1}^{2}\in\mathbb{Z}$ . This is a contradiction for $b_{2}\in\mathbb{B}_{n-1}-\mathbb{B}_{n-2}$ and $n\geq 3$ . Once again, we see that there exists at least one more even $i_{2}$ satisfying $a_{i}\neq 0$ , because otherwise it follows that

\displaystyle(a_{0}^{2}+2a_{i_{1}}^{2})+2a_{0}a_{i_{1}}b_{i_{1}}+a_{i_{1}}^{2}b_{2i_{1}}=(1+2a_{1}^{2})+a_{1}^{2}b_{2}.

Then we have “ $b_{2}=b_{i_{1}}=-b_{2i_{1}}$ ” or “ $b_{2}=b_{i_{1}}$ , $b_{2i_{1}}\in\mathbb{Z}$ ”, that is, “ $i_{1}=2$ , $2i_{1}=2^{n+1}-2$ ” or “ $i_{1}=2$ , $2i_{1}=2^{n}$ ”, which is a contradiction for $n\geq 3$ .

Now $\epsilon$ has at least three non-zero coefficients $a_{1},a_{i_{1}},a_{i_{2}}$ with $2\mid i_{1},i_{2}$ , other than $a_{0}$ . We assume for the contradiction that these are all non-zero ones. In particular (8) becomes

	$\displaystyle(a_{0}^{2}+2a_{i_{1}}^{2}+2a_{i_{2}}^{2})+2a_{0}a_{i_{1}}b_{i_{1}}+2a_{0}a_{i_{2}}b_{i_{2}}+a_{i_{1}}^{2}b_{2i_{1}}+a_{i_{2}}^{2}b_{2i_{2}}+2a_{i_{1}}a_{i_{2}}b_{i_{1}+i_{2}}+2a_{i_{1}}a_{i_{2}}b_{i_{1}-i_{2}}$
	$\displaystyle=(1+2a_{1}^{2})+a_{1}^{2}b_{2}.$		(10)

We consider three cases: “ $i_{1}\equiv i_{2}\equiv 0\bmod 4$ ”, “ $i_{1}\equiv i_{2}\equiv 2\bmod 4$ ”, “ $i_{1}\equiv 2\bmod 4$ , $i_{2}\equiv 0\bmod 4$ ”. First assume that $i_{1}\equiv i_{2}\equiv 0\bmod 4$ . Then we have $2i_{1},2i_{2},i_{1}\pm i_{2}\equiv 0\bmod 4$ . Therefore there does not exist any term in the left-hand side corresponding to $a_{1}^{2}b_{2}$ in the right-hand side, which is a contradiction. Next assume that $i_{1}\equiv i_{2}\equiv 2\bmod 4$ . We have $2i_{1},2i_{2},i_{1}\pm i_{2}\equiv 0\bmod 4$ . Therefore the relation (10) implies

\displaystyle 2a_{0}a_{i_{1}}b_{i_{1}}+2a_{0}a_{i_{2}}b_{i_{2}}=a_{1}^{2}b_{2}.

This follows, for example, by considering the quotient vector space $\mathbb{B}_{n-1}/\mathbb{B}_{n-2}$ . Then we have $i_{1}=i_{2}=2$ , which is a contradiction. Finally assume that $i_{1}\equiv 2\bmod 4$ , $i_{2}\equiv 0\bmod 4$ . We have $i_{1}\pm i_{2}\equiv 2\bmod 4$ , $2i_{1},2i_{2}\equiv 0\bmod 4$ . Similarly as above we obtain

\displaystyle 2a_{0}a_{i_{1}}b_{i_{1}}+2a_{i_{1}}a_{i_{2}}b_{i_{1}+i_{2}}+2a_{i_{1}}a_{i_{2}}b_{i_{1}-i_{2}}=a_{1}^{2}b_{2}.

We have $i_{1}+i_{2}\notin\{\pm 2,\pm(2^{n+1}-2)\}$ by $2\leq i_{1}\leq 2^{n}-2$ , $4\leq i_{2}\leq 2^{n}-4$ . That is, $|b_{2}|\neq|b_{i_{1}+i_{2}}|$ . Then there are two possible cases:

\displaystyle|b_{2}|=|b_{i_{1}}|\neq|b_{i_{1}+i_{2}}|=|b_{i_{1}-i_{2}}|\quad\text{or}\quad|b_{2}|=|b_{i_{1}-i_{2}}|\neq|b_{i_{1}+i_{2}}|=|b_{i_{1}}|.

If the former one holds, then we have $i_{1}=2$ and $i_{1}+i_{2}=-(i_{1}-i_{2})$ , which is a contradiction. If the latter one holds, we have

\displaystyle 2a_{0}a_{i_{1}}b_{i_{1}}+2a_{i_{1}}a_{i_{2}}b_{i_{1}+i_{2}}=0,

which implies $|a_{0}|=|a_{i_{2}}|$ . This is a contradiction for (9). Then the assertion is clear. ∎

Remark 2.4.

(i)

The above proof is independent of any information of the class number $h_{n}$ of $\mathbb{B}_{n}$ . Oppositely, we show that 2.3 implies $h_{3}/h_{2}=1$ in §3.
(ii)

The strategy of the above proof is counting the number of non-zero coefficients $a_{i}$ of a relative unit $\epsilon=\sum_{i=0}^{2^{n}-1}a_{i}b_{i}\in RE_{n}^{+}$ by a combinatorial argument, and showing that the number is greater than or equal to $c_{k}$ if $n\geq k$ , for $k=3$ . The same proof works for $k=4$ , although we used a computer.

For small $n$ or even $n$ , we obtain (a candidate of) the minimal unit $\in RE_{n}^{+}$ explicitly.

Theorem 2.5.

For $n=1,3,5$ , we put

	$\displaystyle u_{1}$	$\displaystyle:=\varepsilon_{1}=3b_{0}+2b_{1},$
	$\displaystyle u_{3}$	$\displaystyle:=\varepsilon_{3}\sigma(\varepsilon_{3})=b_{0}+2(b_{1}+b_{2}+b_{5}+b_{6}),$
	$\displaystyle u_{5}$	$\displaystyle:=\varepsilon_{5}\sigma^{2}(\varepsilon_{5})$
		$\displaystyle=b_{0}+2(b_{11}+b_{12}-b_{14}-b_{15}+b_{17}+b_{18}+b_{19}+b_{20}-b_{22}-b_{23}+b_{25}+b_{26}).$

For $n\in 2\mathbb{N}$ , we put

\displaystyle u_{n}:=b_{0}+(-1)^{\frac{n}{2}}2\sum_{i=\lceil\frac{2^{n+1}}{5}\rceil}^{\lfloor\frac{2^{n+2}}{5}\rfloor}b_{i}.

Here $\lceil\ \rceil$ , $\lfloor\ \rfloor$ denote the ceiling function, the floor function, respectively. Then we have

\displaystyle\mathrm{Tr}\,u_{n}^{2}=2^{n}(1+8c_{n})\quad(n=1,3,5\text{ or }n\in 2\mathbb{N}).

Hence a “half” of 2.2 holds for such $n$ :

\displaystyle\min\{\mathrm{Tr}\,\epsilon^{2}\mid\pm 1\neq\epsilon\in RE_{n}^{+}\}\leq 2^{n}(1+8c_{n})\quad(n=1,3,5\text{ or }n\in 2\mathbb{N}).

Proof.

The cases $n=1,3,5$ follow from a direct calculation, by noting that (7) implies

\displaystyle\mathrm{Tr}\,\biggl{(}\sum_{i=0}^{2^{n}-1}c_{i}b_{i}\biggr{)}^{2}=2^{n}\biggl{(}c_{0}^{2}+2\sum_{i=1}^{2^{n}-1}c_{i}^{2}\biggr{)}\quad(c_{i}\in\mathbb{Z}).

(11)

For even $n$ , easily see that

\displaystyle c_{n}=\lfloor\tfrac{2^{n+2}}{5}\rfloor-\lceil\tfrac{2^{n+1}}{5}\rceil+1.

(12)

It follows that $\mathrm{Tr}\,u_{n}^{2}=2^{n}(1+8c_{n})$ by (11). Hence it suffices to show that $\mathrm{N}_{n/n-1}\,u_{n}=1$ . Let $s:=\lceil\frac{2^{n+1}}{5}\rceil$ , $t:=\lfloor\frac{2^{n+2}}{5}\rfloor$ , $b(n):=b_{n}$ . We can write

	$\displaystyle\mathrm{N}_{n/n-1}\,u_{n}-1$	$\displaystyle=\left(1+(-1)^{\frac{n}{2}}2\sum_{i=s}^{t}b(i)\right)\left(1+(-1)^{\frac{n}{2}}2\sum_{i=s}^{t}(-1)^{i}b(i)\right)-1$
		$\displaystyle=(-1)^{\frac{n}{2}}4\sum_{s\leq 2k\leq t}b(2k)+4\sum_{s\leq 2k\leq t}b(2k)^{2}+8\sum_{s\leq 2k<2l\leq t}b(2k)b(2l)$
		$\displaystyle\quad-4\sum_{s\leq 2k+1\leq t}b(2k+1)^{2}-8\sum_{s\leq 2k+1<2l+1\leq t}b(2k+1)b(2l+1).$

The sum of the second and forth terms in the most right-hand side is equal to

\displaystyle 4\sum_{s\leq 2k\leq t}(b(4k)+2)-4\sum_{s\leq 2k+1\leq t}(b(4k+2)+2)=4\sum_{s\leq 2k\leq t}b(4k)-4\sum_{s\leq 2k+1\leq t}b(4k+2).

since (12) implies that the parities of $s,t$ are even-odd or odd-even. The sum of the third and fifth terms is equal to

	$\displaystyle 8\sum_{s\leq 2k<2l\leq t}(b(2k+2l)+b(2k-2l))-8\sum_{s\leq 2k+1<2l+1\leq t}(b(2k+2l+2)+b(2k-2l))$
	$\displaystyle=8\sum_{s\leq 2k<2l\leq t}b(2k+2l)-8\sum_{s\leq 2k+1<2l+1\leq t}b(2k+2l+2),$

by the parities of $s,t$ again. Hence it suffices to show that

	$\displaystyle(-1)^{\frac{n}{2}}\sum_{s\leq 2k\leq t}b(2k)+\sum_{s\leq 2k\leq t}b(4k)-\sum_{s\leq 2k+1\leq t}b(4k+2)$
	$\displaystyle\quad+2\sum_{s\leq 2k<2l\leq t}b(2k+2l)-2\sum_{s\leq 2k+1<2l+1\leq t}b(2k+2l+2)$		(13)

is equal to $0$ . We divide it into two cases. First assume that $n\equiv 2\mod 4$ . Then $s$ is even and $t$ is odd. Therefore $s\leq 2k+1\leq t$ is equivalent to $s\leq 2k\leq t-1$ , and $s\leq 2k+1<2l+1\leq t$ is equivalent to $s\leq 2k<2l\leq t-1$ , respectively. Then (13) becomes

	$\displaystyle-\sum_{s\leq 2k\leq t-1}b(2k)+\sum_{s\leq 2k\leq t-1}b(4k)-\sum_{s\leq 2k\leq t-1}b(4k+2)$
	$\displaystyle\quad+2\sum_{s\leq 2k<2l\leq t-1}b(2k+2l)-2\sum_{s\leq 2k<2l\leq t-1}b(2k+2l+2).$

Since we have

	$\displaystyle\sum_{s\leq 2k<2l\leq t-1}b(2k+2l)-\sum_{s\leq 2k<2l\leq t-1}b(2k+2l+2)$	$\displaystyle=\sum_{s\leq 2k\leq t-1}b(4k+2)-\sum_{s+t+1\leq 2k\leq 2t}b(2k),$
	$\displaystyle\sum_{s\leq 2k\leq t-1}b(4k)+\sum_{s\leq 2k\leq t-1}b(4k+2)$	$\displaystyle=\sum_{2s\leq 2k\leq 2t}b(2k),$

the problem is reduced to showing that

\displaystyle-\sum_{s\leq 2k\leq t-1}b(2k)+\sum_{2s\leq 2k\leq 2t}b(2k)-2\sum_{s+t+1\leq 2k\leq 2t}b(2k)=0.

Let $c:=\frac{2^{n+2}}{10}$ . For even $n$ (not only for $n\equiv 2\bmod 4$ ), we see that

	$\displaystyle\bullet\text{ $s$ is the least integer $\geq c$, $t$ is the greatest integer $\leq 2c$},$
	$\displaystyle\bullet\text{ $2s$ is the least even integer $\geq 2c$, $2t$ is the greatest even integer $\leq 4c$},$		(14)
	$\displaystyle\bullet\text{ $s+t+1$ is the least even integer $\geq 3c$}.$

Therefore the left-hand side becomes

	$\displaystyle-\sum_{c\leq 2k\leq 2c}b(2k)+\sum_{2c\leq 2k\leq 4c}b(2k)-2\sum_{3c\leq 2k\leq 4c}b(2k)$
	$\displaystyle=-\sum_{c\leq 2k\leq 2c}b(2k)+\sum_{2c\leq 2k\leq 3c}b(2k)-\sum_{3c\leq 2k\leq 4c}b(2k)$

The last sum is equal to $0$ since we have

	$\displaystyle\sum_{c\leq 2k\leq 2c}b(2k)+\sum_{3c\leq 2k\leq 4c}b(2k)=0,$
	$\displaystyle\sum_{2c\leq 2k\leq 3c}b(2k)=\sum_{2c\leq 2k<2^{n}}b(2k)+b(2^{n})+\sum_{2^{n}<2k<3c}b(2k)=0$

by $b(2^{n}+k)=-b(2^{n}-k)$ and $b(2^{n})=0$ .

Next assume that $n\equiv 0\mod 4$ , which implies $s$ is odd and $t$ is even. Then (13) becomes

	$\displaystyle\sum_{s+1\leq 2k\leq t}b(2k)+\sum_{s+1\leq 2k\leq t}b(4k)-\sum_{s-1\leq 2k\leq t-2}b(4k+2)$
	$\displaystyle\quad+2\sum_{s+1\leq 2k<2l\leq t}b(2k+2l)-2\sum_{s-1\leq 2k<2l\leq t-2}b(2k+2l+2).$		(15)

In this case we have

	$\displaystyle\sum_{s+1\leq 2k<2l\leq t}b(2k+2l)-\sum_{s-1\leq 2k<2l\leq t-2}$	$\displaystyle b(2k+2l+2)$
		$\displaystyle=\sum_{s-1\leq 2k\leq t-2}b(4k+2)-\sum_{2s\leq 2k\leq s+t-1}b(2k),$
	$\displaystyle\sum_{s+1\leq 2k\leq t}b(4k)+\sum_{s-1\leq 2k\leq t-2}b(4k+2)$	$\displaystyle=\sum_{2s\leq 2k\leq 2t}b(2k).$

Hence (15) is equal to

\displaystyle\sum_{s+1\leq 2k\leq t}b(2k)+\sum_{2s\leq 2k\leq 2t}b(2k)-2\sum_{2s\leq 2k\leq s+t-1}b(2k).

(16)

By (14), we can rewrite (16) as

	$\displaystyle\sum_{c\leq 2k\leq 2c}b(2k)+\sum_{2c\leq 2k\leq 4c}b(2k)-2\sum_{2c\leq 2k\leq 3c}b(2k)$
	$\displaystyle=\sum_{c\leq 2k\leq 2c}b(2k)+\sum_{3c\leq 2k\leq 4c}b(2k)-\sum_{2c\leq 2k\leq 3c}b(2k),$

which is equal to $0$ by $b(2^{n}+k)=-b(2^{n}-k)$ and $b(2^{n})=0$ . Then the assertion is clear. ∎

We obtain the following corollary by Ineq. (4), Theorems 2.3, 2.5 (and a trivial argument for $n=1$ ).

Corollary 2.6.

2.2 holds true for $n=1,2,3$ .

2.2 should be proved without studying the class number $h_{n}$ of $\mathbb{B}_{n}$ , as we seen above. On the other hand, we have

\displaystyle k_{n}:=\frac{h_{n}}{h_{n-1}}=[RE_{n}^{+}:A_{n}].

(17)

This follows from, for example, [Wa, Theorem 8.2, Proposition 8.11], [H2, (1), (4)]. For a proof, see [Yo, §4.1]. Besides, we have $h_{n}=1$ for $n\leq 6$ , so $RE_{n}^{+}=A_{n}$ for the same $n$ . Since $A_{n}$ is given explicitly, we can verify 2.2 numerically for such $n$ , as follows.

Assume that $u\in RE_{n}^{+}$ satisfies

\displaystyle\mathrm{Tr}\,u^{2}\leq 2^{n}(1+8c_{n}).

(18)

We put

\displaystyle x_{i}:=\log|\sigma^{i}(u)|\in\mathbb{R}\quad(0\leq i\leq 2^{n-1}-1).

Since $\mathrm{N}_{n/n-1}\,\tau(u)=\tau(u)\sigma^{2^{n-1}}(\tau(u))=1$ for $\tau\in G_{n}$ , the inequality (18) turns into

\displaystyle\sum_{i=0}^{2^{n-1}-1}(e^{2x_{i}}+e^{-2x_{i}})\leq 2^{n}(1+8c_{n}).

We consider the logarithmic embedding

\displaystyle RE_{n}^{+}/\{\pm 1\}\hookrightarrow\mathbb{R}^{2^{n-1}},\quad\epsilon\mapsto(\log(|\sigma^{i}(\epsilon)|))_{i=0,1,\dots,2^{n-1}-1}.

Then the square of the length of the image of $u$ is given by

\displaystyle\sum_{i=0}^{2^{n-1}-1}(\log|\sigma^{i}(u)|)^{2}=\sum_{i=0}^{2^{n-1}-1}x_{i}^{2}.

We put

\displaystyle L_{n}:=\max\left\{\sum_{i=0}^{2^{n-1}-1}x_{i}^{2}\,\middle|\,x_{i}\in\mathbb{R},\ \sum_{i=0}^{2^{n-1}-1}(e^{2x_{i}}+e^{-2x_{i}})\leq 2^{n}(1+8c_{n})\right\}.

Namely, the condition (18) implies

\displaystyle\sum_{i=0}^{2^{n-1}-1}(\log|\sigma^{i}(u)|)^{2}\leq L_{n}.

(19)

Now we assume that $RE_{n}^{+}=A_{n}$ , which is equivalent to $k_{n}:=\frac{h_{n}}{h_{n-1}}=1$ by (17) ( $n\leq 6$ is a sufficient condition). Then we may write

\displaystyle u=\prod_{j=0}^{2^{n-1}-1}\sigma^{j}(\varepsilon_{n})^{n_{j}}\quad(n_{j}\in\mathbb{Z}).

Therefore (19) is equivalent to

	$\displaystyle M[\mathbf{n}]:={}^{t}\mathbf{n}M\mathbf{n}\leq L_{n},$		(20)
	$\displaystyle M:=\left[\sum_{k=0}^{2^{n-1}-1}\log\|\sigma^{k+i}(\varepsilon_{n})\|\log\|\sigma^{k+j}(\varepsilon_{n})\|\right]_{i,j=0,1,\dots,2^{n-1}-1},$
	$\displaystyle\mathbf{n}:=[n_{i}]_{i=0,1,\dots,2^{n-1}-1}.$

We can find all such vectors $\mathbf{n}$ by the Fincke-Pohst algorithm (actually, we used the command qfminim of PARI/GP). Here the value of $L_{n}$ is given as follows: Assume that $x_{i}$ satisfies $\sum_{i=0}^{2^{n-1}-1}(e^{2x_{i}}+e^{-2x_{i}})=2^{n}a$ for a fixed $a$ . Note that $a\geq 1$ since $y+y^{-1}\geq 2$ for $y\in\mathbb{R}$ . Then the Lagrange multiplier theorem says that the function $\sum_{i=0}^{2^{n-1}-1}x_{i}^{2}$ takes the maximum value only when $(x_{i})_{i}=\lambda(e^{2x_{i}}-e^{-2x_{i}})_{i}$ for some $\lambda\in\mathbb{R}$ . The solutions of $x=\lambda(e^{2x}-e^{-2x})$ are of the form of $x=\pm b$ with $b\geq 0$ since $e^{2x}-e^{-2x}$ ( $x\geq 0$ ) is a convex function and $x,e^{2x}-e^{-2x}$ are odd functions. It follows that $e^{2x_{i}}+e^{-2x_{i}}$ is constant for all $i$ , that is, $e^{2x_{i}}+e^{-2x_{i}}=2a$ . Namely, $\sum_{i=0}^{2^{n-1}-1}x_{i}^{2}$ takes the maximum value when

\displaystyle x_{i}=\pm\frac{\log(a-\sqrt{a^{2}-1})}{2}.

Therefore we see that

	$\displaystyle L_{n}$	$\displaystyle=\max\left\{2^{n-3}\left(\log\left(a-\sqrt{a^{2}-1}\right)\right)^{2}\,\middle\|\,1\leq a\leq 1+8c_{n}\right\}$
		$\displaystyle=2^{n-3}\left(\log\left(1+8c_{n}-\sqrt{16c_{n}+64c_{n}^{2}}\right)\right)^{2}$

In fact, we have $L_{1}=3.107\ldots$ , $L_{2}=6.214\ldots$ , $L_{3}=17.55\ldots$ , $L_{4}=42.04\ldots$ , $L_{5}=111.0\ldots$ , $L_{6}=291.4\ldots$ , $L_{7}=723.8\ldots$ .

When $n\leq 6$ , we confirmed that $u$ does not satisfies $\mathrm{Tr}\,u^{2}<2^{n}(1+8c_{n})$ for any $\mathbf{n}\neq\mathbf{0}$ satisfying (20): for example, let $n=6$ . Then the number of vectors $\mathbf{n}\neq\mathbf{0}$ satisfying (20) is $290624$ . We computed $\mathrm{Tr}\,(\prod_{i=0}^{2^{5}-1}\sigma(\varepsilon_{n})^{n_{i}})^{2}$ for such $\mathbf{n}$ and checked that the minimal value is equal to $2^{6}(1+8c_{6})$ . To summarize, by numerical computation and by using $k_{n}=1$ , we have the following.

Theorem 2.7.

2.2 holds true for $n\leq 6$ .

Remark 2.8.

(i)

When $n\leq 6$ , all $\epsilon\in RE_{n}^{+}$ satisfying $\mathrm{Tr}\,\epsilon^{2}=2^{n}(1+8c_{n})$ are the conjugates of $u_{n}$ given in 2.5.
(ii)

We can not confirm the case $n>6$ due to the limit of computer power.

3 Relation to $h_{n}=1$ when $n\leq 3$

There are many partial results supporting Weber’s class number problem obtained by studying the unit group. More directly, the second author proved the following. We put $k_{n}:=\frac{h_{n}}{h_{n-1}}$ , where $h_{n}$ denotes the class number of $\mathbb{B}_{n}$ .

Theorem 3.1 ([Yo, Remark in §3.3]).

Ineq. (4) implies $k_{2}=1$ .

In this section, we generalize this result to $n=3$ as follows.

Theorem 3.2.

2.3 implies $k_{3}=1$ .

Proof.

Assume for contradiction that there exists $u\in RE^{+}_{3}-A_{3}$ . Since $[RE^{+}_{3}:A_{3}]<\infty$ , we can write

\displaystyle u:=\prod_{j=0}^{3}\sigma^{j}(\varepsilon_{3})^{x_{j}},\quad(x_{0},x_{1},x_{2},x_{3})\in\mathbb{Q}^{4}-\mathbb{Z}^{4}.

We may replace $(x_{0},x_{1},x_{2},x_{3})$ with $(x_{0}^{\prime},x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime})$ so that $x_{i}\equiv x_{i}^{\prime}\bmod\mathbb{Z}$ . Therefore, by putting

	$\displaystyle T(x_{0},x_{1},x_{2},x_{3}):=\sum_{i=0}^{7}\prod_{j=0}^{3}\|\sigma^{i+j}(\varepsilon_{3})\|^{2x_{j}},$
	$\displaystyle L:=\min_{\alpha_{0}\in\mathbb{R}}\max_{\alpha_{0}\leq x_{0}\leq\alpha_{0}+1}\min_{\alpha_{1}\in\mathbb{R}}\max_{\alpha_{1}\leq x_{1}\leq\alpha_{1}+1}\min_{\alpha_{2}\in\mathbb{R}}\max_{\alpha_{2}\leq x_{2}\leq\alpha_{2}+1}\min_{\alpha_{3}\in\mathbb{R}}\max_{\alpha_{3}\leq x_{3}\leq\alpha_{3}+1}T(x_{0},x_{1},x_{2},x_{3}),$

it suffices to show that

\displaystyle L<2^{3}(1+8c_{3})=264

(21)

since we have

\displaystyle T(x_{0},x_{1},x_{2},x_{3})=\mathrm{Tr}\,u^{2}\geq 264

for $\pm 1\neq u\in RE_{3}^{+}$ by 2.3.

First we show that $T(x_{0},x_{1},x_{2},x_{3})$ is a convex function. In particular, a set

\displaystyle S:=\{(x_{0},x_{1},x_{2},x_{3},y)\mid y\geq T(x_{0},x_{1},x_{2},x_{3})\}

is convex. We can reduce it to the convexity of a function of the form $a^{x_{0}}b^{x_{1}}c^{x_{2}}d^{x_{3}}$ since the sum of convex functions is again convex. Its Hessian matrix is equal to

\displaystyle 4a^{2x_{0}}b^{2x_{1}}c^{2x_{2}}d^{2x_{3}}\begin{pmatrix}(\log a)^{2}&\log a\log b&\log a\log c&\log a\log d\\ \log a\log b&(\log b)^{2}&\log b\log c&\log b\log d\\ \log a\log c&\log b\log c&(\log c)^{2}&\log c\log d\\ \log a\log d&\log b\log d&\log c\log d&(\log d)^{2}\end{pmatrix},

whose eigenvalues and eigenvectors are

	$\displaystyle 0,0,0,4a^{2x_{0}}b^{2x_{1}}c^{2x_{2}}d^{2x_{3}}((\log a)^{2}+(\log b)^{2}+(\log c)^{2}+(\log d)^{2}),$
	$\displaystyle(-\log b,\log a,0,0),(-\log c,0,\log a,0),(-\log d,0,0,\log a),(\log a,\log b,\log c,\log d).$

Therefore $a^{x_{0}}b^{x_{1}}c^{x_{2}}d^{x_{3}}$ is convex since the eigenvalues are non-negative.

By the convexity of $T$ (in particular, the convexity with respect to $x_{3}$ ) we can write

$\displaystyle T_{3}(x_{0},x_{1},x_{2})$	$\displaystyle:=\min_{\alpha_{3}\in\mathbb{R}}\max_{\alpha_{3}\leq x_{3}\leq\alpha_{3}+1}T(x_{0},x_{1},x_{2},x_{3})$
	$\displaystyle=\min_{\alpha_{3}\in\mathbb{R}}\max\{T(x_{0},x_{1},x_{2},\alpha_{3}),T(x_{0},x_{1},x_{2},\alpha_{3}+1)\}$	(22)
	$\displaystyle=T(x_{0},x_{1},x_{2},\alpha)$

for a unique $\alpha$ satisfying

\displaystyle T(x_{0},x_{1},x_{2},\alpha)=T(x_{0},x_{1},x_{2},\alpha+1).

Now we clam that $T_{3}(x_{0},x_{1},x_{2})$ is again convex: namely we have for $t\in[0,1]$

\displaystyle T_{3}(ta_{0}+(1-t)b_{0},ta_{1}+(1-t)b_{1},ta_{2}+(1-t)b_{2})\leq tT_{3}(a_{0},a_{1},a_{2})+(1-t)T_{3}(b_{0},b_{1},b_{2}).

Say

	$\displaystyle T_{3}(a_{0},a_{1},a_{2})=T(a_{0},a_{1},a_{2},\alpha)=T(a_{0},a_{1},a_{2},\alpha+1),$
	$\displaystyle T_{3}(b_{0},b_{1},b_{2})=T(b_{0},b_{1},b_{2},\beta)=T(b_{0},b_{1},b_{2},\beta+1).$

Moreover we put

\displaystyle c_{i}:=ta_{i}+(1-t)b_{i}\ (i=0,1,2),\quad c_{3}:=t\alpha+(1-t)\beta.

Since

	$\displaystyle(a_{0},a_{1},a_{2},\alpha,T_{3}(a_{0},a_{1},a_{2})),$	$\displaystyle(b_{0},b_{1},b_{2},\beta,T_{3}(b_{0},b_{1},b_{2})),$
	$\displaystyle(a_{0},a_{1},a_{2},\alpha+1,T_{3}(a_{0},a_{1},a_{2})),$	$\displaystyle(b_{0},b_{1},b_{2},\beta+1,T_{3}(b_{0},b_{1},b_{2}))$

are elements of the convex set $S$ , so are

	$\displaystyle(c_{0},c_{1},c_{2},c_{3},tT_{3}(a_{0},a_{1},a_{2})+(1-t)T_{3}(b_{0},b_{1},b_{2})),$
	$\displaystyle(c_{0},c_{1},c_{2},c_{3}+1,tT_{3}(a_{0},a_{1},a_{2})+(1-t)T_{3}(b_{0},b_{1},b_{2})).$

Namely we have

\displaystyle tT_{3}(a_{0},a_{1},a_{2})+(1-t)T_{3}(b_{0},b_{1},b_{2})\geq T(c_{0},c_{1},c_{2},c_{3}),T(c_{0},c_{1},c_{2},c_{3}+1).

Hence, by (22), we have

	$\displaystyle tT_{3}(a_{0},a_{1},a_{2})+(1-t)T_{3}(b_{0},b_{1},b_{2})$	$\displaystyle\geq\max\{T(c_{0},c_{1},c_{2},c_{3}),T(c_{0},c_{1},c_{2},c_{3}+1)\}$
		$\displaystyle\geq T_{3}(c_{0},c_{1},c_{2})$

as desired. By repeating the same argument, we can write

	$\displaystyle T_{2}(x_{0},x_{1}):=\min_{\alpha_{2}\in\mathbb{R}}\max_{\alpha_{2}\leq x_{2}\leq\alpha_{2}+1}T_{3}(x_{0},x_{1},x_{2})=T_{3}(x_{0},x_{1},\alpha)$
	$\displaystyle\text{for}\ \alpha\ \text{with}\ T_{3}(x_{0},x_{1},\alpha)=T_{3}(x_{0},x_{1},\alpha+1),$
	$\displaystyle T_{1}(x_{0}):=\min_{\alpha_{1}\in\mathbb{R}}\max_{\alpha_{1}\leq x_{1}\leq\alpha_{1}+1}T_{2}(x_{0},x_{1})=T_{2}(x_{0},\alpha^{\prime})$
	$\displaystyle\text{for}\ \alpha^{\prime}\ \text{with}\ T_{2}(x_{0},\alpha^{\prime})=T_{2}(x_{0},\alpha^{\prime}+1).$

We easily obtain an upper bound of such minimal values as follows. Consider a closed-interval $I=[a,b]$ and divide it into $N+1$ pieces:

\displaystyle A:=\{a,a+(b-a)/N,a+2(b-a)/N,\dots,b-(b-a)/N,b\}.

Then we see that

	$\displaystyle T_{3}(x_{0},x_{1},x_{2})$	$\displaystyle=\min_{\alpha_{3}\in\mathbb{R}}\max\{T(x_{0},x_{1},x_{2},\alpha_{3}),T(x_{0},x_{1},x_{2},\alpha_{3}+1)\}$
		$\displaystyle\leq\min_{\alpha_{3}\in A}\max\{T(x_{0},x_{1},x_{2},\alpha_{3}),T(x_{0},x_{1},x_{2},\alpha_{3}+1)\}.$

By repeating similar arguments, we obtain an upper bound of $L$ as

\displaystyle L\leq\min_{\alpha_{0}\in A}\max_{x_{0}=\alpha_{0},\alpha_{0}+1}\min_{\alpha_{1}\in A}\max_{x_{1}=\alpha_{1},\alpha_{1}+1}\min_{\alpha_{2}\in A}\max_{x_{2}=\alpha_{2},\alpha_{2}+1}\min_{\alpha_{3}\in A}\max_{x_{3}=\alpha_{3},\alpha_{3}+1}T(x_{0},x_{1},x_{2},x_{3}).

Now, we put $[a,b]:=[\frac{-101}{100},\frac{99}{100}]$ , $N:=32$ . Then numerically we have

	$\displaystyle t(\tfrac{-404}{400})=887.4\dots,\ \cdots,\ t(\tfrac{-229}{400})=312.9\dots,t(\tfrac{-204}{400})=260.8\dots,t(\tfrac{-179}{400})=241.1\dots,$
	$\displaystyle\cdots,\ t(\tfrac{171}{400})=239.1\dots,t(\tfrac{196}{400})=259.0\dots,t(\tfrac{221}{400})=308.8\dots,\ \cdots,\ t(\tfrac{396}{400})=1094.5\dots.$

where we put

\displaystyle t(\alpha_{0}):=\min_{\alpha_{1}\in A}\max_{x_{1}=\alpha_{1},\alpha_{1}+1}\min_{\alpha_{2}\in A}\max_{x_{2}=\alpha_{2},\alpha_{2}+1}\min_{\alpha_{3}\in A}\max_{x_{3}=\alpha_{3},\alpha_{3}+1}T(\alpha_{0},x_{1},x_{2},x_{3}).

Hence we obtain $L\leq\max\{t(\frac{-204}{400}),t(\frac{196}{400})\}=260.8\dots$ as desired. ∎

Remark 3.3.

Summarizing the proof of 3.2, we showed that there exists a fundamental domain $D$ of $\mathbb{R}^{2^{n-1}}$ modulo $\mathbb{Z}^{2^{n-1}}$ satisfying

\displaystyle\max\left\{\sum_{i=0}^{2^{n}-1}\prod_{j=0}^{2^{n-1}-1}|\sigma^{i+j}(\varepsilon_{n})|^{2x_{j}}\mid(x_{i})_{i}\in D\right\}<2^{n}(1+8c_{n})

for $n=3$ , by considering the $\mathbb{Z}$ -module structure of $RE_{+}^{n}$ . When $n\geq 4$ , it seems to have to consider its Galois module structure, not only the $\mathbb{Z}$ -module structure, in order to studying the relation between 2.2 and the class number. We provide some partial (and numerical) results in the proceeding sections.

4 $l$ -Indivisibility of $h_{n}$ by numerical calculations

We give a demonstration of numerical checks of the $l$ -indivisibility of $k_{n}$ for several $(l,n)$ , by using 2.2. More powerful results can be seen in [H1, H2, FK1, FK2, FK3, MO1, MO2], including (3). Let $l$ be an odd prime. We put

\displaystyle A_{n}^{\frac{1}{l}}

\displaystyle:=\{x\in\mathbb{R}\mid x^{l}\in A_{n}\}.

Since $A_{n}/\{\pm 1\}$ is a free abelian group generated by $\{\sigma^{i}(\varepsilon_{n})\mid i=0,\dots,2^{n-1}-1\}$ , we may identify the following three $\mathbb{F}_{l}[G_{n}]$ -modules

\displaystyle\begin{array}[]{ccccc}\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)&\cong&A_{n}^{\frac{1}{l}}/A_{n}&\cong&\mathbb{F}_{l}^{2^{n}-1},\\ \rotatebox[origin={c}]{90.0}{$\in$}&&\rotatebox[origin={c}]{90.0}{$\in$}&&\rotatebox[origin={c}]{90.0}{$\in$}\\ \displaystyle\overline{\sum_{i=0}^{2^{n-1}-1}a_{i}x^{i}}&\leftrightarrow&\displaystyle\overline{\prod_{i=0}^{2^{n-1}-1}\sigma^{i}(\varepsilon_{n})^{\frac{a_{i}}{l}}}&\leftrightarrow&(a_{i})_{0\leq i\leq 2^{n-1}-1}.\end{array}

Here $\sigma$ acts on $\mathbb{F}_{l}^{2^{n}-1}$ by

\displaystyle\sigma(a_{0},a_{1},a_{2},\dots,a_{2^{n}-2},a_{2^{n}-1})=(-a_{2^{n}-1},a_{0},a_{1},\dots,a_{2^{n}-3},a_{2^{n}-2}).

$\mathbb{F}_{l}[G_{n}]$ acts on $\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)$ via

\displaystyle\mathbb{F}_{l}[G_{n}]\stackrel{{\scriptstyle\sigma\mapsto x}}{{\cong}}\mathbb{F}_{l}[x]/(x^{2^{n}}-1)\twoheadrightarrow\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1),

and hence we may also consider $\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1),A_{n}^{\frac{1}{l}}/A_{n},\mathbb{F}_{l}^{2^{n}-1}$ are $\mathbb{F}_{l}[x]$ -modules where $x$ acts as $\sigma$ .

By the Chinese remainder theorem, the irreducible decomposition of $\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)$ as a $\mathbb{F}_{l}[x]$ -module is given as

	$\displaystyle\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)=\bigoplus_{f_{i}}M_{f},$
	$\displaystyle M_{f_{i}}:=\tfrac{x^{2^{n-1}}+1}{f_{i}}\cdot\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)\ (\cong\mathbb{F}_{l}^{\deg f_{i}}),$

where $f_{i}$ runs over all irreducible polynomial $f_{i}\in\mathbb{F}_{l}[x]$ dividing $x^{2^{n-1}}+1$ . (Note that $x^{2^{n-1}}+1\bmod l$ has no multiple roots.) Taking a polynomial $g_{f_{i}}\in\mathbb{F}_{l}[x]$ satisfying $\tfrac{x^{2^{n-1}}+1}{f_{i}}\cdot g_{f_{i}}\equiv 1\mod f_{i}$ , the idempotent map is given explicitly as

\displaystyle e_{f_{i}}\colon\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)\twoheadrightarrow M_{f_{i}},\quad h\mapsto\tfrac{x^{2^{n-1}}+1}{f_{i}}g_{f_{i}}h.

Now, we assume that $l\mid k_{n}$ . It follows that there exists $\epsilon\in RE_{n}^{+}$ satisfying

\displaystyle\epsilon\notin A_{n},\quad\epsilon^{l}\in A_{n}.

This element corresponds to a non-trivial element $\overline{g_{\epsilon}}\in\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)\cong A_{n}^{\frac{1}{l}}/A_{n}$ . Then we can take $f_{i}$ so that $e_{f_{i}}(\overline{g_{\epsilon}})\neq\overline{0}$ since $\sum_{i}e_{f_{i}}(\overline{g_{\epsilon}})=\overline{g_{\epsilon}}\neq\overline{0}$ . For such $f_{i}$ , the whole of $M_{f_{i}}$ is contained in $RE_{n}^{+}/A_{n}$ , since we have

\displaystyle\begin{array}[]{cccc}\mathbb{F}_{l}[x]/(x^{2^{n-1}}+1)&\cong&A_{n}^{\frac{1}{l}}/A_{n}\\ \cup&&\cup\\ M_{f_{i}}&\cong&\left\{g\cdot\overline{\varepsilon_{n}}\in A_{n}^{\frac{1}{l}}/A_{n}\mid g\in M_{f_{i}}\right\}\\ \rotatebox[origin={c}]{90.0}{$=$}&&\rotatebox[origin={c}]{90.0}{$=$}\\ \mathbb{F}_{l}[x]\cdot e_{f_{i}}(\overline{g_{\epsilon}})&\cong&\left(\tfrac{x^{2^{n-1}}+1}{f_{i}}g_{f_{i}}\mathbb{F}_{l}[x]\right)\cdot\overline{\epsilon}&\subset RE_{n}^{+}/A_{n}.\end{array}

(28)

For $g=\sum_{i=0}^{m}a_{i}x^{i}\in\mathbb{Z}[x]$ (not only for elements $\in\mathbb{F}_{l}[x]$ ), we put

\displaystyle g\cdot\varepsilon_{n}:=\prod_{i=0}^{m}\sigma^{i}(\varepsilon_{n})^{\frac{a_{i}}{l}}\in A_{n}^{\frac{1}{l}}.

Then the following proposition follows form (28).

Proposition 4.1.

Assume that an odd prime $l$ divides $k_{n}$ . Then there exists an irreducible polynomial $f\in\mathbb{F}_{l}[x]$ dividing $x^{2^{n-1}}+1$ satisfying

\displaystyle\left\{\overline{g\cdot\varepsilon_{n}}\in A_{n}^{\frac{1}{l}}/A_{n}\;\middle|\;g\in\mathbb{Z}[x]\text{ with }g\bmod(x^{2^{n-1}}+1)\in M_{f}\right\}\subset RE_{n}^{+}.

We extend the trace map to

\displaystyle\widetilde{\mathrm{Tr}}\,\big{(}\epsilon^{\frac{2}{l}}\big{)}:=\sum_{i=0}^{2^{n}-1}\left(\sigma^{i}(\epsilon)^{2}\right)^{\frac{1}{l}}\quad(\epsilon\in A_{n}).

By the above proposition, 2.2 can be used for a numerical check of the indivisibility of the class numbers as follows.

Theorem 4.2.

Assume that 2.2 holds true for $n$ . If for each irreducible polynomial $f\in\mathbb{F}_{l}[x]$ dividing $x^{2^{n-1}}+1$ there exists $g\in\mathbb{Z}[x]$ satisfying

	$\displaystyle g\bmod(l,x^{2^{n-1}}+1)\in M_{f}-\{0\},$
	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g\cdot\varepsilon_{n}\right)^{2}\right)<2^{n}(1+8c_{n}),$

then we have $l\nmid k_{n}$ .

4.1 The case $n=4,5$

Example 4.3.

Let $n=4$ , $l<10^{6}$ . For each irreducible polynomial $f\in\mathbb{F}_{l}[x]$ dividing $x^{2^{3}}+1$ , we took the center lift $g$ of a suitable element in $M_{f}$ and confirmed that

\displaystyle\mathrm{Tr}\,\left(\left(g\cdot\varepsilon_{4}\right)^{2}\right)<2^{4}(1+8c_{4})=784.

(29)

Namely, by 4.2, we checked that 2.2 implies $l\nmid k_{4}$ for $l<10^{6}$ .

For example, let $l=3$ . Then the irreducible decomposition of $x^{8}+1\bmod 3$ is given by

\displaystyle x^{8}+1\equiv f_{1}f_{2}\mod 3,\quad f_{1}=x^{4}+x^{2}-1,\quad f_{2}=x^{4}-x^{2}-1.

We choose elements $\frac{x^{8}+1}{f_{i}}\in M_{f_{i}}$ $(i=1,2)$ and take their center lifts $g_{1}:=x^{4}-x^{2}-1$ , $g_{2}:=x^{4}+x^{2}-1$ . Then, by numerical computation, we obtain

	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{1}\cdot\varepsilon_{4}\right)^{2}\right)=\widetilde{\mathrm{Tr}}\,\left(\left(\varepsilon_{4}^{-1}\sigma^{2}(\varepsilon_{4})^{-1}\sigma^{4}(\varepsilon_{4})\right)^{\frac{2}{3}}\right)=95.6\dots,$
	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{2}\cdot\varepsilon_{4}\right)^{2}\right)=\widetilde{\mathrm{Tr}}\,\left(\left(\varepsilon_{4}^{-1}\sigma^{2}(\varepsilon_{4})\sigma^{4}(\varepsilon_{4})\right)^{\frac{2}{3}}\right)=100.1\dots.$

These values satisfy the condition (29) for $3\nmid k_{4}$ .

Next, let $l=7$ . Then we have

	$\displaystyle x^{8}+1\equiv f_{1}f_{2}f_{3}f_{4}\mod 7,$
	$\displaystyle f_{1}=x^{2}+x-1,\quad f_{2}=x^{2}+3x-1,\quad f_{3}=x^{2}-3x-1,\quad f_{4}=x^{2}-x-1.$

First we take center lifts $g_{i}$ of $\frac{x^{8}+1}{f_{i}}\bmod 7$ :

	$\displaystyle g_{1}=x^{6}-x^{5}+2x^{4}-3x^{3}-2x^{2}-x-1,\quad g_{2}=x^{6}-3x^{5}+3x^{4}+2x^{3}-3x^{2}-3x-1,$
	$\displaystyle g_{3}=x^{6}+3x^{5}+3x^{4}-2x^{3}-3x^{2}+3x-1,\quad g_{4}=x^{6}+x^{5}+2x^{4}+3x^{3}-2x^{2}+x-1.$

Then we have

	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{1}\cdot\varepsilon_{4}\right)^{2}\right)=106.5\dots,\quad\widetilde{\mathrm{Tr}}\,\left(\left(g_{2}\cdot\varepsilon_{4}\right)^{2}\right)=546.9\dots,$
	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{3}\cdot\varepsilon_{4}\right)^{2}\right)=840.6\dots,\quad\widetilde{\mathrm{Tr}}\,\left(\left(g_{4}\cdot\varepsilon_{4}\right)^{2}\right)=160.2\dots.$

Note that the case $i=3$ does not satisfy the condition (29). Replacing $g_{3}$ with the center lift $g_{3}^{\prime}=2x^{6}-x^{5}-x^{4}+3x^{3}+x^{2}-x-2$ of $2\cdot\frac{x^{8}+1}{f_{i}}\bmod 7$ , we have

\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{3}^{\prime}\cdot\varepsilon_{4}\right)^{2}\right)=200.7\dots,

which implies $7\nmid k_{4}$ .

Example 4.4.

Let $n=5$ , $l<10^{5}$ , $\neq 97,193,257$ . Then, similarly as in 4.3, the center lift $g$ of a suitable element in $M_{f}$ for each $f$ satisfies the condition of 4.2:

\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g\cdot\varepsilon_{5}\right)^{2}\right)<2^{5}(1+8c_{5})=3104.

We also check the exceptions $97,193,257$ by taking certain non-center lifts.

Let $l=97$ . Then we have

$\displaystyle x^{2^{4}}+1\equiv\prod_{i=1}^{16}f_{i}\mod 97,$
$\displaystyle f_{1}=x+19,$	$\displaystyle f_{2}=x+20,$	$\displaystyle f_{3}=x+28,$	$\displaystyle f_{4}=x+30,$	$\displaystyle f_{5}=x+34,$
$\displaystyle f_{6}=x+42,$	$\displaystyle f_{7}=x+45,$	$\displaystyle f_{8}=x+46,$	$\displaystyle f_{9}=x-46,$	$\displaystyle f_{10}=x-45,$
$\displaystyle f_{11}=x-42,$	$\displaystyle f_{12}=x-34,$	$\displaystyle f_{13}=x-30,$	$\displaystyle f_{14}=x-28,$	$\displaystyle f_{15}=x-20,$
$\displaystyle f_{16}=x-19.$

For $i=1,4,5,6,8,9,11,13,14$ , we put $g_{i}$ to be the center lift of $4\cdot\frac{x^{2^{4}}+1}{f_{i}}$ . Then we have

$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{1}\cdot\varepsilon_{5}\right)^{2}\right)=1123.9\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{4}\cdot\varepsilon_{5}\right)^{2}\right)=1429.9\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{5}\cdot\varepsilon_{5}\right)^{2}\right)=2421.7\dots,$
$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{6}\cdot\varepsilon_{5}\right)^{2}\right)=1632.8\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{8}\cdot\varepsilon_{5}\right)^{2}\right)=2332.6\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{9}\cdot\varepsilon_{5}\right)^{2}\right)=1291.7\dots,$
$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{11}\cdot\varepsilon_{5}\right)^{2}\right)=1537.1\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{13}\cdot\varepsilon_{5}\right)^{2}\right)=1492.2\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{14}\cdot\varepsilon_{5}\right)^{2}\right)=1444.4\dots.$

For $i=2,3,7,10,12,15,16$ , we have to take non-center lifts. Hereinafter, we write a polynomial $\sum_{i=0}^{k}a_{i}x^{i}$ as a vector $[a_{0},\dots,a_{k}]$ for saving pages. We put

	$\displaystyle g_{2}$	$\displaystyle:=[34,8,19,33,42,27,-45,-22,-28,-18,30,\underline{-50},-46,12,-20,1],$
	$\displaystyle g_{3}$	$\displaystyle:=[7,24,13,3,38,\underline{61},29,44,40,-43,5,31,37,16,41,2],$
	$\displaystyle g_{7}$	$\displaystyle:=[41,-16,37,-31,5,43,40,\underline{53},29,36,38,-3,13,-24,7,2],$
	$\displaystyle g_{10}$	$\displaystyle:=[-41,-16,-37,-31,-5,\underline{-54},-40,-44,-29,36,-38,-3,-13,-24,-7,2],$
	$\displaystyle g_{12}$	$\displaystyle:=[17,\underline{49},10,6,23,-25,-15,-9,14,-11,-26,-35,-21,-32,39,4],$
	$\displaystyle g_{15}$	$\displaystyle:=[29,16,-38,-31,\underline{-84},-43,-7,-44,-41,-36,37,-3,-5,24,40,2],$
	$\displaystyle g_{16}$	$\displaystyle:=[5,36,7,31,-29,24,37,\underline{53},13,16,-40,3,41,43,38,2],$

which are lifts of $\frac{x^{2^{4}}+1}{f_{2}},2\cdot\frac{x^{2^{4}}+1}{f_{3}},2\cdot\frac{x^{2^{4}}+1}{f_{7}},2\cdot\frac{x^{2^{4}}+1}{f_{10}},4\cdot\frac{x^{2^{4}}+1}{f_{12}},2\cdot\frac{x^{2^{4}}+1}{f_{15}},2\cdot\frac{x^{2^{4}}+1}{f_{16}}$ respectively. Here components with underlining are not contained in $[-\frac{l-1}{2},\frac{l-1}{2}]$ . Then we have

$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{2}\cdot\varepsilon_{5}\right)^{2}\right)=1492.1\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{3}\cdot\varepsilon_{5}\right)^{2}\right)=1963.0\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{7}\cdot\varepsilon_{5}\right)^{2}\right)=1548.9\dots,$
$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{10}\cdot\varepsilon_{5}\right)^{2}\right)=920.6\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{12}\cdot\varepsilon_{5}\right)^{2}\right)=1831.2\dots,$	$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{15}\cdot\varepsilon_{5}\right)^{2}\right)=2985.0\dots,$
$\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g_{16}\cdot\varepsilon_{5}\right)^{2}\right)=2386.1\dots.$

The other cases $l=193,257$ can be done similarly.

Remark 4.5.

Let $n=6$ . Then, for many $l$ (e.g., $l=31,97,127,193,223,257,449,\dots$ ), the center lift $g$ of any element in $M_{f}$ does not satisfy the condition

\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g\cdot\varepsilon_{6}\right)^{2}\right)<2^{6}(1+8c_{6})=13376.

Moreover searching all non-center lifts is difficult due to the high dimension. We confirmed that 2.2 implies that $l\nmid k_{6}$ only for $l=31$ .

4.2 The case $n=7$ , $l>10^{9}$ , $l\equiv 65\bmod 128$

If $n,l$ are large, it is difficult to check the condition in 4.2. However that becomes relatively easy in some special cases. Let $n=7$ , $l\equiv 65\bmod 128$ . We note that such $l$ are out of the range of (6). Then the irreducible decomposition of $x^{2^{6}}+1\bmod l$ is in the form

\displaystyle x^{2^{6}}+1\bmod l=\prod_{i=1}^{32}(x^{2}+a_{i}).

(30)

In fact, that $l\equiv 65\bmod 128$ is equivalent to that $l$ splits completely in $\mathbb{Q}(\zeta_{6})$ and does not in $\mathbb{Q}(\zeta_{7})$ where $\zeta_{n}:=e^{\frac{2\pi i}{2^{n}}}$ . Then $y^{2^{5}}+1$ , which is a minimal polynomial of $\mathbb{Q}(\zeta_{6})$ , decomposes a product of polynomials of degree $1$ modulo $l$ , and $x^{2^{6}}+1$ of $\mathbb{Q}(\zeta_{7})$ does not. Considering $y=x^{2}$ , we obtain the expression (30). Since half of the coefficients of $\frac{x^{2^{6}}+1}{x^{2}+a}$ are equal to $0$ (that is, $\frac{x^{2^{6}}+1}{x^{2}+a}$ is in the form $\sum_{i=0}^{31}c_{2i}x^{2i}$ ), the value of $\widetilde{\mathrm{Tr}}\,((g\cdot\varepsilon_{7})^{2})$ tends to “small” if we take a center lift $g$ of $\frac{x^{2^{6}}+1}{x^{2}+a}$ multiplied by a constant. For example, let $l=1000000321$ , which is the least prime satisfying $l>10^{9}$ , $l\equiv 65\bmod 128$ . Note that this case is not contained in (6). Then we have

	$\displaystyle x^{2^{6}}+1\bmod l=\prod_{i=1}^{32}(x^{2}+a_{i}),$
	$\displaystyle a_{1}=30063488,\ a_{2}=30912022,\ a_{3}=42483948,\ a_{4}=59955883,\ a_{5}=78186285,$
	$\displaystyle a_{6}=160612070,\ a_{7}=191346380,\ a_{8}=246360387,\ a_{9}=268629094,\ a_{10}=269645956,$
	$\displaystyle a_{11}=280492327,\ a_{12}=303644312,\ a_{13}=311722386,\ a_{14}=424439170,$
	$\displaystyle a_{15}=441230693,\ a_{16}=447503416,\ a_{16+i}=-a_{17-i}\ (1\leq i\leq 16).$

We put $g_{i}$ to be the center lift of $b_{i}\cdot\frac{x^{2^{6}}+1}{x^{2}+a_{i}}$ with

	$\displaystyle b_{1}=231,\ b_{2}=231,\ b_{3}=867,\ b_{4}=125,\ b_{5}=386,\ b_{6}=231,\ b_{7}=100,\ b_{8}=100,$
	$\displaystyle b_{9}=64,\ \ b_{10}=36,\ b_{11}=702,\ b_{12}=771,\ b_{13}=231,\ b_{14}=2069,\ b_{15}=349,\ b_{16}=64,$
	$\displaystyle b_{17}=64,\ b_{18}=64,\ b_{19}=4,\ b_{20}=64,\ b_{21}=686,\ b_{22}=105,\ b_{23}=167,\ b_{24}=64,$
	$\displaystyle b_{25}=100,\ b_{26}=89,\ b_{27}=100,\ b_{28}=100,\ b_{29}=100,\ b_{30}=100,\ b_{31}=100,\ b_{32}=64.$

Then $t_{i}:=\widetilde{\mathrm{Tr}}\,((g_{i}\cdot\varepsilon_{7})^{2})$ are calculated numerically as follows.

	$\displaystyle t_{1}=24947.7\dots,\ t_{2}=15616.7\dots,\ t_{3}=49165.2\dots,\ t_{4}=23454.0\dots,\ t_{5}=46028.1\dots,$
	$\displaystyle t_{6}=41400.4\dots,\ t_{7}=19344.5\dots,\ t_{8}=26943.5\dots,\ t_{9}=42868.4\dots,\ t_{10}=40913.4\dots,$
	$\displaystyle t_{11}=44067.7\dots,\ t_{12}=49457.9\dots,\ t_{13}=18759.3\dots,\ t_{14}=39188.3\dots,$
	$\displaystyle t_{15}=35939.1\dots,\ t_{16}=44713.3\dots,\ t_{17}=41782.1\dots,\ t_{18}=47974.8\dots,$
	$\displaystyle t_{19}=52445.8\dots,\ t_{20}=49841.0\dots,\ t_{21}=43256.3\dots,\ t_{22}=52244.6\dots,$
	$\displaystyle t_{23}=49338.6\dots,\ t_{24}=22229.3\dots,\ t_{25}=36290.0\dots,\ t_{26}=48593.0\dots,$
	$\displaystyle t_{27}=26438.3\dots,\ t_{28}=40208.3\dots,\ t_{29}=23006.2\dots,\ t_{30}=19831.0\dots,$
	$\displaystyle t_{31}=16060.6\dots,\ t_{32}=42470.9\dots.$

These values satisfy the condition in 4.2:

\displaystyle\widetilde{\mathrm{Tr}}\,\left(\left(g\cdot\varepsilon_{7}\right)^{2}\right)<2^{7}(1+8c_{7})=53376.

Namely, 2.2 for $n=7$ implies $l=1000000321\nmid k_{7}$ . Similarly, we checked that 2.2 for $n=7$ implies the $l$ -indivisibility of $k_{7}$ for first $1000$ primes satisfying

\displaystyle 10^{9}<l,\ l\equiv 65\bmod 128,

form $1000000321$ to $1001287361$ .

References

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Minimal relative units of the cyclotomic ℤ2\mathbb{Z}_{2}-extension

Abstract

1 Introduction

Conjecture 1.1 ([MO3, Conjecture 1.1]).

Theorem 1.2 ([MO3, Theorem 6.4]).

Conjecture (2.2).

Theorem (2.5).

Theorem (2.7).

Theorem (2.3).

Theorem (3.2).

2 Minimal relative units

Definition 2.1.

Conjecture 2.2.

Theorem 2.3.

Proof.

Remark 2.4.

Theorem 2.5.

Proof.

Corollary 2.6.

Theorem 2.7.

Remark 2.8.

3 Relation to hn=1h_{n}=1 when n≤3n\leq 3

Theorem 3.1 ([Yo, Remark in §3.3]).

Theorem 3.2.

Proof.

Remark 3.3.

4 ll-Indivisibility of hnh_{n} by numerical calculations

Proposition 4.1.

Theorem 4.2.

4.1 The case n=4,5n=4,5

Example 4.3.

Example 4.4.

Remark 4.5.

4.2 The case n=7n=7, l>109l>10^{9}, l≡65mod128l\equiv 65\bmod 128

References

Minimal relative units of the cyclotomic $\mathbb{Z}_{2}$ -extension

3 Relation to $h_{n}=1$ when $n\leq 3$

4 $l$ -Indivisibility of $h_{n}$ by numerical calculations

4.1 The case $n=4,5$

4.2 The case $n=7$ , $l>10^{9}$ , $l\equiv 65\bmod 128$