The 2-Sylow subgroup of $K_2O_F$ for certain quadratic number fields

Abstract

In this paper, we apply Qin’s theorem for the 4-rank of $K_2O_F$ to establish the relation between the 4-rank of the ideal class group of $F=\mathbb {Q}(\sqrt{d})$ and the 4-rank of $K_2O_F$ provided that all odd prime factors of d are congruent to 1 mod 8. As an application, we give a concise and unified proof of two conjectures proposed by Conner and Hurrelbrink (Acta Arith 73:59–65, 1995).

1 Introduction

Let F be a number field, $O_F$ the ring of integers in F. Let $K_2O_F$ denote the Milnor group of $O_F$, which coincides with the tame kernel of F. As a finite abelian group, $K_2O_F$ contains rich arithmetical information. In particular, the Birch–Tate conjecture and the Lichtenbaum conjecture establish a relation between the order of $K_2O_F$ and the value $\zeta _F^*(-1)$, where $\zeta _F(\cdot )$ is the Dedekind zeta-function of F. To understand $K_2O_F$, we need to know the $p^n$-rank of $K_2O_F$ for any prime p and any positive integer n. If F contains a primitive $p^n$-th root of unity, the $p^n$-rank of $K_2O_F$ formula is given by Tate [13], in particular, the 2-rank of $K_2O_F$ formula is given for any number field. See also Keune [6].

When $F=\mathbb {Q}(\sqrt{d})$ is a quadratic field, Browkin and Schinzel [3] give an explicit formula for the 2-rank of $K_2O_F$. Qin [9,10,11] obtains formulas for the 4-rank of $K_2O_F$ and the 8-rank of $K_2O_F$. See also [8, 12]. In this paper, we apply Qin’s theorem for the 4-rank of $K_2O_F$ to establish a relation between the 4-rank of ideal class group and the 4-rank of $K_2O_F$ provided that all odd prime factors of d are congruent to 1 mod 8. As an application, we give a concise and unified proof of two conjectures proposed by Conner and Hurrelbrink in [4]. Both conjectures are proved by Vazzana [14, 15], but the proofs, which use graph theory, are somewhat involved.

2 Statement of main theorems

We first state main theorems of this paper. For any abelian group A,  let $r_4(A)$ denote the 4-rank of A.

Theorem 2.1

Let $E=\mathbb {Q}(\sqrt{d})$, where $d = p_1\ldots p_k$ with $p_i\equiv 1\pmod {8}$ and let C(E) be the class group of E. Assume that

(i):

the norm of the fundamental unit of E is $-1,$ and

(ii):

an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 32y^2.$ Then

$$\begin{aligned} r_4(K_2O_E)=r_4(C(E)). \end{aligned}$$

Theorem 2.2

Let $F=\mathbb {Q}(\sqrt{2d})$ and $E=\mathbb {Q}(\sqrt{d})$, where $d = p_1\ldots p_k$ with $p_i\equiv 1\pmod {8}$ and let C(E) be the class group of E. Assume that

(i):

the norm of the fundamental unit of E is $-1,$ and

(ii):

an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 64y^2.$ Then

$$\begin{aligned} r_4(K_2O_F)=r_4(C(E)). \end{aligned}$$

Theorem 2.3

Let $F=\mathbb {Q}(\sqrt{2d})$, where $d = p_1\ldots p_k$ with $p_i\equiv 1\pmod {8}$ and let C(F) be the class group of F. Assume that

(i):

the norm of the fundamental unit of F is $-1,$ and

(ii):

an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 64y^2.$ Then

$$\begin{aligned} r_4(K_2O_F)=r_4(C(F))-1. \end{aligned}$$

As corollaries of Theorems 2.1 and 2.2, we have the following two theorems.

Theorem 2.4

The 2-primary part of $K_2(\mathcal {O}_E)$ is elementary abelian if and only if

(i):

the 2-primary part of the ideal class group C(E) is elementary abelian and the norm of the fundamental unit of E is $-1,$ and

(ii):

an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 32y^2.$

Theorem 2.4 gives an explicit set of conditions under which the 2-primary part of $K_2(\mathcal {O}_E)$ is elementary abelian. A similar theorem for the field $F= \mathbb {Q}(\sqrt{2d})$ is as follows.

Theorem 2.5

The 2-primary part of $K_2(\mathcal {O}_F)$ is elementary abelian if and only if

(i):

the 2-primary part of the ideal class group C(E) is elementary abelian and the norm of the fundamental unit of E is $-1,$ and

(ii):

an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 64y^2.$

Both of these two theorems were conjectured by Conner and Hurrelbrink in [4]. Vazzana proved the two theorems in [14] and [15], respectively. In particular, his proof for Theorem 2.5 is complicated. He made use of a graph associated with the primes $p_1,\ldots , p_k$ and studied its relationship to a new graph associated with the primes lying over $p_1,\ldots , p_k$ in $\mathbb {Q}(\sqrt{-2}).$ Our alternative proof is not only unified but also very concise.

3 Prelimilary results

In this section, we review some known results which are useful in this paper. The following result is classical. One can find a proof in [5].

Lemma 3.1

(Legendre’s Theorem) Suppose that a, b, c are square-free,

$$\begin{aligned} (a,\; b)=(a,\;c)=(b,\;c)=1 \end{aligned}$$

and a, b, c do not have the same sign. Then the Diophantine equation

$$\begin{aligned} aX^2+bY^2+cZ^2=0 \end{aligned}$$

has nontrivial solutions if and only if for every odd prime p|abc,  say, p|a,  $\left( \frac{-bc}{p}\right) =1.$

We adopt the notation from [11]. Let $d\ne 0$ be an integer. Put

$$\begin{aligned} \ S(d)= \left\{ \begin{array}{ll} \{\pm 1, \pm 2\} &{} \mathrm{if}\; d > 0, \\ \{1, 2\} &{} \mathrm{if}\; d < 0. \end{array} \right. \end{aligned}$$

(1)

For any abelian group A,  set

$$\begin{aligned} _{2}A=\{x\in A | x^2=1\}. \end{aligned}$$

For a number field F,  we use NF for the set of norms from F over $\mathbb {Q}.$

Let F be a number field with $r_2$ complex places. By Tate [13], every element of order two in $K_2F$ is of the form $\{-1,\ a\}$, where $a\in F^{*}$. Let $\Delta$ denote the Tate kernel of F, i.e., the group of elements $x\in F^{*}$ such that $\{-1,\ x\}=1$. If $\sqrt{2}\not \in F$, then by [13],

$$\begin{aligned} (\Delta :(F^{*})^2)=2^{r_2+1}. \end{aligned}$$

In particular, if $F\not =\mathbb {Q}(\sqrt{2})$ is a real quadratic field, then $\Delta =F^*\cup 2F^*.$

Theorem 3.2

(Browkin and Schinzel) Let $F=\mathbb {Q}(\sqrt{d}), d\in \mathbb {Z}$ square-free. Then $_{2}K_2(O_F)$ can be generated by

$$\begin{aligned} \{-1, m\},\;\; m|d, \end{aligned}$$

together with

$$\begin{aligned} \{-1,u_i+\sqrt{d}\}, \end{aligned}$$

if $\{-1, \pm 2\}\cap NF\ne \emptyset ,$ where $u_i\in \mathbb {Z}$ is such that $u_i^2-d=c_iw_i^2$ for some $w_i\in \mathbb {Z}$ and $c_i\in \{-1, \pm 2\}\cap NF.$

Remark

The general $r_2(K_2O_F)$ formula for an arbitrary number field F is given by Tate [13].

For any integer n,  put $\nabla ^n=\{\alpha \in K_2O_F| \alpha =\beta ^n \;\mathrm{for} \;\mathrm{some} \; \beta \in K_2O_F\}.$

Our main tool is the following theorem of Qin, which completely determines the 4-rank of $K_2O_F$ for any quadratic field (see [9, 10]).

Theorem 3.3

(Qin) Let $F=\mathbb {Q}(\sqrt{d}), d\in \mathbb {Z}$ square-free. Suppose that m|d $(m>0 \;{if} \;d>0,$ ,but m also takes on negative values if $d< 0)$ and write $d=u^2-2w^2$ with $u, w\in \mathbb {Z}$ if $2\in NF.$ Then $\{-1,m\}\in \nabla ^2$ if and only if one can find an $\epsilon \in S(d)$ such that

(i):

$\left( \frac{\frac{d}{m}}{p}\right) =\left( \frac{\epsilon }{p}\right)$ for every odd prime p|m; 

(ii):

$\left( \frac{m}{p}\right) =\left( \frac{\epsilon }{p}\right)$ for every odd prime $p|\frac{d}{m};$and $\{-1,m(u+\sqrt{d})\}\in \nabla ^2$ if and only if one can find an $\epsilon \in S(d)$ such that

(iii):

$\left( \frac{\frac{d}{m}}{p}\right) =\left( \frac{\epsilon (u+w)}{p}\right)$ for every odd prime p|m; 

(iv):

$\left( \frac{m}{p}\right) =\left( \frac{\epsilon (u+w)}{p}\right)$ for every odd prime $p|\frac{d}{m}.$

Corollary 3.4

Let $d = p_1\ldots p_k$ be a product of different primes congruent to 1 mod 8,  and let $F=\mathbb {Q}(\sqrt{d})$ or $F=\mathbb {Q}(\sqrt{2d})$. If $m\mid d$, then $\{-1,m\}\in \nabla ^2$ if and only if

$$\begin{aligned} Z^2=mX^2+\frac{d}{m}Y^2 \end{aligned}$$

is solvable in $\mathbb {Z}$; $\{-1,m(u+\sqrt{d})\}\in \nabla ^2$ if and only if

$$\begin{aligned} (u+w)Z^2=mX^2+\frac{d}{m}Y^2 \end{aligned}$$

is solvable in $\mathbb {Z}.$

Proof

Since for every odd prime p dividing d,  $\left( \frac{\pm 1}{p}\right) =\left( \frac{\pm 2}{p}\right) =1,$ the result follows from Theorem 3.3 and Lemma 3.1. $\square$

4 Proofs of main results

Lemma 4.1

Let $d = p_1\ldots p_k$ be a product of rational primes congruent to 1 mod 8. Assume that $d=u^2-2w^2, v=u+w(u>0).$ Then $v\equiv 3\pmod {4}$ if and only if an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 32y^2.$

Proof

First we assume that $k=1.$ There exist two integers u, w such that $p=u^2-2w^2$. Assume that $u>0$ and write $v=u+w$. Since $u^2-2w^2=(3u+4w)^2-2(2u+3w)^2$ and $p\equiv 1\pmod {8}$, by a suitable choice of u, w,  we can write $p=u^2-2(4w^\prime )^2$. Hence $v=u+4w^\prime \equiv u\pmod {4}$. By [1], we have that $v\equiv 1\pmod {4}$ if and only if $p=x^2 + 32y^2$, equivalently, $v\equiv 3\pmod {4}$ if and only if $p\not =x^2 + 32y^2.$

We consider now the general case. Assume that $d_1=u_1^2-2w_1^2,$ $d_2=u_2^2-2w_2^2$ and $d_1d_2=u^2-2w^2$. Then $d_1d_2=(u_1u_2+2w_1w_2)^2-2(u_1w_2+u_2w_1)^2$. If we assume further that $d_1\equiv d_2\equiv 1\pmod {4}$, then both $w_1$ and $w_2$ are even. We have

$$\begin{aligned} (u_1u_2+2w_1w_2)+(u_1w_2+u_2w_1)\equiv (u_1+w_1)(u_2+w_2)\pmod {4}, \end{aligned}$$

(2)

i.e., $v=u+w\equiv v_1v_2\pmod {4},$ where $v_1=u_1+w_1, v_2=u_2+w_2.$ Now suppose that $p_1\ldots p_k=u^2-2w^2, v=u+w(u>0).$ For any $1\le i\le k$, assume that $p_i=u_i^2-2w_i^2~(u_i>0)$ and write $v_i=u_i+w_i.$ By induction,

$$\begin{aligned} v\equiv v_1\ldots v_k\pmod {4}. \end{aligned}$$

Note that $v_i\equiv 1$ or $3\pmod {4}$ and $v_i\equiv 1\pmod {4}$ if and only if $p_i=x^2 + 32y^2.$ Therefore, $v\equiv 3\pmod {4}$ if and only if an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 32y^2.$ $\square$

Lemma 4.2

Let $p_1,\ldots , p_k$ be rational primes congruent to 1 mod 8. Let $d = p_1\ldots p_k$ or $2p_1\ldots p_k$, let $E=\mathbb {Q}(\sqrt{d}),$ and let C(E) be the class group of E. Write $\varepsilon$ for the fundamental unit of $E=\mathbb {Q}(\sqrt{d})$. Then

(i):

$N\varepsilon =1$ if and only if the following Diophantine equation is solvable for some $m\mid d$, $m\ne \pm 1$ or $\pm d$,

$$\begin{aligned} mX^2-\frac{d}{m}Y^2=\pm 1. \end{aligned}$$

(3)

(ii):

if $N\varepsilon =-1$ then $_2C(E)$ can be generated by the ideals $\wp _1,\ldots ,\wp _k$, where for any i, $\wp _i^2=p_i$.

Proof

  

• (i) We apply the result on the Tate kernel of real quadratic fields to show that the norm of the fundamental unit of $E=\mathbb {Q}(\sqrt{d})$ is 1 if and only if

  $$\begin{aligned} mX^2-\frac{d}{m}Y^2=\pm 1 \end{aligned}$$

  is solvable for some $m\mid d,$ $m\ne \pm 1,\pm d.$ In fact, since $\{-1, \varepsilon \}\in \ _2K_2(O_E),$ by Theorem 3.2, there exists some $m\mid d,$ $m\ne \pm 1,\pm d$ such that $\{-1, \varepsilon \}=\{-1,m\}$ or $\{-1, \varepsilon \}=\{-1,m(u+\sqrt{d})\}.$ It is straightforward to see that

  $$\begin{aligned} \varepsilon m(u+\sqrt{d})\not \in E^{*2}\cup 2E^{*2}. \end{aligned}$$

  Clearly, we have $\varepsilon m\not \in 2E^{*2}$. Hence, $\varepsilon m\in E^{*2}$. Therefore, $m^2=\varepsilon m{{\bar{\varepsilon }} m}=(X^2-dY^2)^2$ for some integers X, Y, so we obtain $\pm m=X^2-dY^2$, i.e., (3) holds.

  Conversely, if $\pm m=X^2-dY^2$, $m\ne \pm 1,\pm d,$ we put

  $$\begin{aligned} \varepsilon =\frac{X+\sqrt{d}Y}{X-\sqrt{d}Y}. \end{aligned}$$

  It is obvious that $(m,Y)=1.$ Hence, for any finite prime P of E, if $P\mid m$, then $v_P(X+\sqrt{d}Y)= v_P(X-\sqrt{d}Y)=1$; if $P\not \mid m$, then $v_P(X+\sqrt{d}Y)= v_P(X-\sqrt{d}Y)=0$. Therefore, $\varepsilon$ is an algebraic integer. Moreover, $\varepsilon$ is not a square, since $\varepsilon (X-\sqrt{d}Y)^2=X^2-dY^2=\pm m,$  $m\ne \pm 1,\pm d,$ is not a square.

• (ii) By genus theory, the 2-rank of C(E) is $k-1$ if $E=\mathbb {Q}(\sqrt{p_1\ldots p_k}),$ and k if $E=\mathbb {Q}(\sqrt{2p_1\ldots p_k})$. For any $m\mid d$, the ideal $(m,\sqrt{d})$ satisfies that $(m,\sqrt{d})^2=(m)$. Moreover, $(m,\sqrt{d})$ is principal in C(E) if and only if

  $$\begin{aligned} \pm m=X^2-dY^2 \end{aligned}$$

  (4)

  is solvable over $\mathbb {Z}$.

  But the assumption that $N\varepsilon =-1$ implies that for any $m\mid d$, $m\ne \pm 1,\pm d,$ (4) has no integer solution. Therefore, the ideals $\wp _1,\ldots ,\wp _k$ generate $_2C(E).$

$\square$

Proof of Theorem 2.1

We apply Theorem 3.3 to compute $r_4(K_2O_E)$.

We first show that the assumption $\mathrm{(ii)}$ implies $\{-1,m(u+\sqrt{d})\}\not \in \nabla ^2.$

Recall that in our case, all prime factors are congruent to 1 mod 8. By Corollary 3.4 to theorem of Qin, the condition that $\{-1,m(u+\sqrt{d}\}\in \nabla ^2$ is equivalent to the fact that there exists a positive factor m of d, so that

$$\begin{aligned} vmZ^2=X^2+dY^2 \end{aligned}$$

(5)

is solvable in $\mathbb {Z}$. Since $d\equiv 1\pmod {8}, m\mid d,$ $X^2+dY^2\equiv 0,1,2\pmod {4}.$ Now the assumption $\mathrm{(ii)}$ and Lemma 4.1 imply that $v\equiv 3\pmod {4}.$ Hence, $vmZ^2\equiv 0,3\pmod {4}$ and so (5) has no solution with $(X,Y,Z)=1$.

Therefore, in $_2K_2O_E$ only $\{-1,m\}$ could belong to $\nabla ^2.$ By sending $\{-1,m\}$ to the ideal $\mathfrak {M} (\mathfrak {M}^2=m),$ we obtain a bijection from the set $\{\{-1,m\},m\mid d\} \subseteq$ $_2K_2O_E$ to $_2C(E).$

By [7], we see that $\mathfrak {M}$ is a square in C(E) if and only if

$$\begin{aligned} mZ^2=X^2-dY^2 \end{aligned}$$

(6)

is solvable in $\mathbb {Z}$. Furthermore, $\{-1,m\}$ is a square in $K_2O_E$, i.e., $\{-1,m\}\in \nabla ^2$ if and only if

$$\begin{aligned} mZ^2=X^2+dY^2 \end{aligned}$$

(7)

is solvable in $\mathbb {Z}.$ As (6) and (7) have the same solvability,

$$\begin{aligned} r_4(K_2O_F)=r_4(C(F)) \end{aligned}$$

holds. $\square$

To prove Theorem 2.2, we need the following Lemma 4.5, which is an analog of Lemma 4.1. First we have

Lemma 4.3

Let $\alpha =2^{\frac{1}{4}}$ and $L=\mathbb {Q}(\alpha )$. Then $\mathbb {Z}[\alpha ]$ is the ring of algebraic integers of L. The class number of L is 1.

Proof

This can be done by the GP calculator. $\square$

Lemma 4.4

Let $p\equiv 1\pmod {8}$ be a prime. Assume that $p=u^2-2w^2$ with $u>0.$ Then $p=x^2+64y^2$ if and only if $u\equiv 1,3\pmod {8}$.

Proof

We need the following result: Let $p\equiv 1\pmod {8}$ be a prime. Then $\gamma ^4 \equiv 2\pmod {p}$ is solvable if and only if $p=x^2+64y^2$. See, for example, Theorem 7.5.2 in [2]. Hence, for a prime $p=x^2+64y^2$, $\gamma ^4 \equiv 2\pmod {p}$ has four distinct solutions. It follows that p splits completely in L. Hence, there are integers a, b, c, d such that

$$\begin{aligned} p=N_{L/\mathbb {Q}}(a+b\alpha +c\alpha ^2+d\alpha ^3)=(a^2+2c^2-4bd)^2-2(b^2+2d^2-2ac)^2. \end{aligned}$$

Since $p\equiv 1\pmod {8}$, we see that a is odd and $2\mid b$. Hence we have $u=a^2+2c^2-4bd\equiv 1,3\pmod {8}.$

Conversely, if $p=u^2-2w^2$ with $|u|\equiv 1,3\pmod {8}$ and w even, then $p=x^2+64y^2.$ In fact, we may write $u=t^2u^\prime , w=2^es^2w^\prime$, where $u^\prime ,w^\prime$ are square-free odd and positive integers. For any prime $l\mid w^\prime$, $(\frac{p}{l})=1$, hence, $w\equiv \alpha ^2\pmod {p}$ since 2 is a square $\pmod {p}$. On the other hand, for any prime $l\mid u^\prime$, $(\frac{-2p}{l})=1$, hence, $|u|\equiv 1,3\pmod {8}$ implies that $u \equiv \alpha ^2\pmod {p}.$ Therefore, $\gamma ^4 \equiv 2\pmod {p}$ is solvable and so $p=x^2+64y^2$. $\square$

Lemma 4.5

Let $d = 2p_1\ldots p_k$ with $p_1\equiv \cdots \equiv p_k \equiv 1\pmod {8}.$ Assume that $p_i=u_i^2-2w_i^2, v_i=u_i+w_i(u_i>0).$ If we write $p_1\ldots p_k=u^2-2w^2,$ and $2p_1\ldots p_k=U^2-2W^2, U>0,$ then $u\equiv u_1\ldots u_k\pmod {8}$ and $U+W\equiv u$ or $3u\pmod {8}$. Moreover, $U+W\equiv 5,$ or $7\pmod {8}$ if and only if an odd number of the primes $p_1,\ldots , p_k$ fail to be represented over $\mathbb {Z}$ by the quadratic form $x^2 + 64y^2.$

Proof

Assume that $a=u^2-2w^2$ and $b=\mu ^2-2\omega ^2$. Then we have

$$\begin{aligned} ab=(u\mu +2w\omega )^2-2(u\omega +\mu w)^2. \end{aligned}$$

(8)

Note that if both $a,b\equiv 1\pmod {8}$, it is easy to see w and $\omega$ are even, so $u\mu +2w\omega \equiv u\mu \pmod {8}$. By induction, we have $u\equiv u_1\ldots u_k\pmod {8}.$ Observe that

$$\begin{aligned} 2a=(2u+2w)^2-2(u+2w)^2. \end{aligned}$$

(9)

Since $(2u+2w)+(-u-2w)=u$ and $(2u+2w)+(u+2w)=3u+4w\equiv 3u\pmod {8},$ $U+W\equiv u$ or $3u\pmod {8}.$

If $k=1$ and $d=2 p_1=U^2-2W^2,$ by Lemma 4.4, $p_1\ne x^2 + 64y^2$ if and only if $U+W\equiv 5,7\pmod {8}.$ Observe that if $u_1\equiv 1,3\pmod {8},u_2\equiv 1,3\pmod {8}$ or $u_1\equiv 5,7\pmod {8},u_2\equiv 5,7\pmod {8},$ then $u_1u_2\equiv 1,3\pmod {8}$, and if $u_1\equiv 1,3\pmod {8},u_2\equiv 5,7\pmod {8},$ then $u_1u_2\equiv 5,7\pmod {8}.$ Therefore, by (8) and (9), if an odd number of the primes $p_1,\ldots , p_k$ fail to be represented by $x^2 + 64y^2,$ then $U+W\equiv 5,$ or $7\pmod {8},$ and vice versa. $\square$

Proof of Theorem 2.2

As in the proof of Theorem 2.1, we apply again Theorem 3.3 to compute $r_4(K_2O_F)$.

Write $2d=u^2-2w^2, ~u>0,~v=u+w$. We claim that for any $m\mid 2d$, $\{-1,m(u+\sqrt{2d})\}\not \in \nabla ^2.$

Indeed, by Theorem 3.3 and Lemma 3.1, if for some positive factor m of 2d, $\{-1,m(u+\sqrt{2d})\}\in \nabla ^2$, then

$$\begin{aligned} vmZ^2=X^2+2dY^2 \end{aligned}$$

(10)

is solvable in $\mathbb {Z}$. By Lemma 3.1, we can assume that m is odd. Since $2d\equiv 2\pmod {8},$ one has $X^2+2dY^2\equiv 1,2,3,6\pmod {8}$ if $(X,Y)=1$. It follows from Lemma 4.5 that the assumption $\mathrm{(ii)}$ implies that $v\equiv 5,7\pmod {4}.$ Hence, $vmZ^2\equiv 0,4,5,7\pmod {8}$ and so (10) has no solution with $(X,Y,Z)=1$.

Note that $\{-1,2m\}=\{-1,m\}$. Therefore, the same discussion as in the proof of Theorem 2.1 shows that

$$\begin{aligned} r_4(K_2O_F)=r_4(C(E)). \end{aligned}$$

$\square$

Proof of Theorem 2.3

The assumption that the norm of the fundamental unit of F is $-1$ implies that there is an ideal $\wp$ of F, which is non-principal with $\wp ^2=(2)$. But in $K_2O_F$, $\{-1,2\}=1$. Then the proof is analogous to that of Theorem 2.2. $\square$

We turn to prove Theorems 2.4 and 2.5. Let $m\mid d$. We introduce the following notation.

$$\begin{aligned} V(m,p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k), \end{aligned}$$

where

$$\begin{aligned} \ \epsilon _i= \left\{ \begin{array}{ll} \left( \frac{\frac{d}{m}}{p_i}\right) &{} \mathrm{if\;} p_i\mid m, \\ \left( \frac{m}{p_i}\right) &{} \mathrm{if\;} p_i\not \mid m. \end{array} \right. \end{aligned}$$

(11)

It is easy to check that $\sqcap _{i=1}^{k}\epsilon _i=1$. Similarly, we put

$$\begin{aligned} V(m(u+\sqrt{d}),p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k), \end{aligned}$$

where

$$\begin{aligned} \ \epsilon _i= \left\{ \begin{array}{ll} \left( \frac{\frac{vd}{m}}{p_i}\right) &{} \mathrm{if\;} p_i\mid m, \\ \left( \frac{vm}{p_i}\right) &{} \mathrm{if\;} p_i\not \mid m. \end{array} \right. \end{aligned}$$

(12)

By Theorem 3.3, $\{-1,m\}\in \nabla ^2$ if and only if $V(m,p_1,\ldots ,p_k)=(1,\ldots ,1)$. If $m,n\mid d$ and $(m,n)=t$, we write $V(mn,p_1,\ldots ,p_k)=V(mn/t^2,p_1,\ldots ,p_k)$. Moreover, if $V(m,p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k)$ and $V(n,p_1,\ldots ,p_k)=(\delta _1,\ldots ,\delta _k),$ then $V(mn,p_1,\ldots ,p_k)=(\epsilon _1\delta _1,\ldots ,\epsilon _k\delta _k).$ Similarly, $V(m(u+\sqrt{d}),p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k),$ and $V(n,p_1,\ldots ,p_k)=(\delta _1,\ldots ,\delta _k),$ then $V(mn(u+\sqrt{d}),p_1,\ldots ,p_k)=(\epsilon _1\delta _1,\ldots ,\epsilon _k\delta _k).$

Proof of Theorem 2.4

If both (i) and (ii) are satisfied, then by Theorem 2.1, we have $r_4(K_2(\mathcal {O}_E))=0$, i.e., the 2-primary part of $K_2(\mathcal {O}_E)$ is elementary abelian.

Conversely, if the 2-primary part of $K_2(\mathcal {O}_E)$ is elementary abelian, then by Lemma 4.2, Lemma 3.1 and Theorem 3.3, we see that the norm of the fundamental unit of E is $-1$. We show that $\{-1,m(u+\sqrt{d})\}\not \in \nabla ^2$ implies condition $\mathrm{(ii)}$, which in turn shows that $v\equiv 3\pmod {4}$ as we can see from Lemma 4.1.

Since $r_4(K_2O_E)=0$, for any $m\mid d$,

$$\begin{aligned} mZ^2=X^2+dY^2 \end{aligned}$$

and equivalently,

$$\begin{aligned} mZ^2=X^2-dY^2 \end{aligned}$$

has no integer solutions. Hence, we have proved that $r_4(C(E))=0$.

Since $r_4(K_2O_E)=0$, for any $m\mid d$, $m\ne \pm 1,\pm d$, $V(m,p_1,\ldots ,p_k)\not =(1,\ldots ,1)$. By a suitable choice of integers $m_i$ dividing d, we may assume that $\{-1,-1\},\{-1,m_1\},$ $\ldots ,\{-1,m_{k-1}\},\{-1,u+\sqrt{d}\},\{-1,u_{-1}+\sqrt{d}\}$ generate $_2K_2O_E$, where $u_{-1}^2-d=-w_{-1}^2$, and the $m_j's$ are chosen in such a way that, for $1\le i\le k-1$, $\epsilon _k=\epsilon _i=-1$ and for $j\not =i,k,\epsilon _j=1$.

Suppose that $v\equiv 1\pmod {4}$. We have $(u+w)(u-w)-w^2=d$ since $u^2-2w^2=d$, hence $\left( \frac{-d}{v}\right) =1$ and so $\left( \frac{v}{d}\right) =1.$ On the other hand, if $m\mid d$ with $m\equiv 1\pmod {8},$ then $\left( \frac{\frac{d}{m}}{m}\right) \left( \frac{m}{\frac{d}{m}}\right) =1.$ For any odd and positive $m\mid d$, assuming that $V(m(u+\sqrt{d}),p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k),$ we have $\sqcap _{i=1}^{k}\epsilon _i=\left( \frac{\frac{vd}{m}}{m}\right) \left( \frac{vm}{\frac{d}{m}}\right) =\left( \frac{v}{d}\right) \left( \frac{\frac{d}{m}}{m}\right) \left( \frac{m}{\frac{d}{m}}\right) =\left( \frac{v}{d}\right) =1$. Therefore, there exists $n\mid d$ such that $V(n(u+\sqrt{d}),p_1,\ldots ,p_k)=(1,\ldots ,1).$ This contradicts the assumption that $r_4(K_2O_E)=0$. Hence, $v\equiv 3\pmod {4}$, and so we have (ii). $\square$

Proof of Theorem 2.5

The proof is analogous to that of Theorem 2.4.

Applying Theorem 2.2, we see that (i) and (ii) imply that $r_4(K_2(\mathcal {O}_F))=0$.

Conversely, the assumption that $r_4(K_2(\mathcal {O}_F))=0$ implies that $r_4(C(E))=0$ and the norm of the fundamental unit of E is $-1$, i.e., (i) holds. To prove (ii), by Lemma 4.5, it is sufficient for us to show that $V=U+W\equiv 5,~7\pmod {8}.$

Since $U^2-2W^2=2d$, $\left( \frac{-2d}{V}\right) =1$. Hence, $\left( \frac{V}{d}\right) =1$ if and only if $V\equiv 1,~3\pmod {8}.$ If $V\equiv 1,~3\pmod {8},$ then for any odd and positive $m\mid d$, assuming that $V(m(U+\sqrt{d}),p_1,\ldots ,p_k)=(\epsilon _1,\ldots ,\epsilon _k),$ we have $\sqcap _{i=1}^{k}\epsilon _i=\left( \frac{V}{d}\right) =1$. Therefore, one can find $n\mid d$ such that $V(n(U+\sqrt{d}),p_1,\ldots ,p_k)=(1,\ldots ,1),$ i.e., $r_4(K_2(\mathcal {O}_F))>0$, a contradiction!

This completes the proof. $\square$