数理解析研究所講究録
第 1574 巻 2007 年 1-21

1

On some infinite totally real extensions of$\mathbb{Q}$

鹿児島国際大学国際文化学部 福崎賢治 (Kenji Fukuzaki)
Faculty of Intercultural Studies,
The international University of Kagoshima

Abstract

Every number fields are known to be undecidable. Nevertheless the only
known undecidable infinite algebraic extensions of the rationals are fields whose
descriptions depend on non-recursive sets. No ‘natural’ such fields seem to be
known until now.
Let be a prime greater than 5 and $l\equiv-1(mod 4)$ . We prove that asubset ofsuch thatis definable in the ring language using
the formula of Julia Robinson in [8] and give some propert-es of , aiming to
$l$

$A$

$K_{l}$$\mathbb{Z}\subseteq A\subseteq O_{k_{l}}$$A$

prove the undecidability of1$K_{l}= \bigcup_{n}\mathbb{Q}(\cos(2\pi/l^{n}))$.

Introduction

In 1959 Julia Robinson [8] proved that any number field, as well as the corresponding
ring of algebraic integers, is undecidable, by showing that is -definable (in the ringlanguage) in the ring, and the ring is -definable in its number field. The formulas
which she used depend on number fields. Later she [9] showed that there is a uniform
way of defining in the ring of algebraic integers of a number field. Hence, the theoryof the ring of algebraic integers of number fields is undecidable.
These results were extended by R. Rumely [12] to prove that the theory of global
fields is undecidable. His formula is independent of global fields. (J. Robinson usedthe Hasse-Minkowski theorem on quadratic forms. On the other hand, R. Rumely
used Hasse’s Norm Theorem.) Recently’ B. Poonen [7] extended the results. He
proved that the theory of infinite finitely generated fields is undecidable.
As for undecIdable infinite algebraic extensions of the rationals, the only known
such fields are fields whose descriptions depend on non-recursive sets. For example,
if we adjoin to the rationals the square roots of a non-recursive set of prime numbers,
then the resulting field is certainly undecidable. (See [10].) On the other hand, J.
Robinson [9] proved that the theory of the ring of all totally real algebraic integers
is undecidable. We say that an algebraic number is totally real iff and an its
conjugates are real. Then she conjectured that the fieldof all totally real algebraic$N$$\emptyset$$\emptyset$

$N$

$a$

$a$

$\mathbb{Q}^{tr}$2numbers was undecidable. But in 1994 Fried, Haran, and V\"olklein [1] proved thatis decidable. So it remains open whether or not there are ‘natural’ undecidable
infinite algebraic extensions of .
In order to define the ring of algebraic integers in a given number field, J. Robinson
constructed a formula which includes but excludes non-algebraic integers, which
only depends on the ramification index of prime ideals of a number field which divides
be such a formula. J. Robinson defined the ring
2. Let $F$ be a number fleld and$O$ of $F$
be an integral
in $F$ in the following way. Letof algebraic integers
be the minimal polynomial of over (hencebasis ofD $(s=[F:\mathbb{Q}])$ , and let$F$over ) for each . Then in$\mathbb{Q}^{tr}$$\mathbb{Q}$$\mathbb{Z}$$\psi(t)$$a_{1},$$P_{i}(x)$$\ldots a_{\delta}$$\mathbb{Q}$$a_{i}$$i$$\mathbb{Z}$$t\in O\Leftrightarrow\exists x_{1},$$\ldots x_{e},$ $y_{1},$ $\ldots y_{\delta}(t=x_{1}y_{1}+\cdots+x_{s}y_{e}\wedge\bigwedge_{i}P_{i}(y_{i})\wedge\bigwedge_{i}\psi(x_{i}))$holds. Note that this formula depends on .
includes and excludes non-algebraic integers
In this article we will show thatwith an odd prime.
also in
Unfortunately we cannot define the ring of algebraic integers in the same way as initself defines the ring of algebraic
number fields. Nevertheless we conjecture that
integers in
if $l>5$ is a prime and $-1$ mod 4. If this conjecture is true, then it$F$$\psi(t)$$K_{l}= \bigcup_{n}\mathbb{Q}(\cos(2\pi/l^{n}))$$\mathbb{Z}$$l$$\psi(t)$$K_{l}$

by the results of J. Robinson [9].
follows that is definable in suchin [8]. which we need in section
In section 2, we describe the construction ofincludes and excludes non-algebraic integers
3. In section 3, we will prove thatwithan
odd prime. In section 4 we will prove somealso infacts on quadratic charcters with polynomial arguments. In section 5 we will give.
some properties of$N$$K_{l}$$\psi(t)$$\psi(t)$$K_{l}= \bigcup_{n}\mathbb{Q}(\cos(2\pi/l^{n}))$$\mathbb{Z}$$l$$\psi(K_{l})$2Construction of$\psi(t)$be a number field (a finite algebraic extension of the rationals ) and let
be the ring of algebraic integers of $F$ . By we denote a valuation of $F$ and by
the completion of $F$ with respect to . Since non-Archemedean valuations of $F$ are-adic valuations for some prime ideal of $F$ , we use the same letter for both thewevaluation and the prim ideal. Let be a prime ideal of $F$ and $a\in F$ . By$b\inF^{*}$,,
wesymboluse Hilbert
which isdenote the order of at . Given
defined to $be+1$ if $ax^{2}+w^{2}=1$ is solvable in , otherwise defined to be-l.
The following lemma is well-known:

Let

$F$

$0$

$\mathbb{Q}$$F_{\mathfrak{p}}$$\mathfrak{p}$$\mathfrak{p}$$\mathfrak{p}$$\mathfrak{p}$$\mathfrak{p}$$\nu_{\mathfrak{p}}(a)$$\mathfrak{p}$$a$$\mathfrak{p}$$(a, b)_{\mathfrak{p}}$$a,$$F_{\mathfrak{p}}$Lemma 1 $h\in F$ “ can be represented by the the temary quadratic form$x^{2}-ay^{2}-bz^{2}iff-ab/h\not\in F_{\mathfrak{p}}^{*2}$.
for any valuation such that$\mathfrak{p}$$(a, b)_{\mathfrak{p}}=-1$

3This follows from the property of quaternary quadratic forms and the HasseMinkowski theorem on quadratic forms. See [6, p. 187] and [14, p. 111].
Using this lemma, J. Robinson proved the following:(\dagger ) Let $m$ be a positive integer such that$\mathfrak{p}^{m}\#$for all prime ideals . Let$\mathfrak{p}$$\varphi(s, u, t)$$be$$\exists x,y,$For$t\not\in 0$, there are$z(1-sut^{2m}=x^{2}-sy^{2}-uz^{2})$$a,$ $b\in 0$.

such that,
1.
2. $F\models\forall c(\varphi(a, b, c)arrow\varphi(a, b, c+1))$ .
$F\models\neg\varphi(a, b, t)$

Then we can use inductive form: Let$\forall s,$$\psi(t)$be$u(\forall c(\varphi(s, u, c)arrow\varphi(s,u, c+1))arrow\varphi(s,u, t))$,then the solution set ofin $F,$ $\psi(F)$ , includes but excludes non-algebraic integers, that is,. Sinceholds for every $u\in F$ , the inductive
form insures that every positive integer satisify . Since $\varphi(s, u, t)rightarrow\varphi(s,u, -t)$ , every
rational integer also satisfies . The above statement (\dagger ) shows that non-algebraic
integers fail to satisfy . Note that for(and for $t\in 0$ ), it is not so difficult tofind $b\in F$ such that 1 holds, but difficult to findsuch that both 1 and 2 hold.
J. Robinson proved the above statement from two lemmas. We state these twolemmas in a little bit different forms for our sake. Before stating these lemmas, weneed some lemmas. The following two lemmas are special cases of a theorem provedin [4, p. 166].
$\psi(t)$$\mathbb{Z}$$\varphi(s, u, 0)$$\mathbb{Z}\subseteq\psi(F)\subseteq 0$$s,$$\psi$$\psi$$t\not\in 0$$\psi$

$a,$

$a,$

$b$

Lemma 2 There are infinitely many prime ideals in every ideal class.

Lemma 3 If $a\in 0$ is prime to an ideal$p\in 0$such that$p\equiv a$, there are infinitely many prime elements(mod m).

Lemma 4 Let $a\in 0$ and$(a, b)_{\mathfrak{p}}=-1$$\mathfrak{m}$$\nu_{\mathfrak{p}}(a)=1$.

.

Then there is $b\in 0$ with$\mathfrak{p}A^{b}$such that

Proof. It is proved in [6, pp. 161-165] that there is a unit in a local field, wherethat it is congruent to a square $(mod 40)$ but not
integers and a prime ideal of $I/I$ . And if is such a unit,
element . Take such a unit
. There is a unit
such thatis congruent to a square $(mod 40)$ but not.
We state two basic lemmas due to J. Robinson [8, Lemma 8,9].
$(mod 4\mathfrak{p})$$\mathfrak{p}$$a$$\epsilon_{0}$$(a, \epsilon)_{\mathfrak{p}}=-1$$\epsilon$$\epsilon\in F_{\mathfrak{p}}$

$0$

such
is the ring offor a prime$AI$$\epsilon_{0}\in F$$(mod 4\mathfrak{p})$$\epsilon_{0}\equiv\epsilon(mod 4\mathfrak{p})$.

$\square$4

Lemma 5 Given a prime idealof and an odd prime number , there are relatively prime elements and in 0* such that$F$$\mathfrak{p}_{1}$

$l$

$b$

$a$

, where,are distinct prime ideals which include every
Prime ideals which divides 2, and dose not divide $forj=2,$ $\ldots 2k$ , and1.

$(a)=\mathfrak{p}_{1}\cdot\cdot \mathfrak{p}_{2k}$$\mathfrak{p}_{1},$$\ldots$$\mathfrak{p}_{2k}$$l$$\mathfrak{p}_{j}$

2.

$b$

is a totally positive prime element such that$(a, b)_{\mathfrak{p}}=-1$

iff$\mathfrak{p}|a$.

be a set of disticnt prime ideals such that it includes every
prime idals dividing 2 and, $2k-1$ . Let Sl be thedose not divide for $j=2,$
ideal class which contains the product. By Lemma 2 we can choose aprime ideal, $2k-1$ and within the ideal classwithfor $i=1,$

Proof. Let$\mathfrak{p}_{1},$$\ldots \mathfrak{p}_{2k-1}$$l$$\mathfrak{p}_{j}$$\ldots$$\mathfrak{p}_{1}\cdots \mathfrak{p}_{2k-1}$$R^{-1}$$\mathfrak{p}_{2k}$$\mathfrak{p}_{2k}\parallel(l)$$\mathfrak{p}_{2k}\neq \mathfrak{p}_{i}$$\ldots$.

, $2k$ , by Lemma 4 we can choose
For $i=1,$
prime to so that$-1$ . Let $m$ be a positive integer such thatfor every prime ideal . Consider
the simultaneous system of congruences$b_{i}\in 0$$\ldots$$(a, b_{i})_{\mathfrak{p}:}=$$\mathfrak{p}_{i}$$\mathfrak{p}^{m}\parallel 2$$(mod \mathfrak{p}_{i}^{2m})$$x\equiv b_{i}$$\mathfrak{p}$for $i=1,$$\ldots$, $2k$ .

By the Chinese Remainder Theorem, there is a solution $c\in 0$ and so is every element
which is congruent to. Since is prime to the modulus, by Lemma
3 there are infinitely many totally positive prime elements such that$c(mod \mathfrak{p}_{1}^{2m}\cdots \mathfrak{p}_{2k}^{2m})$

$c$

$p$

$(mod \mathfrak{p}_{1}^{2m}\cdots \mathfrak{p}_{2k}^{2m})$$p\equiv c$.

be one of such elements. is coprime to .
We claim thatfor each ; sinceand is prime to ,, then we apply Newton’s method of iteration [4, p. 42]: “Let$f(x)$ be a polynomoial with coefficients in. If there is an element of$f(x)$
such that, thenhas a root in. Letting $f(x)=x^{2}-b_{i}/b$, we get thatand. Hencefor each . On the other
hand, $(a, b)_{\mathfrak{p}}=+1$ for all Archimedean valuations since is totally positive. It
is easy to see that ifthen is an Archimedean valuation or the prime
ideal dividing 2ab (see [6, p. 166]). Then the only other other valuation for which$(a, b)_{\mathfrak{p}}=-1$ could hold would be; but, by the product formula for the
Hilbert symbol ([6, p. 190]),
for an even number of valuations. $Therefore\square$
iff $P|a$ .

Let

$b$

$b$

$a$

$b\equiv b_{i}(mod \mathfrak{p}_{i}^{2m})$$i$$b_{i}/b\in F_{\mathfrak{p}_{i}}^{2}$$b_{i}$$\mathfrak{p}_{i}$$\nu_{\mathfrak{p}_{i}}(1-b_{i}/b)>\nu_{\mathfrak{p}_{\{}}(4)$$O_{F_{\mathfrak{p}:}}$$\alpha_{0}$$|f(\alpha_{0})|<|f’(\alpha_{0})^{2}|$$\alpha_{0}=1$$O_{F_{\mathfrak{p}_{i}}}$$O_{F_{\mathfrak{p}:}}$$(a, b)_{\mathfrak{p}_{i}}=-1$$b_{i}/b\in F_{\mathfrak{p}_{1}}^{n2}$$b$$\mathfrak{p}$$(a, b)_{\mathfrak{p}}=-1$$i$$\mathfrak{p}$$\mathfrak{p}$$\mathfrak{p}=(b)$$(a, b)_{\mathfrak{p}}=-1$$(a, b)_{\mathfrak{p}}=-1$

Lemma 6 Let
such thatare distinct prime ideals which
in clude every prime ideals which divides 2, and let
be coprime to such thatand $m$ be a positive integer such that
ifffor every prime ideal$(a)=\mathfrak{p}_{1}\cdots \mathfrak{p}_{2k}$$\mathfrak{p}_{1},$$\ldots \mathfrak{p}_{2k}$$b\in 0^{*}$$(a, b)_{\mathfrak{p}}=-1$$\mathfrak{p}$$\mathfrak{p}|a_{f}$$a$$\mathfrak{p}^{m}\beta$. Then,$1-abc^{2m}=x^{2}-ay^{2}-bz^{2}$ is solvable

for$x,$ $y$and$z$

in $F$ iff$\nu_{\mathfrak{p}:}(c)\geq 0$

for each .
$i$5Proof. Let $h=1-abc^{2m}$ . Then$h\neq 0$

since$\nu_{\mathfrak{p}_{1}}(abc^{2m})\neq 0$. Suppose thatfor each . By Lemma$\nu_{\mathfrak{p}:}(c)\geq 0$

for each . Sinceand
1 and the assumption, $h=x^{2}-ay^{2}-bz^{2}$ is solvable for $x,$ and in $F$ .
Now suppose thatfor some . Let. We show that $-ab/h\in$. Since, applying again Newton’s method of iteration[4, p. 42] with $x^{2}-(-ab/h)$ and, we get that
. It follows that$h=x^{2}-ay^{2}-bz^{2}$ is not solvable forand in $F$ .
$i$$\nu_{\mathfrak{p}_{i}}(-ab)=1,$$\nu_{\mathfrak{p}_{1}}(h)=0$$y$$\nu_{\mathfrak{p}:}(c)<0$$i$$h/(-ab)\not\in F_{\mathfrak{p}_{1}}^{*2}$

$i$

$z$

$\nu_{\mathfrak{p}_{*0}}.(c)<0$$\nu_{\mathfrak{p}_{i_{0}}}(1-(-ab/h))>\nu_{\mathfrak{p}_{*0}}.(4)$$F_{\mathfrak{p}_{1_{0}}}^{2}$$\alpha_{0}=1$$-ab/h\in F_{\mathfrak{p}:_{0}}^{*2}$$x,$ $y$$\square$$z$

It is easy to derive the statement (\dagger ) from the above two lemmas. For, take$b\inD$such thatandas in Lemma , then the statement (\dagger ) holds,
notingiffor every prime ideal .
$t\not\in O$$\nu_{\mathfrak{p}_{1}}(t)<0$$\mathfrak{p}_{1}$$\nu_{\mathfrak{p}}(c+1)\geq 0$3$\psi(t)$

in

$a,$

$\nu_{\mathfrak{p}}(c)\geq 0$$\mathfrak{p}$$K_{l}$

The following lemma on cyclotomic fields is well-known and proved in [2, pp. 256-258].
We denote by Euler’s function.
$\phi$

Lemma 7 Let$\Lambda,\prime f=\mathbb{Q}(\zeta_{m})$m-th root of unity. Then:

1.

$[\Lambda\prime I:\mathbb{Q}]=\phi(m)$, where $m$ is an positive integer and$\zeta_{m}$is a primitive.

2. The only ramified prime ideals in $M$ are those dividing $m$ .
If $m=l^{n}$ uyith odd pnme, then there is only one prime$l$$\mathfrak{p}=(1-\zeta_{m})$lying above , and it is totally ramified.

of

$M$

$l$

3. Let$p$

be a prime unth$p^{f}\equiv 1(mod m)$degree.

, and let bethe smallest positive integer such that
Then in, where each has residue
we have$f$$p\parallel m$$\Lambda l$$p=\mathfrak{p}_{1}\cdots \mathfrak{p}_{g}$$\mathfrak{p}_{i}$and $fg=\phi(m)$ .

$f$Lemma 8 Let $F=\mathbb{Q}(\cos(2\pi/m))$ and$A\cdot\prime I=\mathbb{Q}(\zeta_{m})$be as above. Then:

1. 2 $\cos(2k\pi/m)$ with $0\leq k\leq m$ are algebraic integers, and
2 $\cos(2k\pi/m)$ with $0\leq k\leq m/2$ and $(k, m)=1$ form a set of conjugates.

2. $M\supset F$, [A :

$F$

$f$

$\phi(m)/2)$.

] $=2$ (hence$\Lambda f$

is abelian extension of$\mathbb{Q}$, and $[F : \mathbb{Q}]=$are those dividing $m$ .
If $m=l^{n}$ with odd prime, then there is only one prime $\mathfrak{p}=(2-2\cos(2\pi/m))$ oflying above , and it is totally ramified, and 2 $\cos(2k\pi/m)$ with $0\leq k\leq m/2$
and $(k, m)=1$ are units in the ring of algebmic integers.

3. The only$ru$

mified prime ideals in$l$$\Lambda,I$

$l$

$M$

6

Proof. Since 2 $\cos(2k\pi/m)=e^{2k\pi/m}+1/e^{2k\pi/m},$ $2\cos(2k\pi/m)$ are algebraic integers.

$1/e^{2k\pi/m}$ with $0\leq k\leq m/2$ and $(k, m)=1$ are primitive roots
Noting thatof unity, we have that 2 $\cos(2k\pi/m)$ with $0\leq k\leq m/2$ and $(k, m)=1$ form a set of
conjugates. It follows that $M\supset F[M:F]=2$ , and the only ramified prime ideals
inare those dividing $m$ .
Let $m=l^{n}$ with odd prime. Then,$e^{2k\pi/m},$$\Lambda/I$$l$$(x^{l^{n-1}})^{l-1}+(x^{l^{n-1}})^{l-2}+\cdots+x^{l^{n-1}}+1$$=0 \leq k\leq\iota^{n}/2(kl)=1\prod_{1}(x-e^{2k\pi/l^{n}})(x-1/e^{2k\pi/l^{n}})$.
$0 \leq k\leq l’/2\prod_{(k,l)=1}(x^{2}-2\infty s(2k\pi/l^{n})x+1)$

Letting $x=1$ , we have,$l= \prod_{0\leq k\leq t’/2}(2-2\cos(2k\pi/l^{n}))$

.

Since
$\frac{2-2\cos(2k_{1}\pi/l^{n})}{2-2\cos(2k_{2}\pi/l^{n})}=\frac{(1-e^{2k_{1}\pi/l^{n}})(1-1/e^{2k_{1}\pi/l^{n}})}{(1-e^{2k_{2}\pi/l^{n}})(1-e^{2k_{2}\pi/l^{n}})}$$(2-2 \cos(2k_{1}\pi/l^{n}))/(2-2\cos(2k_{2}\pi/l^{n}))$are units if$k_{1}\neq k_{2}$$(l)=(2-2\cos(2\pi/l^{n}))^{\phi(l^{n})}$. Hence,.

It follows that there is only one prime $\mathfrak{p}=(2-2\cos(2\pi/l^{n}))$ of $M$ lying above , and
it is totally ramified.
Letting $x=\sqrt{-1}$, we have,$l$$\pm 1=\prod_{0\leq k\leq t’/2}2\cos(2k\pi/l^{n})$.

Therefore 2 $\cos(2k\pi/m)$ with $0\leq k\leq m/2$ and $(k,m)=1$ are units in the ring ofalgebraic integers.
It is proved in [11] that 2 $\cos(2k\pi/m)$ with $0\leq k\leq m/2$ and $(k, m)=1$ are
algebraic units iff $m\neq 1,2,4$ and is not of the form
with prime.
$F_{n}=\mathbb{Q}(\cos(2\pi/l^{n}))$, where is an odd prime, and let $K_{l}=$Rom now on, let. We denote by
the ring of algebraic integers in.
.
integersinThenand bythe ring of algebraic
From Lemma 8, we easily see that,$\square$$4p^{n}$$p$$l$$\bigcup_{n}\mathbb{Q}(\cos(2\pi/l^{n}))(F_{0}=\mathbb{Q})$$O_{K_{l}}$$O_{n}$$K_{l}$$F_{n}$$O_{K_{l}}= \bigcup_{n}O_{n}$7Lemma 9 Let $0<i<j$ and1. If$\mathfrak{p}\Lambda^{l}$$[F_{j} :2. If$\mathfrak{p}|l$, then in.

be a Prime ideal of$P$$F_{j;}\mathfrak{p}=\mathfrak{P}_{1}\cdots \mathfrak{P}_{k}$, where$F_{i}$. Then:are Primes in$\mathfrak{P}_{r}$and $kdi$ 例 des$F_{j}$

F_{i}]=t^{j-i}$, then in$\mathfrak{p}=\mathfrak{P}^{l^{j-1}}$$F_{j},$, where $\mathfrak{p}=(2-2\cos(2\pi/l^{i})),\mathfrak{P}=(2-2\cos(2\pi/l^{j}))$ .

The next lemma is also proved in [2, p. 272].

Lemma 10 Let
, let
For
$\alpha\in K_{\mathfrak{P}}^{*}$$K\supset k$number fields andbe primes of $K$ and.
. Then,and$\mathfrak{P}\supset \mathfrak{p}$$a=N_{K\varphi/k_{p}}(\alpha)$$k$respectively.

$(\alpha, b)_{\mathfrak{P}}=(a,b)_{\mathfrak{p}}$$b\in k_{\mathfrak{p}}$The next lemma follows from Lemma 10.

Lemma 11 Let $0<i<j,$. Then for$a,$ $b\in F_{1}^{*},$$\mathfrak{p}$Proof. Since$\mathfrak{p}$a prime ideal of

iff$(a,b)_{\mathfrak{p}}=1$$F_{i}$$(a,b)_{\mathfrak{P}}=1$.

and

be a prime in$\mathfrak{P}$$F_{j}$

lying overis an abelian extension, the local degree at
divides the degree, that is,of(see [6, p. 32].) Let be the local degree at$N_{K\varphi/k,}(a)=a^{u}$. Then. Since is odd, it follows
and
i旺
that
.
We now extend J. Robinson’s result [8] tofor. Note that in eachevery prime ideal in .
$F_{j}/F_{i}$$F_{j}/F_{i}$$\mathfrak{P}$$[(F_{j})_{\mathfrak{P}} :(F_{i})_{\mathfrak{p}}]|[F_{j} :$u$

F_{i}]$$u$$(a, b)_{\mathfrak{P}}=(a^{u}, b)_{\mathfrak{p}}=(a, b)_{\mathfrak{p}}^{u}$$\mathfrak{P}$$(a, b)_{\mathfrak{p}}=1$$(a, b)_{\mathfrak{P}}=1$$\square$$F_{n},$$K_{l}$$\mathfrak{p}^{2}\Lambda^{2}$$F_{n}$

Theorem 12 Let$\varphi(s, u, t)$$\exists x,$

and$\psi(t)$be$y,$$z(1-abt^{4}=x^{2}-sy^{2}-uz^{2})$be$\forall s,$$u(\forall c(\varphi(s, u, c)arrow\varphi(s,u, c+1))arrow\varphi(s,u, t))$then the solution set of
integers, that is,$\psi(t)$in$K_{l},$$\mathbb{Z}\subseteq\psi(K_{l})\subseteq O_{k_{l}}$Proof. It is clear thatare$a,$ $b\in K_{l}$$\mathbb{Z}\subseteq\psi(K_{l})$$\psi(K_{l})$

.

. Let

, includes$t\in K_{l}\backslash O_{K_{l}}$$\mathbb{Z}$but excludes non-algebraic. For this , we show that there$t$such that$K_{1}\models\neg\varphi(a, b, t)\wedge\forall c(\varphi(a, b, c)arrow\varphi(a, b, c+1))$

We fixsuch that $t\in F_{m}$ and $m>1$ . Then
By Lemma 2, there are relatively prime elements$F_{m}$

1.

, whereevery prime ideals in$\ldots$.

for some primeinsuch that$\nu_{\mathfrak{p}_{1}}(t)<0$$(a)=\mathfrak{p}_{1}$

$2,$

,

$\mathfrak{p}_{2k}$, $2k$ , and$p_{1},$$F_{m}$

$a$

and

$b$

$\mathfrak{p}_{1}$in$F_{m}$.

$O_{m}$,
which include
are distinct prime ideals indose not divide for $j=$which divides 2, and$\ldots$$F_{m}$$\mathfrak{p}_{2k}$$l$$\mathfrak{p}_{j}$

8

2.

$b$

is a totally positive prime element in$F_{m}$

such that$(a, b)_{\mathfrak{p}}=-1$

iff$\mathfrak{p}|a$.

By Lemma 6, $1-abt^{4}=x^{2}-ay^{2}-bz^{2}$ is not solvable forand in, and forevery $c\in F_{m}$ , ifthen $F_{m}\models\varphi(a, b, c+1)$ .
, it is enough to show that for every $s>m$ such that $s-m$ is even,
For this$1-abt^{4}=x^{2}-ay^{2}-bz^{2}$ is not solvable for $x,y$ and, ifin , and for every
.
then
$a,b$
relatively
prime
Note thatarealso in .
$x,$ $y$$F_{m}$$z$$F_{m}\models\varphi(a, b, c)$

$a,$

$b$

$F_{8}$$z$$F_{s}\models\varphi(a, b, c)$$c\in F_{\delta}$$F_{\delta}\models\varphi(a, b, c+1)$$O_{\delta}$

Case 1: Pl$\int l$.

By Lemma 9, the decomposition of the ideal $(a)$ in
is given by,
where
are mutually distinct prime ideals and include every prime ideals
iff. We letwhich devides 2. By Lemma 11,. Sinoe, we have that
. By Lemma 6, we $\infty nclude$ that $1-abt^{4}=x^{2}-ay^{2}-bz^{2}$
is not solvable for $x,$ and in , and for every
, if
then
$(a)=\mathfrak{P}_{1}\cdots \mathfrak{P}_{2r}$$F_{\epsilon}$$\mathfrak{P}_{1},$$\ldots \mathfrak{P}_{2r}$$(a, b)_{\mathfrak{P}}=-1$$0$$\mathfrak{P}|a$$\nu_{\mathfrak{p}_{1}}(t)<$$\mathfrak{p}_{1}\subset \mathfrak{P}_{1}$$\nu_{\varphi_{1}}(t)<0$

$z$

$y$

$F_{\epsilon}\models\varphi(a, b, c+1)$$F_{l}\models\varphi(a, b, c)$$c\in F_{\epsilon}$$F_{l}$.

.
Case 2:
By Lemma 9, the decomposition of the ideal $(a)$ in$\mathfrak{p}_{1}|l$$(a)=\mathfrak{P}^{l}i^{-m}\cdots \mathfrak{P}_{2r’}$$F_{\delta}$

is given by

,

,
where
are mutually distinct prime ideals and include every prime ideals
which devides 2, and $\mathfrak{p}_{1}=(2-2\cos(2\pi/l^{m})),\mathfrak{P}_{1}=(2-2\cos(2\pi/l^{\theta}))$ .
Let $a’=a/(2-2\cos(2\pi/l^{\delta}))^{l^{-m}-1}$ . Then $a’\in O_{s}$ andin .
Since $a=a’((2-2\cos(2\pi/l^{\delta}))^{(l^{\epsilon-m}-1)/2})^{2},$for each . Hence wehave that $(a’,b)_{\mathfrak{P}}=-1$ iff.
Suppose that $1-abt^{4}=x^{2}-ay^{2}-bz^{2}$ were solvable forand in $F,$ . Then$\mathfrak{P}_{1},$$\ldots$$\mathfrak{P}_{2r’}$$(a’)=\mathfrak{P}_{1}\cdots \mathfrak{P}_{2r’}$$(a, b)_{\mathfrak{P}_{i}}=(a’, b)_{\mathfrak{P}:}$$F_{\iota}$$i$$\mathfrak{P}|a’$$x,$ $y$$z$$1-a’b(t(2-2\cos(2\pi/l^{\epsilon}))^{(l^{-m}-1)/4})^{4}=x^{2}-a’((2-2\cos(2\pi/l^{\delta}))^{(l^{-m}-1)/2}y)^{2}-bz^{2}$

is solvable for $x,$ and in , noting that $(l^{e-m}-1)/4$ is a positive integer since$l-m$ is even. But $\nu_{\mathfrak{P}1}(t(2-2\cos(2\pi/l^{s}))^{(l^{-n}-1)/4})<0$ since. We have acontradiction by Lemma 6.
Next we show that ifthen $F_{\delta}\models\varphi(a, b, c+1)$ . Suppose that, that is, $1-abc^{4}=x^{2}-ay^{2}-bz^{2}$ is solvable for $x,$ and in . Then
$y$

$z$

$F_{l}$$\mathfrak{p}_{1}=\mathfrak{P}^{l}i^{-m}$$F_{s}\models\varphi(a, b, c)$$F_{\epsilon}\models\varphi(a, b, c)$

$y$

$z$

$F_{\delta}$$1-a’b(c(2-2\cos(2\pi/l^{\delta}))^{(l^{-n}-1)/4})^{4}=x^{2}-a’((2-2\cos(2\pi/l^{\epsilon}))^{(l^{-n}-1)/2}y)^{2}-bz^{2}$

is solvable for $x,y$ and in . By Lemma 6,
for each . It follows that $\nu_{\mathfrak{P}_{i}}((c+1)(2-2\cos(2\pi/l^{\delta}))^{(l^{-m}-1)/4})\geq 0$ for each .
Therefore we have that $F,$ $\models\varphi(a, b, c+1).\ovalbox{\tt\small REJECT}$$z$$F_{f}$$\mathfrak{P}\iota$

Remark 13 We can easily show the following.

$\nu_{\mathfrak{P}*}(c(2-2\infty s(2\pi/l^{\delta}))^{(l^{-m}-1)/4})\geq 0$$\mathfrak{P}_{i}$9

1. For every

2. for everyS.

$n\in \mathbb{Z},$$m\in \mathbb{Z}$$t\in\psi(K_{l})$iff $t+n\in\psi(K_{l})$ ,, if $t\in\psi(K_{l})$ , then $mt\in\psi(K_{l})_{f}$is closed under automorophism, thatare also in .

$\psi(K_{l})$

of

$a$

$\dot{w}$, if$a\in\psi(K_{l})_{f}$then all conjugates$K_{l}$For 2., we use the equivalence$\varphi(a, b, mc)rightarrow\varphi(m^{2}a, m^{2}b, c)$.

holds also for towers of cyclotomics similarly. Let
Remark 14 The result forroot of unity, and, where is an odd prime andis a primitive
the ring of algebraic integers in .
let
. We denote by$K_{l}$$M_{n}=\mathbb{Q}(\zeta_{l^{n}})$$l$$N_{l}= \bigcup_{n}\mathbb{Q}(\zeta_{l^{n}})(AI_{0}=\mathbb{Q})$

Then,

4

$l^{n_{-}}th$$\zeta_{l^{n}}$$\mathbb{Z}\subseteq\psi(N_{l})\subseteq O_{N_{l}}$$N_{t}$$O_{N_{l}}$.

quadratic characters with polynomial arguments

In this section we will prove some facts on some character sums of finite fields which
be a finite field with elements, and $q=p^{f}$ where is
we will use later. We let
an odd prime. We let be the quadratic character of , that is, $\eta(0)=0,\eta(c)=1$
and $\eta(c)=-1$ otherwise.
if
We consider the following character sum$F_{q}$$p$$q$$F_{q}$$\eta$$c\in F_{q}^{r2}$$I_{n}(a)= \sum_{c\in F_{q}}\eta(c^{n}+a)$,where $a\in F_{q}$ . Moreover we use the following character sum$H_{n}(a)= \sum_{c\in F_{q}}\eta(c^{n+1}+ac)$,

which is caned a Jacobsthal sum. Using these character sums, we will show that if$\eta(a)=-1,$ $p\equiv 3(mod 4)$ and $p>3$ , then there are $b\in F_{q}$ and $i\in F_{p}$ such that$\eta(b^{4}+d)\eta((b+i)^{4}+d)=-1$.

Lemma 15 Let $p\equiv 3(mod 4),$ $q=p^{f}$ , and$a\in \mathbb{F}_{q}$

1. If

$f$

is odd, $I_{4}(a)=-1$ .

2. If

$f$

is even and $\eta(a)=-1,$ $I_{4}(a)=-1$ .

. Then:is odd and $q\equiv 1(mod 4)$ if is even.
For 1., it is proved in [5, pp. 231-232] that $I_{2}(a)=-1$ for all $a\in F_{q},$ $I_{2n}=$, then$I_{n}(a)+H_{n}(a)$ , and if the largest power of 2 dividing $q-1$ also divides$H_{n}(a)=0$ . Therefore we get that $H_{2}(a)=0$ and $I_{4}(a)=-1$ for all.

Proof. We first note that $q\equiv 3(mod 4)$ if

$f$

$f$

$n$

$a\in \mathbb{F}_{q}$10

For 2., we use the following formula [5, p. 231].
$I_{n}(a)= \eta(a)\sum_{j=1}^{d-1}\lambda^{j}(-a)J(\lambda^{j},\eta)$

where is a multicative character ofsum, that is,$\lambda$$F_{q}$,of order $d=(n, q-1)$ and$J( \lambda^{j}, \eta)=\sum_{c_{1}+c_{2}=1}\lambda^{j}(c_{1})\eta(c_{2})$$J(\lambda^{j}, \eta)$

is a Jacobi.

Letting $n=4$ , we see that is a multiplicative character of order 4, hence
Therefore we see by [5, p. 207] that$\lambda$$J( \lambda^{2}, \eta)=-\frac{1}{q}G(\eta, \chi_{1})^{2}$

where$G(\eta, \chi_{1})$$\eta=\lambda^{2}$.

,

is a Gaussian sum. Further we know by [5, p. 199] that
$G(\eta, \chi_{1})=(-1)^{f-1}i^{f}q^{1/2}$.

Therefore we get$I_{4}(a)=\eta(a)(\lambda(-a)J(\lambda, \eta)-\lambda^{2}(-a)(-1)^{f}+\lambda^{3}(-a)J(\lambda^{3}, \eta))$.

It is easy to see that $(q-1)/4$ is even, and $\lambda(-1)=-1$ iff $(q-1)/4$ is odd, hence we
see that $\lambda(-1)=1$ . Together with $\eta(-1)=(-1)^{(q-1)/2}=1$ and, we have$\lambda^{3}=\overline{\lambda}$$I_{4}(a)=\lambda^{3}(a)J(\lambda, \eta)+(-1)^{f+1}+\lambda(a)\overline{J(\lambda,\eta)}$.

Here we have that $\lambda(a)=\pm i$ since $\eta(a)=-1$ . Then
$I_{4}(a)=-1\pm 2{\rm Im} J(\lambda, \eta)$.

We now calculateof . Let $J(\lambda,\eta)=A+Bi.$
and are rational$\pm1and\pmi$
integers since assumes only the values. By [5, p. 209], we know that$A^{2}+B^{2}=p^{f}$, hence we have that
. It is well-known that for such
$A=\pmp^{f/2}$that $p\equiv 3(mod 4)$ , it is the case thatand $B=0$ , or vice versa. However$A=p^{f/2}$$A=-p^{f/2}$$f/2$
we can show thatifis odd,if $f/2$ is even, and $B=0$ by thesimilar way in $[5, p233]$ , from which $I_{4}(a)=-1$ follows. It is proved in [5, p. 232]${\rm Im} J(\lambda, \eta)$$\lambda$$A$$F_{q}$$B$$0,$$|J(\lambda, \eta)|=q^{1/2}$

$p$

that
$H_{n}(a)= \eta(a)\lambda(-1)\sum_{j=0}^{d-1}\lambda^{2j+1}(a)J(\lambda^{2j+1}, \eta)$,

11

where $d=(n, q-1)$ andformula we get$\lambda$

is a multiplicative character of$F_{q}$of order $2d$ . From this$H_{2}(1)=\lambda(-1)(J(\lambda, \eta)+J(\lambda^{3}, \eta))=\lambda(-1)(J(\lambda, \eta)+\overline{J(\lambda,\eta)})=2{\rm Re} J(\lambda, \eta)$

hence
Let

. We will now show that $\frac{1}{2}H_{2}(1)\equiv-1(mod 4)$ .
be a primitive element ofand let $q=4k+1$ . Since $\eta(-1)=1$ and${\rm Re} J( \lambda, \eta)=\frac{1}{2}H_{2}(1)$$g$$-1=g^{2k}$,$F_{q}$, we can write$H_{2}(1)$$= \sum_{i=1}^{4k}\eta(g^{i})\eta((g^{i})^{2}+1)$$\sum_{i=1}^{2k}\eta(g^{i})\eta((g^{i})^{2}+1)+\sum_{i=1}^{2k}\eta(-g^{i})\eta((-g^{i})^{2}+1)$2$\sum_{i=1}^{2k}\eta(g^{i})\eta((g^{i})^{2}+1)$,

so that$\frac{1}{2}H_{2}(1)=\sum_{:f=1}^{2k}\eta(g^{i})\eta((g^{i})^{2}+1)$.

From $I_{2}(1)=-1$ we get
一$1=1+ \sum_{i=1}^{4k}\eta((g^{i})^{2}+1)=1+2\sum_{i=1}^{2k}\eta((g^{i})^{2}+1)$

,

hence
$-1= \sum_{i=1}^{2k}\eta((g^{*})^{2}+1)$.

By subtraction, we obtain$\frac{1}{2}H_{2}(1)+1=\sum_{i=1}^{2k}(\eta(g^{i})-1)\eta((g^{i})^{2}+1)$.

For $1\leq i\leq 2k$ , we have$(\eta(g^{i})-1)(\eta((g^{i})^{2}+1)-1)\equiv 0$$(mod 4)$ whenever $\eta((g^{i})^{2}+1)\neq 0$

.

12

Thus,
$(\eta(g^{i})-1)\eta((g^{i})^{2}+1)\equiv\eta(g^{i})-1$$(mod 4)$ whenever $\eta((g^{i})^{2}+1)\neq 0$.

Now $\eta((g^{i})^{2}+1)=0$ if and only if $i=k$ or $3k$ . Consequently,$\frac{1}{2}H_{2}(1)+1$$\equiv\sum_{i=1}^{2k}(\eta(g^{i})-1)-(\eta(g^{k})-1)$$\equiv\sum_{i=1}^{2k}\eta(g^{i})-(2k-1)-\eta(g^{k})$$(mod 4)$.

Furthermore,$0= \sum_{i=1}^{4k}\eta(g^{i})=2\sum_{i=1}^{2k}\eta(g^{i})$and $\eta(g^{k})=\lambda^{2}(g^{k})=\lambda(-1)=-1$ , so that$\frac{1}{2}H_{2}(1)+1\equiv-2k$

Since

$k$

$(mod 4)$.

is even, we see that$\frac{1}{2}H_{2}(1)+1\equiv 0$$(mod 4)$ ,as claimed.

$\square$even, and $\eta(a)=1$ . Then from theRemark 16 Let $p\equiv 3(mod 4),$ $q=p^{f},$proof of the above lemma, we see that $I_{4}(a)=-1+2{\rm Re} J(\lambda, \eta)$ if order of inis mod 4, $I_{4}(a)=-1-2{\rm Re} J(\lambda, \eta)$ if order of is 2 mod 4. Note that the value ofis independent of the choice of . Therfore $I_{4}(a)=-1\pm 2p^{f/2}$ .
$f$

$a$

$0$

$F_{q}^{*}$$a$$I_{4}(a)$$\lambda$

Lemma 17 Let be an odd prtme such that $p\equiv 3(mod 4)$ , and $q=p^{f}$ . Let $a\in F_{q}$and $\eta(a)=-1$ . Then:
$p$

1. If

$f$

2. If

$f$

is even, there are $b\in F_{q}$ and $j\in F_{p}$ such that $\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ .

is odd and $p>3$ , there are $b\in F_{q}$ and $j\in F_{p}$ such that $\eta(b^{4}+a)\eta((b+$$j)^{4}+a)=-1$.

3. If $f>1$ is odd, $p=3_{f}$ and$\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$$a\not\in F_{3}$.

, there are $b\in F_{q}$ and $j\in F_{p}$ such that

13

Proof. For 1., we first note that $x^{4}+a=0$ has no solutions insince $\eta(-1)=1$ and$\eta(-a)=-1$ . Suppose not. Then, for any $c\in F_{q},$ $\eta(x^{4}+a)$ assumes the same value
for $\{c,c+1, \ldots , c+p-1\}$ . Therefore,
must be mod , a contradiction.
, sinceFor 2., we first note that $x^{4}+a=0$ has exactly two solutions in , say,$\eta(-1)=-1$ and $\eta(a)=-1$ .
Suppose not. Then, for any $c\in F_{q}$ such that $c\pm e\not\in F_{p},$ $\eta(x^{4}+a)$ assumes thesame value for $\{c, c+1, \ldots , c+p-1\}$ .
, $e+$
If $e-(-e)=2e\not\in F_{p}$ , then $\eta(x^{4}+a)$ assumes the same value for { $e+1,$$p-1\}$ except , and similarly for $\{-e, -e+1, \ldots , -e+p-1\}$ except $-e$ . Notingthat $\eta(-e+j)=-\eta(e-j),$must be mod . Thus we get a contradiction
since $I_{4}(a)=-1$ .
$a\in F_{p}$ .
If $2e\in F_{p)}$ then it follows that
Let be the quadratic character of. Then we see that $\eta(c)=\eta’(c)$ for all $c\in F_{p}$ since is odd. Therfore we have$0$

$I_{4}(a)$$F_{q}$$p$$\pm e$$F_{q}$$e,$$\ldots$$e$$0$$I_{4}(a)$$p$$\pm e,$$\eta’$$f$$F_{p}$$\sum_{c\in F_{p}}\eta(c)=\sum_{c\in P_{p}}\eta’(c)=-1$

So it is not the case that $\eta(x^{4}+a)$ assumes the same value for $\{0,1, \ldots p-1\}$ except
slnce $p\geq 7$ . Hence there areandsuch that $\eta(b^{4}+a)\eta((b+i)^{4}+a)=$$\pm e$$-1$$i\in F_{p}$$b\in F_{q}$.

For 3., noting $that\pm e\not\in F_{3}$ , we can prove the assertion.

$\square$

There are no elements $b,j\in F_{3}$ such that $\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ , for 2 isthe only element such that $\eta(2)=-1$ and $\eta(1^{4}+2)=\eta(2^{4}+2)=0$ . For F3 with
odd and $a=2$ , we cannot say whether or not there are $b\in F_{q}$ and $j\in F_{p}$ such that$\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ in general.
On the other hand, we cannot establish an explicit formula for $q=p^{f}$ with$(mod 4)$ . For example, in F5, $\eta(2)=-1$ and $I_{4}(2)=-5,$ $\eta(3)=-1$ and $I_{4}(3)=3$ .
However we are interested in residue fields of $F_{n}=\mathbb{Q}(\cos(2\pi/l^{n})$ with $l>5$ . Letof isbe a prime oflying above a rational prime . Then the residue degree$p\neql$, letdetermined as follows:
for lying above , and for lying above$p^{f}\equiv1(modl^{n})$, then is either or $f/2$ .
be the smallest positive integer such that
From now on, we let be a prime greater than 5 and let be an odd prime other tha, and denote. We denote by one of primes of such thatby, wherethe residue degree of
denotes the ringat , that is,.
of algebraic integers of . We denoteby . Obviously,
Since is either the order of mod or its half, hence $(f_{1}, l)=1$ . Let $p^{f}1=1+kl$ .
We easily see that if $(k, l)=1$ , then $f_{n}=f_{1}l^{n-1}$ for all , and if $k=l^{b}g$ with$gcd(g, l)=1$ and $b>1$ , then $f_{1}=f_{2}=\cdots=f_{b+1}$ and $f_{b+h}=f_{1}l^{h-1}$ if $h>1$ . In
either case, there issuch that $f_{m+1}=f_{m}l$ for all $m\geq n_{0}$ .
for all
We again easily see that if $\eta(c)=1$ inwith $n\geq 1$ , then $\eta(c)=1$ in. So we use the same symbol for character$m>n$ , and similarly for $\eta(c)=-1$ in
$f$

$f$

$p\equiv 1$$F_{n}$$\mathfrak{p}$$p$$f_{\mathfrak{p}}=1$$f_{\mathfrak{p}}$$l$$\mathfrak{p}$$f$$\mathfrak{p}$$f$$f_{\mathfrak{p}}$

$l$

$p$

$l$

$F_{n}$$\mathfrak{p}_{n}$$f_{n}$$F_{n}$$p\subset \mathfrak{p}_{1}\subset \mathfrak{p}_{2}\subset \mathfrak{p}_{3}\subset\cdots$$F_{p^{f_{\hslash}}}$$p$$O_{n}$$F_{p^{f_{n}}}=O_{n}/\mathfrak{p}_{n}$$\mathfrak{p}_{n}$$F_{n}$$f_{1}$$\mathfrak{p}$$F_{p}\subset\overline{F}_{1}\subset\overline{F}_{2}\subset\cdots$$F_{n}$$l$$n$$n_{0}$$\overline{F}_{m}$$F_{n}$$F_{n}$$\eta$

14

with $n\geq 1$ . We denote by the quadratic character of . Note that
if $l\equiv-1(mod 4)$ , then $\eta’(c)=\eta(c)$ for all $c\in F_{p}$ since $[F_{1} : \mathbb{Q}]=(l-1)/2$ is odd,
and in case of $l\equiv 1(mod 4),$ $\eta’(c)=1$ for all $c\in F_{p}$ if is even and $\eta’(c)=\eta(c)$ for
all $c\in F_{p}$ since $[F_{1} : \mathbb{Q}]=(l-1)/2]$ is even.
We first deal with the case of F3 with odd and $a=2$ . Note that 2 is the only
element of F3 with $\eta(a)=-1$ .

sums of all$F_{n}$$\eta’$$\mathbb{F}_{p}$$f_{1}$$f$$f$

Lemma 18 Let $p=3$ and $l\equiv-1(mod 4)$ . Then there is $N$ such that there are
and $j\in F_{3}$ such that $\eta(b^{4}+2)\eta((b+j)^{4}+2)=-1$ for all $n\geq N$ .

$b\in\overline{F}_{n}$Proof. We will show that if $f\geq 3$ is odd, then there aresuchthat $\eta(b^{4}+2)\eta((b+j)^{4}+2)=-1$ . Suppose not. Since $I_{4}(2)=-1,$ $\eta(2)=-1$ , and$\eta(1^{4}+2)=\eta(2^{4}+2)=0$ , we have
. Let $q=3^{f}$ . Since thesolution of $x^{4}+2=0$ inare {1, 2}, the number of the set $A=\{c\in F_{q}\backslash F_{S}$ :$\eta(c^{4}+2)=1\}$ is $(q-3)/2$ .
Now we consider the following system of equations.
$b\in F_{3J}$and$j\in \mathbb{F}_{3}$$\sum_{c\in F_{3^{f}}\backslash Fs}\eta(c^{4}+2)=0$$F_{q}$$y^{2}-x^{4}+1$$=0$$z^{2}-(x+1)^{4}+1$$=0$$w^{2}-(x+2)^{4}+1$$=0$

We consider the number of common solutions of these equations in . By assumption
we have $\eta(c^{4}+2)=\eta((c+1)^{4}+\cdot 2)=\eta((c+2)^{4}+2)=1$ or $\eta(c^{4}+2)=\eta((c+1)^{4}+2)=$$\eta((c+2)^{4}+2)=-1$ for any. Therefore the number of common solutions
is $(q-1)/2\cross(q-1)^{3}=(q-1)^{4}/2$ .
On the other hand, the equation$\mathbb{F}_{q}^{4}$$c\in F_{q}\backslash F_{3}$$(y^{2}-x^{4}+1)(z^{2}-(x+1)^{4}+1)(w^{2}-(x+2)^{4}+)=0$

has at most 12 solutions inby [5, p. 275]. Hence we get $(q-1)^{4}<12q^{3}$ , a$q\geq3^{3}=27$
contradiction since.
, every is odd since $l\equiv-1(mod 4)$ , and obviously there is
As for$3\leqf_{N}\leq
f_{N+1}\leq f_{N+2}\leq\cdots$ .
that
such
$q^{3}$$\overline{F}_{n}=F_{3^{f_{n}}}$$F_{q}^{3}$$f_{n}$$N\square$

Lemma 19 We let $p\equiv 1(mod 4)$ . For any $n\geq 1$ and for any
wzth $\eta(a)=$$-1$ , there is $N>n$ such that there are
and $j\in F_{p}$ such that $\eta(b^{4}+a)\eta((b+$$a\in\overline{F}_{n}$$b\in\overline{F}_{N}$$j)^{4}+a)=-1$ .

Proof. We first note that $\eta(-1)=1$ in all$x^{4}+a=0$ has no solutions in all$\overline{F}_{m}$.

$\overline{F}_{m}$since every power of is 1 mod 4, hence
$p$

15

Fix anddose not divide. Here$a\in\overline{F}_{n}$

$n$

$p$

$a)$

with $\eta(a)=-1$ . Take an integer $n’$ such that $n’> \max(n, n_{0})$ . If$\eta(c^{4}+a)$ , we are done. Suppose that
divides. Let $q=p^{f_{n’}}$ . We again use the formula$p$$\sum_{c\in P}.,$$\overline{F}_{n’}=F_{p^{f_{n}}},$$I_{n}(a)= \eta(a)\sum_{j=1}^{d-1}\lambda^{j}(-a)J(\lambda^{j}, \eta)$$\sum_{c\in F_{n}},$$\eta(c^{4}+$,We have that
$J( \lambda^{2}, \eta)=-\frac{1}{q}G(\eta, \chi_{1})^{2}$,

as before. But this time we have by [5, p. 199] that
$G(\eta, \chi_{1})=(-1)^{f-1}q^{1/2}$.

Therefore we get$I_{4}(a)=\eta(a)(\lambda(-a)J(\lambda, \eta)-\lambda^{2}(-a)+\lambda^{3}(-a)J(\lambda^{3},\eta))$

Since$\eta=\lambda^{2}$.

and $\eta(-1)=1$ , we have.

$I_{4}(a)=\lambda^{3}(-a)J(\lambda, \eta)-1+\lambda(-a)\overline{J(\lambda,\eta)}$

Here we have that $\lambda(-a)=\pm i$ since $\eta(-a)=-1$ . Then
$I_{4}(a)=\{\begin{array}{ll}-1+2{\rm Im} J(\lambda, \eta) if \lambda(-a)=i-1-2{\rm Im} J(\lambda, \eta) if \lambda(-a)=-i\end{array}$in the same way as before. We also canfor any $b\in F_{q}$ with $\eta(d)=-1$ similarly. Notethat $\lambda(-1)=\pm 1$ since $\eta(-1)=1$ . We see that $\lambda(-1)=1$ if $q\equiv 1(mod 8)$ , and$\lambda(-1)=-1$ if $q\equiv 5(mod 8)$
At the same time We can show that $\frac{1}{2}H_{2}(1)\equiv-1(mod 4)$ in the similar way
as before. Further we can show that $\frac{1}{2}H_{2}(d)\equiv-2k(mod 4)$ with $k=(q-1)/4$
similarly.
, there are
We will show that inand $j\in F_{p}$ such that$a)\eta((b+j)^{4}+a)=-1$ . It is enough to show thatis not divisible by .
It is proved in [5, p. 210] that

We can show thatshow that${\rm Re} J( \lambda, \eta)=\frac{1}{2}\lambda(-1)H_{2}(1)$${\rm Im} J( \lambda, \eta)=\frac{1}{2}\lambda(-1)H_{2}(d)$$\overline{F}_{n’+1}=F_{q^{l}}$$\eta(b^{4}+$$b\in F_{q^{l}}$$p$$\sum_{c\in F_{q^{l}}}I_{4}(a)$$J(\lambda_{1}’, \ldots\lambda_{k}’)=(-1)^{(\epsilon-1)(k-1)}J(\lambda_{1}, \ldots\lambda_{k})^{\epsilon}$

,

where
are multiplicative characters of , not all of which are trivial, and, respectively, of .
which are lifted to characters$\lambda_{1},$$\ldots\lambda_{k}$$\mathbb{F}_{q}$$\lambda_{1}’,$$\ldots\lambda_{k}’$$F_{q}\iota$

16

for all
We say that is lifted to if. The quadratic
is lifted to the quadratic character of , and characters of order 4 ofcharacter ofare lifted to characters of order 4 of , since$\eta’(c)=\eta(c)$ as stated before. Now
and $(q^{l}-1)/(q-1)$ is odd. Note that for
we consider characters of order 4. Note that there are two characters of order 4 in
if $q\equiv 1(mod 4)$ , and there are none if $q\equiv-1(mod 4)$ . Let be a quadratic. Obviously, forcharacter ofand be lifted to ofwith $\lambda(c)=\pm 1$ , wehave that $\lambda’(c)=\pm 1$ , respectively. Since $(q^{l}-1)/(q-1)\equiv l(mod 4)$ , we have that,
if $l\equiv 1(mod 4)$ respectively, andforwith $\lambda(c)=\pm i,$if$l\equiv-1(mod 4)$ respectively.
$\lambda’(-a)=\lambda(-a)$ if $l\equiv 1(mod 4)$ ,Consequently, we have thatand
if $l\equiv-1(mod 4)$ .
, we haveOn the other hand, also in$\lambda_{j}$$c\in F_{q^{I}}$$\lambda’(c)=\lambda(N_{F_{q^{l}}/F_{q}}(c))$$\lambda_{j}’$$F_{q}$$F_{q^{l}}$$F_{q}$$N_{F_{q^{l}}/F_{q}}(c)=cc^{q}\cdots c^{q^{\iota-1}}=c^{(q^{l}-1)/(q-1)}$$F_{q^{l}}$$c\in F_{q},$$\lambda$$F_{q}$$F_{q}$$\lambda’$$\lambda$$c\in F_{q}$$F_{q^{l}}$$\lambda’(c)=\pm i$$c\in F_{q}$$\lambda’(c)=\mp i$$J(\lambda’,\eta)=J(\lambda, \eta)^{l},$$\lambda’(-a)=\overline{\lambda(-a)}$$F_{q^{l}}$$I_{4}(a)=\{\begin{array}{ll}-1+2{\rm Im} J(\lambda’, \eta) if \lambda’(-a)=i-1-2{\rm Im} J(\lambda’, \eta) if \lambda’(-a)=-i\end{array}$similarly.
We first let $l\equiv-1(mod 4)$ . Let$J(\lambda, \eta)=A+Bi,$ $J(\lambda’,\eta)=A’+B’i$$I_{4}’(a)=-1\mp 2B’$Hence we get, respectively. Since

denote the character sum inand let$\lambda(-a)=\pm
i$, then $I_{4}(a)=-1\pm 2B$ andIf, we have $A’+B’i=(A+Bi)^{l}$ .

$I_{4}’(a)$.

$F_{q^{l}}$$J(\lambda’, \eta)=J(\lambda,\eta)^{l}$$B’=(\begin{array}{l}l1\end{array})A^{l-1}B-(\begin{array}{l}l3\end{array})A^{l-3}B^{3}+\cdots+(-1)^{(j-1)/2}(\begin{array}{l}lj\end{array})A^{l-j}B^{j}+\cdots-B^{l}$.

Let $\lambda(-a)=i$ . By the assumption that $I_{4}(a)\equiv 0(mod p)$ , we have $B\equiv 1/2$$(mod p)$ . On the other hand, by $|J(\lambda, \eta)|=q^{1/2}$ , we have $A^{2}\equiv-1/4(mod p)$ .
Hence we get $B^{j}\equiv-1(mod p)$ and $I_{4}’(a)=-1-2B’\equiv 1(mod p)$ . In case of$\lambda(-a)=i$ , we have that $B^{j}\equiv 0(mod p)$ and $I_{4}’(a)=-1-2B’\equiv-1(mod p)$ . Thus
we are done.
$I_{4}(a)=-1\pm 2B$ andSecondly, we let $l\equiv 1(mod 4)$ . Then, if$I_{4}’(a)=-1\pm 2B’$ , respectively. Similarly, we have that $B’\equiv-1(mod p)$ and$I_{4}’(a)=-1+2B’\equiv-3(mod p)1f\lambda(-a)=i$ , and $B’\equiv 0(mod p)$ and$-1-2B’\equiv-1(mod p)$ if $\lambda(-a)=i$ . Thus we are done since $p\equiv 1(mod 4)$ .
In , there iswith $\eta’(a)=-1$ such that there are no and no $j\in F_{5}$ suchthat $\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ : take $a=2$ , then $\eta’(2)=\eta’(1+2)=\eta’(2^{4}+2)=$$...=\eta’(4^{4}+2)=-1$ . In case of $l\equiv 1(mod 4)$ and $\eta(c)=1$ for all $c\in$ F5 (forexample, let $l=13$ )) we dont know that whether or not there is $N\geq 1$ in which thereareand $j\in F_{5}$ such that $\eta(b^{4}+2)\eta((b+j)^{4}+2)=-1$ . However for primes
greater than 5, we have the following:$\lambda(-a)\cdot=\pm i,$$I_{4}’(a)=\square$$\mathbb{F}_{5}$$b\in\overline{F}_{N}$$a\in F_{S}$

$b$

17

Lemma 20 Let

be a prime greater than 5, then there are $b,j\in F_{p}$ such that$\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ for any $a\in F_{p}$ .
$p$Proof. From the formula$I_{n}(a)= \eta(a)\sum_{j=1}^{d-1}\lambda^{j}(-a)J(\lambda^{j}, \eta)$,

we get $|I_{4}(a)|\leq(d-1)p^{1/2}$ , where $d=(4,p-1)$ . Hence $|I_{4}(a)|\leq 3\sqrt{p}$ if $p\equiv 1$$(mod 4)$ , andif $p\equiv-1(mod 4)$ . Therefore $|I_{4}(a)|<p-2$ , and the$x^{4}+a$assertion follows sincehas possively two solutions in case of $p\equiv-1(mod 4)\square$
and $\eta’(a)=-1$ .
$|I_{4}(a)|\leq\sqrt{p}$Some properties of5$\psi(K_{l})$In this section we let is a prime greater than 5 and $-1$ mod 4, and we keep thenotation of section 2. Note that under the assumption ofis odd for all .
We will give some properties of. We recall thatais formula
$l$

$l,$

$\psi(t)$$\varphi(s, u,t)$$n$$\psi(t)$,is a formula$\exists x,$For\mathbb{Q}]$$u(\forall c(\varphi(s, u, c)arrow\varphi(s,u, c+1))arrow\varphi(s, u, t))$$\forall s,$

and$[F_{n} :$y,$ $z$(1–abt $4=x^{2}-sy^{2}-uz^{2}$ ).

, we let $S_{n}=$ {spots on: $(a,$}, and let $H_{n}=\{(a,b)\in$: all spots individe 2}.
Furthermore we recall that there are $b\in K_{l}$ such that$a,$ $b\in F_{n}$$\mathfrak{p}prime$$F_{\mathfrak{n}}\cross F_{n}$$b)_{\mathfrak{p}}=-1$$F_{n}$$S_{n}$$a,$$K_{l}$$\models\forall c(\varphi(a, b, c)arrow\varphi(a, b, c+1))arrow\varphi(a, b, t))$$K_{l}$$\models\exists x,$$y,$ $z(1-ab\alpha^{4}=x^{2}-sy^{2}-uz^{2})$andfor any$\alpha\in O_{k_{l}}$,and insuch that $b\in F_{n},$for allandis odd if
by the proof of Theorem 12.
Generally we can prove the following proposition. From now on the ring of integersofis denoted by, its maximal ideal is also denoted by , its residue fieldby, and the group of units inby. For, we denote by
its residue class in. FUrther we let lie above a rational prime .
$F_{n}$$\nu_{\mathfrak{p}}(-ab)\geq 1$$a,$$(F_{n})_{\mathfrak{p}}$$\mathfrak{p}\in S_{n}$$\nu_{\mathfrak{p}}(-ab)$$\mathfrak{p}|2$$(0_{n})_{\mathfrak{p}}$$\mathfrak{p}$$\overline{(F_{n})_{\mathfrak{p}}}$$(0_{n})_{\mathfrak{p}}/\mathfrak{p}$$(0_{n})_{\mathfrak{p}}$$\overline{\alpha}$$\alpha\in \mathbb{F}_{n}$$(U_{n})_{\mathfrak{p}}$$\overline{(F_{n})_{\mathfrak{p}}}$$p$$\mathfrak{p}$Proposition 21 Let$a,$ $b\in K_{l}^{*}$and suppose that

$K_{l}\models\forall c(\varphi(a, b,c)arrow\varphi(a, b, c+1))$

and insuch that
Then we have$F_{n}$$a,$ $b\in F_{n},$$\nu_{\mathfrak{p}}(-ab)\geq 1$$K_{l}\models\varphi(a, b, \alpha)$

for all

for all$\alpha\in O_{K_{l}}$.

$\mathfrak{p}\in S_{n}$and$\nu_{\mathfrak{p}}(-ab)$is odd$if\mathfrak{p}|2$.

18

Proof. Suppose thatfor all$\varphi(a, b, j)$such that$\alpha,$. Fix such . Since
for some, we have. We easily see that $1-ab\alpha^{4}\neq 0$ . Take
, then we have $n_{0}>0$ and$\alpha\in O_{K_{l}}$$K_{l}\models\neg\varphi(a, b, \alpha)$$j\in \mathbb{Z}$$\alpha$$\alpha\not\in \mathbb{Z}$$a,$ $b\in F_{n_{0}}$$F_{n0}\models\neg\varphi(a, b, \alpha)$$K_{l}\models$be$n_{0}$.

By Lemma 1, we have$(1-ab\alpha^{4})/(-ab)=\alpha^{4}-1/ab\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{\alpha 2}$. Fix such .
such thatfor some
is Archimedian. Then
is not Archimedian. Suppose that
We claim that
. We can take $n_{1}>no$ such that
there is in $\in N$ such thatlying above
be a valuation ofsince $K_{l}\models\varphi(a, b, m)$ . Let, we have. Since. Then we have, a contradiction.
. Hence by Lemma 1, we haveis not Archimedian.
Therfore, and for $n>n_{0},$ consists of primes of which lie above each
We have, by Lemma 10 and by the above argument for Archimedian ones. We
prime inis not Archimedian similarly.
see that every prime in$(a, b)_{\mathfrak{p}_{0}}=-1$$\mathfrak{p}_{0}$$\mathfrak{p}_{0}$$\mathfrak{p}_{0}$$\mathfrak{p}_{0}$$m^{4}-1/ab\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$F_{n_{1}}\models\varphi(a, b, m)$$F_{n_{1}}$$\mathfrak{p}_{0}’$$(F_{no})_{\mathfrak{p}0}=(F_{n_{1}})_{\mathfrak{p}_{0}’}\simeq \mathbb{R}$$m^{4}-1/ab\in(F_{n_{1}})_{\mathfrak{p}_{0}}^{*2}$$\mathfrak{p}_{0}$$F_{n_{1}}\models\neg\varphi(a, b, m)$$(a, b)_{\mathfrak{p}_{\acute{0}}}=-1$$\mathfrak{p}_{0}$$F_{n}$$S_{n}$$S_{n0}\neq\emptyset$$S_{no}$$S_{no}$.
caselying abovefor
We claim that if $n\geq n_{0},$$n\geqn_{0}$
and for all
for alland . We note that
. We can take $n’>n$ such that
. Let be a prime of lying above . Then we have
$1:\mathfrak{p}_{0}p$$\nu_{\mathfrak{p}}(-ab)=0$$\mathfrak{p}$$\mathfrak{p}\in S_{n}$$-ab\not\in(F_{n})_{\mathfrak{p}}^{*2}$$n$. Fix such$\mathfrak{p}\in S_{n}$, since$F_{n’}\models\varphi(a, b, 1)$$(a, b)_{\mathfrak{p}}=(a, -ab)_{\mathfrak{p}}=-1$$K_{l}\models\varphi(a, b, 1)$$\mathfrak{p}_{0}$$F_{n’}$$\mathfrak{p}’$$(1-ab)/(-ab)=1-1/ab\not\in(F_{n’})_{\mathfrak{p}}^{s2}$.

( , pp. 163]). Hencefor
It is known that
. On the other hand, we have
$1+\mathfrak{p}=(1+\mathfrak{p})^{2}$since$\mathfrak{p}$$\mathfrak{p}\parallel 2$$[6$$\nu_{\mathfrak{p}’}(-1/ab)\leq 0$, so$\nu_{\mathfrak{p}’}(-ab)\geq 0$$(1-ab\alpha^{4})/(-ab)=\alpha^{4}-1/ab\in(F_{n’})\mathfrak{p}^{\prime r2}$.
since
, thenIf
contradiction. Therefore we have$(F_{n0})_{\mathfrak{p}_{0}}\subseteq(F_{n’})_{\mathfrak{p}’}$$\nu_{\mathfrak{p}’}(-ab)>0$.
Case
We first note that

$1-ab\alpha^{4}\in(F_{n’})_{\mathfrak{p}}^{*2}$$\nu_{\mathfrak{p}’}(-ab)=0$

since
, and

$\alpha\in O_{n’}$, a, hence, a contradiction.
$-ab\in F_{\mathfrak{p}}^{*2}$$\nu_{\mathfrak{p}}(-ab)=0$$2:\mathfrak{p}_{0}|2$$\nu_{\mathfrak{p}_{0}}(2)=1$

since$\mathfrak{p}_{0}$is unramified. Similarly as before, we have$-1/ab\not\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$(1-ab)/(-ab)$$(1-ab\alpha^{4})/(-ab)$$=$$1-1/ab\not\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$=\alpha^{4}-1/ab\in(F_{n0})_{\mathfrak{p}_{0}}^{s2}$

(1)

(2)
(3)

19

( , pp. 163]). So we have
It is known that $(1 +p^{r})^{2}=1+2p^{r}$ ifbyby (2) and. Hence we have(3). It follows that $-3<\nu_{\mathfrak{p}_{0}}(-ab)<3$ . Further we see thatby
, a contradiction since(3). If, then we have
. Therefore we have $\nu_{Po}(-1/ab)\neq-1$ .
,. In caseWe will show that. Suppose that
we have, henceby (3), a contradiction. In case.
, let $A=1-1/ab,$ $B=\alpha^{4}-1/ab$ . Then we have
$A\neq0$, hence, we have
Notingandand, letting $A=-1/ab$ we would, a contradiction. In case
so
,
.
see
have
that
a contradiction. Thus we
Let. Every elements ofbe the group ofroots of unity inare squares in. Let C’ $=C\cup\{0\}$ . Let. We can wright$\delta=c_{0}+c_{1}2+c_{2}2^{2}+\cdots$ , for some $c_{i}\in C’$ with. We easily see that, then we
iff $c_{1}=0$ and $c_{2}/c_{0}\equiv c(c+1)(mod \mathfrak{p}_{0})$ for some $c\in C’$ . If$C$
havein . Hence we see
since
for somethatby (3). Accordingly $\nu_{\mathfrak{p}_{0}}(-1/ab)=0or\pm 2,hence\nu_{0}(-ab)=0\square$$or\pm 2$ , a contradiction.
FVrthermore we can prove the following.
$[6$$\mathfrak{p}^{r}\subseteq 2\mathfrak{p}$$\nu_{\mathfrak{p}_{0}}(-ab)<3$$\nu_{\mathfrak{p}_{0}}(-1/ab)<3$$1+\mathfrak{p}_{\mathfrak{p}_{0}}^{3}=(1+\mathfrak{p}_{\mathfrak{p}_{0}}^{2})^{2}$$0\leq\nu_{\mathfrak{p}_{0}}(\alpha)<2$$\nu_{\mathfrak{p}_{0}}(-1/ab)=-1$$\nu_{\mathfrak{p}_{0}}(\alpha^{4}-1/ab)=-1$$\alpha^{4}-1/ab\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$\nu_{\mathfrak{p}_{0}}(-1/ab)<0$$\nu_{\mathfrak{p}0}(\alpha)=1$$\nu_{\mathfrak{p}0}(\alpha)=0$$-ab\in(F_{n0})_{\mathfrak{p}0}^{*2}$$1-ab\alpha^{4}\in(F_{n0})_{\mathfrak{p}0}^{*2}$$A\equiv B(mod \mathfrak{p}_{0}^{3})$$\nu_{\mathfrak{p}_{0}}(-1/ab)>0$$B/A\equiv 1(mod \mathfrak{p}_{0}^{3})$$\nu_{\mathfrak{p}_{0}}(A)=0$$\nu_{\mathfrak{p}0}(-1/ab)=0$$A\in(F_{n0})_{\mathfrak{p}_{0}}^{*2}$$\nu_{\mathfrak{p}_{0}}(\alpha)=0$$A\in(F_{n_{0}})_{\mathfrak{p}0}^{*2}$$C$$(N\mathfrak{p}-1)^{th}$$C$$B/A\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$(F_{n0})_{\mathfrak{p}_{0}}$$\delta\in(U_{n_{0}})_{\mathfrak{p}_{0}}$$(F_{n0})_{\mathfrak{p}_{0}}$$c_{0}\neq 0$$\delta\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{2}$$\nu_{\mathfrak{p}0}(-1/ab)=1$$c_{0}\neq 0$$\alpha^{4}\equiv c_{0}^{4}(mod \mathfrak{p}_{0}^{3})$$\alpha^{4}-1/ab\not\in(F_{no})_{\mathfrak{p}_{0}}^{*2}$$\nu_{\mathfrak{p}_{0}}(-1/ab)\neq 1$

Proposition 22 Let$a,$ $b\in K_{l}^{*}$and suppose that

$K_{l}\models\forall c(\varphi(a, b, c)arrow\varphi(a, b,c+1))$

and insuch that
Then we have$F_{n}$$a,$ $b\in F_{n},$$\nu_{\mathfrak{p}}(-ab)=0$$K_{l}\models\varphi(a, b, \alpha)$

Proof. Suppse thatfor all$K_{l}\models\neg\varphi(a, b, \alpha)$

andfor all$\mathfrak{p}\beta$$\alpha\in O_{K_{l}}$.

for some
anda spot$\alpha\in O_{K_{l}}$$\mathfrak{p}\in S_{n}$.

. Fix such alpha. Then we

haveof. We see that is afor some
prime ofas before.
It is known that for.
iffin
in caseHence we see that $\eta(\overline{-1/ab})=-1$ in. Let $d=-1/ab$ .
with $n\geq n_{0}$ and
By Lemma 17, 18, 19, there are $n_{1}\geq n0,$,with
a prime ofand $j_{0}\in\{1, \ldots p-1\}$ such thatin
We mayassume thatwithout of loss of generality.
andWe can take
. Let $S_{n_{1}}=$such thatsince. By the Chinese Remainder Theorem, there aresuch that$\alpha^{4}-1/ab\not\in(F_{n_{0}})_{\mathfrak{p}_{0}}^{*2}$$n_{0}$$\mathfrak{p}_{0}$$F_{n_{0}}$$\mathfrak{p}_{0}$$\mathfrak{p}_{0}$$F_{n0}$$\alpha\in(U_{n})_{\mathfrak{p}},$$\overline{(F_{n})_{\mathfrak{p}}}$$\eta(\overline{\alpha})=1$$\alpha\in(F_{n})_{\mathfrak{p}}^{*2}$$\mathfrak{p}|\mathfrak{p}_{0}$$(F_{n})_{\mathfrak{p}^{2}}^{*}$$F_{n_{1}}$$\mathfrak{P}_{1}$$\mathfrak{P}_{1}|\mathfrak{p}_{0},$$\eta(\overline{b}^{4}+\overline{d})\eta((\overline{b}+\overline{j}_{0})^{4}+\overline{d})=-1$$\eta(\overline{b}^{4}+\overline{d})=-1$$\overline{\beta}=\overline{b}$$O_{n_{1}}/\mathfrak{P}_{1}\simeq 0_{n_{1}}/\mathfrak{P}_{1}$$\gamma\in O_{n}$$\{\mathfrak{P}_{1}, \ldots , \mathfrak{P}_{k}\}$$\gamma\equiv\beta$$\gamma$$\equiv$$\gamma\equiv$$(mod \mathfrak{P}_{1})$1$(mod \mathfrak{P}_{i})$$3$$(mod \mathfrak{P}_{j}^{3})$ifif$\mathfrak{P}_{i}\#$$\mathfrak{P}_{J}|2$.

,$\overline{b}\in\overline{(F_{n\iota})_{\mathfrak{P}_{1}}}$$\overline{(F_{\mathfrak{n}_{1}})_{\mathfrak{P}}1}$$\eta((\overline{b}+\overline{j}_{0})^{4}+\overline{d})=1$$\beta\in O_{n_{1}}$$\mathfrak{p}\Lambda^{2}$

20, we claim that
. SinceinLetwith $i\neq 1$ . Take $n’>n_{1}$ so thatlying above . Then we havebe a prime ofFor$\mathfrak{P}_{i}\beta$$\gamma^{4}-1/ab\not\in(F_{n_{1}})_{\mathfrak{P}}^{r2}1$$\mathfrak{P}’$$\overline{\gamma}=\overline{\beta}$$\gamma^{4}-1/ab\not\in(F_{n_{1}})_{\mathfrak{P}_{i}}^{*2}$$\mathfrak{P}_{i}\parallel 2$$F_{n’}$$\overline{(F_{n_{1}})_{\mathfrak{P}_{1}}}$, we have that

$F_{n’}\models\varphi(a,b, 1)$. Let$\mathfrak{P}_{i}$$(1-ab)/(-ab)=1-1/ab\not\in F_{\mathfrak{P}’}^{*2}$. Since,, we are done.
, we see that
since
and
, hence. Consequently we have that.
, we see thatin
Now since. We claim thathence we have that
. It isenough to show thatfor all $n’>n_{1}$ . Take $n’>n_{1}$ and a prime
lying aboveof. Then since, we know
also in, hence we have thatthat.
, a contradiction, since
Therefore we get
hence
For

$1-1/ab\not\in F_{\mathfrak{P}:}^{*2}$$\mathfrak{P}j|2$$\overline{\gamma}=\overline{1}$in$\overline{(F_{n_{1}})_{\mathfrak{P}:}}$$3+\mathfrak{P}_{j}^{3}=1+2+\mathfrak{P}_{j}^{3}\not\in$$\nu_{\mathfrak{P}_{j}}(2)=1$$\gamma\not\in(F_{n_{1}})_{\mathfrak{P}_{j}}^{*2}$$K_{l}\models\varphi(a, b,\gamma)$$F_{n\iota}\models\varphi(a, b, \gamma)$$(U_{n_{1}})_{\mathfrak{P}_{j}}^{2}$$\eta((\overline{b}+j_{0})^{4}+\overline{d})=1$$\overline{(F_{n_{1}})_{\mathfrak{P}_{1}}}$$(\gamma+j_{0})^{4}-1/ab\in(F_{n_{1}})_{\mathfrak{P}}^{*2}j$$F_{n_{1}}\models\neg\varphi(a, b, \gamma+j_{0})$$K_{l}\models\neg\varphi(a, b,\gamma+j_{0})$$F_{n’}\models\neg\varphi(a, b, \gamma+j_{0})$$\mathfrak{P}’$$F_{n’}$$\eta((\overline{\gamma}+j_{0})^{4}+\overline{d})=1$$\mathfrak{P}_{1}$$(\gamma+j)^{4}-1/ab\in(F_{n’})_{\mathfrak{P}}^{s2},$$F_{n’}\models\neg\varphi(a, b, \gamma+j_{0})$$K_{l}\models\varphi(a, b, \gamma)\wedge\neg\varphi(a, b, \gamma+j_{0})$$\forall c(\varphi(a, b, c)arrow\varphi(a, b, c+1))$.

$\overline{(F_{n’})_{\mathfrak{P}’}}$$K_{l}\models$口

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