independence notions in model theor Citation 数理解析研究所講究録 (2010), 1718:

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1 Uniformly definable subrings som Titleextensions the rationals (New de independence notions in model theor Author(s) Fukuzaki Kenji Citation 数理解析研究所講究録 (2010) 1718: Issue Date URL Right Type Departmental Bulletin Paper Textversion publher Kyoto University

2 using Uniformly definable subrings some infinite algebraic extensions the rationals (Kenji Fukuzaki) Faculty Intercultural Studies The international University Kagoshima Abstract We consider the formulas used by Julia Robinson in her pro that number fields are first order undecidable We extend the result [1] We prove that it defines subrings in some infinite algebraic extensions the rationals As an application we dcuss undecidablities those infinite algebraic extensions 1 Introduction In 1959 Julia Robinson [8] proved that any number field as well as the corresponding $\mathbb{n}$ ring algebraic integers undecidable by showing that $\emptyset$-definable (in the ring language) in the ring the ring $\emptyset$-definable in its number field She first considered the formula $\varphi_{m}(s u t)$ : $x$ $y$ $z(1-sut^{2m}=x^{2}-sy^{2}-uz^{2})$ $\mathfrak{p}^{m}\wedge 2$ where $m$ a positive integer that for all prime ideals a given number field $F$ that $m$ an integer greater than all the ramification indices prime ideals $F$ $\mathfrak{p}_{1}$ which divide 2 Then she proved that for a given prime $F$ there are $a$ $b\in F$ that $\varphi_{m}(a b t)$ defines a finite intersection valuation $\triangle$ $\bigcap_{\mathfrak{p}\in\delta}o_{\mathfrak{p}}$ rings where a finite set primes $F$ $\mathfrak{p}_{1}$ containing (We actually $\mathfrak{p}_{0}$ can define the valuation ring two with some choice those $\varphi_{m}(s t u)$ parameters ) We denote by the solution set in $\varphi_{m}(a b F)$ $\varphi_{m}(a b t)$ $F$ that $\varphi_{m}(a b F)=\{\alpha\in F : F\models\varphi_{m}(a b \alpha)\}$ It easy to see that $\bigcap_{ab\in F}\varphi_{m}(a b F)=0$ Therefore in order to define the ring algebraic integers in a given number field $0_{F}$ $F$ J Robinson considered the intersection all $\mathbb{z}$ $\varphi_{m}(a b F)$ containing which defined by : $\psi_{m}(t)$ $\forall s$ $u(\forall c(\varphi_{m}(s u c)arrow\varphi_{m}(s u c+1))arrow\varphi_{m}(s u t))$ Note that We denote by the solution set $\varphi_{m}(s u t)rightarrow\varphi_{m}(s u -t)$ $\psi_{m}(f)$ $\psi_{m}(t)$ in as $F$ before It possible to define since $0_{F}$ $\mathbb{z}\subseteq\psi_{m}(f)\subseteq 0_{F}$ $F$ has an integral bas over the rationals (The defining formula $0_{F}$ depends on $F$ ) In th paper we calculate the solution set in some infinite algebraic $\psi_{2}(t)$ extensions

3 $F_{\mathfrak{p}}$ $\mathfrak{p}_{1}$ $F_{\mathfrak{p}_{i}}^{*2}$ for we for the ) be Construction $\psi(t)$ Let be a number field $F$ (a finite algebraic extension the rationals let $0_{F}$ $F^{*}$ be the ring algebraic integers F will denote the set non-zero elements $F$ By denote a place $F$ $F_{\mathfrak{p}}$ by completion $F$ with respect to Since non-archimedean places are $F$ p-adic valuations for some prime ideal $F$ we use the same letter for both the place the prime ideal The ring integers $F_{\mathfrak{p}}$ $(0_{F})_{\mathfrak{p}}$ denoted by its maximal ideal also denoted by Let a prime ideal $F$ $a\in F$ $\nu_{\mathfrak{p}}(a)$ By we denote the order at $a$ Given $a$ $b\in F^{*}$ we $(a b)_{\mathfrak{p}}$ use Hilbert symbol which defined to be $+1$ if $ax^{2}+by^{2}=1$ solvable in otherwe defined to be-l For $a$ we $b\in F^{*}$ denote by $S_{F}(a b)$ the set places $F$ that We know that it contains even number places $(a b)_{\mathfrak{p}}=-1$ $F$ The following lemma well-known Lemma 1 A nonzero element $h$ form $x^{2}-ay^{2}-bz^{2}$ in $F$ if only if $h/(-ab)\not\in $(a b)_{\mathfrak{p}}=-1$ can $F$ be represented by the the temary quadmtic for any place F_{\mathfrak{p}}^{*2}$ that Th follows from the properties quaternary quadratic forms the Hasse- Minkowski theorem on quadratic forms See [7 p 187] $\mathfrak{p}_{1}$ Lemma 2 Given even number dtinct prime ideals $\ldots$ $F$ there are $\mathfrak{p}_{2k}$ $b$ $a$ in that for $F^{*}$ $S_{F}(a b)=\{\mathfrak{p}_{1} \ldots \mathfrak{p}_{2k}\}$ $\nu_{\mathfrak{p}_{i}}(a)=1$ $\nu_{\mathfrak{p}_{i}}(b)=0$ $i=1$ $\ldots$ $2k$ Pro By weak approximation we get an element $a$ $F^{*}$ with $\nu_{\mathfrak{p}_{i}}(a)=1$ for all $i$ We know by [7 71:19 Theorem p 203] that there $b\in F^{*}$ that $S_{F}(a b)=$ $\{\mathfrak{p}_{1} \ldots \mathfrak{p}_{2k}\}$ $b$ In the pro [7 71:19 Theorem p 203] we can take with $\nu_{\mathfrak{p}_{i}}(b)=0$ for $i=1$ $\ldots$ $2k$ $\mathfrak{p}_{1}$ J Robinson actually proved in [8 Lemma 9] that given a prime ideal there are relatively prime elements $a$ $b$ in that where $0_{F}$ $(a)=\mathfrak{p}_{1}\cdots \mathfrak{p}_{2k}$ $b$ $\ldots$ are dtinct prime ideals that include every prime ideal dividing 2 $\mathfrak{p}_{2k}$ a totally positive prime element that iff $(a b)_{\mathfrak{p}}=-1$ $\mathfrak{p} a$ $F$ Lemma 3 Let $a$ $b$ $c\in F$ $b$ $a$ If satfy Lemma 2 $m$ be a positive integer $\mathfrak{p}^{m}\int 2$ that for every prime ideal Then $1-abc^{2m}=x^{2}-ay^{2}-bz^{2}$ solvable for in iff $x$ $y$ $z$ $F$ $\nu_{\mathfrak{p}_{i}}(c)\geq 0$ $i$ for each Pro By Lemma 1 $h=1-abc^{2m}$ can be represented by $x^{2}-ay^{2}-bz^{2}$ iff $h/(-ab)\not\in$ $1\leq i\leq 2k$ $\nu_{\mathfrak{p}_{i}}(c)\geq 0$ $i$ If for each $i$ $F_{\mathfrak{p}_{i}}$ square each then we have hence $\nu_{\mathfrak{p}_{i}}(h/(-ab))=-1$ $h/(-ab)$ not a

4 since $\mathfrak{p}_{2}$ $\mathfrak{p}_{3}$ we that for since with 104 $i$ $F_{\mathfrak{p}}$ $\nu_{\mathfrak{p}_{i}}(c)<0$ Suppose for some We know in if $(1+\mathfrak{p}^{r})^{2}=1+2\mathfrak{p}^{r}$ $\mathfrak{p}^{r}\subseteq 2\mathfrak{p}$ by [7 p 163] Noting $h/(-ab)=c^{2m}(1-1/(abc^{2m}))$ we see that $h/(-ab)$ a square $F_{\mathfrak{p}_{i}}$ $\nu_{\mathfrak{p}:}(1/(abc^{2m}))\geq 2m-1$ $\mathfrak{p}^{2m-1}\subseteq 2\mathfrak{p}$ $\square$ Thus we have that if $a$ $b$ satfy Lemma 2 $\varphi_{m}(a b F)=\bigcap_{1\leq i\leq 2k}O_{\mathfrak{p}_{i}}$ $\forall c(\varphi_{m}(a b c)arrow\varphi_{m}(a b c+1)$ holds in $F$ since a ring containing $\varphi_{m}(a b F)$ $\mathbb{z}$ For a given $c\in F^{*}$ there are $a$ $b\in F^{*}$ that since we can $c\not\in\varphi_{m}(a b F)$ construct $a$ that $b\in F^{*}$ $F_{\mathfrak{p}}$ $1-1/(abc^{2m})$ $(a b)_{\mathfrak{p}}=$ a square some $-1$ Noting for all $0\in\varphi_{m}(a b F)$ $a$ we have $b$ $\bigcap_{ab\in F}\varphi_{m}(a b F)=0$ Nevertheless we have that a subset $\psi_{m}(f)$ $\mathbb{z}$ $0_{F}$ containing $\psi_{m}(f)$ the intersection all the solution set $\forall c(\varphi_{m}(a b c)arrow\varphi_{m}(a b c+1))arrow\varphi_{m}(a b t)$ If the preme the above formula fails the solution set $F$ We don t know what But we can show what if $\psi_{m}(f)$ $\psi_{2}(k)$ $K$ a certain infinite algebraic extension $\mathfrak{p}_{1}$ $\mathfrak{p}_{1}$ Remark 4 For a given prime ideal can define the valuation ring Take $\mathfrak{p}_{1}$ three prime ideal $F$ $a$ $b$ $c$ $d\in 0_{F}$ that $S_{F}(a b)=\{\mathfrak{p}_{1} \mathfrak{p}_{2}\}$ $S_{F}(c d)=\{\mathfrak{p}_{1} \mathfrak{p}_{3}\}$ then we easily see that defines $\varphi_{m}(a b F)+\varphi_{m}(c d F)$ $O_{\mathfrak{p}_{1}}$ 3 The solution set $\psi(t)$ in some nfinite algebraic extensions Let $F$ $\mathscr{p}$ be a number field let be an infinite set finite Galo extensions $M$ $F$ that $[M : F]$ odd every prime ideal $M$ dividing 2 unramified in (We say that 2 unramified in Note $M/\mathbb{Q}$ $M/\mathbb{Q}$ $\mathfrak{p}^{2}\parallel 2$ for all prime ideals $)$ M Let $K$ $\mathscr{p}$ be the composite field all fields in Then $K$ an infinite Galo extension $F$ every finite Galo subextension $M$ has odd extension degree over $\mathfrak{o}_{k}$ We denote by the ring algebraic integers $K$ In th section we will prove that the solution set in $\psi_{2}(k)$ $\psi_{2}(t)$ $K$ a subset $\mathbb{z}$ $J\supset_{K}$ containing We need the following lemma which proved in [2 pp ] Lemma 5 Let $M$ $L$ be number fields with $L\supset M$ let $\mathfrak{p}\supset $M$ respectively For $\alpha\in $(\alpha b)_{\mathfrak{p}}=(a b)_{\mathfrak{p}}$ L_{\mathfrak{P}}^{*}$ let $a=n_{l_{\mathfrak{p}}/m_{\mathfrak{p}}}(\alpha)$ The next lemma follows from Lemma 5 \mathfrak{p}$ be $b\in M_{\mathfrak{p}}$ primes $L$ Then we have

5 be be a 105 Lemma 6 Let $L$ be a finite Galo extension a number field $M$ with $[L:M]$ odd Let a prime ideal $M$ $\mathfrak{p}$ let $a$ $b\in M^{*}$ we have $(a b)_{\mathfrak{p}}=1$ iff $(a b)_{\mathfrak{p}}=1$ a prime $L$ lying over Then for Pro Since $L/M$ a Galo extension the local degree at as divides the degree $L/M$ that $[(L)_{\mathfrak{P}} : (M)_{\mathfrak{p}}] [L:M]$ (see [7 p 32]) Let be the local degree at $u$ $\mathfrak{p}$ Then $N_{(L)_{\mathfrak{P}}/(M)_{\mathfrak{p}}}(a)=a^{u}$ $(a b)_{\mathfrak{p}}=(a^{u} b)_{\mathfrak{p}}=(a b)_{\mathfrak{p}}^{u}$ Since odd it follows $u$ that iff $(a b)_{\mathfrak{p}}=1$ $(a b)_{\mathfrak{p}}=1$ $\square$ $\varphi_{2}(s u t)$ We recall that $\exts x$ $y$ $z(1-sut^{4}=x^{2}-sy^{2}-uz^{2})$ $\psi_{2}(t)$ $\forall s$ $u(\forall c(\varphi(s u c)arrow\varphi(s u c+1))arrow\varphi_{2}(s u t))$ Lemma 7 Let $M$ be a subfield $K$ with $M/F$ finite Galo Let $a$ with $ab\neq 0$ Then $M\models\varphi(a b \alpha)$ iff $K\models\varphi(a b \alpha)$ Pro If $M\models\varphi(a b \alpha)$ then we have trivially $K\models\varphi(a b \alpha)$ $b$ $\alpha\in M$ If then for some $M\models\neg\varphi(a b \alpha)$ $(1-ab\alpha^{4})/(-ab)\in M_{\mathfrak{p}}^{*2}$ place $M$ that Let be any subfield $(a b)_{\mathfrak{p}}=-1$ $L$ $K$ with $L/M$ finite Galo let as be a place $M$ lying above Since $[L : M]$ odd we have $(a b)_{\mathfrak{p}}=-1$ $(1-ab\alpha^{4})/(-ab)\in L_{\mathfrak{P}}^{*2}$ Hence Note that for $L\models\neg\varphi(a b \alpha)$ $K\models\neg\varphi(a b \alpha)$ $\mathfrak{p}\subset$ archimedean places as it is also true that $(a$ iff $b)_{\mathfrak{p}}=1$ $(a$ $b)$ Pt $=1$ Theorem 8 The solution set $\psi_{2}(k)$ $\psi_{2}(t)$ $\mathbb{z}(\mathbb{z}\subseteq\psi_{2}(k)\subseteq \mathfrak{o}_{k})$ in $K$ a subset $\mathfrak{o}_{k}$ containing Pro We have trivially Let We show that there are $\mathbb{z}\subseteq\psi_{2}(k)$ $t\in K\backslash 0_{K}$ $a$ $b\in K$ that $K\models\neg\varphi_{2}(a b t)\wedge\forall c(\varphi_{2}(a b c)arrow\varphi_{2}(a b c+1))$ We fix a subfield $M$ $K$ that $[M : F]$ finite $t\in M$ Then we have $\nu_{\mathfrak{p}_{1}}(t)<0$ $\mathfrak{p}_{1}$ for some prime $M$ $\mathfrak{p}_{2}\neq \mathfrak{p}_{1}$ Take a prime $M$ Lemma 2 there By are $a$ $b$ in that for $i=12$ $M^{*}$ $\nu_{\mathfrak{p}_{i}}(a)=1$ $\nu_{\mathfrak{p}_{i}}(b)=0$ $(a b)_{\mathfrak{p}_{i}}=-1$ $t\not\in\varphi_{2}(a b M)$ By Lemma 7 $1-abt^{4}=x^{2}-ay^{2}-bz^{2}$ not solvable for $x$ $y$ in $z$ $K$ Let in $c$ $K$ suppose $K\models\varphi_{2}(a b c)$ Take a subfield $L$ $K$ that $L$ contains $L/M$ a finite Galo extension then we have by $c$ $L\models\varphi_{2}(a b c)$

6 $L_{\mathfrak{P}}$ since $\prime \mathfrak{p}$ \mathfrak{p}$ divides in $h /(-ab)$ \mathfrak{p}_{i}$ are be $\square$ 106 Lemma 7 Let $h=1-abc^{4}$ $\mathfrak{p}_{1}$ $\mathfrak{p}_{k}$ $h =1-ab(c+1)^{4}$ Let $\ldots$ all the primes lying above $L$ $\mathfrak{p}_{k+1}$ $\mathfrak{p}_{1}$ $\ldots$ be all the primes lying above $L$ $\mathfrak{p}k+s$ $\mathfrak{p}_{2}$ By Lemma 5 we have that $S_{L}(a b)=\{\mathfrak{p}_{1} \ldots \mathfrak{p}k+s\}$ $\prime \mathfrak{p}$ $\sigma all the primes $L$ that $(a b)_{\mathfrak{p}}=-1$ $k$ are odd since $L/M$ Galo with odd extension $s$ $L^{\mathfrak{P}_{i}}$ $\mathfrak{p}_{i}$ degree We will show that for all not a square assuming $\mathfrak{p}=\mathfrak{p}_{i}$ $h/(-ab)$ not Take one We will break into cases according to whether or 2 not $\zeta $\mathfrak{p}\parallel Case 1: As mentioned before we have if $(1+\mathfrak{p}^{r})^{2}=1+2\mathfrak{p}^{r}$ $\mathfrak{p}^{r}\subseteq 2\mathfrak{p}$ by [7 p 163] Hence we have If then $h =1-ab(c+1)^{4}$ a square $(1+\mathfrak{P})^{2}=1+\mathfrak{P}$ $\nu_{\mathfrak{p}}(c)\geq 0$ 2$ $\nu_{\mathfrak{p}}(-ab(c+1)^{4})>0$ Since $(a b)_{\mathfrak{p}}=(a -ab)$ as $=-1$ we $have-ab$ not a $L_{\mathfrak{P}}$ square hence $h /(-ab)$ also not We consider the case Since $h/(-ab)=c^{4}(1-1/(abc^{4})$ it follows that $\nu_{\mathfrak{p}}(c)<0$ $\mathfrak{p}_{i}$ $\nu_{\mathfrak{p}}(-abc^{4})\geq 0$ Let Ep lie above let be the ramification index $e=e(\mathfrak{p}/\mathfrak{p}_{i})$ $e$ must be odd since $L/M$ Galo with odd extension degree Hence we have $\nu_{\mathfrak{p}}(-abc^{4})>0$ Then we have $\nu_{\mathfrak{p}}(-ab(c+1)^{4})=\nu_{\mathfrak{p}}(-ab)+4\nu_{\mathfrak{p}}(c)=\nu_{\mathfrak{p}}(-abc^{4})>0$ $L_{\mathfrak{P}}$ hence $h =1-ab(c+1)^{4}$ a square $h /(-ab)$ not Case 2: as$ 2$ $L/\mathbb{Q}$ Since 2 unramified in we have Furthermore $\nu_{\mathfrak{p}}(2)=1$ $\nu_{\mathfrak{p}}(-ab)=1$ we know by [7 p 163] If then $h/(-ab)=c^{4}(1-1/(abc^{4})$ $(1+\mathfrak{P})^{2}=1+\mathfrak{P}^{3}$ $\nu_{\mathfrak{p}}(c)<0$ $L^{\mathfrak{P}}$ would be a square hence we have It follows that $\nu_{\mathfrak{p}}(c)\geq 0$ $\nu_{\mathfrak{p}}(h /(-ab))=-1$ $L_{\mathfrak{P}}$ $h /(-ab)$ not a square Example 9 1 Let $F=\mathbb{Q}((\zeta_{l}))$ $\mathscr{p}$ be a set all where $M_{n}=\mathbb{Q}(\zeta_{l^{n}})(n>1)$ $\zeta_{l^{n}}$ an odd integer $>1$ a primitive $l^{n}$-th root unity $K= \bigcup_{n}m_{n}$ 2 Let $F=\mathbb{Q}$ $\mathscr{p}$ be a set all where $\mathbb{q}(\cos(2\pi/l^{n}))$ $n\in \mathbb{n}$ an odd prime with $l\equiv-1(mod 4)$ $K=\mathbb{Q}(\{\cos(2\pi/l^{n})$ : $n\in N$ a prime $l\equiv-1$ $(mod 4)\})$ Here $O_{\mathfrak{p}_{\mathfrak{i}}}^{M}$ $\varphi_{2}(a b M)=O_{\mathfrak{p}_{1}}^{M}\cap O_{\mathfrak{p}_{2}}^{M}$ Remark 10 In the pro Theorem 8 we have $\mathfrak{p}_{i}$ denotes the valuation ring $M$ But it not necessarily true that $\varphi_{2}(a b L)=$ $\bigcap_{i}o_{\mathfrak{p}_{i}}^{l}$ Actually we have $\varphi_{2}(a b M)\subseteq\bigcap_{i}\mathcal{O}_{\mathfrak{P}_{i}}^{L}\subseteq\varphi_{2}(a b L)$ Nevertheless we can prove for where $\varphi_{2}(a b L)=\bigcap_{i}O_{\mathfrak{P}_{i}}^{L}$ $K= \bigcup_{n}\mathbb{q}(\zeta_{l^{n}})$ an $\zeta_{l^{n}}$ odd prime a primitive $l^{n}$-th root unity 4 The structure $\psi(k)$ In th section we let that let $F=\mathbb{Q}$ $K$ $\mathscr{p}_{0}$ be the composite all fields in where $\mathscr{p}_{0}$ a set infinitely many finite Galo extensions $M$ that odd $[M : \mathbb{q}]$

7 lie unramified in We let $M/\mathbb{Q}$ $\mathscr{p}$ be the family all finite Galo subextensions $K$ Then every $M$ also has odd extension degree over 2 unramified in $M/\mathbb{Q}$ $\psi$ $\psi_{2}$ $\varphi$ $\varphi_{2}$ We write instead respectively We shall investigate what $\psi(k)$ For $a$ $b\in K$ we let $T_{ab}$ be the set elements $\alpha$ $K$ that $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))arrow\varphi(a b \alpha)$ $ab=0$ Then we have We easily see $\psi(\mathfrak{o}_{k})=\bigcap_{ab\in K}T_{ab}$ $T_{ab}=K$ for $a$ with $b$ So we shall investigate what for $T_{ab}$ $a$ $b\in K^{*}$ We recall that for $a$ $b\in M^{*}$ $M\models\neg\varphi(a b \alpha)$ iff for some Hence we easily see $\alpha^{4}-1/ab\in M_{\mathfrak{p}}^{*2}$ $\mathfrak{p}\in S_{M}(a b)$ the following: for $a$ $b\in K^{*}$ if $S_{M}(a b)=\emptyset$ for some with $M\in \mathscr{p}$ $a$ $b\in M$ then $\varphi(a b K)=T_{ab}=K$ by Lemma 6 So we shall investigate what for $T_{ab}$ $a$ $b\in K^{*}$ that for some with $M\in \mathscr{p}$ $a$ $b\in M$ $S_{M}(a b)\neq\emptyset$ From now on we use the following notation For a number field $M$ the ring $M_{\mathfrak{p}}$ $(0_{M})_{\mathfrak{p}}$ integers denoted by its maximal ideal also denoted by its residue $(0_{M})_{\mathfrak{p}}/\mathfrak{p}$ $(\overline{m})_{\mathfrak{p}}$ $(0_{M})_{\mathfrak{p}}$ $(U_{M})_{\mathfrak{p}}$ $\alpha\in \mathcal{m}$ field by the group units by For we $(\overline{m})_{\mathfrak{p}}$ denote by a its residue class in Furthermore we usually let above a rational $(\overline{m})_{\mathfrak{p}}\simeq 0_{M}/\mathfrak{p}\simeq F_{p^{f}}$ prime $p$ Note that where the residue degree $f$ $M$ at Lemma 11 Let a $b\in K^{*}$ that $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ holds Then for every $M\in \mathscr{p}$ with $a$ $b\in M$ every $\mathfrak{p}\in S_{M}(a b)$ not archimedean Th proved similarly as Lemma 14 in [1] Lemma 12 Let Let $M\in \mathscr{p}$ $a$ $b\in M^{*}$ $\alpha\in 0_{M}$ $\mathfrak{p}_{0}\in S_{M}(a b)$ $\mathfrak{p}_{0}\parallel with that 1 $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ 2$ 2 $\alpha^{4}-1/ab\in M_{\mathfrak{p}_{0}}^{*2}$ hold Then $\nu_{\mathfrak{p}_{0}}(-ab)=0$ $\nu_{\mathfrak{p}_{0}}(\alpha)=0$ Th also proved similarly as Lemma 15 in [1] Now we will prove the following lemma on finite fields a finite field with $q$ elements Lemma 13 Let $p$ be an odd prime $q=p^{f}$ $F_{q}$ Let be $F_{q}$ other than F3 $\eta$ F5 We let be the quadmtic chamcter that $\eta(0)=0$ $\eta(c)=1$ if $c\in F_{q}^{*2}$ $\eta(c)=-1$ otherwe Then for all $a\in F_{q}^{*}$ with $\eta(a)=-1$ $(\dagger$ $)$ there are $b\in F_{q}$ $j\in F_{p}$ that $\eta(b^{4}+a)\eta((b+j)^{4}+a)=-1$ Exceptional cases are F3 $a=2$ F5 $a=2$

8 $N_{\mathfrak{p}_{0}}^{*2}$ $\square$ 108 Pm We will first prove the following; for all with sufficiently large $a\in F_{q}^{*}$ $q$ we can $(\dagger$ $)$ $s$ take $j=1$ in the statement We use Weil Theorem [5 p 225 Theorem 541] from which we have that for $a\in F_{q}^{*}$ $ \sum_{c\in F_{q}}\eta\{(c^{4}+a)((c+1)^{4}+a)\} \leq 7q^{1/2}$ Thus if $q$ satfies inequality $7q^{\iota/2}<q-8$ then for all there that $a\in F_{q}^{*}$ $b\in F_{q}$ $F_{q}$ $\eta(b^{4}+a)\eta((b+1)^{4}+a)=-1$ Hence for all with $q>64$ the assertion holds For the small values $q\leq 64$ we can check the assertion directly $(\dagger$ $)$ Note that in the statement we cannot always take $j=1$ if $q\leq 64$ ; for $b$ example in F7 there no that $\eta(b^{4}+5)\eta((b+1)^{4}+5)=-1$ but in F7 $\eta(1^{4}+5)\eta((1+2)^{4}+5)=-1$ holds Note also that we need the assumption $\eta(a)=-1$ $(\dagger$ $)$ for F9 since for $a=12$ for which $\eta(a)=1$ the statement dose not hold Lemma 14 Let Let $M\in \mathscr{p}$ $a$ $b\in M^{*}$ Suppose that $S_{M}(a b)$ contains a nonarchimedean place that $\mathfrak{p}_{0}\sqrt 2$ $(\overline{m})_{\mathfrak{p}_{0}}\neq F_{3}$ $\mathscr{p}_{5}$ $\mathfrak{p}_{0}$ $\nu_{\mathfrak{p}_{0}}(-ab)=0$ Then $K\models\neg\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ The pro similar to that Lemma 16 in [1] Proposition 15 Let For $M\in \mathscr{p}$ $a$ $b\in M^{*}$ if $S_{M}(a b)$ contains no primes dividing $J\supset_{K}\subseteq T_{ab}$ 2 then we have that $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))arrow\varphi(a b \alpha)$ for all $\alpha\in l\supset_{k_{l}}$ Pro We first note the following; if we take $N\in \mathscr{p}$ that $a$ $b\in N^{*}$ then $\alpha\in also contains no primes dividing 2 by Lemma 6 Suppose not Then there that $S_{N}(a b)$ 1\supset_{K}$ $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ but $K_{l}\models\neg\varphi(a b \alpha)$ Take that $N\in \mathscr{p}$ $a$ $b$ $\alpha\in N$ We have by Lemma 7 $N\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ but $N\models\neg\varphi(a b \alpha)$ Then there a that $\mathfrak{p}_{0}\in S_{N}(a b)$ $\alpha^{4}-1/ab\in N_{\mathfrak{p}_{0}}^{*2}$ $\mathfrak{p}_{0}$ We see that not archimedean by Lemma 11 that $\nu_{\mathfrak{p}_{o}}(-ab)=0$ $\nu_{\mathfrak{p}_{0}}(\alpha)=0$ $(\overline{n})_{\mathfrak{p}_{0}}\neq F_{3}$ $\mathscr{p}_{5}$ by Lemma 12 If we get a contradiction by Lemma 14 $(\overline{n})_{\mathfrak{p}_{0}}=\mathscr{p}_{5}$ Suppose that Since $(a b)_{\mathfrak{p}_{0}}=-1$ $N\models\psi(1)$ we $have-1/ab\in$ 1 hence $-1/ab\in N_{\mathfrak{p}_{0}}^{*2}$ $-1/ab\equiv 2(mod \mathfrak{p}_{0})$ $\nu_{\mathfrak{p}_{0}}(\alpha)=0$ Since we have

9 must 109 $\alpha^{4}\equiv 1(mod \mathfrak{p}_{0})$ Then we have $\alpha^{4}-1/ab\equiv 3(mod \mathfrak{p}_{0})$ hence $\alpha^{4}-1/ab\not\in N_{\mathfrak{p}_{0}}^{*2}$ acontradiction $(\overline{n})_{\mathfrak{p}_{0}}=\mathscr{p}_{3}$ Suppose $\mathfrak{p}_{0}$ that We first deal with the case where not ramified in $N/\mathbb{Q}$ $N_{\mathfrak{p}_{0}}$ Then 3 a prime element we can $write-1/ab=2+s_{1}3+s_{2}3^{2}+\cdots$ where $s_{i}\in\{012\}$ We note that $N\models\varphi(a b n)$ for all $n\in \mathbb{n}$ If $s_{1}=0$ then $2^{4}-1/ab=(s_{2}+2)3^{2}+\cdots$ $7^{4}-1/ab=s_{2}3^{2}+\cdots$ $11^{4}-1/ab=(s_{2}+1)3^{2}+\cdots$ $N_{\mathfrak{p}_{0}}^{*2}$ Thus we have one these three must be contained in a contradiction Likewe if $s_{1}=1$ then $4^{4}-1/ab=(s_{2}+2)3^{2}+\cdots$ $13^{4}-1/ab=s_{2}3^{2}+\cdots$ $5^{4}-1/ab=$ $(s_{2}+1)3^{2}+\cdots$ And if $s_{1}=2$ then $1^{4}-1/ab=(s_{2}+1)3^{2}+\cdots$ $8^{4}-1/ab=s_{2}3^{2}+\cdots$ $10^{4}-1/ab=(s_{2}+1)3^{2}+\cdots$ $\mathfrak{p}_{0}$ Thus in the case where not ramified in $N/\mathbb{Q}$ we get contradictions $\mathfrak{p}_{0}$ Secondly We deal with the case where ramified in Let $N/\mathbb{Q}$ $\nu_{\mathfrak{p}_{0}}(3)=e$ $\pi$ $N_{\mathfrak{p}_{0}}$ let be a prime element We can write $-1/ab=2+s_{1}\pi+s_{2}\pi^{2}+\cdots$ where $s_{i}\in\{012\}$ We may write $\alpha=1+c_{1}\pi+c_{2}\pi^{2}+\cdots$ where $c_{i}\in\{012\}$ since if then Since we have $\alpha\equiv 2(mod \mathfrak{p}_{0})$ $-\alpha\equiv 1(mod \mathfrak{p}_{0})$ $N\models\neg\varphi(a b \alpha)$ $N\models\neg\varphi(a b \alpha-n)$ for all $n\in \mathbb{n}$ But $(\alpha-1)^{4}-1/ab\equiv 2(mod \mathfrak{p}_{0})$ hence there must be another prime with $\mathfrak{p}_{1}\in S_{N}(a b)$ $(\alpha-1)^{4}-1/ab\in N_{\mathfrak{p}_{1}}^{*2}$ $\mathfrak{p}_{1}$ be a prime lying above 3 And we have $\alpha\equiv 2(mod \mathfrak{p}_{1})$ $(\alpha-(3k+1))^{4}-1/ab\equiv 2$ $(mod \mathfrak{p}_{0})$ $(\alpha-(3k+2))^{4}-1/ab\equiv 0(mod \mathfrak{p}_{0})$ Likewe $(\alpha-(3k+1))^{4}-1/ab\equiv 0$ $(mod \mathfrak{p}_{1})$ $(\alpha-(3k+2))^{4}-1/ab\equiv 2(mod \mathfrak{p}_{1})$ Since there are finitely many primes in $S_{N}(a b)$ we must have for some $k(\alpha-(3k+2))^{4}-1/ab\equiv 0(mod \mathfrak{p}_{0})$ $(\alpha-(3k+2))^{4}-1/ab\in N_{\mathfrak{p}_{0}}^{*2}$ We have since $s_{1}+c_{1}\equiv 0(mod \mathfrak{p}_{0})$ $\alpha^{4}-1/ab=(s_{1}-c_{1})\pi+\cdots$ And we have $s_{1}-c_{1}\equiv 0(mod \mathfrak{p}_{0})$ since $(\alpha-(3k+2))^{4}-1/ab=(s_{1}-c_{1})\pi+\cdots$ Thus we have $s_{1}\equiv 0(mod \mathfrak{p}_{0})$ $c_{1}\equiv 0(mod \mathfrak{p}_{0})$ Likewe we have $s_{2}\equiv 0(mod \mathfrak{p}_{0})$ $c_{2}\equiv 0$ $(mod \mathfrak{p}_{0})$ We can proceed to $\pi^{e-1}$ It follows $that-1/ab=2+s_{e}\pi^{e}+s_{e+1}\pi^{e+1}+\cdots$ Then we have $2^{4}-1/ab=(s_{e}+2)3^{2}+\cdots$ $7^{4}-1/ab=s_{e}3^{2}+\cdots$ $11^{4}-1/ab=\square$ $(s_{e}+1)3^{2}+\cdots$ a contradiction We will deal with primes dividing 2 Lemma 16 Let Let $M\in \mathscr{p}$ $a$ $b\in M^{*}$ $\alpha\in 0_{M}$ $\mathfrak{p}_{0}\in S_{M}(a b)$ $\mathfrak{p}_{0} 2$ with that 1 $K\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ 2 $\alpha^{4}-1/ab\in M_{\mathfrak{p}_{0}}^{*2}$ hold Then $\nu_{\mathfrak{p}_{0}}(-ab)=\pm 2$ The pro similar to that Lemma 18 in [1] We shall prove a similar result to Lemma 14

10 with 110 Lemma 17 Let $M\in \mathscr{p}$ $a$ $b\in M^{*}$ Suppose that contains a $S_{n}(a b)$ $\mathfrak{p}_{0}$ that $\mathfrak{p}_{0} 2$ $\nu_{\mathfrak{p}_{0}}(-ab)=-2$ Then $K_{l}\models\neg\forall c(\varphi(a b c)arrow\varphi(a b c+1))$ The pro similar to that Lemma 19 in [1] Thus we get the following proposition The pro similar to that Proposition 15 Proposition 18 Let be $S_{n}(a b)$ contains no primes that an odd prime that $l\equiv-1(mod 4)$ For $a$ $b\in F_{n}^{*}$ if that $\mathfrak{p} 2$ $\nu_{\mathfrak{p}}(-ab)=2$ then we have $1\supset_{K_{l}}\subseteq $K_{l}\models\forall c(\varphi(a b c)arrow\varphi(a b c+1))arrow\varphi(a b \alpha)$ for all $\alpha\in \mathfrak{o}_{k_{l}}$ T_{ab}$ Since $\psi(k)=\bigcap_{ab\in K^{*}}T_{ab}\subseteq \mathfrak{o}_{k}$ Proposition 18 implies $\psi(k)=\bigcap_{(ab)\in\delta}t_{ab}$ $\triangle$ where the set $(a b)\in K^{*}\cross K^{*}$ that for some $M$ with $a$ $b\in M$ $S_{M}(a b)$ $\mathfrak{p} 2$ contains a prime $\nu_{\mathfrak{p}}(-ab)=2$ Such $a$ $b$ ext for example let $a=2$ $b=10$ Let $M\in \mathscr{p}$ $=\mathfrak{p}_{1}\cdots \mathfrak{p}_{k}$ (2) in $M$ Put $P_{M}= \bigcap_{i}((1+\mathfrak{p}_{i})\cup \mathfrak{p}_{i})$ Then $P_{M}$ a subring containing 1 Let $0_{M}$ $P_{K}=\cup\{P_{M} : M\in \mathscr{p}\}$ $P_{K}$ $\mathfrak{o}_{k}$ a subring containing 1 Theorem 19 $\psi(k)=p_{k}$ The pro similar to that Proposition 20 in [1] Example 20 1 $K_{l}=U_{n}\mathbb{Q}(\cos(2\pi/l^{n}))$ with a prime with $l\equiv-1(mod 4)$ 2 $K_{W}= \prod_{l\in W}K_{l}$ $(W=\{l a prime: l\equiv-1(mod 4)\})$ 3 $K_{0}=\mathbb{Q}(\{\cos(2\pi/l)$ : 5 Undecidability results a prime $l\equiv-1(mod 4)\})$ Let In [1] we proved that if $K_{l}= \bigcup_{n}\mathbb{q}(\cos(2\pi/l^{n}))$ a prime that $l\equiv-1$ $\mathfrak{o}_{k_{l}}$ $K_{l}$ $(mod 4)$ 2 a prime then undecidable But in 2000 CR Videla $K_{l}$ [12] proved that undecidable for every prime He considered $K/F$ a pro-p Galo extension over a number field $F$ using Rumely s formula in [6] he proved $\mathfrak{o}_{k_{l}}$ that definable with parameters Then he also used the results Kronecker J Robinson Kronecker [3] determined all sets conjugate algebraic integers in the interval $c-2\leq x\leq c+2$ provided that a rational integer; they have the form $c$ $x=c+2\cos(2k\pi/m)$ with $0\leq k\leq m/2$ $(k m)=1$

11 inside if 111 Note that if $m=1234$ then $x=c+2$ $c-2$ $c\pm 1$ $c$ respectively Furthermore it known that an interval length less than 4 can contain only finitely many complete sets conjugate algebraic integers (See [11]) Therefore we see that the interval $(04)$ contains infinitely many complete conjugate sets totally real algebraic integers that no sub-interval does These facts are used by J Robinson in [9] Her results concerns the integral closure $\mathbb{z}$ totally real fields not necessarily finite over She calls a ring a totally real algebraic integer ring In 1962 she proved the following: The natural numbers can be defined arithmetically in any totally real algebraic integer ring $A$ that there a smallest interval $(0 s)$ with real or $s$ $\infty$ which contains infinitely many complete conjugate sets numbers But we can say more We recall that $A$ $\mathbb{z}^{tr}$ denotes the ring all totally real algebraic integers $\mathbb{z}^{tr}$ $\mathbb{z}$ Theorem 21 $LteR$ be a subring containing that there a smallest interval $(0 s)$ with $s$ $\infty$ real or which contains infinitely many complete conjugate sets numbers R Here $s$ In particular a ring undecidable need not be in R Then $N$ definable in $R$ The pro J Robinson just works See [9 pp ] $\mathfrak{o}_{k_{l}}$ Thus it follows that for every positive integer $l>1$ undecidable from $K_{l}$ which Videla proved that undecidable Note that even if the defining formula contains parameters it possible to define N See [12] We give alternative pro th fact in the case where a prime with $l\equiv-1$ $(mod 4)$ We know that a subring $\psi(k_{l})$ $\mathbb{z}^{tr}$ $\mathbb{z}$ containing a prime that $l\equiv-1(mod 4)$ Furthermore we know by [11 p 312] that 2 2 $+$ $\cos(2\pi/l^{n})$ $\mathfrak{o}_{k_{l}}$ are units in that 1 2 $+$ $\cos(2\pi/l^{n})$ $\mathfrak{o}_{k_{l}}$ are units in if $l\neq 3$ $ N_{F_{n}/\mathbb{Q}}(1+$ $2\cos(2\pi/3^{n})) =3$ for $n\geq 2$ Hence we see that 2 2 are $+$ $\cos(2\pi/l^{n})$ not in $\psi(k_{l})$ if $ln\neq 3$ On the other h 4 4 are $+$ $\cos(2\pi/l^{n})$ $\bigcap_{i}\mathfrak{p}_{i}^{(2)}-$ in hence in Thus $\psi(k_{l})$ we see that the interval (08) contains infinitely many complete conjugate sets numbers the interval $\psi(k_{l})$ $(04)$ does not We show that $(08)$ actually a smallest interval for $\psi(k_{l})$ Lemma 22 Let be an odd prime that $l\equiv-1(mod 4)$ Then $(08)$ a smallest interval the $(0 c)$ form which contains infinitely many complete conjugate sets numbers $\psi(k_{l})$ Pm We know that $K_{l}$ has only finitely many primes lying above 2 (See Lemma $)$ 13 in [1] Thus where $\psi(k_{l})=p_{k_{l}}=\bigcap_{i}((1+\mathfrak{p}_{i})-\cup \mathfrak{p}_{i})-$ $\mathfrak{p}_{1}$ $\ldots \mathfrak{p}_{k}$ $K_{l}$ are primes lying above 2 We easily see that a union $\psi(k_{l})$ $2^{k}$ $\mathfrak{o}_{k_{l}}/20_{k_{l}}$ cosets Suppose that $(08)$ $(0 \delta)$ not a smallest interval Then some interval with $\delta<8$ contains infinitely many complete conjugate sets numbers $\psi(k_{l})$ Then we have that $\alpha+2\mathfrak{o}_{k_{l}}$ some coset say contains infinitely many complete conjugate

12 $\square$ 112 sets numbers It follows that an interval length less than 4 contains infinitely many complete conjugate sets algebraic integers a contradiction Let where $K_{\Delta}= \prod_{l\in\delta}k_{l}$ $\triangle$ a finite set primes From the result Videla we $\Delta$ $K_{\Delta}$ deduce that undecidable If a finite set primes with $l\equiv-1(mod 4)$ then we can give another pro similarly Nevertheless we can give a new undecidable infinite algebraic extension by our method Let be a set Sophie Germain primes that a prime $V$ $p$ that $2p+1$ again a prime It considered that there are infinitely many Sophie Germain primes but it not proved Let $K_{V}=\mathbb{Q}(\{\cos(2\pi/l)$ : $l\in V\})$ Then we have $\psi(k_{v})=(1+2\mathfrak{o}_{k_{v}})\cup 0_{K_{V}}$ hence undecidable $K_{V}$ References [1] K Fukuzaki Undecidable infinite totally real extensions RIMS 1602 (2008) pp Kokyuroku [2] S Iyanaga(Editor) The Theory Numbers North-Holl Publhing Company 1975 [3] L Kronecker Zwei S\"atze \"uber Gleichungen mit ganzzahligen Coefficienten Reine Angew Math 53 (1857) pp [4] S Lang Algebraic Number Theory 2nd ed Graduate Texts in Mathematics vol 211 Springer-Verlag New York 1994 [5] R Lidl H Niederreiter Finite fields Encyclopedia Mathematics its Applications vol 20 Cambridge University Press 1997 [6] R Rumely The undecidability algebraic rings fields Proc Amer Math Soc 262 (1980) pp [7] OT $O$ Meara Introduction to Quadratic Forms Springer-Verlag Berlin Heidelberg New York 1973 [8] J Robinson The undecidability algebraic rings fields Proc Amer Math Soc 10 (1959) pp [9] J Robinson On the decion problem for algebraic rings Studies in Mathematical Analys Related Topics no 42 Stanford Univ Press Stanford Calif 1962 pp [10] J Robinson The decion problem for fields The Theory Models: Proceedings the 1963 International Symposium at Berkeley (J W Addon et al eds) North-Holl Amsterdam 1965 pp

13 113 [11] RM Robinson Intervals Containing Infinitely Many Sets Conjugate Algebraic Integers Studies in Mathematical Analys Related Topics no 43 Stanford Univ Press Stanford Calif 1962 pp [12] CR Videla Definability the ring integers in pro-p Galo extensions number fields Israel J Math 118 (2000) pp 1-14