On Fields of Totally $ S $-Adic Numbers

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Proof. If v(y) = 0, then v(t−1y) < 0 = v(y−1), so v(y−1 + t−1y) < 0. If v(y) < 0, then v(y−1) > 0 and v(t−1y) < 0, so v(y−1 + t−1y) < 0. If v(y) > 0, then v(y−1) < 0 and v(t−1y) ≥ 0 since t is a uniformizer, so again v(y−1 + t−1y) < 0. Thus, in each case, v(γ(y, t)) = −v(y−1 + t−1y) > 0.

Lemma 3. Let F be a ﬁeld and t ∈ F r {0, −1}. If char(F )=2, assume in addition that t is not a square in F . Then f(X,γ(Y, t)) is irreducible over F (Y ).

Proof. If char(F ) =6 2, then f(X,γ(Y, t)) is reducible if and only if the discriminant 1+4γ(Y, t)2 is a square in F (Y ). This is the case if and only if (Y 2 + t)2 + 4(tY )2 is a square. Writing (Y 2 + t)2 + 4(tY )2 =(Y 2 + aY + b)2 and comparing coeﬃcients we get that a = 0, b2 = t2, and a2 +2b = 2t(1+2t). Hence, t =0 or t = −1. If char(F ) = 2, then f(X,γ(Y, t)) is irreducible if and only if

g(X) := f(X + γ(Y, t),γ(Y, t)) = X2 + X + γ(Y, t) is irreducible. If v denotes the normalized valuation on F (Y ) corresponding to the irreducible polynomial Y 2 + t ∈ F [Y ], then v(γ(Y, t)) = −1. This implies that a zero x of 1 g(X) in F (Y ) would satisfy v(x) = − 2 , so g(X) has no zero in F (Y ) and is therefore irreducible.

Proof of Theorem 1. Without loss of generality assume that S =6 ∅ and that the absolute values in S are pairwise inequivalent. Let F = KS. The weak approximation theorem gives an element t ∈ Kr{0, −1} that is a uniformizer for each of the ultrametric absolute values in S. Clearly, if S contains an ultrametric discrete absolute value (in particular if char(K) = 2), then t is not a square in F . Hence, by Lemma 3, f(X,γ(Y, t)) is irreducible over F (Y ). Assume, for the purpose of contradiction, that F is Hilbertian. Then there exists y ∈ F such that f(X,γ(y, t)) is deﬁned and irreducible over F . Let |·|∈ S. If |·| is archimedean (this means we are in the case char(K) =6 2), let ≤ be an ordering corresponding to an extension of |·| to F , and let E be a real closure of (F, ≤). Since γ(y, t)2 ≥ 0, there exists x ∈ E such that f(x, γ(y, t)) = 0 (note that 2 the map E≥0 → E≥0, ξ 7→ ξ + ξ is surjective). If |·| is ultrametric and v is a discrete valuation corresponding to an extension of |·| to F , let E be a Henselization of (F, v). Since v(γ(y, t)) > 0 by Lemma 2, f(X,γ(y, t)) ∈ Ov[X] and

f(X,γ(y, t)) = X(X + 1)

has a simple root, so by Hensel’s Lemma there exists x ∈ E with f(x, γ(y, t)) = 0. Thus in each case, f(X,γ(y, t)) has a root in E, so since it is of degree 2 all of its roots are in E. Since F is the intersection over all such E, all roots of f(X,γ(y, t)) lie in F , contradicting the irreducibility of f(X,γ(y, t)).

AÈÈEÆDÁX: THE TOTALLY S-ADIC IS NOT HILBERTIAN

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Let K be an arbitrary field, and S be a finite set of orderings and/or non-trivial valuations of K. We denote by KS | K the maximal subextesion of a separable closure sep sep K |K of K in which all v ∈ S are totally split. For v ∈ S we will denote by Kv ⊂ K a fixed real closure of K with respect to v in the case v is an ordering, respectively a fixed sep Henselization of K with respect to v in the case v is a valuation. Recall that Kv ⊂ K is unique up to GK -conjugation, where GK is the absolute Galois group of K. One has: S σ 1) K = ∩v∈S ∩σ∈GK Kv . sep S In particular, if Kv0 = K for some v0 ∈ S, then K does not depend on v0. Therefore, sep without loss of generality, we suppose that Kv =6 K for all v ∈ S. Further, for S σ S polynomials r(X) ∈ K [X] and their GK-conjugates r (X) ∈ K [X] one has: S σ 2) r(X) has all its roots in K iff r (X) has all its roots in Kv for all v ∈ S, σ ∈ GK . sep sep Let L|K be all the finite Galois subextensions of K |K. Then K = ∪LL, and since sep Kv ⊂ K is a strict inclusion, there exits L|K finite Galois such that L is not contained in Kv. In particular, since the family (L|K)L is filtered, there exists L|K such that L is not contained in any Kv, v ∈ S. Translated into the language of polynomials, we have the following: Let p(X) ∈ K[X] be a monic polynomial having splitting field L|K and degree deg p(X) = [L : K]. Then L = k[α] for every root α of p(X). Hence the fact that L is not contained in Kv translates into:

3) There exist non-constant polynomials p(X) ∈ K[X] having no roots in Kv, v ∈ S. Equivalently, by general decomposition theory for valuations and orderings, it follows that p(Kv) is bounded away from zero, see e.g., [10], i.e., there exists a v-neighborhood Uv of 0 ∈ Kv such that Uv ∩ p(Kv) is empty. In particular, for every v ∈ S there exists × 1 tv ∈ K such that v(tv) < vp(x) for all x ∈ Kv. Taking into account that the non-zero × elements t ∈ K approximate 0 ∈ Kv simultaneously for v ∈ S, we get: × 4) Let t ∈ K satisfy v(t) < v(tv), v ∈ S. Then v(t) < vp(x) for all x ∈ Kv, v ∈ S. We next recall the theorem on the continuity of roots in the following form, see e.g., [10] for details: Let q(Y ) ∈ K[Y ] be a polynomial of degree n> 0 which has n distinct roots y1,...,yn in K. For polynomials qv(Y ) ∈ Kv[Y ], we define v(qv −q) := max{v(avi −ai)}i, where (avi)i and (ai)i are the coefficients of qv(Y ), respectively q(Y ). Then for every × v ∈ S there exists δv ∈ K such that all polynomials qv(Y ) ∈ Kv[Y ] of degree n satisfy: If v(qv − q) < v(δv), then the roots yv1,...,yvn of qv(Y ) are distinct and lie in Kv. We view polynomialsq ˜(Y ) ∈ KS[Y ] of degree n and their conjugatesq ˜σ(Y ) ∈ KS[Y ] S × :as polynomials in Kv[Y ] via K ֒→ Kv, v ∈ S. Then for δ˜ ∈ K one has S σ 5) Suppose that v(δ˜) ≤ v(δv), v ∈ S, and q˜(Y ) ∈ K [Y ] satisfy v(˜q − q) < v(δ˜) for S S all v ∈ S, σ ∈ GK . Then all the roots of q˜(Y ) ∈ K [Y ] lie in K . Key Lemma. In the above notations, let δ ∈ K× satisfy v(δ) ≤ v(δ˜) if v is a valuation, and v(2δ) ≤ v(δ˜) if v is an ordering. Then f(X,Y ) := p(X)q(Y ) − tδ ∈ K[X,Y ] is S S absolutely irreducible, and for all x ∈ K the specialization fx(Y ) := f(x, Y ) ∈ K [Y ] splits in linear factors in KS[Y ]. Therefore, the field KS is not Hilbertian.

* Variants of 1990/2013. Last supported by the NSF grant DMS-1101397. 1 We write v(ab)= v(a)v(b) for valuations, and v(a) = max{a, −a} if v is an ordering. 3 S σ σ Proof. Let x ∈ K be given. Then x ∈ Kv for all v ∈ S, σ ∈ GK , thus p(x ) ∈ p(Kv). Hence by the definition of t we have v(t) < vp(xσ), and in particular, p(xσ) =6 0. Further, setting a := 1/p(x) and u := at, it follows that aσ =1/p(xσ) and uσ = aσt lie in S σ S K and v(u ) < v(1) for all v ∈ S, σ ∈ GK . Setq ˜(Y ) := afx(Y )= q(Y )+ uδ ∈ K [Y ]. Then the G-conjugates ofq ˜(Y )areq ˜σ(Y )= q(Y )+uσδ ∈ KS[Y ], thusq ˜σ −q =(uσ −u)δ. On the other hand, one has that v(uσ − u) ≤ v(uσ)+ v(u) < v(1)+ v(1) = v(2) if v is an absolute value, respectively v(uσ −u) ≤ max{v(uσ), v(u)} < v(1) if v is a valuation. Thus using the definition of δ, one has that v(˜q σ − q) = v(uσ − u) v(δ) < v(δ˜) for all v ∈ S, S σ ∈ GK. Therefore, by point 5) above it follows thatq ˜(Y ) has all its roots in K and therefore, so does fx(Y ). To conclude the proof of Key Lemma, notice that tδ =6 0, and q(Y ) is separable. Therefore f(X,Y ) is absolutely irreducible, see e.g., [8] for a proof. Remarks. 1) With a proof which is virtually identical with the one above, one proves that the intersection of all the v-topological Henselizations of K, v ∈ S, is not Hilbertian. 2) One can “axiomatize” the proof of the Key Lemma and make it work for infinite families of orderings and/or valuations, satisfying some obvious approximation conditions. References [1] Pierre Dèbes and Dan Haran. Almost Hilbertian fields. Acta Arithmetica, 88:269–287, 1999. [2] Ido Efrat. Absolute Galois groups of p-adically maximal PpC fields. Forum Math., 3:437–460, 1991. [3] Michael D. Fried and Moshe Jarden. Field Arithmetic. Ergebnisse der Mathematik III 11. Springer, 2008. 3rd edition, revised by M. Jarden. [4] Michael D. Fried, Dan Haran, and Helmut Völklein. The absolute Galois group of the totally real numbers. Comptes Rendus de l’Académie des Sciences Paris, 317:95–99, 1993. [5] Michael D. Fried, Dan Haran, and Helmut Völklein. Real Hilbertianity and the field of totally real numbers. In N. Childress and J. W. Jones, editors, Arithmetic Geometry, Contemporary Mathe- matics 174, pages 1–34. American Mathematical Society, 1994. [6] Dan Haran, Moshe Jarden, and Florian Pop. The absolute Galois group of the field of totally S-adic numbers. Nagoya Mathematical Journal, 194:91–147, 2009. [7] Dan Haran, Moshe Jarden, and Florian Pop. The absolute Galois group of subfields of the field of totally S-adic numbers. Functiones et Approximatio Commentarii Mathematici, 2012. [8] B. Heinemann and A. Prestel. Fields regularly closed with respect to finitely many valuations and orderings. Can. Math. Soc. Conf. Proc. 4 (1984), pp.297–336. [9] Florian Pop. Embedding problems over large fields. Annals of Mathematics, 144(1):1–34, 1996. [10] Alexander Prestel and Martin Ziegler. Model theoretic methods in the theory of topological fields. Journal reine angew. Math. 299/300:318–341, 1978. [11] Richard Taylor. Galois representations. Annales de la Facultédes Sciences de Toulouse, 8(1):73–119, 2004.

School of Mathematical Sciences, Tel Aviv University, Ramat Aviv, Tel Aviv 69978, Israel E-mail address: [email protected] Fachbereich Mathematik und Statistik, University of Konstanz, 78457 Konstanz, Ger- many E-mail address: [email protected] Department of Mathematics, University of Pennsylvania, DRL, 209 S 33rd Street, Philadelphia, PA 19104, USA E-mail address: [email protected] URL: http://math.penn.edu/~pop