The Journal of Symbolic Logic Volume 80, Number 1, March 2015
EXISTENTIAL∅-DEFINABILITY OF HENSELIAN VALUATION RINGS
ARNO FEHM
Abstract. In [1], Anscombe and Koenigsmann give an existential∅-definition of the ring of formal power seriesF[[t]] in its quotient field in the case whereF is finite. We extend their method in several directions to give general definability results for henselian valued fields with finite or pseudo-algebraically closed residue fields.
§1. Introduction. The question of first order definability of valuation rings in their quotient fields has a long history. Given a valued field K, one is interested in whether there exists a first order formulaϕ in the languageL ={+,−,·,0,1} of rings such that the setϕ(K) defined byϕinKis precisely the valuation ring, and what complexity such formula must have.
Many results of this kind are known forhenselian valued fields, like fields of formal power seriesK =F((t)) over a fieldF, and their valuation ringF[[t]]. In this setting, a definition going back to Julia Robinson gives an existential definition of the valuation ring using the parametert. Later, Ax [2] gave a definition of the valuation ring, which uses no parameters, but is not existential.
Recently, Anscombe and Koenigsmann [1] succeeded to give an existential and parameter-free definition of F[[t]] inF((t)) in the special case where F = Fq is a finite field. Their proof uses the fact that Fq can be defined in Fq((t)) by the quantifier-free formulaxq−x= 0. In particular, their result does not apply to any infinite fieldF, and their formula depends heavily onq.
In this note we simplify and extend their method. As a first application we get the following general definability result for henselian valued fields with finite or pseudo-algebraically closed residue fields (Theorem 2.6 and Theorem 3.5), which generalizes [1, Theorem 1.1] on Fq((t)) and [5, Theorem 6] on finite extensions ofQp:
Theorem1.1. LetKbe a henselian valued field with valuation ringOand residue fieldF. IfF is finite or pseudo-algebraically closed and the algebraic part ofF is not algebraically closed, then there exists an∃-∅-definition ofOinK.
As a further application, in Section 4, we find definitions of the valuation ring which are uniform for large (infinite) families of finite residue fields, like the following one for finite prime fields (Theorem 4.3):
Received August 22, 2013.
Key words and phrases. Henselian valuation, existential definability, finite fields, pseudo-algebraically closed fields.
c2015, Association for Symbolic Logic 0022-4812/15/8001-0015 DOI:10.1017/jsl.2014.13
301
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https://dx.doi.org/10.1017/jsl.2014.13
Theorem1.2. For every >0there exists an∃-∅-formulaϕand a setPof prime numbers of Dirichlet density at least1−such that for any henselian valued fieldK with valuation ringOand residue fieldF with|F| ∈P, the formulaϕdefinesOinK.
In particular, this applies to power series fields Fp((t)) andp-adic fields Qp. Theorem 1.2 is in a sense optimal, see the discussion at the end of this note.
§2. Defining subsets of the valuation ring. LetKbe a henselian valued field with valuation ringO ⊆K, maximal idealm⊆ Oand residue fieldF =O/m. Fora ∈ O we let ¯a=a+m∈F be its residue class and write ¯f∈F[X] for the reduction of a polynomialf∈ O[X].
We start by simplifying the key lemma of [1], thereby generalizing it to arbitrary henselian valuations. This proof follows Helbig [10]. Here, and in what follows, by f(K)−1we mean the set{f(x)−1:x∈K}and implicitly claim thatf(x)= 0 for allx∈K.
Lemma2.1. Letf ∈ O[X]be a monic polynomial such thatf¯has no zero inF, and leta∈K. LetUf,a:=f(K)−1−f(a)−1. Then the following holds:
a) f(K)−1⊆ O, b) Uf,a ⊆ O,
c) If in additiona∈ Oandf(a)∈/ m, thenm⊆Uf,a.
Proof. a) We have thatf(K)∩m =∅: Ifx ∈ Kwithf(x)∈m, thenx ∈ O sinceOis integrally closed andfis monic, and hence ¯f( ¯x) = 0, contradicting the assumption that ¯fhas no zero inF. Therefore,f(K)−1⊆(Km)−1=O.
b) From a) we get that f(K)−1 ⊆ O, and in particularf(a)−1 ∈ O. Thus, Uf,a⊆ O.
c) Now assume thata ∈ Oandf(a)∈/ m. Letx ∈ m. Sincea ∈ Owe have f(a)∈ O, hencef(a)∈ O×. Defineg(X) =f(X)−(f(a) +x)∈ O[X]. Then g(a) =−x∈mandg(a) =f(a)∈/ m, so by the assumption thatOis henselian there existsb ∈ Owithg(b) = 0, i.e.f(a) +x=f(b). Hence,f(a) +m⊆f(K).
Sincef(a)∈ O×we get thatf(a)−1+m= (f(a)+m)−1 ⊆f(K)−1, and therefore
m⊆Uf,a.
We observe that one can get rid of the element a even if it is not in the (model theoretic) algebraic closure of the prime field:
Lemma2.2. Letf∈ O[X]be a monic polynomial such thatf¯has no zero inF, and a∈ Osuch thatf(a)∈/m. ThenU:=f(K)−1−f(K)−1satisfiesm⊆U⊆ O.
Proof. By Lemma 2.1a, f(K)−1 ⊆ O, hence U ⊆ O. Since a ∈ O and f(a)∈/ m, Lemma 2.1c implies thatm⊆Uf,a ⊆U.
Clearly,Ucan be defined inKby the∃-formula
ϕf(x)≡(∃y, z, y1, z1)(x=y1−z1∧y1f(y) = 1∧z1f(z) = 1).
Note that iff∈Z[X], thenϕfis an∃-∅-formula.
Lemma 2.3. IfU, T ⊆ Oare such thatm ⊆ U andT meets all residue classes (i.e.T¯ =F), thenO=U+T.
Proof. If forx∈ Owe lett ∈T with ¯t= ¯x, thenx=u+twithu:=x−t∈
m⊆U.
∅-DEFINABILITY OF HENSELIAN VALUATION RINGS
Thus, ifϕdefinesUanddefinesT, then
(x)≡(∃u, t)(x=u+t∧ϕ(u)∧(t)) definesO. Note that ifϕandare∃-∅-formulas, then so is.
We now give a first generalization of [1, Theorem 1.1]. We denote by F0 the prime field ofF and byFalgthe algebraic closure ofF0inF. By abuse of notation we will consider polynomials f ∈ Z[X] as elements of O[X] via the canonical homomorphismZ→ O.
Lemma 2.4. For every prime p and positive integer m, there exists f ∈ Fp[X] monic, separable, and irreducible of degreemwithf(0)= 0.
Proof.Letq=pm. SinceFq/Fpis Galois it has a normal basis, i.e. there exists α ∈Fq such that the conjugates ofαform anFp-basis ofFq. In particular,αhas degreemand nonzero trace overFp. Letf∈Fp[X] be the minimal polynomial of α−1. Thenfis irreducible of degreemandf(0) =±TrFq/Fp(α)/NFq/Fp(α)= 0.
Lemma 2.5. If F is finite, then there exist f ∈ F0[X] monic, separable, and irreducible which has no zero inF, anda∈F withf(a)= 0.
Proof.Identify F0 = Fp, let m be any positive integer that does not divide [F :F0], choosefof degreemas in Lemma 2.4, and leta= 0.
Theorem2.6. LetKbe a henselian valued field with valuation ringOand residue fieldF. IfF is finite, then there exists an∃-∅-definition ofOinK.
Proof.IfF = Fq, let g = Xq −X ∈ Z[X] and (x) ≡ (g(x) = 0). Since
¯
g =−1, the assumption thatOis henselian gives thatT :=(K)⊆ Ois a set of representatives ofF. In particular, it meets all residue classes. Choosef ∈F0[X] as in Lemma 2.5 and let ˜f∈ Z[X] be a monic lift off. Since there existsa ∈ F withf(a)= 0, a lift ˜a ∈ Oofasatisfies ˜f( ˜a)∈/ m. Letϕ≡ϕf. By Lemma 2.2, U :=ϕ(K) satisfiesm⊆U ⊆ O. Therefore, Lemma 2.3 shows that(K) =O.
The special case of Theorem 2.6 whereF is a finite field andK = F((t)) was proven by Anscombe and Koenigsmann in [1, Theorem 1.1]. The special case where Kis a finite extension ofQpwas proven by Cluckers, Derakhshan, Leenknegt, and Macintyre in [5, Theorem 6].
§3. Pseudo-algebraically closed residue fields. We now consider assumptions on the residue fieldF under which we can define a setT as in Lemma 2.3. For basics on pseudo-algebraically closed (PAC) fields we refer to [8, Chapter 11]. Ford ∈N we fix the constantc(d) = (2d−1)4.
Lemma 3.1. Let f ∈ F[X]be nonconstant and square-free (over the algebraic closure). ThenF =f(F)f(F)∪{0}ifF is PAC orF is finite with|F|> c(deg(f)).
Proof.Let 0=c∈F. One checks that the polynomialf(X)f(Y)−c∈F[X, Y] is absolutely irreducible, cf. [9, Proposition 1.1]. Thus, ifF is PAC we can conclude that there existx, y ∈F withf(x)f(y)−c = 0, i.e.c∈f(F)f(F). IfF is finite with|F|> c(deg(f)) we come to the same conclusion by applying the Hasse-Weil
bound, cf. [8, Corollary 5.4.2].
Lemma 3.2. Let f ∈ O[X]be monic such thatf¯is square-free and has no zero inF. ThenT :=f(K)−1f(K)−1∪ {0} ⊆ O. If in additionF is PAC or finite with
|F|> c(deg(f)), thenT meets all residue classes.
Proof. By Lemma 2.1a,f(K)−1⊆ O, henceT ⊆ O. IfF is PAC or finite with
|F|> c(deg(f)), then, sinceF×⊆f(F¯ ) ¯f(F) by Lemma 3.1, also F×⊆( ¯f(F) ¯f(F))−1⊆(f(O)·f(O))−1⊆f(K)−1f(K)−1,
henceT satisfiesF =T.
Clearly, the setTcan be defined inKby the∃-formula
f(x)≡(∃y, z, y1, z1)(x= 0∨(x=y1z1∧y1f(y) = 1∧z1f(z) = 1)).
Let
f(x)≡(∃u, t)(x=u+t∧ϕf(u)∧f(t)).
Proposition3.3. Letf∈ O[X]be monic such thatf¯is square-free and has no zero inF. Thenf(K)⊆ O. If in addition there existsa ∈ Osuch thatf(a)∈/ m andF is PAC or finite with|F|> c(deg(f)), thenf(K) =O.
Proof. Let U = ϕf(K), so U ⊆ O by Lemma 2.1. By Lemma 3.2, T :=
f(K)⊆ O, sof(K) =U +T ⊆ O. If in addition there existsa ∈ Osuch that f(a)∈/ mandF is PAC or finite with|F|> c(deg(f)), then Lemma 2.2 gives that m⊆U, and Lemma 3.2 gives thatTmeets all residue classes, hencef(K) =Oby
Lemma 2.3.
Lemma 3.4. IfF is infinite andFalg is not algebraically closed, then there exist f∈F0[X]monic, separable, and irreducible which has no zero inF, anda∈F with f(a)= 0.
Proof. SinceFalg is not algebraically closed, there exists a monic irreducible f∈F0[X] which has no zero inFalg, hence inF. SinceF0is perfect,fis separable, hencef = 0. Therefore, sinceF is infinite, there existsa∈F withf(a)= 0.
Theorem3.5. LetKbe a henselian valued field with valuation ringOand residue fieldF. IfF is pseudo-algebraically closed andFalgis not algebraically closed, then there exists an∃-∅-definition ofOinK.
Proof. Choosef∈F0[X] as in Lemma 3.4 and let ˜f∈Z[X] be a monic lift of f. Since there existsa ∈F withf(a)= 0, a lift ˜a ∈ Oofa satisfies ˜f( ˜a)∈/ m.
By Proposition 3.3,f˜(K) =O.
Corollary3.6. LetKbe a henselian valued field with valuation ringOand residue fieldF. IfF is pseudo-real closed andFalgis neither real closed nor algebraically closed, then there exists an∃-∅-definition ofOinK.
Proof. LetK =K(√
−1). Then the residue fieldF =F(√
−1) ofKis PAC by [11], andFalg = Falg(√
−1) is not algebraically closed by the Artin-Schreier theorem. By Theorem 3.5 there exists an∃-∅-definition of the unique prolongation O ofOinK. By interpretingKinK we get an∃-∅-definition ofO=O∩K in
K.
Remark 3.7. Note that as soon as F is infinite we cannot hope to have an
∃-∅-definition of aset of representativesT ⊆ OofF: For example, ifK =F((t)), thenF is never∃-∅-definable inKunless it is finite, cf. [7, Corollary 9]. This explains why we rather define a setT ⊆ Othatmeets all residue classes.
Remark3.8. We point out that the assumption thatFalgis not algebraically closed in Theorem 3.5 is indeed necessary. For example, letKbe the field of generalized
∅-DEFINABILITY OF HENSELIAN VALUATION RINGS
power seriesF((Q)) over a fieldF. IfFalg is algebraically closed, then so isK :=
Falg((Q)), cf. [6, 18.4.3]. Therefore,K is existentially closed in K. So, ifϕ is an
∃-∅-definition of the valuation ring inK, thenϕ(K) =ϕ(K)∩Kis a nontrivial valuation ring, contradicting the fact that definable subsets of an algebraically closed field are finite or cofinite.
§4. Uniform definitions. We now deal with definitions which are uniform over certain families of finite residue fields. We start with an example in fixed residue characteristicp:
Theorem4.1. Given a prime numberp and a positive integer mthere exists an
∃-∅-formulaϕ such thatϕ(K) =Ofor all henselian valued fieldsK with valuation ringOand residue fieldF =Fpnwithm |n.
Proof.Assume thatF =Fpnwithm |n. Choosef∈Fp[X] irreducible of degree mas in Lemma 2.4. Thenfhas no zero inF and there existsa∈F withf(a)= 0.
Let ˜f∈Z[X] be a monic lift off. By Proposition 3.3,f˜(K)⊆ O, andf˜(K) =O forpn> c(m). Fork∈Nwithm |kletk(x)≡(xpk −x= 0) and let
k(x)≡(∃u, t)(x=u+t∧ϕf˜(u)∧k(t)).
As in the proof of Theorem 2.6 we see thatk(K)⊆ O, andk(K) =Oifn=k. Therefore, withM ={k ∈N:m |kandpk≤c(m)},
ϕ(x)≡f˜(x)∨
k∈M
k(x)
satisfiesϕ(K) =Ofor allnwithm |n.
Remark 4.2. The condition m |n in Theorem 4.1 is indeed necessary:
If a ∃-∅-formulaϕ definesFpn[[t]] inFpn((t)) for alln in a set M, then there is somem∈Nsuch thatm |nfor alln ∈M: Otherwise,
n∈MFpn would equal the algebraic closure ofFp, so since every finite extension ofFpn((t)) is isomorphic to Fpn((t)) for somen|n, we would get a definition of a nontrivial valuation ring in the algebraic closure ofFp((t)), which is impossible. For details the reader may consult [5, Theorem 4], where it is shown that no∃-∅-formula can define the valuation ring uniformly for all finite extensions of a fixed henselian valued fieldK.
We now turn to uniformity inp. LetPdenote the set of all odd prime numbers.
For a subsetP ⊆P, we denote by(P) the Dirichlet-density ofP, if it exists. For a formulaϕlet
P(ϕ) ={p∈P:ϕ(Qp) =Zp}
and letP(ϕ) be the set ofp∈Psuch thatϕ(K) =Ofor all henselian valued fields Kwith valuation ringOand residue fieldF =Fp. We have thatP(ϕ)⊆P(ϕ), and it is known thatP(ϕ) has a Dirichlet-density for every formulaϕ, cf. [3, Theorem 16], [8, Theorem 20.9.3]. It is also known thatP(ϕ) differs from{p∈P:ϕ(Fp((t))) = Fp[[t]]} only by a finite set, see [4, p. 606], so for all results concerning Dirichlet density we could as well useFp((t)) instead ofQp.
Theorem4.3. For every >0there exists an∃-∅-formulaϕsuch that(P(ϕ))>
1−.
Proof. Forn∈Nletfn=X2−n∈Z[X] and Pn=
p∈P:
n p
=−1
=
p∈P:Fp|= (∃y)(y2 =n) .
Note that ifKis henselian valued with residue fieldF =Fp,p >2, thenp ∈Pn
if and only ifK |= (∃y)(y2 = n). If p ∈ Pn with p > c(2), then p ∈ P(fn) by Proposition 3.3. By the quadratic reciprocity law and Dirichlet’s theorem, there existsN ∈Nsuch that forP=N
n=2Pnwe have(P)>1−. Let ϕn(x)≡(∃y)(y2 =n)∨fn(x)
andϕ(x)≡N
n=2ϕn(x). Letp∈Pwhich lies in the open intervalI := (c(2),∞). If p∈Pn, thenϕn(K) =O, otherwiseϕn(K) =K. Thus,ϕ(K) =N
n=2ϕn(K) =O ifp ∈ P, andϕ(K) = K otherwise. So, ifp ∈ P, thenp ∈ P(ϕ) ⊆ P(ϕ), and if p /∈ P, thenp /∈ P(ϕ). Thus,P(ϕ)∩I = P(ϕ)∩I = P∩I, and therefore
(P(ϕ)) =(P(ϕ)) =(P)>1−.
On the other hand, it is well known that there is no such formula that works uniformly for almost allp:
Proposition4.4. LetP⊆Pbe a cofinite set of prime numbers. Then there exists no∃-∅-formulaϕsuch thatP ⊆P(ϕ).
A proof of this can be found in [5, Theorem 5]. In fact, the proof given there shows the following stronger statement:
Proposition4.5. Let P be a set of prime numbers with(P) = 1. Then there exists no∃-∅-formulaϕsuch thatP⊆P(ϕ).
This also explains that Theorem 4.3 cannot be strengthened to give a uniform
∃-∅-definition foreverysetPwith(P)<1:
Proposition4.6. There exists a setPof prime numbers with(P) = 0for which there exists no∃-∅-formulaϕsuch thatP⊆P(ϕ).
Proof. List all∃-∅-formulas asϕ1, ϕ2, . . . and letN ={1, 2, . . .} ⊆ Pbe any infinite set with (N) = 0. Proposition 4.4 implies that for each i,P(ϕi) is not cofinite inP. Therefore, we can choose somepi ∈Pwithpi > i andpi ∈/ P(ϕi).
ThenP={p1, p2, . . .}has(P)≤(N) = 0, butP⊆P(ϕi) for eachi.
§5. Acknowledgements. The author would like to thank Will Anscombe, Patrick Helbig, Franziska Jahnke, and Jochen Koenigsmann for interesting and productive discussions on this subject, and Moshe Jarden for helpful comments on a previous version of the manuscript.
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∅-DEFINABILITY OF HENSELIAN VALUATION RINGS
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FACHBEREICH MATHEMATIK UND STATISTIK UNIVERSITY OF KONSTANZ
78457 KONSTANZ, GERMANY E-mail: arno.fehm@uni-konstanz.de
