Valued fields with finitely many defect extensions of prime degree
aa r X i v : . [ m a t h . A C ] J a n VALUED FIELDS WITH FINITELY MANY DEFECTEXTENSIONS OF PRIME DEGREE
F.-V. KUHLMANN
Abstract.
We prove that a valued field of positive characteristic p that has onlyfinitely many distinct Artin-Schreier extensions (which is a property of infiniteNTP fields) is dense in its perfect hull. As a consequence, it is a deeply ramifiedfield and has p -divisible value group and perfect residue field. Further, we provea partial analogue for valued fields of mixed characteristic and observe an openproblem about 1-units in this setting. Finally, we fill a gap that occurred in a proofin an earlier paper in which we first introduced a classification of Artin-Schreierdefect extensions. Introduction
It is shown in [4, Theorem 3.1] that an infinite field of positive characteristic thatis definable in an NTP theory has only finitely many Artin-Schreier extensions.NTP is a large class of first order theories (“without the tree property of thesecond kind”) defined by S. Shelah generalizing simple and NIP theories. Algebraicexamples of NTP structures are given by ultraproducts of p -adic fields and certainvalued difference fields. For the precise definition (which we will not need in thispaper) and for further examples, we refer the reader to [4].In [4, Proposition 3.2] it is shown that the value group of a valued field of charac-teristic p > p -divisible. Inthis note, assuming throughout that the valuation is nontrivial, we prove a strongerresult, namely, that such a field is dense in its perfect hull. For a valued field ofcharacteristic p >
0, this is equivalent to being a deeply ramified field in the sense of[7]. The density also implies that the value group is p -divisible and the residue field Date : October 1, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Valued field, deeply ramified field, Artin-Schreier extension, Kummerextension, defect.The author is supported by Opus grant 2017/25/B/ST1/01815 from the National Science Centreof Poland. This paper was started while the author was on a research visit to the Departmentof Mathematics at the Federal University of S˜ao Carlos, supported by the S˜ao Paulo ResearchFoundation (FAPESP), grant number 2017/17835-9. The author would like to thank the membersof the department for their hospitality, and Josnei Novacoski for drawing his attention to the gapin an earlier paper that is now filled at the end of this paper. He also thanks the referee for thecareful reading of the manuscript and many helpful corrections and comments. is perfect. By Theorem 2.1 in Section 2.1, every deeply ramified field of positivecharacteristic is a semitame field, which we shall define now.Take a valued field (
K, v ) of arbitrary characteristic. Its value group will bedenoted by vK , and its residue field by Kv . Accordingly, the value of an element a ∈ K will be denoted by va , and its residue by av . We say that ( K, v ) is semitame if either char Kv = 0, or char Kv = p > (DRst) the value group vK is p -divisible, (DRvr) the homomorphism O K c /p O K c ∋ x x p ∈ O K c /p O K c is surjective, where O K c denotes the valuation ring of the completion K c of ( K, v ).Note that this condition implies that the residue field Kv is perfect.The following result will be proven in Section 3.2: Theorem 1.1.
Take a valued field ( K, v ) of characteristic p > . If K admits onlyfinitely many distinct Artin-Schreier extensions, then ( K, v ) is a semitame field. Corollary 1.2.
A nontrivially valued field of positive characteristic that is definablein an NTP theory is a semitame field. We will prove Theorem 1.1 by showing that if (
K, v ) is not dense in its perfecthull, then it admits infinitely many distinct Artin-Schreier extensions; see Proposi-tion 3.4. However, we will show more than this. We are interested in Galois defectextensions of degree p a prime. These are immediate Galois extensions ( L | K, v )of degree p of valued fields for which v has a unique extension from K to L (in thiscase we must have that p = char Kv ). Here, ( L | K, v ) denotes an extension L | K offields with v a valuation on L and K endowed with its restriction; the extension issaid to be immediate if the canonical embeddings of vK in vL and of Kv in Lv are onto. For more details on the defect see Section 2.4, and for further background,see [3, 9, 11].In [3] a classification of Galois defect extensions E = ( L | K, v ) of prime degree p is given as follows. We show that the setΣ σ := (cid:26) v (cid:18) σf − ff (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) f ∈ L × (cid:27) is independent of the choice of a generator σ of Gal ( L | K ), and we denote it by Σ E .We say that E has independent defect ifΣ E = { α ∈ vK | α > H E } for some proper convex subgroup H E of vK ; otherwise we say that E has dependentdefect . If vK is archimedean ordered, or in other words, ( K, v ) is of rank
1, then H E can only be equal to { } .In Section 3.2, we will prove: INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 3
Theorem 1.3.
Take a valued field ( K, v ) of characteristic p > . If it admitsan Artin-Schreier extension with dependent defect, then it admits infinitely manydistinct Artin-Schreier extensions with dependent defect. In [11] the classification was originally introduced only for valued fields of pos-itive characteristic. An Artin-Schreier defect extension ( L | K, v ) was said to havedependent defect if in a certain way it depends on immediate purely inseparableextensions of degree p which do not lie in the completion of ( K, v ). It was thenshown in Section 4.2 of [11] that an Artin-Schreier extension has dependent defectif it is obtained from such a purely inseparable defect extension by a certain trans-formation of an inseparable minimal polynomial which makes it separable. We willdescribe this transformation in Section 3.1 and use it for the proof of Theorem 1.3.In [3] we have made the above more precise by showing that every
Artin-Schreierextension with dependent defect can be obtained by such a transformation. It isalso shown that the new definition of the classification given in [3] is compatiblewith the one given in [11].The new definition of the classification became necessary in order to generalize theoriginal definition to the case of valued fields (
K, v ) of characteristic 0 with residuefield Kv of positive characteristic p ( mixed characteristic ), where we cannot relyon nontrivial purely inseparable extensions. We will now present a partial analogueof Theorem 1.3 in the mixed characteristic case. To avoid a number of technicaldetails in the present paper, we will restrict our scope to valued fields ( K, v ) of rank1. We will also assume that K contains a primitive p -th root of unity. We thenconsider Kummer defect extensions of degree p , that is, Galois defect extensions of( K, v ) generated by some η / ∈ K such that η p ∈ K . Because such extensions areimmediate, we can show that we can assume η to be a 1 -unit , i.e., v ( η − > vη = 0); see Section 2.7.We need some more preparation. Take an extension ( L | K, v ) and a ∈ L . Wedefine: v ( a − K ) := { v ( a − c ) | c ∈ K } ;for details on this set, see Section 2.2. For α in the divisible hull f vK of the valuegroup vK , we will write v ( a − K ) < α if v ( a − c ) < α for all c ∈ K , and similarlyfor “ ≤ ” in place of “ < ”. We write v ( a − K ) < | α if v ( a − K ) < α and v ( a − K ) is bounded away from α in f vK , i.e., there is β ∈ f vK such that v ( a − K ) ≤ β < α .A Kummer defect extension of degree p generated by a 1-unit η as above alwayssatisfies(1.1) v ( η − K ) < vpp − , see Section 2.7. In the case of archimedean value groups (where H E cn only be equalto { } ), the definition we have given for the extension to have dependent defect is F.-V. KUHLMANN equivalent to the condition that v ( η − K ) < | vpp − (see our discussion at the end ofSection 2.7).Note that in [3] in place of v ( η − K ) a more flexible invariant dist ( η, K ) is usedfor our work with defect extensions. We will give its definition in Section 2.3 toenable the reader to relate the results in [3] and [11] to those in the present paper.In view of Theorem 1.3 it seems reasonable to conjecture that if ( K, v ) admits aKummer extension with dependent defect, then it admits infinitely many distinctKummer extensions with dependent defect. By [3, part 2) of Theorem 1.5], thiswould imply that if (
K, v ) admits only finitely many Kummer defect extensions,then it is semitame, in perfect analogy to the equal positive characteristic case.However, so far we have only been able to prove a weaker result. We will say thata Kummer defect extension as above has super-dependent defect if(1.2) v ( η − K ) < | vpp . Theorem 1.4.
Take a valued field ( K, v ) of mixed characteristic and rank 1 whichcontains a primitive p -th root of unity. If it admits a Kummer extension of de-gree p with super-dependent defect, then it admits infinitely many distinct Kummerextensions of degree p with super-dependent defect. This leads us to the following
Open questions:
Does the above theorem also hold with dependent defect in placeof super-dependent defect? What is (possibly) special about a Kummer extensionwith dependent defect that is not super-dependent?We have observed in earlier work already that the threshold vpp plays a certain rolewhen dealing with 1-units in mixed characteristic (see [12, Corollary 2.11 d)]), butit is not yet sufficiently understood what exactly this role is.In the last section of this paper we extend our study of Artin-Schreier extensionsof valued fields in order to fill a gap that occurred in the proof of Lemma 2.31 of[11], which gives a criterion for such extensions to have nontrivial defect.For general background on valuation theory, we refer the reader to [5, 6, 13, 14].2.
Preliminaries
Deeply ramified and semitame fields.
For a field K of characteristic p > K /p ∞ the perfect hull of K . Further, we set K p = { a p | a ∈ K } and K /p = { a /p | a ∈ K } ; then K p is a subfield of K , and K /p | K is a field extensionwhich is trivial if and only if K is perfect.In [3], the following is proven: Theorem 2.1.
Take a nontrivially valued field ( K, v ) of characteristic p > . Thenthe following statements are equivalent:a) ( K, v ) is a semitame field, INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 5 b) ( K, v ) is a deeply ramified field,c) ( K, v ) satisfies (DRvr),d) the completion of ( K, v ) is perfect,e) ( K, v ) is dense in its perfect hull, i.e., K /p ∞ ⊂ K c ,f ) K p is dense in ( K, v ) . Here, the property “dense” refers to the topology induced by the valuation.See [3] for more details and the connection of semitame and deeply ramified fieldswith the classification of defect extensions.2.2.
The set v ( a − K ) . Take a totally ordered set (
T, < ). For a nonempty subset S of T and an element t ∈ T we will write S < t if s < t for every s ∈ S . A set S ⊆ T is called an initial segment of T if for each s ∈ S every t < s also lies in S .Similarly, S ⊆ T is called a final segment of T if for each s ∈ S every t > s alsolies in S .If ( T, < ) is an ordered abelian group, n ∈ N , α ∈ T , and S is an initial segmentof T , then nS + α := { nβ + α | β ∈ S } is an initial segment of nT .Take a valued field extension ( K ( a ) | K, v ). We will now collect various propertiesof the sets v ( a − K ) and v ( a − K ) ∩ vK . Proposition 2.2.
Take a valued field extension ( K ( a ) | K, v ) .1) If ( K ( a ) | K, v ) is immediate, then v ( a − K ) has no largest element.2) If v ( a − K ) has no largest element, then v ( a − K ) ⊆ vK .3) The set v ( a − K ) ∩ vK is an initial segment of vK .4) The set v ( a − K ) \ ( v ( a − K ) ∩ vK ) has at most one element.5) For every c, d ∈ K , v ( da + c − K ) = v ( a − K ) + vd .
6) If a, b are elements in some valued field extension of ( K, v ) such that v ( a − b ) >v ( a − K ) , then v ( a − c ) = v ( b − c ) for all c ∈ K and v ( a − K ) = v ( b − K ) .Proof. c ∈ K ; we wish to show that v ( a − c ) ∈ vK . By assumption there is d ∈ K such that v ( a − d ) > v ( a − c ). Hence v ( a − c ) = min { v ( a − c ) , v ( a − d ) } = v ( c − d ) ∈ vK .3): Take α ∈ v ( a − K ) ∩ vK and β ∈ vK such that β < α . Choose b, c ∈ K with vb = β and v ( a − c ) = α . Then c − b ∈ K and vb = min { v ( a − c ) , vb } = v ( a − ( c − b )) ∈ v ( a − K ).4): Assume that v ( a − c ) , v ( a − d ) ∈ v ( a − K ) \ ( v ( a − K ) ∩ vK ). If they weredistinct, say v ( a − d ) > v ( a − c ), then as in the proof of 2) we would obtain that v ( a − c ) ∈ vK , contradiction. F.-V. KUHLMANN v ( a + c − K ) = v ( a − K ) and v ( da − K ) = v ( a − K ) + vd , which are straightforward to prove.6): For all c ∈ K , from v ( a − b ) > v ( a − c ) we obtain that v ( a − c ) = v ( b − c ) asin the proof of 2); hence v ( a − c ) ∈ v ( b − K ) and v ( b − c ) ∈ v ( a − K ), which showsthat v ( a − K ) = v ( b − K ). (cid:3) We will mostly work with immediate extensions, in which case we have that v ( a − K ) ⊆ vK ( a ) = vK for every element a / ∈ K in the extension. However,several of our results hold more generally for extensions that are not necessarilyimmediate. Some of them, such as the following one, remain true when the set v ( a − K ) is replaced by v ( a − K ) ∩ vK . Lemma 2.3.
Take a valued field extension ( K ( a ) | K, v ) . Take α ∈ v ( a − K ) andassume that d ∈ K is such that (2.1) α + vd > v ( a − K ) (note that d always exists when v ( a − K ) is bounded by some element from vK ).Then the sets v ( a − K ) − vd n , n ∈ N , are pairwise distinct.Proof. Since α ∈ v ( a − K ) and α + vd > v ( a − K ), we know that vd >
0. Take anynatural numbers m < n . Then α − mvd ≥ α + vd − nvd > v ( a − K ) − nvd , which shows that α − mvd / ∈ v ( a − K ) − nvd = v ( a − K ) − vd n . But α − mvd ∈ v ( a − K ) − mvd = v ( a − K ) − vd m , so the two sets are distinct. (cid:3) Distances.
Take again a totally ordered set (
T, < ). A pair (Λ L , Λ R ) of subsetsof T is called a cut in T if Λ L is an initial segment of T and Λ R = T \ Λ L ; it thenfollows that Λ R is a final segment of T . To compare cuts in ( T, < ) we will use thelower cut sets comparison. That is, for two cuts Λ = (Λ L , Λ R ) , Λ = (Λ L , Λ R ) in T we will write Λ < Λ if Λ L Λ L , and Λ ≤ Λ if Λ L ⊆ Λ L . This defines a linearorder on the set of all cuts in T .For a given subset S of T we define S + to be the cut (Λ L , Λ R ) in T such that Λ L is the least initial segment containing S , that is, S + := ( { t ∈ T | ∃ s ∈ S : t ≤ s } , { t ∈ T | t > S } ) . Likewise, we denote by S − the cut (Λ L , Λ R ) in T such that Λ L is the largest initialsegment disjoint from S , i.e., S − := ( { t ∈ T | t < S } , { t ∈ T | ∃ s ∈ S : t ≥ s } ) . For s ∈ T , we set s + := { s } + and s − := { s } − . INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 7
We note that(2.2) S + ≤ s − ⇔ S < s and S + < s − ⇔ ∃ t ∈ T : S ≤ t < s . Indeed, S + ≤ s − means that the smallest initial segment of T containing S is asubset of the initial segment { t ∈ T | t < s } , and this is equivalent to S < s .Likewise, S + < s − means that the smallest initial segment of T containing S is aproper subset of { t ∈ T | t < s } , and as both are initial segments, this is equivalentto the existence of some t ∈ T such that S ≤ t .We embed T in the linearly ordered set of all cuts in T by identifying each s ∈ T with s − .The set v ( a − K ) ∩ vK is an initial segment of vK and thus the lower cut set ofa cut in vK . However, in order to be able to compare v ( a − K ) with v ( a − L ) when L | K is algebraic, it is more convenient to work with the cut v ( a − K ) induces inthe divisible hull f vK of vK . Indeed, f vK is equal to the value group of the algebraicclosure K ac of K , and if L | K is algebraic, then K ac = L ac and therefore, f vK = f vL .Hence we define:dist ( a, K ) := ( v ( a − K ) ∩ f vK ) + in the divisible hull f vK of vK .We call this cut the distance of z from K . The distance replaces the use ofsuprema in the case of non-archimedean value groups, which are not contained in R . Lemma 2.4.
Let the situation be as above and assume that v ( a − K ) ⊆ f vK .Take some α ∈ f vK . Then dist ( a, K ) ≤ α − is equivalent to v ( a − K ) < α , and dist ( a, K ) < α − is equivalent to v ( a − K ) < | α .Proof. We have: dist ( a, K ) ≤ α − ⇔ v ( a − K ) ∩ f vK < α ⇔ v ( a − K ) < α , where the first equivalence holds by (2.2) and the second equivalence holds since v ( a − K ) ⊆ f vK . Similarly,dist ( a, K ) < α − ⇔ ∃ β ∈ f vK : v ( a − K ) ∩ f vK ≤ β < α ⇔ ∃ β ∈ f vK : v ( a − K ) ≤ β < α ⇔ v ( a − K ) < | α . (cid:3) F.-V. KUHLMANN
The fundamental equality and the defect.
Take an extension ( L | K, v ) ofvalued fields such that the extension of v from K to L is unique. Assume thatchar Kv = p >
0. Then(2.3) [ L : K ] = p ν · ( vL : vK )[ Lv : Kv ] , where by the Lemma of Ostrowski ν is a nonnegative integer (see [13, Th´eor`eme 2,p. 236] or [14, Corollary to Theorem 25, p. 78]). The factor d ( L | K, v ) = p ν is calledthe defect of the extension ( L | K, v ).2.5.
Criteria for defect extensions.Lemma 2.5.
Take an extension ( K ( a ) | K, v ) of valued fields of degree p = char( Kv ) and such that the extension of v from K to K ( a ) is unique. If v ( a − K ) has nomaximal element, then ( K ( a ) | K, v ) is immediate with defect p .Proof. By [11, part (1) of Theorem 2.21], ( K ( a ) | K, v ) is immediate. As the extensionof v from K to K ( a ) is assumed to be unique, it follows that the defect is equal tothe degree of the extension. (cid:3) Lemma 2.6.
Assume that ( K ( a ) | K, v ) is a normal extension of prime degree p .1) If K h is some henselization of ( K, v ) and a / ∈ K h , then the extension of v from K to K ( a ) is unique.2) Assume that ( K, v ) is of rank 1. If v ( a − K ) has no maximal element and isbounded in vK , then ( K ( a ) | K, v ) is immediate and has defect p .Proof. K ( a ) | K is normal and of degree p , it is linearly disjoint from everyother algebraic extension L over K in which it is not contained. This is seen asfollows. If K ( a ) | K is inseparable, then it is purely inseparable, and so is L ( a ) | L ;hence the latter extension can only be of degree 1 or p . If K ( a ) | K is separable,then it is Galois with a Galois group of order p . Then also L ( a ) | L is Galois. Asrestriction embeds its Galois group in that of K ( a ) | K , again the degree of L ( a ) | L can only be 1 or p .Since a / ∈ K h by assumption, from what we have just shown we see that K ( a )must be linearly disjoint from K h over K , which by [1, Lemma 2.1] implies ourassertion.2): Since ( K, v ) is of rank 1, K lies dense in its henselization K h . By assumption, v ( a − K ) is bounded in vK , which implies that a / ∈ K h and thus by part 1),the extension of v from K to K ( a ) is unique. Now our assertion follows fromLemma 2.5. (cid:3) INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 9
Artin-Schreier extensions of valued fields.
Throughout this section, welet K be a field of characteristic p >
0. Recall that L | K is called an Artin-Schreierextension if it is generated by a root of a polynomial of the form X p − X − b with b ∈ K . Note that every Artin-Schreier extension of K is a Galois extension of degree p , and vice versa. We call ϑ an Artin-Schreier generator of an Artin–Schreierextension L | K if L = K ( ϑ ) with ϑ p − ϑ ∈ K . Lemma 2.7.
Assume that ϑ is an Artin-Schreier generator of an Artin–Schreierextension L | K . Then ϑ ′ is another Artin-Schreier generator of L | K if and only if ϑ ′ = iϑ + c for some i ∈ F p and c ∈ K .Proof. If ϑ and ϑ ′ are roots of the same polynomial X p − X − b , then ϑ − ϑ ′ is a rootof X p − X , whose roots are exactly the elements of F p . Hence, ϑ + i , i ∈ F p , are allroots of X p − X − b . Pick a nontrivial σ ∈ Gal L | K . We then have that σϑ − ϑ = j for some j ∈ F × p .If ϑ ′ is another Artin-Schreier generator of L | K such that σϑ − ϑ = j = σϑ ′ − ϑ ′ ,then we have σ ( ϑ − ϑ ′ ) = ϑ − ϑ ′ . Since σ is a generator of Gal L | K ≃ Z /p Z , itfollows that τ ( ϑ − ϑ ′ ) = ϑ − ϑ ′ for all τ ∈ Gal L | K , that is, ϑ − ϑ ′ ∈ K .If ϑ ′ is another Artin-Schreier generator of L | K and σϑ ′ − ϑ ′ = j ′ ∈ F × p , thenthere is some i ∈ F × p such that ij = j ′ . Since σi = i , we then have that σiϑ − iϑ = i ( σϑ − ϑ ) = ij = j ′ . Then by what we have shown before, ϑ ′ = iϑ + c for some c ∈ K .Conversely, if ϑ is an Artin-Schreier generator of L | K and if i ∈ F × p and c ∈ K ,then ( iϑ + c ) p − ( iϑ + c ) = i ( ϑ p − ϑ ) + c p − c ∈ K . But iϑ + c cannot lie in K ,so K ( iϑ + c ) = L since [ L : K ] is a prime. This shows that also iϑ + c is anArtin-Schreier generator of L | K . (cid:3) Corollary 2.8.
Let ( L | K, v ) be an Artin-Schreier extension of valued fields. Then v ( ϑ − K ) is independent of the choice of the Artin-Schreier generator ϑ of the ex-tension, so it is an invariant of the extension.Proof. Take two Artin-Schreier generators ϑ, ϑ ′ of L | K . By Lemma 2.7 we can write ϑ ′ = iϑ + c for some i ∈ F × p and c ∈ K . Since vi = 0 for every valuation, it followsfrom assertion 5) of Proposition 2.2 that v ( ϑ ′ − K ) = v ( ϑ − K ). (cid:3) Proposition 2.9.
Assume that ( K ( η ) | K, v ) is a purely inseparable extension and ( K ( ϑ ) | K, v ) is an Artin-Schreier extension of valued fields. Then K ( η, ϑ ) | K ( ϑ ) ispurely inseparable and so the extension of v from K ( ϑ ) to K ( η, ϑ ) is unique. If (2.4) v ( η − ϑ ) > v ( ϑ − K ) , then the extension of v from K to K ( ϑ ) is unique. If in addition, ( K ( η ) | K, v ) isimmediate, then ( K ( ϑ ) | K, v ) is immediate with defect p .Proof. Assume that (2.4) holds. Then ϑ / ∈ K h since otherwise it would followfrom Theorem 2 of [10] that K ( η ) | K is not purely inseparable, contradicting our assumption. Now our first assertion follows from part 1) of Lemma 2.6. By part 6)of Proposition 2.2, (2.4) also implies that v ( ϑ − K ) = v ( η − K ).Now assume in addition that ( K ( η ) | K, v ) is immediate. Then by part 1) of Propo-sition 2.2, v ( ϑ − K ) = v ( η − K ) has no maximal element. Therefore, our secondassertion follows from Lemma 2.5. (cid:3) Kummer extensions of prime degree of valued fields.
Take a valuedfield (
K, v ) of mixed characteristic, that is, char K = 0 while char Kv = p > p . Such an extension is generated by anelement η such that η p ∈ K . If ( K ( η ) | K, v ) is immediate, then it can be assumedthat η and hence also η p is a 1-unit, i.e., v ( η − >
0. Indeed, since ( K ( η ) | K, v ) isimmediate, we have that vη ∈ vK ( η ) = vK , so there is c ∈ K such that vc = − vη .Then vηc = 0, and since ηcv ∈ K ( η ) v = Kv , there is d ∈ K such that dv = ( ηcv ) − .Then v ( ηcd ) = 0 and ( ηcd ) v = 1. Hence ηcd is a 1-unit. Furthermore, K ( ηcd ) = K ( η ) and ( ηcd ) p = η p c p d p ∈ K . Thus we can replace η by ηcd and assume from thestart that η is a 1-unit.The next proposition follows from [3, Corollary 3.6 and Proposition 3.7]: Proposition 2.10.
Take a valued field of mixed characteristic and a Kummer defectextension as detailed above. Then the distance dist ( η, K ) does not depend on thechoice of the generator η of the extension ( K ( η ) | K, v ) as long as it is a -unit andsatisfies η p ∈ K . Moreover, (2.5) 0 < dist ( η, K ) ≤ (cid:18) vpp − (cid:19) − . Under the above conditions, the Kummer extension is immediate, hence we havethat v ( η − K ) ⊆ vK . From the definition of the distance it then follows that v ( η − K )is the intersection of the lower cut set of dist ( η, K ) with vK , hence the distancedetermines uniquely the set v ( η − K ), showing that this set does not depend onthe choice of the generator η . Moreover, from Lemma 2.4 we obtain that inequality(1.1) holds.In [3, Proposition 3.7] we show that a Kummer extension ( K ( η ) | K, v ), where η isa 1-unit with η p ∈ K , has dependent defect if and only ifdist ( a, K ) = vpp − H − , for some proper convex subgroup H of f vK . If ( K, v ) has rank 1, then f vK isarchimedean ordered and therefore H = { } . In this case, the above equationjust becomes dist ( η, K ) < (cid:18) vpp − (cid:19) − . In view of Lemma 2.4, this translates to v ( η − K ) < | vpp − , as mentioned in theIntroduction. INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 11
Differences of p -th powers in mixed characteristic.Lemma 2.11. Take η and a to be two elements of some valued field of characteristic0 with residue field characteristic p > . If (2.6) v ( η − a ) < vpp − vη , then (2.7) v ( η p − a p ) = pv ( η − a ) . Proof.
Let ζ be a primitive p -th root of unity. We have that η p − a p = p Y i =1 ( ζ i η − a ) = p Y i =1 ( ζ i η − η + η − a ) . Using the well known fact that v ( ζ i −
1) = vpp − v ( ζ i −
1) + vη = vpp − vη > v ( η − a ) . Hence we have that v ( ζ i η − a ) = v ( ζ i η − η + η − a ) = min { v ( ζ i −
1) + vη, v ( η − a ) } = v ( η − a ) , which yields equation (2.7). (cid:3) The case of valued fields of characteristic p >
A basic transformation.
We describe a transformation that was introducedin [11]. Take a valued field (
K, v ) of characteristic p > K ( η ) | K, v ) of degree p of valued fieldssuch that η p ∈ K and that v ( η − K ) is bounded from above by an element in vK ( η )(and hence also by an element in vK ). We are starting with the minimal polynomial Y p − η p of η over K and turn it into the separable polynomial(3.1) Y p − d p − Y − η p , where d = 0. Setting Y = dX and then dividing the resulting polynomial by d p , wetransform this polynomial into the Artin-Schreier polynomial(3.2) X p − X − η p d p . Under the condition that vd is large enough, the following lemma describes thebehaviour of the set v ( η − K ) when η is replaced by roots of these two polynomials. Lemma 3.1.
Take d ∈ K such that (3.3) vd p − > pv ( η − K ) − vη . Let ˜ ϑ d be a root of the polynomial (3.1). Then ϑ d := ˜ ϑ d d is a root of (3.2) and K ( ϑ d ) | K is an Artin-Schreier extension with a unique extensionof v from K to K ( ϑ d ) . Furthermore, we have that (3.4) v ( ˜ ϑ d − K ) = v ( η − K ) , (3.5) v (cid:16) ηd − ϑ d (cid:17) > v ( ϑ d − K ) , and (3.6) v ( ϑ d − K ) = v ( η − K ) − vd . Proof.
Once we prove that(3.7) v ( η − ˜ ϑ d ) > v ( η − K ) , we obtain equation (3.4) by part 6) of Proposition 2.2, which in turn implies equa-tion (3.6) by part 5) of Proposition 2.2. Further, again using part 5) of Proposi-tion 2.2 again, we obtain v (cid:16) ηd − ϑ d (cid:17) = v ( η − ˜ ϑ d ) − vd > v ( η − K ) − vd = v ( ˜ ϑ d − K ) − vd = v ( ϑ d − K ) , which proves (3.5). Hence we will now prove (3.7). We compute:(3.8) ( η − ˜ ϑ d ) p = η p − ˜ ϑ pd = η p − η p − d p − ˜ ϑ d = − d p − ˜ ϑ d . We set Y = dX to obtain that ϑ d is a root of the polynomial d p X p − d p − dX − η p and hence also of the Artin-Schreier polynomial (3.2).Since vη ∈ v ( η − K ), from (3.3) we obtain:( p − vd = vd p − > pvη − vη = ( p − vη , so that v η p d p = p ( vη − vd ) < . Hence we have that vϑ d < vϑ pd = pvϑ d < vϑ d and pv ηd = v η p d p = min { vϑ pd , vϑ d } = vϑ pd = pvϑ d , which yields that v ˜ ϑ d = vd + vϑ d = vη . From this together with (3.3) and (3.8) we obtain: v ( η − ˜ ϑ d ) = 1 p v ( d p − ˜ ϑ d ) = 1 p ( vd p − + vη ) > v ( η − K ) , INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 13 as desired.It remains to prove that K ( ϑ d ) | K is nontrivial (and hence an Artin-Schreier ex-tension), and that the extension of v from K to K ( ϑ d ) is unique. Since v ( η − K )is bounded from above in vK by assumption, the same holds for v ( η − K ) − vd which by (3.6) is equal to v ( ϑ d − K ). This implies in particular that ϑ d / ∈ K , so theextension K ( ϑ d ) | K is nontrivial. Since η is purely inseparable over K by assump-tion, the same holds for η/d . Thus our assertion follows from (3.5) together withProposition 2.9, where we replace η by η/d and ϑ by ϑ d . (cid:3) Proof of Theorems 1.1 and 1.3.
Throughout, let (
K, v ) be a valued field ofcharacteristic p >
0. First, we prove: if K admits only finitely many distinct Artin-Schreier extensions, then ( K, v ) is asemitame field, which is the assertion of Theorem 1.1. Take a purely inseparable (not necessarilyimmediate) extension K ( η ) | K of degree p such that η p ∈ K and that v ( η − K ) isbounded from above by an element in vK . Proposition 3.2.
Assume that α ∈ v ( η − K ) and d ∈ K are such that α + vd >v ( η − K ) (that is, (2.1) holds for η in place of a ) and vd p − > pv ( η − K ) − vη (that is, (3.3) holds). Then with the notation from Lemma 3.1, the Artin-Schreierextensions K ( ϑ d n ) | K , n ∈ N , are pairwise distinct, and the extensions of v from K to K ( ϑ d n ) are unique for all n ∈ N .Proof. It follows from Lemma 3.1 that for each n ∈ N , K ( ϑ d n ) | K is an Artin-Schreierextension and the extension of v from K to all K ( ϑ d n ) is unique. From Lemma 2.3 weinfer that the sets v ( ϑ d n − K ) = v ( η − K ) − vd n are distinct. Thus by Corollary 2.8,the extensions K ( ϑ d n ) | K are distinct. (cid:3) Lemma 3.3.
If the perfect hull of K does not lie in the completion of ( K, v ) , then ( K, v ) admits a purely inseparable extension K ( η ) | K of degree p such that v ( η − K ) is bounded from above by an element in vK .Proof. Assume that the perfect hull K /p ∞ does not lie in the completion K c of( K, v ), and take an element ˜ η ∈ K /p ∞ \ K c . We may assume that ˜ η p ∈ K c (oth-erwise, we replace ˜ η by ˜ η p ν for a suitable ν ≥ η / ∈ K c , we have that v (˜ η − K ) is bounded from above by some α ∈ vK and v (˜ η p − K p ) = pv (˜ η − K )is bounded from above by pα . On the other hand, since ˜ η p ∈ K c , there is some b ∈ K such that v (˜ η p − b ) > pα . We choose η ∈ K /p such that η p = b .Then pv (˜ η − η ) = v (˜ η p − η p ) = v (˜ η p − b ) > pα ≥ pv (˜ η − K ), which yields that v (˜ η − η ) > v (˜ η − K ). From this it follows by part 6) of Proposition 2.2 that v ( η − K ) = v (˜ η − K ), showing that also v ( η − K ) is bounded from above by α . (cid:3) Since v ( η − K ) is assumed to be bounded from above by an element in vK , thereis some d ∈ K which satisfies vd p − > pv ( η − K ) − vη . Then the same is true forevery d ′ ∈ K with vd ′ ≥ vd . Hence α ∈ v ( η − K ) and d ∈ K can be chosen suchthat α + vd > v ( η − K ). Thus we can use Lemma 3.3 together with Proposition 3.2to obtain: Proposition 3.4.
If the perfect hull of K does not lie in the completion of ( K, v ) ,then K admits infinitely many Artin-Schreier extensions. By the equivalence of assertions a) and e) of Theorem 2.1, this proposition provesTheorem 1.1.We will now prove the assertion of Theorem 1.3, which states: if ( K, v ) admits an Artin-Schreier extension with dependent defect, then it admitsinfinitely many distinct Artin-Schreier extensions with dependent defect. We assume that (
K, v ) admits an Artin-Schreier extension with dependent defect,which means that it must be obtained via the transformation described in Section 3.1from a purely inseparable defect extension ( K ( η ) | K, v ) of degree p . This is onlypossible if v ( η − K ) is bounded from above by an element in vK . Also, since theextension is of degree p with nontrivial defect, the defect must be p and the extensionmust be immediate. Hence by part 1) of Proposition 2.2, v ( η − K ) has no largestelement and consequently, the same holds for the sets v ( η − K ) − vd n = v ( ϑ d n − K ).Therefore, since the extension of v from K to K ( ϑ d n ) is unique by Proposition 3.2,it follows from Lemma 2.5 that ( K ( ϑ d n ) | K, v ) is immediate with defect p . As thisextension is obtained from a purely inseparable defect extension of degree p bythe transformation described in Section 3.1, this defect is dependent by definition.Finally, Proposition 3.2 shows that the extensions K ( ϑ d n ) | K , n ∈ N , are distinct.This completes our proof.4. The case of valued fields of mixed characteristic
Throughout this section, we assume that (
K, v ) is a valued field of rank 1 andof characteristic 0 with residue field of characteristic p >
0. Further, we assumethat K contains a primitive p -th root of unity, and that ( K ( η ) | K, v ) is a Kummerextension, where η is a 1-unit with η p ∈ K .4.1. A basic transformation in the mixed characteristic case.
Given d ∈ K ,we transform the minimal polynomial X p − η p of η over K into the polynomial(4.1) f d ( X ) := X p + h d ( X ) − η p with h d ( X ) := p − X i =1 (cid:18) pi (cid:19) d p − i X i . INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 15
Lemma 4.1.
Assume that vd < and that (4.2) v ( η − K ) < vp + ( p − vdp . Take ˜ ϑ d to be a root of the polynomial (4.1). Then v ( η − K ) = v ( ˜ ϑ d − K ) . Proof.
We compute the value of the coefficients of h d , using our assumption (4.2):(4.3) v (cid:18)(cid:18) pi (cid:19) d p − i (cid:19) = vp + ( p − i ) vd ≥ vp + ( p − vd > pv ( η − K ) . Since 0 = v ( η − ∈ v ( η − K ), this shows that all coefficients of h d have positivevalue, while vη p = 0; this forces ˜ ϑ d to have value 0 by the ultrametric triangle law.Consequently,(4.4) vh d ( ˜ ϑ d ) > pv ( η − K ) . Suppose that there is c ∈ K such that v ( ˜ ϑ d − η ) ≤ v ( η − c ). Combined with ourassumption (4.2), this implies that assumption (2.6) of Lemma 2.11 holds for η andfor ˜ ϑ d in place of a since vη = 0. Hence by Lemma 2.11 we have that v ( ˜ ϑ pd − η p ) = pv ( ˜ ϑ d − η ) . Using this together with (4.4) and the fact that ˜ ϑ pd = η p + h d ( ˜ ϑ d ) by definition of˜ ϑ d , we compute: v ( ˜ ϑ d − η ) = 1 p v ( ˜ ϑ pd − η p ) = 1 p v ( η p + h d ( ˜ ϑ d ) − η p )= 1 p vh d ( ˜ ϑ d ) > v ( η − K ) , contradicting our assumption. This shows that v ( ˜ ϑ d − η ) > v ( η − c ) for all c ∈ K ,whence v ( ˜ ϑ d − η ) > v ( η − K ). Hence by part 6) of Proposition 2.2, v ( η − K ) = v ( ˜ ϑ d − K ). (cid:3) Proof of Theorem 1.4.
We let (
K, v ) be a valued field of mixed characteristicand rank 1 which contains a primitive p -th root of unity. We have to prove theassertion of Theorem 1.4, which states: If ( K, v ) admits a Kummer extension of degree p with super-dependent defect, then itadmits infinitely many distinct Kummer extensions of degree p with super-dependentdefect. We assume that ( K ( η ) | K, v ) is a Kummer extension of degree p with super-dependentdefect, and that η is a 1-unit with η p ∈ K . Since 0 = vη ∈ v ( η − K ) and v ( η − K ) doesnot contain a maximal element, it must contain positive elements. Since ( K ( η ) | K, v ) is a super-dependent defect extension, we know that (1.2) holds. Since by assump-tion, ( K, v ) is of rank 1 and v ( η − K ) does not contain a maximal element but isbounded, vK must be dense in f vK . Hence there is some d ∈ K such that vd < v ( η − K ) < vpp + 2 vd . Then inequality (4.2) of Lemma 4.1 is satisfied, because vpp + 2 vd < vp + ( p − vdp . We obtain from Lemma 4.1 that for a root ˜ ϑ d of the polynomial f d ( X ) defined inthat lemma, v ( η − K ) = v ( ˜ ϑ d − K ) . Now we set X = dY ; dividing the resulting polynomial f d ( dY ) by d p , we obtainthe polynomial g d ( Y ) := Y p + p − X i =2 (cid:18) pi (cid:19) Y i − η p d p . We observe that ϑ d := ˜ ϑ d d is a root of g d ( Y ) and that by part 5) of Proposition 2.2,(4.6) v ( ϑ d − K ) = v ( ˜ ϑ d − K ) − vd . Now we set Y = X −
1, which turns the polynomial g d ( Y ) into the polynomial X p − (cid:18) η p d p + 1 (cid:19) with root η d := ϑ d + 1 . Again by part 5) of Proposition 2.2, we have that v ( η d − K ) = v ( ϑ d − K ) = v ( ˜ ϑ d − K ) − vd = v ( η − K ) − vd = v ( η − K ) , where the last inequality holds since vd = 0 and v ( η − K ) is a bounded subset of anarchimedean ordered abelian group. As vd < vη , η pd = η p d p + 1is a 1-unit.Since vK is dense, there are infinitely many d ′ ∈ K with vd ′ < d . With the same argument as above, we see that v ( η − K ) − vd = v ( η − K ) − vd ′ if vd = vd ′ . In this way we obtain infinitely many Kummer extensions( K ( η d ) | K, v ) with distinct sets v ( η d − K ), which by Proposition 2.10 shows that theseextensions are pairwise distinct. INITELY MANY DEFECT EXTENSIONS OF PRIME DEGREE 17
Since ( K ( η ) | K, v ) is a nontrivial immediate extension, part 1) of Proposition 2.2shows that the set v ( η − K ) has no maximal element, while being bounded byassumption. The same is consequently true for the sets v ( η − K ) − vd = v ( η d − K ). Asthe rank of ( K, v ) is assumed to be 1, part 2) of Lemma 2.6 shows that ( K ( η d ) | K, v )is immediate and has defect p . Finally, this defect is super-dependent since v ( η d − K ) = v ( η − K ) − vd < vpp + 2 vd − vd = vpp + vd . Filling a gap in [11]In [11, Lemma 2.31] the following is stated:
Lemma 5.1.
Assume that char K = p > and ( K ( ϑ ) | K, v ) is an Artin-Schreierextension with Artin-Schreier generator ϑ . If dist ( ϑ, K ) ≤ − and v ( ϑ − K ) has nomaximal element, then the extension of v from K to K ( ϑ ) is unique and ( K ( ϑ ) | K, v ) is immediate with defect p . In the proof it is written: “In [10] we show that the assumption that dist ( ϑ, K ) < v from K to K ( ϑ ) is unique.” Since v ( ϑ − K )has no maximal element, Lemma 2.5 shows that this assertion indeed implies that( K ( ϑ ) | K, v ) is immediate. In addition, dist ( ϑ, K ) < ∞ implies that ϑ / ∈ K , so theextension is nontrivial and thus has defect p .However, the assertion was never proven in [10]. In order to complete the proofof Lemma 5.1, we will prove it here. As v ( ϑ − K ) has no maximal element, weknow from part part 2) of Proposition 2.2 that v ( a − K ) ⊆ vK ⊆ f vK . Hence byLemma 2.4 the assumption dist ( ϑ, K ) ≤ − is equivalent to v ( ϑ − K ) < ϑ / ∈ K h . Suppose otherwise.Then by [10, Theorem 1], there are α ∈ vK and a convex subgroup H = { } of vK such that the coset α + H is cofinal in v ( ϑ − K ). Since v ( ϑ − K ) <
0, we must havethat α + H <
0, which yields that − α > H and − α >
0. As α ∈ α + H , there issome c ∈ K such that v ( ϑ − c ) ≥ α . Since ϑ − c is also an Artin-Schreier generatorof the extension and v ( ϑ − c − K ) = v ( ϑ − K ), we may assume w.l.o.g. that vϑ ≥ α .Let X p − X − b with b ∈ K be the minimal polynomial of ϑ over K , and take d ∈ K with vd = − α . Then by part 5) of Proposition 2.2, H is cofinal in v ( dϑ − K ).Further, dϑ is a root of d − p X p − d − X − b and hence also of X p − d p − X − d p b . Let η be the root of X p − d p b . We wish to show that v ( dϑ − K ) = v ( η − K ) . We compute: pv ( dϑ − η ) = v (( dϑ ) p − η p ) = v ( d p ϑ + d p b − d p b ) = vd p ϑ ≥ − pα + α ≥ − α > H . Since H is a convex subgroup of vK , this implies that v ( dϑ − η ) ≥ − p α > H ,and as H is cofinal in v ( dϑ − K ), we obtain: v ( dϑ − η ) > v ( dϑ − K ) . By construction, K ( η ) | K is purely inseparable, hence by Theorem 2 of [10], dϑ cannot lie in the henselization K h . We have thus shown that ϑ / ∈ K h , as desired. References [1] Blaszczok, A. – Kuhlmann, F.-V. :
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Institute of Mathematics, University of Szczecin, ul. Wielkopolska 15 70-451Szczecin, Poland
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