Algebraic independence of elements in immediate extensions of valued fields
aa r X i v : . [ m a t h . A C ] A p r ALGEBRAIC INDEPENDENCE OF ELEMENTS IN IMMEDIATEEXTENSIONS OF VALUED FIELDS
ANNA BLASZCZOK AND FRANZ-VIKTOR KUHLMANN
Abstract.
Refining a constructive combinatorial method due to MacLaneand Schilling, we give several criteria for a valued field that guarantee thatall of its maximal immediate extensions have infinite transcendence degree.If the value group of the field has countable cofinality, then these criteriagive the same information for the completions of the field. The criteria haveapplications to the classification of valuations on rational function fields. Wealso apply the criteria to the question which extensions of a maximal valuedfield, algebraic or of finite transcendence degree, are again maximal. In thecase of valued fields of infinite p -degree, we obtain the worst possible examplesof nonuniqueness of maximal immediate extensions: fields which admit analgebraic maximal immediate extension as well as one of infinite transcendencedegree. Introduction
In this paper, we denote a valued field by (
K, v ), its value group by vK , and itsresidue field by Kv . When we talk of a valued field extension ( L | K, v ) we meanthat (
L, v ) is a valued field, L | K a field extension, and K is endowed with therestriction of v . For the basic facts about valued fields, we refer the reader to[2, 3, 9, 14, 16, 17].One of the basic problems in valuation theory is the description of the possibleextensions of a valuation from a valued field ( K, v ) to a given extension field L .The case of an algebraic extension L | K is taken care of by ramification theory.Another important case is given when L | K is an algebraic function field. Val-uations on algebraic function fields appear naturally in algebraic geometry, realalgebraic geometry and the model theory of valued fields, to name only a few areas.Local uniformization, the local form of resolution of singularities, is essentially aproperty of valued algebraic function fields (cf. [6]). This problem, which is stillopen in positive characteristic, requires a detailed knowledge of all possible valua-tions on such function fields. The same is true for corresponding problems in themodel theory of valued fields.By means of ramification theory, the problem of describing all appearing valua-tions is reduced to the case of rational function fields. The case of a single variableattracted many authors; see the references in [7] for a selection from the exten-sive literature on this case. The case of higher transcendence degree was treated Date : 4. 4. 2013.2010
Mathematics Subject Classification. in [7]. What at first glance appeared to be problem easily solvable by induction,turned out to tightly connected with the intricate question whether the maximalimmediate extensions of a given valued field have finite or infinite transcendencedegree.An extension ( L | K, v ) of valued fields is called immediate if the canonical em-beddings vK ֒ → vL and Kv ֒ → Lv are onto. It was shown by W. Krull [11] thatmaximal immediate extensions exist for every valued field. The proof uses Zorn’sLemma in combination with an upper bound for the cardinality of valued fields withprescribed value group and residue field. Krull’s deduction of this upper bound ishard to read; later, K. A. H. Gravett [4] gave a nice and simple proof.A valued field ( K, v ) is called henselian if it satisfies Hensel’s Lemma, or equiva-lently, if the extension of its valuation v to its algebraic closure, which we will denoteby ˜ K , is unique. A henselization of ( K, v ) is an extension which is henselian andminimal in the sense that it can be embedded over K , as a valued field, in everyother henselian extension field of ( K, v ). Therefore, a henselization of (
K, v ) can befound in every henselian extension field, and henselizations are unique up to valua-tion preserving isomorphism over K (this is why we will speak of the henselizationof ( K, v ). The henselization is an immediate separable-algebraic extension.A valued field is called maximal if it does not admit any proper immediateextension; clearly, maximal immediate extensions are maximal fields. I. Kaplan-sky ([5]) characterized the maximal field as those in which every pseudo Cauchysequence admits a pseudo limit. From this result it follows that power series fieldsare maximal fields. For example, for any field k the Laurent series field k (( t )) withthe t -adic valuation is a maximal immediate extension of k ( t ), and it is well knownthat k (( t )) is of infinite transcendence degree over k ( t ). This can be shown by acardinality argument (and some facts about field extensions in case k is not count-able). A constructive proof was given by MacLane and Schilling in Section 3 of[12]. Our main theorem is a far-reaching generalization of their result. A part ofthis theorem has already been applied in [7] to the problem described above. Theorem 1.1.
Take a valued field extension ( L | K, v ) of finite transcendence degree ≥ , with v nontrivial on L . Assume that one of the following four cases holds:valuation-transcendental case: vL/vK is not a torsion group, or Lv | Kv is tran-scendental;value-algebraic case: vL/vK contains elements of arbitrarily high order, or thereis a subgroup Γ ⊆ vL containing vK such that Γ /vK is an infinite torsion groupand the order of each of its elements is prime to the characteristic exponent of Kv ;residue-algebraic case: Lv contains elements of arbitrarily high degree over Kv ;separable-algebraic case: L | K contains a separable-algebraic subextension L | K such that within some henselization of L , the corresponding extension L h | K h isinfinite.Then each maximal immediate extension of ( L, v ) has infinite transcendence degreeover L . If the cofinality of vL is countable (which for instance is the case if vL contains an element γ such that γ > vK ), then already the completion of ( L, v ) hasinfinite transcendence degree over L . Note that the cofinality of vL is equal to the cofinality of vK unless there issome γ in vL which is larger than every element of vK . In that case, because L | K LGEBRAIC INDEPENDENCE 3 has finite transcendence degree, vL will have countable cofinality, no matter whatthe cofinality of vK is.Note further that the condition in the residue-algebraic case always holds when Lv | Kv contains an infinite separable-algebraic subextension; this is a consequenceof the Theorem of the Primitive Element. There is no analogue of this theoremin abelian groups; therefore, the first condition in the value-algebraic case doesnot follow from the second. As an example, take q to be a prime different fromchar Kv and consider the case where vL/vK is an infinite product of Z /q Z . Underthe second condition, however, the result can easily be deduced from the separable-algebraic case by passing to a henselization L h of L and using Hensel’s Lemma toshow that L h | K admits the required subextension. For the details, see the proof ofTheorem 1.1 in Section 3.The key assumption in the separable-algebraic case is that the separable-algebraicsubextension remains infinite when passing to the respective henselizations. Weshow that this condition is crucial. Take a valued field ( k, v ) which has a tran-scendental maximal immediate extension ( M, v ). We know that (
M, v ) is henselian(cf. Lemma 2.1). Take a transcendence basis T of M | k and set K := k ( T ). Thenfrom Lemma 2.3 it follows that the henselization K h of K inside of ( M, v ) is aninfinite separable-algebraic subextension of ( M | K, v ). But M is a maximal imme-diate extension of L := K h and M | L is algebraic. Hence the assertion of Theorem1.1 does not necessarily hold without the condition that L h | K h is infinite.An interesting special case is given when ( K, v ) is itself a maximal field. In thiscase, it is well known that if ( L | K, v ) is a finite extension, then (
L, v ) is itself amaximal field. So we ask what happens if ( L | K, v ) is infinite algebraic or tran-scendental of finite transcendence degree. Under which conditions could (
L, v ) beagain a maximal field? This question will be addressed in Section 4, where all ofthe following theorems will be proved.
Theorem 1.2.
Take a maximal field ( K, v ) and an infinite algebraic extension ( L | K, v ) . Assume that L | K contains an infinite separable subextension or that (1) ( vK : pvK )[ Kv : Kv p ] < ∞ , where p is the characteristic exponent of Kv . Then every maximal immediate ex-tension of ( L, v ) has infinite transcendence degree over L . As an immediate consequence, we obtain:
Corollary 1.3.
Take a maximal field ( K, v ) of characteristic and an algebraicextension ( L | K, v ) . Then ( L, v ) is maximal if and only if L | K is a finite extension. It remains to discuss the case where L | K is an infinite extension, its maximalseparable subextension K ′ | K is finite, and condition (1) fails. Since then also( K ′ , v ) is maximal, we can replace K by K ′ and simply concentrate on the casewhere L | K is purely inseparable.Note that if the maximal field K is of characteristic p , then condition (1) impliesthat the p -degree of K is finite, as it is equal to ( vK : pvK )[ Kv : Kv p ]. If condition(1) does not hold, then the purely inseparable extension K /p | K is infinite; since vK /p = p vK and K /p v = ( Kv ) /p , we then have that vK /p /vK is of exponent p , every element in K /p v \ Kv has degree p over Kv , and at least one of the twoextensions is infinite. Since ( K /p , v ) is again maximal (regardless of the p -degree ANNA BLASZCZOK AND FRANZ-VIKTOR KUHLMANN of K , see Lemma 4.1), this case shows that the assertion of Theorem 1.1 may faileven when vL/vK is an infinite torsion group or Lv | Kv is an infinite algebraicextension. In fact, all possible cases can appear for infinite p -degree: Theorem 1.4.
Take a maximal field ( K, v ) of characteristic p > for which con-dition (1) fails (which is equivalent to K having infinite p -degree). Take κ to be themaximum of ( vK : pvK ) and [ Kv : Kv p ] , considered as cardinals. Then:a) The valued field ( K /p , v ) is again maximal, although vK /p /vK is an infinitetorsion group or K /p v | Kv is an infinite algebraic extension.b) For every n ∈ N and every infinite cardinal λ ≤ κ , there are subextensions ( L n | K, v ) and ( L λ | K, v ) of ( K /p | K, v ) such that ( K /p | L λ , v ) is an immediatealgebraic extension of degree λ and ( K /p | L n , v ) is an immediate algebraic extensionof degree p n .c) There is a purely inseparable extension ( L | K, v ) with • vL = p vK and Lv = Kv if ( vK : pvK ) = ∞ , • vL = vK and Lv = ( Kv ) /p if [ Kv : Kv p ] = ∞ ,such that every maximal immediate extension of ( L, v ) has transcendence degree atleast κ . In both cases, L can also be taken to simultaneously satisfy vL = p vK and Lv = ( Kv ) /p .If the cofinality of vK is countable, then in b), K /p can be replaced by thecompletion of L λ or L n , respectively, and in c), “maximal immediate extension”can be replaced by “completion”. Case b) of this theorem is a generalization of Nagata’s example ([13, Appendix,Example (E3.1), pp. 206-207]). Similar to that example, the valued fields in b) arenonmaximal fields admitting an algebraic maximal immediate extension. We notethat the field L of part c) is not contained in K /p .It was shown by Kaplansky that if a valued field satisfies “hypothesis A”, then itsmaximal immediate extensions are unique up to isomorphism ([5, Theorem 5]; seealso [10]). Kaplansky also gives an example for a valued field for which uniquenessfails ([5, Section 5]). The question whether uniqueness always fails when hypothesisA is violated is open. Different partial answers were given in [10] and in [15]. Tothe best knowledge of the authors, the next theorem presents, for the first time inthe literature, the worst case of nonuniqueness: Theorem 1.5.
Take a maximal field ( K, v ) of characteristic p > satisfying oneof the following conditions:i) vK/pvK is infinite and vK admits a set of representatives of the cosets modulo pvK which contains an infinite bounded subset, orii) the residue field extension Kv | ( Kv ) p is infinite and the value group vK is notdiscrete.Then there is an infinite purely inseparable extension ( L, v ) of ( K, v ) which admits ( K /p , v ) as an algebraic maximal immediate extension, but also admits a maximalimmediate extension of infinite transcendence degree. Let us mention that the condition of i) always holds when vK/pvK is infinite and vK is of finite rank, or Γ /p Γ is infinite for some archimedean component Γ of vK .For example, if F p denotes the field with p elements and G is an ordered subgroup LGEBRAIC INDEPENDENCE 5 of the reals of the form L i ∈ N r i Z , then the power series field F p (( G )) satisfies thecondition of i). If t i , i ∈ N , are algebraically independent over F p , then the powerseries field F p ( t i | i ∈ N )(( Q )) with residue field F p ( t i | i ∈ N ) satisfies the conditionof ii).Finally, let us discuss the case of transcendental extensions ( L, v ) of a maximalfield (
K, v ). In view of the valuation-transcendental case of Theorem 1.1, it remainsto consider the valuation-algebraic case where vL/vK is a torsion group and Lv | Kv is algebraic. Here is a partial answer to our above question: Theorem 1.6.
Take a maximal field ( K, v ) and a transcendental extension ( L, v ) of ( K, v ) of finite transcendence degree. Assume that Lv | Kv is separable-algebraicand vL/vK is a torsion group such that the characteristic of Kv does not dividethe orders of its elements. Then Lv | Kv or vL/vK is infinite and every maximalimmediate extension of ( L, v ) has infinite transcendence degree over L . We do not know an answer in the case where the conditions on the value group andresidue field extensions fail. 2.
Preliminaries By va we denote the value of an element a ∈ K , and by av its residue. Givenany subset S of K , we define vS = { va | = a ∈ S } and Sv = { av | a ∈ S, va ≥ } . The valuation ring of (
K, v ) will be denoted by O K .2.1. The fundamental inequality.
Every finite extension ( L | K, v ) of valued fieldssatisfies the fundamental inequality (cf. (17.5) of [2] or Theorem 19 on p. 55 of[17]):(2) [ L : K ] ≥ ( vL : vK )[ Lv : Kv ] . The nature of the “missing factor” on the right hand side is determined by the
Lemma of Ostrowski which says that whenever the extension of v from K to L is unique, then(3) [ L : K ] = p ν · ( vL : vK ) · [ Lv : Kv ] with ν ≥ , where p is the characteristic exponent of Lv , that is, p = char Lv if this ispositive, and p = 1 otherwise. For the proof, see [14, Theoreme 2, p. 236]) or [17,Corollary to Theorem 25, p. 78]).The factor d = d( L | K, v ) = p ν is called the defect of the extension ( L | K, v ). Ifd = 1, then we call ( L | K, v ) a defectless extension . Note that ( L | K, v ) is alwaysdefectless if char Kv = 0.We call a henselian field ( K, v ) a defectless field if equality holds in the fun-damental inequality (2) for every finite extension L of K . Theorem 2.1.
Every maximal field is henselian and a defectless field.Proof.
The henselization of a valued field is an immediate extension. Therefore,a maximal field is equal to its henselization and thus henselian. For a proof thatmaximal fields are defectless fields, see [16, Theorem 31.21]. (cid:3)
ANNA BLASZCZOK AND FRANZ-VIKTOR KUHLMANN
Some facts about henselian fields and henselizations.
Let (
K, v ) beany valued field. If a ∈ ˜ K \ K is not purely inseparable over K , we choose someextension of v from K to ˜ K and definekras( a, K ) := max { v ( τ a − σa ) | σ, τ ∈ Gal ( ˜ K | K ) and τ a = σa } ∈ v ˜ K and call it the Krasner constant of a over K . Since all extensions of v from K to˜ K are conjugate, this does not depend on the choice of the particular extension of v .For the same reason, over a henselian field ( K, v ) our Krasner constant kras( a, K )is equal to max { v ( a − σa ) | σ ∈ Gal ( ˜ K | K ) and a = σa } . Lemma 2.2.
Take an extension ( K ( a ) | K, v ) of henselian fields, where a is anelement in the separable-algebraic closure of K with va ≥ . Then (4) va ≤ kras( a, K ) , and for every polynomial f = d m X m + . . . + d ∈ K [ X ] of degree m < [ K ( a ) : K ] , (5) vf ( a ) ≤ vd m + m kras( a, K ) . Proof.
Since (
K, v ) is henselian, vσa = a and therefore v ( a − σa ) ≥ va for all σ .This yields inequality (4).Take any element b in the separable-algebraic closure of K with [ K ( b ) : K ] < [ K ( a ) : K ]. Then v ( a − b ) ≤ kras( a, K ) since otherwise, Krasner’s Lemmawould yield that a ∈ K ( b ) and [ K ( b ) : K ] ≥ [ K ( a ) : K ]. If we write f ( X ) = d m Q mi =1 ( X − b i ), then [ K ( b i ) : K ] ≤ deg( f ) < [ K ( a ) : K ]. Hence, vf ( a ) = vd m + m X i =1 v ( a − b i ) ≤ vd m + m kras( a, K ) . This proves inequality (5). (cid:3)
Lemma 2.3.
Take a nontrivially valued field ( k ( T ) , v ) , where T is a nonempty setof elements algebraically independent over k . Then the henselization of ( k ( T ) , v ) inside of any henselian valued extension field is an infinite extension of k ( T ) .Proof. Set F := k ( T ) and take a henselization F h of F inside of some henselianvalued extension field. Pick an arbitrary t ∈ T . Without loss of generality we canassume that vt >
0. By Hensel’s Lemma, F h contains a root ϑ of the polynomial X − X − t such that vϑ >
0. We proceed by induction. Once we have constructed ϑ i with vϑ i > i ∈ N , we again use Hensel’s Lemma to obtain a root ϑ i +1 ∈ F h of the polynomial X − X − ϑ i with vϑ i +1 > F ( ϑ i | i ∈ N ) | F is infinite. To thisend, we consider the t − -adic valuation w on F = k ( T \ { t } )( t − ) which is trivialon k ( T \ { t } ). We note that wF = Z . Since wt <
0, we obtain that wϑ = wt and by induction, wϑ i = i wt . Therefore, the 2-divisible hull of Z is containedin wF ( ϑ i | i ∈ N ). In view of the fundamental inequality (2), this shows that F ( ϑ i | i ∈ N ) | F cannot be a finite extension. (cid:3) Lemma 2.4.
Assume ( L, v ) to be henselian and K to be relatively separable-algebraically closed in L . Then Kv is relatively separable-algebraically closed in Lv . If in addition Lv | Kv is algebraic, then the torsion subgroup of vL/vK is a p -group, where p is the characteristic exponent of Kv . LGEBRAIC INDEPENDENCE 7
Proof.
Take ζ ∈ Lv separable-algebraic over Kv . Choose a monic polynomial g ( X ) ∈ K [ X ] whose reduction gv ( X ) ∈ Kv [ X ] modulo v is the minimal polynomialof ζ over Kv . Then ζ is a simple root of gv . Hence by Hensel’s Lemma, there isa root a ∈ L of g whose residue is ζ . As all roots of gv are distinct, we can liftthem all to distinct roots of g . Thus, a is separable-algebraic over K . From theassumption of the lemma, it follows that a ∈ K , showing that ζ ∈ Kv . This provesthat Kv is relatively separable-algebraically closed in Lv .Now assume in addition that Lv | Kv is algebraic. Then Kv is relatively separable-algebraically closed in Lv , by what we have proved already. Take α ∈ vL and n ∈ N not divisible by p such that nα ∈ vK . Choose a ∈ L and b ∈ K such that va = α and vb = nα . Then v ( a n /b ) = 0. Since Lv | Kv is a purely inseparable extension,there exists m ∈ N such that (( a n /b ) v ) p m ∈ Kv . We choose c ∈ K satisfying vc = 0 and cv = (( a n /b ) v ) p m , to obtain that ( a np m /cb p m ) v = 1. So the reductionof the polynomial X n − a np m /cb p m modulo v is X n −
1. Since n is not divisibleby p , 1 is a simple root of this polynomial. Hence by Hensel’s Lemma, there isa simple root d ∈ L of the polynomial X n − a np m /cb p m with dv = 1, whence vd = 0. Consequently, a p m /d is a simple root of the polynomial X n − cb p m andthus is separable algebraic over K . Since K was assumed to be relatively separable-algebraically closed in L , we find that a p m /d ∈ K . As n is not divisible by p , thereare k, l ∈ Z such that 1 = kn + lp m . This yields: α = knα + lp m α = knα + l ( p m va − vd ) = k ( nα ) + lv (cid:18) a p m d (cid:19) ∈ vK. (cid:3) Valuation independence.
For the easy proof of the following lemma, see [1,chapter VI, § Lemma 2.5.
Let ( L | K, v ) be an extension of valued fields. Take elements x i , y j ∈ L , i ∈ I , j ∈ J , such that the values vx i , i ∈ I , are rationally independent over vK , and the residues y j v , j ∈ J , are algebraically independent over Kv . Then theelements x i , y j , i ∈ I , j ∈ J , are algebraically independent over K .Moreover, if we write f = X k c k Y i ∈ I x µ k,i i Y j ∈ J y ν k,j j ∈ K [ x i , y j | i ∈ I, j ∈ J ] in such a way that for every k = ℓ there is some i s.t. µ k,i = µ ℓ,i or some j s.t. ν k,j = ν ℓ,j , then (6) vf = min k v c k Y i ∈ I x µ k,i i Y j ∈ J y ν k,j j = min k vc k + X i ∈ I µ k,i vx i . That is, the value of the polynomial f is equal to the least of the values of itsmonomials. In particular, this implies: vK ( x i , y j | i ∈ I, j ∈ J ) = vK ⊕ M i ∈ I Z vx i K ( x i , y j | i ∈ I, j ∈ J ) v = Kv ( y j v | j ∈ J ) . Moreover, the valuation v on K ( x i , y j | i ∈ I, j ∈ J ) is uniquely determined by itsrestriction to K , the values vx i and the residues y j v . ANNA BLASZCZOK AND FRANZ-VIKTOR KUHLMANN
Conversely, if ( K, v ) is any valued field and we assign to the vx i any values in anordered group extension of vK which are rationally independent, then (6) definesa valuation on L , and the residues y j v , j ∈ J , are algebraically independent over Kv . Corollary 2.6.
Let ( L | K, v ) be an extension of valued fields. Then (7) trdeg L | K ≥ trdeg Lv | Kv + rr ( vL/vK ) . If in addition L | K is a function field and if equality holds in (7), then the extensions vL | vK and Lv | Kv are finitely generated.Proof. Choose elements x , . . . , x ρ , y , . . . , y τ ∈ L such that the values vx , . . . , vx ρ are rationally independent over vK and the residues y v, . . . , y τ v are algebraicallyindependent over Kv . Then by the foregoing lemma, ρ + τ ≤ trdeg L | K . This provesthat trdeg Lv | Kv and the rational rank of vL/vK are finite. Therefore, we maychoose the elements x i , y j such that τ = trdeg Lv | Kv and ρ = dim Q Q ⊗ ( vL/vK )to obtain inequality (7).Set L := K ( x , . . . , x ρ , y , . . . , y τ ) and assume that equality holds in (7). Thismeans that the extension L | L is algebraic. Since L | K is finitely generated, itfollows that this extension is finite. By the fundamental inequality (2), this yieldsthat vL | vL and Lv | L v are finite extensions. Since already vL | vK and L v | Kv are finitely generated by the foregoing lemma, it follows that also vL | vK and Lv | Kv are finitely generated. (cid:3) The algebraic analogue to the transcendental case discussed in Lemma 2.5 is thefollowing lemma (see [17] for a proof):
Lemma 2.7.
Let ( L | K, v ) be an extension of valued fields. Suppose that η , . . . , η k ∈ L such that vη , . . . , vη k ∈ vL belong to distinct cosets modulo vK . Further, assumethat ϑ , . . . , ϑ ℓ ∈ O L such that ϑ v, . . . , ϑ ℓ v are Kv -linearly independent. Then theelements η i ϑ j , ≤ i ≤ k , ≤ j ≤ ℓ , are K -linearly independent, and for everychoice of elements c ij ∈ K , we have that v X i,j c ij η i ϑ j = min i,j vc ij η i ϑ j = min i,j ( vc ij + vη i ) . If the elements η i ϑ j form a K -basis of L , then vL = vK + M ≤ i ≤ k Z vη i and Lv = Kv ( ϑ j v | ≤ j ≤ ℓ ) . For any element x in a field extension of K and every nonnegative integer n , weset K [ x ] n := K + Kx + . . . + Kx n . Since dim K K [ x ] n ≤ n + 1, we obtain the following corollary from Lemma 2.7: Corollary 2.8.
Take a valued field extension ( K ( x ) | K, v ) . Then for every n ≥ ,a) the elements of vK [ x ] n lie in at most n + 1 many distinct cosets modulo vK ,b) the Kv -vector space K [ x ] n v is of dimension at most n + 1 . LGEBRAIC INDEPENDENCE 9
Immediate extensions.
We will assume some familiarity with the basicproperties of pseudo Cauchy sequences; we refer the reader to Kaplansky’s pa-per “Maximal fields with valuations” ([5]). In particular, we will use the followingtwo main theorems:
Theorem 2.9. (Theorem 2 of [5])
For every pseudo Cauchy sequence ( a ν ) ν<λ in ( K, v ) of transcendental type thereexists an immediate transcendental extension ( K ( x ) , v ) such that x is a pseudo limitof ( a ν ) ν<λ . If ( K ( y ) , v ) is another valued extension field of ( K, v ) such that y is apseudo limit of ( a ν ) ν<λ , then y is also transcendental over K and the isomorphismbetween K ( x ) and K ( y ) over K sending x to y is valuation preserving. Theorem 2.10. (Theorem 3 of [5])
Take a pseudo Cauchy sequence ( a ν ) ν<λ in ( K, v ) of algebraic type. Choose apolynomial f ( X ) ∈ K [ X ] of minimal degree whose value is not fixed by ( a ν ) ν<λ ,and a root z of f . Then there exists an extension of v from K to K ( z ) such that ( K ( z ) | K, v ) is an immediate extension and z is a pseudo limit of ( a ν ) ν<λ .If ( K ( z ′ ) , v ) is another valued extension field of ( K, v ) such that z ′ is also a rootof f and a pseudo limit of ( a ν ) ν<λ , then the field isomorphism between K ( a ) and K ( b ) over K sending a to b will preserve the valuation. We will need a few more results that are not in Kaplansky’s paper.
Lemma 2.11.
Take an algebraic algebraic field extension ( K ( a ) | K, v ) , where a is apseudo limit of a pseudo Cauchy sequence ( a ν ) ν<λ in ( K, v ) without a pseudo limitin K . Then ( a ν ) ν<λ does not fix the value of the minimal polynomial of a over K .Proof. We denote the minimal polynomial of a over K by f ( X ) = Q ni =1 ( X − σ i a )with σ i ∈ Gal ( ˜ K | K ). Since a is a pseudo limit of ( a ν ) ν<λ , the values v ( a ν − a ) areultimately increasing. If v ( a − σ i a ) > v ( a ν − a ) for all ν < λ , then also the values v ( a ν − σ i a ) = min { v ( a ν − a ) , v ( a − σ i a ) } = v ( a ν − a ) are ultimately increasing. Ifon the other hand, v ( a − σ i a ) ≤ v ( a ν − a ) for some ν < λ , then for ν < ν < λ , thevalue v ( a ν − σ i a ) = min { v ( a ν − a ) , v ( a − σ i a ) } = v ( a − σ i a ) is fixed. We concludethat the values vf ( a ν ) = P ni =1 v ( a ν − σ i a ) are ultimately increasing. (cid:3) Lemma 2.12.
Take a henselian field ( K, v ) of positive characteristic p and a pseudoCauchy sequence ( a ν ) ν<λ in ( K, v ) without a pseudo limit in K . If ( K ( a ) | K, v ) isa valued field extension of degree p such that a is a pseudo limit of ( a ν ) ν<λ , then ( K ( a ) | K, v ) is immediate.Proof. By the previous lemma, ( a ν ) ν<λ does not fix the value of the minimal poly-nomial f of a over K . On the other hand, we will show that ( a ν ) ν<λ fixes thevalue of every polynomial of degree less than deg f = p . We take g ∈ K [ X ] to be apolynomial of smallest degree such that ( a ν ) ν<λ does not fix the value of g . Since( a ν ) ν<λ admits no pseudo limit in ( K, v ), the polynomial g is of degree at least 2.Take a root b of g . By Theorem 2.10, there is an extension of the valuation v from K to K ( b ) such that ( K ( b ) | K, v ) is immediate. Since [ K ( b ) : K ] ≥ K, v ) ishenselian, the Lemma of Ostrowski implies that [ K ( b ) : K ] ≥ p . This shows that f is a polynomial of smallest degree whose value is not fixed by ( a ν ) ν<λ . Henceagain by Theorem 2.10, there is an extension of the valuation v from K to K ( a )such that ( K ( a ) | K, v ) is immediate. Since (
K, v ) is henselian, this extension coin-cides with the given valuation on K ( a ) and we have thus proved that the extension( K ( a ) | K, v ) is immediate. (cid:3)
The following result is Proposition 4.3 of [8]:
Proposition 2.13.
Take a valued field ( F, v ) of positive characteristic p . Assumethat F admits an immediate purely inseparable extension F ( η ) of degree p suchthat the element η does not lie in the completion of ( F, v ) . Then for each element b ∈ F × such that (8) ( p − vb + vη > pv ( η − c ) holds for every c ∈ F , any root ϑ of the polynomial X p − X − (cid:16) ηb (cid:17) p generates an immediate Galois extension ( F ( ϑ ) | F, v ) of degree p with a uniqueextension of the valuation v from F to F ( ϑ ) . Characteristic blind Taylor expansion.
We need a Taylor expansion thatworks in all characteristics. For polynomials f ∈ K [ X ], we define the i -th formalderivative of f as(9) f i ( X ) := n X j = i (cid:18) ji (cid:19) c j X j − i = n − i X j =0 (cid:18) j + ii (cid:19) c j + i X j . Then regardless of the characteristic of K , we have the Taylor expansion of f at c in the following form:(10) f ( X ) = n X i =0 f i ( c )( X − c ) i . Algebraic independence of elements in maximal immediateextensions
This section is devoted to the proof of Theorem 1.1. Our first goal is a basicindependence lemma.Take i ∈ N , any field K and a polynomial f ∈ K [ X , . . . , X i ]. With respect tothe lexicographic order on Z i , let ( µ , . . . , µ i ) be maximal with the property thatthe coefficient of X µ · · · X µ i i in f is nonzero. Then define c f to be this coefficientand call ( µ , . . . , µ i ) the crucial exponent of f .For our basic independence lemma, we consider the following situation. Wechoose a function ϕ : N × N −→ N such that ϕ ( k, ℓ ) > max { k, ℓ } and ϕ ( k + 1 , ℓ ) > ϕ ( k, ℓ ) for all k, ℓ ∈ N ,and for each i ∈ N a strictly increasing sequence ( E i ( k )) k ∈ N of integers ≥ k ≥ i ≥ E ( k + 1) ≥ ϕ ( k, E ( k )) + 1 ,E i ( k + 1) ≥ E i − ( ϕ ( k, E i ( k )) + 1) . (cid:27) Then for i, k ∈ N ,(12) E i ( k ) > k and E i ( k + 1) ≥ ϕ ( k, E i ( k )) + 1 > E i ( k ) + 1 . Further, we take an extension ( L | K, v ) of valued fields, elements a j ∈ L and α j ∈ vL for all j ∈ N , LGEBRAIC INDEPENDENCE 11 and K -subspaces S j ⊆ L , j ∈ N . We assume that for all i, k, ℓ ∈ N , the following conditions are satisfied: (A1) ≤ va k ≤ α k < va k +1 and kα E i ( k ) ≤ α ϕ ( k,E i ( k )) , (A2) a , . . . , a k ∈ S k and S k ⊆ S k +1 , (A3) if d , . . . , d k ∈ S k and u ∈ S ℓ , then d + d u + . . . + d k u k ∈ S ϕ ( k,ℓ ) , (A4) if m ≤ k and d , . . . , d m ∈ S k , then v ( d + d a k +1 + . . . + d m a mk +1 ) ≤ vd m + mα k +1 . Now we choose any maximal immediate extension (
M, v ) of (
L, v ). For each i ,we take an arbitrary pseudo limit y i ∈ M of the pseudo Cauchy sequence k X j =1 a E i ( j ) k ∈ N . In this situation, we can prove the following basic independence lemma:
Lemma 3.1.
Suppose that k ≥ is an integer and f ∈ L [ X , . . . , X i ] is a polyno-mial with coefficients in S k − ∩ O L such that α E i ( k ) ≥ vc f and that f has degreeless than k in each variable. Then (13) vf ( y , . . . , y i ) < va E i ( k +1) . Proof.
We shall prove the lemma by induction on i . We start with i = 1 and set u := k X j =1 a E ( j ) and z := y − u . Then u ∈ S E ( k ) because of (A2) and the fact that the S k are vector spaces. By(A1), the definition of E and our assumption that α E ( k ) ≥ vc f ,(14) vz = va E ( k +1) ≥ va ϕ ( k,E ( k ))+1 > α ϕ ( k,E ( k )) ≥ kα E ( k ) ≥ vc f + ( k − α E ( k ) . We use the Taylor expansion(15) f ( y ) = f ( u + z ) = f ( u ) + zf ( u ) + z f ( u ) + . . . where f j ( X ) ∈ O L [ X ] is the j -th formal derivative of f as defined in (9). We havethat f j ( u ) ∈ O L for all j . Hence,(16) v ( zf ( u ) + z f ( u ) + . . . ) ≥ vz . We wish to prove that vf ( u ) < vz . We set u ′ := k − X j =1 a E ( j ) ∈ S E ( k − so that u = u ′ + a E ( k ) . We use the Taylor expansion f ( u ) = f ( u ′ + a E ( k ) ) = f ( u ′ ) + f ( u ′ ) a E ( k ) + . . . + f m ( u ′ ) a mE ( k ) where m = deg f < k . By definition, c f is the leading coefficient of f , which in turnis equal to the constant f m ( u ′ ) = f m ( X ) ∈ L . Since f has coefficients in the vector space S k − , we know from (9) that also all f j have coefficients in S k − . Thus, (A3)and (A2) show that f ( u ′ ) , f j ( u ′ ) ∈ S ϕ ( k − ,E ( k − ⊆ S E ( k ) − for each j . Further, m ≤ k − ≤ E ( k ) −
1. Hence by (A4) and (14), vf ( u ) ≤ vf m ( u ′ ) + mα E ( k ) ≤ vc f + ( k − α E ( k ) < vz . From this together with (15) and (16), we deduce that vf ( y ) = vf ( u ) < vz , which gives the assertion of our lemma for the case of i = 1.In the case of i > i − i , and we set u := k X j =1 a E i ( j ) ∈ S E i ( k ) , u ′ := k − X j =1 a E i ( j ) ∈ S E i ( k − , and z := y i − u . Then by (12), (A1) and our assumption that α E i ( k ) ≥ vc f , vz = va E i ( k +1) ≥ va ϕ ( k,E i ( k ))+1 > α ϕ ( k,E i ( k )) ≥ kα E i ( k ) ≥ vc f + ( k − α E i ( k ) . We use the Taylor expansion f ( y , . . . , y i − , u + z ) = f ( y , . . . , y i − , u ) + zf ( y , . . . , y i − , u )+ z f ( y , . . . , y i − , u ) + . . . where f j ∈ O L [ X , . . . , X i ] is the j -th formal derivative of f with respect to X i .We obtain the analogue of inequality (16); hence it will suffice to prove that(17) vf ( y , . . . , y i − , u ) < vz . We set g ( X , . . . , X i − ) := f ( X , . . . , X i − , u )so that g ( y , . . . , y i − ) = f ( y , . . . , y i − , u ). Viewing f as a polynomial in thevariables X , . . . , X i − with coefficients in L [ X i ], we denote by h ( X i ) the coefficientof X µ · · · X µ i − i − in f . Note that h has coefficients in S k − , its leading coefficient is c f and its degree is µ i < k . Again, since h has coefficients in S k − , (9) shows thatthe same is true for the j -th formal derivative h j of h , for all j . Thus, (A3), (12)and (A2) imply that h ( u ′ ) , h j ( u ′ ) ∈ S ϕ ( k − ,E i ( k − ⊆ S E i ( k ) − for each j . As in the first part of our proof we find that vh ( u ) ≤ vh µ i ( u ′ ) + µ i α E i ( k ) = vc f + µ i α E i ( k ) since h µ i ( u ′ ) = c f . In particular, this shows that h ( u ) = 0. Hence if ( µ , . . . , µ i ) isthe crucial exponent of f , then ( µ , . . . , µ i − ) is the crucial exponent of g , and c g = h ( u ) . We set k ′ := ϕ ( k, E i ( k )) > max { k, E i ( k ) } . Since µ i ≤ k − vc f ≤ α E i ( k ) by assumption, and by virtue of (A1) and (12),it follows that vc g = vh ( u ) ≤ vc f + ( k − α E i ( k ) ≤ kα E i ( k ) ≤ α k ′ < α E i − ( k ′ ) . LGEBRAIC INDEPENDENCE 13
Since every coefficient of g is of the form h ( u ) with h a polynomial of degree lessthan k and coefficients in S k − , we know from (A3), our conditions on ϕ and (A2)that the coefficients of g lie in S ϕ ( k − ,E i ( k )) ⊆ S ϕ ( k,E i ( k )) − = S k ′ − . Also, its degree in each variable is less than k , hence less than k ′ . Therefore, wecan apply the induction hypothesis to the case of i −
1, with k ′ in place of k . Weobtain, by (A1) and our choice of the numbers E i ( k ): vf ( y , . . . , y i − , u ) < va E i − ( ϕ ( k,E i ( k ))+1) ≤ va E i ( k +1) = vz . This establishes our lemma. (cid:3)
By (A2), S ∞ := [ k ∈ N S k contains a k for all k . We set K ∞ := K ( S ∞ ) . Further, we note that condition (A1) implies thatΓ := { α ∈ vK ∞ | − va k ≤ α ≤ va k for some k } is a convex subgroup of vK ∞ . Corollary 3.2.
Assume that every element of K ∞ with value in Γ can be writtenas a quotient r/s with r, s ∈ S ∞ such that ≤ vs ∈ Γ . Then the elements y i , i ∈ N , are algebraically independent over K ∞ .Proof. We have to check that g ( y , . . . , y i ) = 0 for all i and all nonzero polynomi-als g ( X , . . . , X i ) ∈ K ∞ [ X , . . . , X i ]. After division by some coefficient of g withminimal value we may assume that g has integral coefficients in K ∞ and at leastone of them has value 0 ∈ Γ. We write all its coefficients which have value in Γ inthe form as given in our assumption. We take ˜ s to be the product of all appearingdenominators. Then v ˜ s ∈ Γ. After multiplication with ˜ s , all coefficients of g withvalue in Γ are elements of S ∞ , and there is at least one such coefficient. Now wewrite g ( X , . . . , X i ) = f ( X , . . . , X i ) + h ( X , . . . , X i ) where every coefficient of f is in S ∞ and has value less than va k for some k , and every coefficient of h hasvalue bigger than va k for all k (we allow h to be the zero polynomial). Since g hascoefficients of value v ˜ s , the polynomial f is nonzero. Since vy i ≥ i , wehave that vh ( y , . . . , y i ) is bigger than va k for all k .We choose k such that the assumptions of Lemma 3.1 hold; note that k existssince by our definition of f , the coefficient c f has value less than va k for some k .We obtain that vf ( y , . . . , y i ) < va E i ( k +1) < vh ( y , . . . , y i ) . This gives that vg ( y , . . . , y i ) = v ( f ( y , . . . , y i )+ h ( y , . . . , y i )) = vf ( y , . . . , y i ) < va E i ( k +1) < ∞ , that is, g ( y , . . . , y i ) = 0. (cid:3) Now we are able to give the
Proof of Theorem 1.1:
In all cases of the proof, we will choose functions ϕ that have the previously requiredproperties. We will choose a suitable sequence ( b k ) k ∈ N of elements in L and asequence ( c k ) k ∈ N in K . Then we will set a k := c k b k and choose some values α k ≥ va k .First, let us consider the valuation-transcendental case. We set ϕ ( k, ℓ ) := k + kℓ , and note that equations (11) now read as follows: E ( k + 1) ≥ k + kE ( k ) + 1 ,E i ( k + 1) ≥ E i − ( k + kE i ( k ) + 1) . Further, we will work with a suitable element t ∈ O L transcendental over K andset, after a suitable choice of the sequence ( c k ) k ∈ N , a k := c k t k ,α k := va k ,S k := K + Kt + . . . + Kt k . Conditions (A2) and (A3) are immediate consequences of our choice of S k as theset of all polynomials in K [ t ] of degree at most k .Suppose that vL/vK is not a torsion group. Then we pick t ∈ O L such that vt isrationally independent over vK (that is, nvt / ∈ vK for all integers n > k we set b k = t k and c k = 1 so that a k = t k . Then condition (A1) is satisfiedsince we have that0 ≤ va k = α k = vt k = kvt < ( k + 1) vt = vt k +1 = va k +1 and kα E i ( k ) = kva E i ( k ) = kvt E i ( k ) = kE i ( k ) vt < ( k + kE i ( k )) vt = α ϕ ( k,E i ( k )) . Suppose now that vL/vK is a torsion group. In this case, Kv | Lv is transcen-dental by assumption, and we note that since v is assumed nontrivial on L , it mustbe nontrivial on K . We pick t ∈ O L such that vt = 0 and tv is transcendental over Kv . Further, we choose a sequence ( c k ) k ∈ N in O K such that vc k +1 ≥ kvc k for all k . Since va k = vc k + kvt = vc k , we obtain that va k = α k < va k +1 and(18) kα k = kva k ≤ va k +1 . Then by (12),(19) kα E i ( k ) < E i ( k ) α E i ( k ) ≤ va E i ( k )+1 ≤ va ϕ ( k,E i ( k )) ≤ α ϕ ( k,E i ( k )) . Hence again, condition (A1) is satisfied.Now we have to verify (A4), simultaneously for all of the above choices for a k .Take d , . . . , d m ∈ S k , m ≤ k , and write d j = P kν =0 d jν t ν with d jν ∈ K . Then d + d a k +1 + . . . + d m a mk +1 = m X j =0 k X ν =0 c jk +1 d jν t j ( k +1)+ ν . LGEBRAIC INDEPENDENCE 15
In this sum, each power of t appears only once. So we have, by Lemma 2.5, v ( d + d a k +1 + . . . + d m a mk +1 ) = min j,ν vc jk +1 d jν t j ( k +1)+ ν := β . If this minimum is obtained at j = j and ν = ν , then β = vc j k +1 d j ν t j ( k +1)+ ν = min ν vc j k +1 d j ν t j ( k +1)+ ν = (min ν vd j ν t ν ) + va j k +1 = vd j a j k +1 , where the last equality again holds by Lemma 2.5. For all j , β ≤ min ν vc jk +1 d jν t j ( k +1)+ ν = (min ν vd jν t ν ) + va jk +1 = vd j a jk +1 . This gives that v ( d + d a k +1 + . . . + d m a mk +1 ) = β = min j vd j a jk +1 ≤ vd m + mva k +1 = vd m + mα k +1 , as required. Finally, we have to verify the assumption of Corollary 3.2. Eachelement in K ∞ can be written as a quotient r/s of polynomials in t with coefficientsin K . After multiplying both r and s with a suitable element from K we may assumethat s has coefficients in O K and one of them is 1. If this is the coefficient of t i ,say, then it follows by Lemma 2.5 that 0 ≤ vs ≤ vt i ≤ va i and thus, vs ∈ Γ.Now we take any maximal immediate extension (
M, v ) and y i as defined pre-ceeding to Lemma 3.1. Then we can infer from Corollary 3.2 that the elements y i are algebraically independent over K ∞ ; that is, the transcendence degree of M over K ∞ is infinite. Since the transcendence degree of L over K and thus also thatof L over K ∞ is finite, we can conclude that the transcendence degree of M over L is infinite.Next, we consider the value-algebraic case and the residue-algebraic case. Wewill assume for now that there is an algebraic subextension L | K of L | K such that vL /vK contains elements of arbitrarily high order, or L v contains elements ofarbitrarily high degree over Kv . The remaining cases will be treated at the end ofthe proof of our theorem.For the present case as well as the separable-algebraic case, we work with anyfunction ϕ that satisfies the conditions outlined in the beginning of this section,and with S k := K ( a , . . . , a k ) . Then S ∞ is a field and the assumption of Corollary 3.2 are trivially satisfied (taking s = 1). Further, condition (A2) is trivially satisfied. To prove that condition (A3)holds, take any u ∈ S ℓ = K ( a , . . . , a ℓ ) . If n = max { k, ℓ } , then d , . . . , d k , u ∈ K ( a , . . . , a n ) = S n and therefore, d + d u + . . . + d k u k ∈ S n ⊆ S ϕ ( k,ℓ ) . This shows that (A3) holds.By induction, we define a k ∈ L as follows, and we always take α k = va k .We start with a = 1 and α = 0. Suppose that a , . . . , a k are already de-fined. Since K ( a , . . . , a k ) | K is a finite extension, also vK ( a , . . . , a k ) /vK and K ( a , . . . , a k ) v | Kv are finite. Hence by our assumption in the algebraic case, thereis some b k +1 ∈ L such that0 , vb k +1 , vb k +1 , . . . , kvb k +1 lie in distinct cosets modulo vK ( a , . . . , a k ), or(20) 1 , b k +1 v, ( b k +1 v ) , . . . , ( b k +1 v ) k are K ( a , . . . , a k ) v -linearly independent.(21)If L v contains elements of arbitrarily high degree over Kv , we always choose b k +1 such that (21) holds; in this case, vb k +1 = 0 and we choose the elements c k asin the residue-transcendental case above. Otherwise, vL /vK contains elements ofarbitrarily high order, and we always choose b k +1 such that (20) holds. In this case,we choose c k +1 such that for a k +1 := c k +1 b k +1 we obtain kα k = kva k ≤ va k +1 ;this is possible since the values of b k and hence of all a k lie in the convex hull of vK in vL . As in the residue-transcendental case above, we obtain (18) and (19),showing that condition (A1) is satisfied.To prove that (A4) holds, take any k ≥ d , . . . , d k ∈ S k = K ( a , . . . , a k ).By Lemma 2.7 applied to b k +1 , v ( d + d a k +1 + . . . + d k a kk +1 ) = v ( d + d c k +1 b k +1 + . . . + d k c kk +1 b kk +1 )= min i vd i c ik +1 b ik +1 = min i vd i a ik +1 . This shows that (A4) holds.As in the valuation-transcendental case, we can now deduce our assertion aboutthe transcendence degree.Next, we consider the separable-algebraic case. In this case, we can w.l.o.g.assume that (
K, v ) is henselian. Indeed, each maximal immediate extension of(
L, v ) contains a henselization L h of ( L, v ) and hence also a henselization K h of( K, v ), and our assumption on L implies that the subfield L .K h of L h is an infiniteseparable-algebraic extension of K h . (Here, L .K h denotes the field compositum,i.e., the smallest subfield of L h which contains L and K h .)We take S k and ϕ ( k, ℓ ) as in the previous case, so that again, (A2), (A3) andthe assumption of Corollary 3.2 hold. Then we take a = b to be any elementin O L \ K and choose some α ∈ vK such that α ≥ kras( a , K ) ∈ v ˜ K ; this ispossible since vK is cofinal in its divisible hull, which is equal to v ˜ K . Inequality(4) of Lemma 2.2 shows that kras( a , K ) ≥ va , so that α ≥ va . Suppose wehave chosen a , . . . , a k ∈ O L . Since L | K is infinite and separable-algebraic, thesame is true for L | K ( a , . . . , a k ). By the Theorem of the Primitive Element, wecan therefore find an element b k +1 ∈ L such that[ K ( a , . . . , a k , b k +1 ) : K ( a , . . . , a k )] ≥ k + 1 . We choose c k +1 ∈ K such that for a k +1 := c k +1 b k +1 we have that kα k ≤ va k +1 .Finally, we choose α k +1 ∈ vK such that α k +1 ≥ kras( a k +1 , K ) ≥ va k +1 . Again, we obtain that (19) and (A1) hold.It only remains to show that (A4) holds. But this follows readily from inequality(5) of Lemma 2.2, where we take K ( a , . . . , a k ) in place of K and a = a k +1 , togetherwith the fact that kras( a k +1 , K ( a . . . , a k )) ≤ kras( a k +1 , K ).As before, we now obtain our assertion about the transcendence degree. LGEBRAIC INDEPENDENCE 17
It remains to prove the value-algebraic case and the residue-algebraic case fortranscendental valued field extensions ( L | K, v ) of finite transcendence degree. Weassume that vL/vK is a torsion group containing elements of arbitrarily high orderor the extension Lv | Kv is algebraic and such that Lv contains elements of arbitrarilyhigh degree over Kv .Take any subextension E | K of L | K . Then ( L | K, v ) satisfies the above assump-tion if and only if at least one of the extensions ( L | E, v ) and ( E | K, v ) satisfies theassumption. Choose a transcendence basis ( x , . . . , x n ) of L | K and set F := K ( x , . . . , x n ) . Then L | F is algebraic. By what we have already proved, if vL/vF contains elementsof arbitrarily high order or Lv contains elements of arbitrarily high degree over F v ,then any maximal immediate extension of (
L, v ) has infinite transcendence degreeover L .Suppose now that ( F | K, v ) satisfies the assumption on the value group or theresidue field extension. Take s ∈ N minimal such that vK ( x , . . . , x s ) /vK containselements of arbitrarily high order or K ( x , . . . , x s ) v contains elements of arbitrarilyhigh degree over Kv . Then the assertion holds also for the value group or theresidue field extension of ( K ( x , . . . , x s ) | K ( x , . . . , x s − ) , v ). We can replace K by K ( x , . . . , x s − ) and we will write x in place of x s so that now we have asubextension ( K ( x ) | K, v ) that satisfies the assertion for its value group or its residuefield extension.In both the value-algebraic and the residue-algebraic case we define b k ∈ K [ x ]by induction on k and set S k := K [ x ] N k with N k := deg b k .Assume that vK ( x ) contains elements of arbitrarily high order modulo vK . Thensuch elements can be already chosen from vK [ x ]. We set b = 1. Suppose that b , . . . , b k are already chosen with deg b i − < deg b i for 1 < i ≤ k . From Corol-lary 2.8 we know that vS k contains only finitely many values that represent dis-tinct cosets modulo vK . Since all of these values are torsion modulo vK , thesubgroup h vS k i of vK ( x ) generated by vS k satisfies ( h vS k i : vK ) < ∞ . By as-sumption, there is b k +1 ∈ K [ x ] for which the order of vb k +1 modulo vK is at least( k + 1)( h vS k i : vK ); this forces 0 , vb k +1 , vb k +1 , . . . , kvb k +1 to lie in distinct cosetsmodulo h vS k i . Since b k +1 / ∈ K [ x ] N k , we have that N k +1 = deg b k +1 > N k .Assume now that K ( x ) v contains elements of arbitrarily high degree over Kv .Without loss of generality we can assume that vK ( x ) /vK is then a torsion groupwith a finite exponent N . Otherwise, vL/vK is not a torsion group and we arein the valuation-transcendental case or vK ( x ) /vK contains elements of arbitrarilyhigh order and we are in the value-algebraic case.The elements of arbitrarily high degree over Kv can be chosen from K [ x ] v .Indeed, suppose there is m ∈ N such that [ Kv ( f v ) : Kv ] ≤ m for every polynomial f of nonnegative value. Take any r = hg , where g, h ∈ K [ x ] and vr = 0. By theassumption on vK ( x ) /vK we have that nvh = vd for some natural number n ≤ N and d ∈ K . Then r = d − h n d − h n − g and vd − h n − g = vd − h n = 0, since vh = vg . Therefore we may assume that vh = vg = 0. Hence,[ Kv ( rv ) : Kv ] ≤ [ Kv ( rv, gv ) : Kv ] = [ Kv ( hv, gv ) : Kv ] ≤ m for every r ∈ K ( x ) with vr = 0, a contradiction to our assumption.As in the value-algebraic case, we set b = 1. Suppose that b , . . . , b k are alreadychosen with deg b i − < deg b i for 1 < i ≤ k . By Corollary 2.8, there are atmost N N k + 1 many Kv -linearly independent elements in K [ x ] NN k v , and as allof them are algebraic over Kv , it follows that the extension Kv ( K [ x ] NN k v ) | Kv isfinite. By assumption, there is b k +1 ∈ K [ x ] such that vb k +1 = 0 and the degree of b k +1 v over Kv is at least ( k + 1)[ Kv ( K [ x ] NN k v ) : Kv ], which forces the elements1 , b k +1 v, ( b k +1 v ) , . . . , ( b k +1 v ) k to be Kv ( K [ x ] NN k v )-linearly independent. Since b k +1 / ∈ K [ x ] NN k , we have that N k +1 = deg b k +1 > N N k ≥ N k .For the value-algebraic as well as for the residue-algebraic case we set ϕ ( k, l ) := N k + N k N l . Since in both cases ( N k ) k ∈ N is a strictly increasing sequence of natural numbers, ϕ has the required properties. As in the first part of the proof of the value-algebraicand the residue-algebraic case, one can show that the elements c k ∈ K can bechosen in such a way that condition (A1) holds for a k := c k b k and α k := va k .Since ( N k ) k ∈ N is strictly increasing, condition (A2) is trivially satisfied. Moreover, N k ≥ k for every k ∈ N . Hence for any d , . . . , d k ∈ S k and u ∈ S l ,deg( d + d u + · · · + d k u k ) ≤ N k + kN l ≤ ϕ ( k, l ) ≤ N ϕ ( k,l ) . Thus, d + d u + · · · + d k u k ∈ S ϕ ( k,l ) . This shows that (A3) holds.To verify (A4), we take any k, m ∈ N with m ≤ k , and d , . . . , d m ∈ S k . Wewish to estimate the value of the element d + d a k +1 + · · · d m a mk +1 . We discussfirst the value-algebraic case. Note that the values v ( d i a ik +1 ), 0 ≤ i ≤ m , lie indistinct cosets modulo vK . Indeed, vd i a ik +1 = vd i c ik +1 + ivb k +1 , where d i c ik +1 ∈ S k .Therefore, if vd i a ik +1 + vK = vd j a jk +1 + vK for some 0 ≤ i ≤ j ≤ m , then also ivb k +1 + h vS k i = jvb k +1 + h vS k i , which by our choice of b k +1 yields that i = j . Hence, from Lemma 2.7 it followsthat v ( d + d a k +1 + · · · d m a mk +1 ) = min i vd i a ik +1 ≤ vd m + mva k +1 . We obtain the same assertion also in the residue-algebraic case. If d i = 0 for all i , then it is trivially satisfied. If not, take i so that vd i c i k +1 = min i vd i c ik +1 = min i vd i a ik +1 . We have that vd Ni = vc for some c ∈ K . Setting d := c − c − i k +1 d N − i , we obtain that v ( d + d a k +1 + · · · d m a mk +1 ) = − vd + vξ with ξ := dd + dd c k +1 b k +1 + · · · + dd m c mk +1 b mk +1 . Note that dd i ∈ K [ x ] NN k for0 ≤ i ≤ m , and that vdd i c ik +1 ≥ vdd i c i k +1 = vc − d Ni = 0 . LGEBRAIC INDEPENDENCE 19
In particular, vξ ≥
0, and ξv = ( dd ) v + ( dd c k +1 ) vb k +1 v + · · · + ( dd m c mk +1 ) v ( b k +1 v ) m is a linear combination of 1 , b k +1 v, ( b k +1 v ) , . . . , ( b k +1 v ) m with coefficients from Kv ( K [ x ] N · N k v ). Since at least one of them, the element dd i c i k +1 v , is nonzero,also the linear combination is nontrivial by our choice of b k +1 . Hence vξ = 0 and v ( d + d a k +1 + · · · d m a mk +1 ) = − vd = vd i c i k +1 ≤ vd m + mva k +1 . Therefore, condition (A4) is satisfied in both cases.It suffices now to verify the assumptions of Corollary 3.2. Take any element hg of K ∞ = K ( x ), where g, h ∈ S ∞ = K [ x ]. In both the value-algebraic and the residue-algebraic case we assumed that vK ( x ) /vK is a torsion group. Therefore, as in theresidue-algebraic case above one can multiply h and g by a suitable polynomial toobtain that vg = 0 ∈ Γ. Hence the assumptions of the corollary are satisfied.Since the transcendence degree of the extension L | K ( x ) is finite, we can nowdeduce the assertion about about the transcendence degree as in the previous cases.In the value-algebraic case, we still have to deal with the case where there is asubgroup Γ ⊆ vL containing vK such that Γ /vK is an infinite torsion group and theorder of each of its elements is prime to the characteristic exponent of Kv . We mayassume that Lv | Kv is algebraic since otherwise, the assertion of our theorem followsfrom the valuation-transcendental case. Since every maximal immediate extensionof ( L, v ) contains a henselization of (
L, v ), we may assume that both (
L, v ) and(
K, v ) are henselian. We take L ′ to be the relative separable-algebraic closure of K in L . Then by Lemma 2.4, vL/vL ′ is a p -group, which yields that Γ ⊆ vL ′ . Inview of the fundamental inequality, we find that L ′ | K must be an infinite extension.Now the assertion of our theorem follows from the separable-algebraic case.Finally, we have to deal with our additional assertion about the completion.Since the transcendence degree of L | K is finite, we know that vL/vK has finiterational rank. Therefore, vK is cofinal in vL or there exists some α ∈ vL such thatthe sequence ( iα ) i ∈ N is cofinal in vL . In the latter case (which always holds if vL contains an element γ such that γ > vK ), we are in the value-transcendental caseand we choose the element t such that vt = α . In the former case, provided thatthe cofinality of vL is countable, we choose the elements c i such that the sequence( vc i b i ) i ∈ N is cofinal in vL . In all of these cases, the sequence ( va i ) i ∈ N will be cofinalin vL and the elements y i will lie in the completion of ( L, v ). (cid:3) Extensions of maximal fields
We start with the
Proof of Theorem 1.2 :Take a maximal field (
K, v ) which satisfies (1), and denote by p the characteristicexponent of Kv . Further, take an infinite algebraic extension ( L | K, v ). Denote therelative separable-algebraic closure of K in L by L ′ . Assume that L ′ | K is infinite.Since K is henselian, the separable-algebraic case of Theorem 1.1 shows that anymaximal immediate extension of ( L, v ) has infinite transcendence degree over L .Assume now that L ′ | K is a finite extension. Then the field ( L ′ , v ) is maximal,( vL ′ : pvL ′ )[ L ′ v : L ′ v p ] = ( vK : pvK )[ Kv : Kv p ] < ∞ , and L | L ′ is an infinite purely inseparable extension. Therefore at least one of the extensions vL | vL ′ or Lv | L ′ v is infinite. Indeed, suppose that ( vL : vL ′ ) and [ Lv : L ′ v ] were finite. Takeany finite subextension E | L ′ of L | L ′ such that [ E : L ′ ] > ( vL : vL ′ )[ Lv : L ′ v ]. Then[ E : L ′ ] > ( vL : vL ′ )[ Lv : L ′ v ] ≥ ( vE : vL ′ )[ Ev : L ′ v ] , which contradicts the fact that L ′ as a maximal field is defectless by Theorem 2.1.If vL/vL ′ contains elements of arbitrarily high order or Lv contains elements ofarbitrarily high degree over L ′ v , then from the value-algebraic or residue-algebraiccase of Theorem 1.1 we deduce that any maximal immediate extension of L isof infinite transcendence degree over L . Otherwise, vL/vL ′ is a p -group of finiteexponent, Lv | L ′ v is a purely inseparable extension with ( Lv ) p n ⊆ L ′ v for somenatural number n , and vL/vL ′ or Lv | L ′ v is infinite. But this is not possible if[ L ′ v : L ′ v p ]( vL ′ : pvL ′ ) < ∞ . (cid:3) For the proof of Theorem 1.4 we will need the following result:
Lemma 4.1. If ( K, v ) is a maximal field of characteristic p > , then also K /p with the unique extension of the valuation v is a maximal field.Proof. If ( a ν ) is a pseudo Cauchy sequence in L , then ( a pν ) is a pseudo Cauchysequence in K . Since ( K, v ) is maximal, it has a pseudo limit b ∈ K . But then, a = b /p ∈ L is a pseudo limit of ( a ν ). (cid:3) After this preparation, we can give the
Proof of Theorem 1.4 :Part a) follows immediately from Lemma 4.1.To prove assertions b) and c) we consider the following subsets of K . We take A to be a set of elements of K such that the cosets p va + vK , a ∈ A , form a basisof the Z /p Z -vector space p vK/vK . Similarly, we take B to be a set of elements ofthe valuation ring of ( K, v ) such that the residues ( bv ) /p , b ∈ B , form a basis of( Kv ) /p | Kv . Then1 p vK = vK + X a ∈ A p va Z and ( Kv ) /p = Kv (( bv ) /p | b ∈ B ) . In order to prove assertion b) of our theorem, we set L ∞ := K ( a /p , b /p | a ∈ A, b ∈ B ) ⊆ K /p and obtain that vL ∞ = p vK and L ∞ v = ( Kv ) /p . So the extension ( K /p | L ∞ , v )is immediate. Lemma 4.1 shows that ( K /p , v ) is a maximal immediate extensionof ( L ∞ , v ). Our goal is now to show that under the assumptions of the theorem,this extension is of degree at least κ . Once this is proved, we can take X ⊆ K /p tobe a minimal set of generators of the extension K /p | L ∞ . Then the elements of X are p -independent over L ∞ . Take any natural number n . As X is infinite, we canchoose x , . . . , x n ∈ X and set L n := L ∞ ( X \ { x , . . . , x n } ). Then K /p | L n is animmediate extension of degree p n . Similarly, for λ any infinite cardinal ≤ κ , take Y ⊆ X of cardinality λ and set L λ := L ∞ ( X \ Y ). Then K /p | L λ is an immediatealgebraic extension of degree λ . LGEBRAIC INDEPENDENCE 21
We assume first that κ = ( vK : pvK ), so the set A is infinite. Then we take apartition of A into κ many countably infinite sets A τ , τ < κ . We choose enumera-tions A τ = { a τ,i | i ∈ N } . For every µ < κ we set A µ := S τ<µ A τ and K µ := K ( a /p | a ∈ A µ ) . Note that A = ∅ and K = K . We claim that(22) vK µ = vK + X a ∈A µ p va Z and K µ v = Kv .
The inclusions “ ⊇ ” are clear. For the converses, we observe that value groupand residue field of K µ are the unions of the value groups and residue fields ofall finite subextensions of K µ | K . Such subextensions can be written in the form F = K ( a , . . . , a k ) with distinct a , . . . , a k ∈ A µ , and we have that p k ≥ [ F : K ] ≥ ( vF : vK )[ F v : Kv ] ≥ p k · , so equality holds everywhere. Consequently, vF = vK + P ki =1 va i Z and F v = Kv .This proves our claim.For every τ < κ we choose a sequence ( c τ,i ) i ∈ N of elements in K such that thesequence of values(23) ( vc τ,i a /pτ,i ) i ∈ N is strictly increasing. If the cofinality of vK is countable, then the elements c τ,i canbe chosen in such a way that the sequence (23) is cofinal in p vK . For every n ∈ N ,we set(24) ξ τ,n := n X i =1 c τ,i a /pτ,i ∈ K τ +1 . Then ( ξ τ,n ) n ∈ N is a pseudo Cauchy sequence, hence it admits a pseudo limit ξ τ in the maximal field ( K /p , v ). In order to show that the degree of K /p | L ∞ isat least κ , we prove by induction that for every µ < κ and each K ′ such that K µ +1 ⊆ K ′ ⊆ L ∞ , the pseudo Cauchy sequence ( ξ µ,n ) n ∈ N admits no pseudo limitin K ′ ( ξ τ | τ < µ ) and the extension(25) ( K ′ ( ξ τ | τ ≤ µ ) | K ′ , v )is immediate.Take µ < κ and assume that our assertions have already been shown for all µ ′ < µ . If µ = µ ′ + 1 is a successor ordinal, then from (25) we readily get that theextension(26) ( K ′ ( ξ τ | τ < µ ) | K ′ , v )is immediate for every K ′ such that K µ ⊆ K ′ ⊆ L ∞ . If µ is a limit ordinal, then(26) follows from the induction hypothesis since K µ ′ ⊆ K µ ⊆ K ′ for each µ ′ < µ andsince the union over the increasing chain of immediate extensions K ′ ( ξ τ | τ ≤ µ ′ ), µ ′ < µ , of ( K ′ , v ) is again an immediate extension of ( K ′ , v ). In order to prove the induction step, suppose towards a contradiction that( ξ τ,n ) n ∈ N admits a pseudo limit η µ in K ′ ( ξ τ | τ < µ ) for some K ′ such that K µ +1 ⊆ K ′ ⊆ L ∞ . Then η µ lies already in a finite extension(27) E := K µ ( ξ τ | τ < µ )( a /p , . . . , a /pk , b /p , . . . , b /pℓ )of K µ ( ξ τ | τ < µ ) in L ∞ ( ξ τ | τ < µ ), with distinct elements a , . . . , a k ∈ A \ A µ and b , . . . , b ℓ ∈ B . We claim that(28) vE = vK µ + k X i =1 p va i Z and Ev = K µ v (( b v ) /p , . . . , ( b ℓ v ) /p ) . As the extension (26) is immediate for K µ in place of K ′ , the inclusions “ ⊇ ” areclear. Conversely, from these inclusions together with the equations in (22) and ourassumption on the a i , it follows that ( vE : vK ) ≥ p k as well as [ Ev : Kv ] ≥ p ℓ .Therefore, we have that p k · p ℓ ≥ [ E : K ] ≥ ( vE : vK )[ Ev : Kv ] ≥ p k · p ℓ , so equalityholds everywhere. Consequently, ( vE : vK ) = p k and [ Ev : Kv ] = p ℓ , which provesthat the inclusions are equalities.Now we take n to be the minimum of all i ∈ N such that a µ,i is not among the a , . . . , a k . We set ξ E := 0 if n = 1, and ξ E := ξ µ,n − otherwise. Then η µ − ξ E ∈ E .In contrast, the fact that η µ is a pseudo limit, together with the first equation of(28), yields that v ( η µ − ξ E ) = v ( ξ µ,n − ξ E ) = vc µ,n + 1 p va µ,n / ∈ vK µ + k X i =1 p va i Z = vE . This contradiction proves that ( ξ µ,n ) n ∈ N admits no pseudo limit in K ′ ( ξ τ | τ < µ ).Thus in particular, ξ µ / ∈ K ′ ( ξ τ | τ < µ ). Since K µ +1 ⊆ K ′ , ( ξ µ,n ) n ∈ N is a pseudoCauchy sequence in ( K ′ ( ξ τ | τ < µ ) , v ). As [ K ′ ( ξ τ | τ < µ )( ξ µ ) : K ′ ( ξ τ | τ < µ )] = p is a prime, Lemma 2.12 shows that the extension ( K ′ ( ξ τ | τ ≤ µ ) | K ′ ( ξ τ | τ < µ ) , v )is immediate. As also the extension (26) is immediate, we find that the extension( K ′ ( ξ τ | τ ≤ µ ) | K ′ , v ) is immediate. This completes our induction step. Becauseevery extension L ∞ ( ξ τ | τ ≤ µ ) | L ∞ ( ξ τ | τ < µ ) is nontrivial, it follows that thedegree of K /p | L ∞ is at least κ .A simple modification of the above arguments allows us to show the assertionof part b) of the theorem in the case of κ = [ Kv : ( Kv ) p ], in which the set B isinfinite. Let us describe these modifications.We take a partition of B into κ many countably infinite sets B τ , τ < κ , andchoose enumerations B τ = { b τ,i | i ∈ N } . For every µ < κ we set B µ := S τ<µ B τ and K µ := K ( b /p | b ∈ B µ ) . Similarly as before, it is shown that(29) vK µ = vK and K µ v = Kv (( bv ) /p ) | b ∈ B µ ) . We choose a sequence ( c i ) i ∈ N of elements in K with strictly increasing values.Again, if the cofinality of vK is countable, then the elements c i can be chosen in LGEBRAIC INDEPENDENCE 23 such a way that the sequence of their values is cofinal in vK . For every τ < κ and n ∈ N , we set(30) ξ τ,n := n X i =1 c i b /pτ,i ∈ K τ +1 . Now the only further part of the proof that needs to be modified is the one thatshows that η µ ∈ E , where η µ is a pseudo limit of ( ξ µ,n ) n ∈ N , leads to a contradiction.In the present case, we take n to be the minimum of all i ∈ N such that b µ,i is notamong the b , . . . , b ℓ . As before, we set ξ E := 0 if n = 1, and ξ E := ξ µ,n − otherwise. Then η µ − ξ E ∈ E . In contrast, the fact that η µ is a pseudo limit,together with the second equation of (28), yields that c − n ( η µ − ξ E ) v = c − n ( ξ µ,n − ξ E ) v = ( b /pµ,n ) v = ( b µ,n v ) /p / ∈ K µ v (( b v ) /p , . . . , ( b ℓ v ) /p ) = Ev , a contradiction. This completes our modification and thereby the proof that theextension ( K /p | L ∞ , v ) is of degree at least κ .We now turn to part c) of the theorem. Again, we consider separately the casesof κ = ( vK : pvK ) and of κ = [ Kv : ( Kv ) p ].We assume first that κ = ( vK : pvK ) and take a partition of A as in the proofof part b). Further, we set s (1) = 0 and s ( m ) = 1 + 2 + · · · + ( m −
1) for m > τ < µ and every m ∈ N , we set z τ,m := m X i =1 d τ,s ( m )+ i a p − i τ,s ( m )+ i ∈ K /p ∞ , where d τ,j are elements from K such that for every m ∈ N ,1) the sequence ( vd τ,s ( m )+ i a p − i τ,s ( m )+ i ) ≤ i ≤ m is strictly increasing,2) vd τ,s ( m )+ m a p − m τ,s ( m )+ m < vd τ,s ( m +1)+1 a p − τ,s ( m +1)+1 .If the cofinality of vK is countable, then the elements d τ,i can be chosen in such away that the sequence that results from the above is cofinal in vK .We note that z p m τ,m ∈ K with(31) [ K ( z τ,m ) : K ] = p m and 1 p va τ,s ( m )+1 , . . . , p va τ,s ( m )+ m ∈ vK ( z τ,m ) . We set L µ := K ( z τ,m | τ < µ, m ∈ N ) for µ ≤ κ , and L := L κ . Further, we fix a maximal immediate extension (
M, v ) of (
L, v ). We claim that(32) vL µ = vK + X a ∈A µ p va and L µ v = Kv .
In particular, this shows that(33) vL = 1 p vK and Lv = Kv .
To prove our claim, we observe that the first inclusion “ ⊇ ” in (32) follows from(31). We choose any µ < κ , k ∈ N , τ , . . . , τ k < µ and m , . . . , m k ∈ N such thatthe pairs ( τ i , m i ), 1 ≤ i ≤ k , are distinct. Then we compute, using (31): p m · . . . · p m k ≥ [ K ( z τ ,m , . . . , z τ k ,m k ) : K ] ≥ ( vK ( z τ ,m , . . . , z τ k ,m k ) : vK )[ K ( z τ ,m , . . . , z τ k ,m k ) v : Kv ] ≥ ( vK ( z τ ,m , . . . , z τ k ,m k ) : vK ) ≥ ( vK + k X j =1 m j X i =1 p va τ j ,s ( m j )+ i Z : vK ) ≥ p m · . . . · p m k , showing that equality holds everywhere. Therefore, vK ( z τ ,m , . . . , z τ k ,m k ) = vK + k X j =1 m j X i =1 p va τ j ,s ( m j )+ i Z ⊆ vK + X a ∈A µ p va and K ( z τ ,m , . . . , z τ k ,m k ) v = Kv .
Since the value group and residue field of L µ are the unions of the value groups andresidue fields of all subfields of the above form, this proves our claim.For every τ < κ and n ∈ N , we set ζ τ,n := n X m =1 z τ,m ∈ L. Then ( ζ τ,n ) n ∈ N is a pseudo Cauchy sequence in ( L, v ), hence it admits a pseudo limit ζ τ in the maximal field ( M, v ). In order to show that the transcendence degree of M | L is at least κ , we prove by induction that for every µ < κ and every field L ′ suchthat L µ +1 ⊆ L ′ ⊆ L , the pseudo Cauchy sequence ( ζ µ,n ) n ∈ N is of transcendentaltype over L ′ ( ζ τ | τ < µ ), so that the extension ( L ′ ( ζ τ | τ ≤ µ ) | L ′ ( ζ τ | τ < µ ) , v ) isimmediate and transcendental and then also the extension(34) ( L ′ ( ζ τ | τ ≤ µ ) | L ′ , v )is immediate.Take µ < κ and assume that our assertions have already been shown for all µ ′ < µ . If µ = µ ′ + 1 is a successor ordinal, then from (34) we readily get that theextension(35) ( L ′ ( ζ τ | τ < µ ) | L ′ , v )is immediate for every L ′ such that L µ ⊆ L ′ ⊆ L . If µ is a limit ordinal, then (35)follows from the induction hypothesis since L µ ′ ⊆ L µ ⊆ L ′ for each µ ′ < µ andsince the union over an increasing chain of immediate extensions of ( L ′ , v ) is againan immediate extension of ( L ′ , v ).In order to prove the induction step, take any L ′ such that L µ +1 ⊆ L ′ ⊆ L .Suppose towards a contradiction that the pseudo Cauchy sequence ( ζ µ,n ) n ∈ N in L µ +1 is of algebraic type over ( L ′ ( ζ τ | τ < µ ) , v ) (which includes the case whereit has a pseudo limit in L ′ ( ζ τ | τ < µ )). Then by Theorem 2.10 there exists animmediate algebraic extension ( L ′ ( ζ τ | τ < µ )( d ) | L ′ ( ζ τ | τ < µ ) , v ) with d a pseudolimit of the sequence. The element d is also algebraic over L µ ( ζ τ | τ < µ ). Onthe other hand, we will now show that from the fact that d is a pseudo limit of LGEBRAIC INDEPENDENCE 25 ( ζ µ,n ) n ∈ N it follows that the value group vL µ ( ζ τ | τ < µ )( d ) is an infinite extensionof vL µ ( ζ τ | τ < µ ). Take n ∈ N and define η µ,n := ζ p n − µ,n − d p n − µ,s ( n )+ n a /pµ,s ( n )+ n = ζ p n − µ,n − + n − X i =1 d p n − µ,s ( n )+ i a p n − − i µ,s ( n )+ i ∈ K .
Since d is a pseudo limit of the pseudo Cauchy sequence ( ζ µ,n ) n ∈ N , we deduce that(36) v ( d − ζ µ,n ) = vz µ,n +1 = vd µ,s ( n +1)+1 a /pµ,s ( n +1)+1 > vd µ,s ( n )+ n a p − n µ,s ( n )+ n . Therefore, v (cid:16) d p n − − η µ,n (cid:17) = v (cid:16) d p n − − ζ p n − µ,n + d p n − µ,s ( n )+ n a /pµ,s ( n )+ n (cid:17) = p n − v (cid:16) d − ζ µ,n + d µ,s ( n )+ n a p − n µ,s ( n )+ n (cid:17) = p n − min n v ( d − ζ µ,n ) , v (cid:16) d µ,s ( n )+ n a p − n µ,s ( n )+ n (cid:17)o = p n − v (cid:16) d µ,s ( n )+ n a p − n µ,s ( n )+ n (cid:17) = p n − vd µ,s ( n )+ n + 1 p va µ,s ( n )+ n , which shows that 1 p va µ,s ( n )+ n ∈ vL µ ( ζ µ | µ < τ )( d )for all n ∈ N . In view of (32), these values are not in vL µ . Since the extension (35) isimmediate for L µ in place of L ′ , they are also not in vL µ ( ζ τ | τ < µ ). It follows thatthe index ( vL µ ( ζ τ | τ < µ )( d ) : vL µ ( ζ τ | τ < µ )) is infinite. This contradicts thefact that the extension L µ ( ζ τ | τ < µ )( d ) | L µ ( ζ τ | τ < µ ) is finite. This contradictionproves that the pseudo Cauchy sequence ( ζ µ,n ) n ∈ N is of transcendental type over L ′ ( ζ τ | τ < µ ). From Theorem 2.9 it follows that ( L ′ ( ζ τ | τ ≤ µ ) | L ′ ( ζ τ | τ < µ ) , v )is an immediate transcendental extension. Since the extension (35) is immediate,we obtain that also ( L ′ ( ζ τ | τ ≤ µ ) | L ′ , v ) is immediate.This completes our induction step. By induction on µ we have therefore shownthat ( L ( ζ τ | τ < µ ) , v ) is an immediate extension of ( L, v ) for each µ < κ , whichyields that also the union ( L ( ζ τ | τ < κ ) , v ) of these fields is an immediate extensionof ( L, v ). As every extension L ( ζ τ | τ ≤ µ ) | L ( ζ τ | τ < µ ) is transcendental, thetranscendence degree of L ( ζ τ | τ < κ ) over L is at least κ .A simple modification of the above arguments allows us to show the assertion ofpart c) of the theorem in the case of κ = [ Kv : ( Kv ) p ]. We take the partition of B as in the proof of part b). We now list the modifications.Since the vb = 0 for all b ∈ B , the only requirement for the elements d τ,i thatwe need is that vd τ,i < vd τ,j for i < j . If the cofinality of vK is countable, thenthe elements d τ,i can be chosen in such a way that the sequence of their values iscofinal in vK . We set z τ,m := m X i =1 d τ,s ( m )+ i b p − i τ,s ( m )+ i ∈ K /p ∞ , Equation (31) is replaced by(37) [ K ( z τ,m ) : K ] = p m and ( b τ,s ( m )+1 v ) /p , . . . , ( b τ,s ( m )+ m v ) /p ∈ K ( z τ,m ) v . One proves in a similar way as before that(38) vL µ = vK and L µ v = Kv (( bv ) /p | b ∈ B µ ) . In particular, this shows that(39) vL = vK and Lv = ( Kv ) /p . Now the only further part of the proof that needs to be modified is the one thatshows that the extension L µ ( ζ τ | τ < µ )( d ) | L µ ( ζ τ | τ < µ ) cannot be finite. Wedefine η µ,n as before, with “ b ” in place of “ a ”. Also (36) holds with “ b ” in place of“ a ”, whence v d − p n − µ,s ( n )+ n (cid:16) d p n − − ζ p n − µ,n (cid:17) = p n − vd − µ,s ( n )+ n ( d − ζ µ,n ) > . This leads to d − p n − µ,s ( n )+ n (cid:16) d p n − − η µ,n (cid:17) v = d − p n − µ,s ( n )+ n (cid:16) d p n − − ζ p n − µ,n + d p n − µ,s ( n )+ n b /pµ,s ( n )+ n (cid:17) v = (cid:16) d − p n − µ,s ( n )+ n ( d p n − − ζ p n − µ,n ) + b /pµ,s ( n )+ n (cid:17) v = ( b /pµ,s ( n )+ n ) v = ( b µ,s ( n )+ n v ) /p , which shows that ( b µ,s ( n )+ n v ) /p ∈ L µ ( ζ τ | τ < µ )( d ) v for all n ∈ N . In view of (38), these residues are not in L µ v . As before, thisis shown to contradict d being algebraic over L µ ( ζ τ | τ < µ ). This completes ourmodification and thereby the proof that ( M | L, v ) is of transcendence degree at least κ . (cid:3) We now come to the proof of
Proof of Theorem 1.5 :Note that a field (
K, v ) which satisfies the assumptions of Theorem 1.5 also satisfiesthe assumptions of Theorem 1.4. We choose the sets
A, B ⊆ K /p and define L := L ∞ as in the proof of part b) of Theorem 1.4. Then, as we have already seen,( K /p , v ) is a maximal immediate extension of ( L, v ).To show the existence of an immediate extension of L of infinite transcendencedegree over L , we consider separately the cases i) and ii) of the theorem. We assumefirst that the conditions of case i) hold. Then the set A can be chosen so as to containan infinite countable subset A ′ such that the set of values S = { va | a ∈ A ′ } isbounded. It must contain a bounded infinite strictly increasing or a bounded infinitestrictly decreasing sequence. If it does not contain the former, we replace A ′ by { a − | a ∈ A ′ } , thereby passing from S to − S . Note that in our proof we will notneed that A ′ ⊆ A ; we will only use that A ′ ⊆ L . Now we can choose a sequence( a j ) j ∈ N of elements in A ′ such that the sequence ( va j ) j ∈ N is strictly increasing andbounded by some γ ∈ vK . We partition the sequence ( a j ) j ∈ N into countably manysubsequences ( a N,i ) i ∈ N ( N ∈ N ) . As in the proof of Theorem 1.4, we define K N := K ( a n,i | n < N , i ∈ N ) ⊆ K /p .For every N ∈ N we consider the pseudo Cauchy sequence ( ξ N,m ) m ∈ N defined by ξ N,m := m X i =1 a /pN,i ∈ K N +1 . LGEBRAIC INDEPENDENCE 27 and the pseudo limits ξ N of the sequences in the maximal immediate extension( K /p , v ) of ( L, v ). We show that for every N the pseudo limit ξ N does not lie inthe completion L c of ( L, v ). Fix N ∈ N and take any d ∈ L . Then d lies already insome finite extension E := K ( a /p , . . . , a /pk , b /p , . . . , b /pl )of K in L . Choosing ξ E as in the proof of Theorem 1.4, we obtain that ξ E − d ∈ E .But from equalities (28) with µ = 0 it follows that v ( ξ N − ξ E ) = p va N,n / ∈ vE .Thus, v ( ξ N − d ) = min { v ( ξ N − ξ E ) , v ( ξ E − d ) } ≤ p va N,n < p γ . Hence the values v ( ξ N − d ), d ∈ L , are bounded by p γ and consequently, ξ N / ∈ L c .Again from the proof of Theorem 1.4 it follows that for every field K ′ such that K ⊆ K ′ ⊆ L the extension ( K ′ ( ξ ) | K ′ , v ) is immediate and purely inseparableof degree p . Since ξ / ∈ L c , from Proposition 2.13 we deduce that for an element d ∈ K × satisfying inequality (8) with η = ξ , a root ϑ of the polynomial f := X p − X − (cid:18) ξ d (cid:19) p generates an immediate Galois extension ( L ( ϑ ) | L, v ) of degree p with a uniqueextension of the valuation v from L to L ( ϑ ). Take any field K ′ such that K ⊆ K ′ ⊆ L . Then ξ / ∈ K ′ c and the element d satisfies inequality (8) with everyelement c ∈ K ′ . Therefore also ( K ′ ( ϑ ) | K ′ , v ) is an immediate extension of degree p with a unique extension of v from K ′ to K ′ ( ϑ ).Take any m >
1. Suppose that we have shown that for every l < m there is d l ∈ K × such that a root ϑ l of the polynomial f l := X p − X − (cid:18) ξ l d l (cid:19) p generates, for any field K ′ with K l +1 ⊆ K ′ ⊆ L , an immediate Galois exten-sion ( K ′ ( ϑ , . . . , ϑ l ) | K ′ ( ϑ , . . . , ϑ l − ) , v ) of degree p with a unique extension ofthe valuation v from K ′ ( ϑ , . . . , ϑ l − ) to K ′ ( ϑ , . . . , ϑ l ). Then in particular, theextension ( K m +1 ( ϑ , . . . , ϑ m − ) | K m +1 , v ) is immediate. Take any field K ′ suchthat K m +1 ⊆ K ′ ⊆ L . Replacing in the argumentation of the proof of part b)of Theorem 1.4 the field K ′ ( ξ l | l < m ) by K ′ ( ϑ l | l < m ), we deduce that( K ′ ( ϑ , . . . , ϑ m − )( ξ m ) | K ′ ( ϑ , . . . , ϑ m − ) , v ) is an immediate purely inseparable ex-tension of degree p . Since ξ m / ∈ L c and L ( ϑ , . . . , ϑ m − ) c = L c ( ϑ , . . . , ϑ m − ) isa separable extension of L , linearly disjoint from the purely inseparable extension L c ( ξ m ) | L c , we obtain that[ L c ( ϑ , . . . , ϑ m − )( ξ m ) : L c ( ϑ , . . . , ϑ m − )] = p. Therefore, ξ m does not lie in L ( ϑ , . . . , ϑ m − ) c . Thus, from Proposition 2.13 itfollows that for an element d m ∈ K × satisfying inequality (8) with η = ξ m , a root ϑ m of the polynomial f m := X p − X − ( ξ m /d m ) p generates an immediate Galoisextension ( L ( ϑ , . . . , ϑ m ) | L ( ϑ , . . . , ϑ m − ) , v ) of degree p with a unique extension ofthe valuation v from L ( ϑ , . . . , ϑ m − ) to L ( ϑ , . . . , ϑ m ). As in the case of m = 1 wededuce that also the extension ( K ′ ( ϑ , . . . , ϑ m ) | K ′ ( ϑ , . . . , ϑ m − ) , v ) is immediateand the valuation v of K ′ ( ϑ , . . . , ϑ m − ) extends uniquely to K ′ ( ϑ , . . . , ϑ m ). By induction, we obtain an infinite immediate separable-algebraic extension F := L ( ϑ m | m ∈ N ) of L with the unique extension of the valuation v of L to F . Thus,the extension F | L is linearly disjoint from L h | L . From the separable algebraic caseof Theorem 1.1 it follows that each maximal immediate extension ( M, v ) of (
F, v )has infinite transcendence degree over F . Since ( F | L, v ) is immediate, M is also amaximal immediate extension of L .Similar arguments allow us to prove the assertion in the case of an infinite residuefield extension Kv | ( Kv ) p when the value group vK is not discrete. Let us describethe modifications.Take an infinite countable subset B ′ of B and an infinite partition of B intoinfinite sets B N = { b N,i | i ∈ N } ( N ∈ N ) . Since vK is not discrete, we can choose elements c i ∈ K such that the sequence( vc i ) i ∈ N of their values is strictly increasing and bounded by some element γ ∈ vK .For every N we consider the pseudo Cauchy sequence ( ξ N,m ) m ∈ N defined by (30).The only further part of the proof that needs to be modified is the one thatshows that ξ N / ∈ L c . More precisely, we need to show that for any element d ∈ E we have that v ( ξ N − d ) < γ . Take ξ E as in the second case of the proof of part b) ofTheorem 1.4. From the equalities (28) with µ = 0 we deduce that c − n ( ξ N − ξ E ) v =( b N,n v ) /p / ∈ Ev . Suppose that v ( ξ N − d ) > v ( ξ n − ξ E ). Then v (cid:0) c − n ( ξ N − ξ E ) − c − n ( ξ E − d ) (cid:1) > vc − n ( ξ N − ξ E ) = 0 . It follows that c − n ( ξ N − ξ E ) v = c − n ( ξ E − d ) v ∈ Ev , a contradiction. Consequently, v ( ξ N − d ) ≤ v ( ξ N − ξ E ) = vc n < γ . This completes our modification and thereby the proof that (
L, v ) admits a maximalimmediate extension of infinite transcendence degree over L . (cid:3) Finally, we give the
Proof of Theorem 1.6 :Take an extension ( L | K, v ) as in the assumptions of the theorem. In view of thevalue-algebraic and residue-algebraic cases of Theorem 1.1, it suffices to show thatat least one of the extensions vL | vK or Lv | Kv is infinite.Take K ′ to be the relative algebraic closure of K in L h . By the assumptions onthe residue field and value group extensions of ( L | K, v ), it follows from Lemma 2.4that vK ′ = vL h = vL and K ′ v = L h v = Lv . Therefore, ( L h | K ′ , v ) is an immediatetranscendental extension.Suppose that the value group extension and the residue field extension of ( L | K, v )and hence of ( K ′ | K, v ) were finite. But since K is henselian and a defectless fieldby Theorem 2.1, the degree [ K ′ : K ] is equal to ( vK ′ : vK )[ K ′ v : Kv ] and hencewould be finite, so ( K ′ , v ) would again be a maximal field, which contradicts thefact that ( L h | K ′ , v ) is a nontrivial immediate extension. (cid:3) LGEBRAIC INDEPENDENCE 29
References [1] Bourbaki, N. :
Commutative algebra , Paris (1972)[2] Endler, O. :
Valuation theory , Springer, Berlin (1972)[3] Engler, A.J. – Prestel, A.:
Valued fields , Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2005[4] Gravett, K. A. H. :
Note on a result of Krull , Cambridge Philos. Soc. Proc. (1956), 379[5] Kaplansky, I. : Maximal fields with valuations I , Duke Math. Journ. (1942), 303–321[6] Kuhlmann, F.–V. : Valuation theoretic and model theoretic aspects of local uniformization ,in: Resolution of Singularities - A Research Textbook in Tribute to Oscar Zariski. HerwigHauser, Joseph Lipman, Frans Oort, Adolfo Quiros (eds.), Progress in Mathematics Vol. ,Birkh¨auser Verlag Basel (2000), 381–456[7] Kuhlmann, F.-V. :
Value groups, residue fields and bad places of rational function fields ,Trans. Amer. Math. Soc. (2004), 4559-4600[8] Kuhlmann, F.-V.:
A classification of Artin-Schreier defect extensions and a characterizationof defectless fields , Illinois J. Math. (2010), 397–448[9] Kuhlmann, F.–V. : Valuation Theory , book in preparation. Preliminary versions of severalchapters are available on the web site: http://math.usask.ca/˜fvk/Fvkbook.htm.[10] Kuhlmann, F.-V., Pank, M., Roquette, P.:
Immediate and purely wild extensions of valuedfields , Manuscripta math., (1986), 39–67[11] Krull, W. : Allgemeine Bewertungstheorie , J. reine angew. Math. (1931), 160–196[12] MacLane, S. – Schilling, O.F.G. :
Zero-dimensional branches of rank 1 on algebraic varieties ,Annals of Math. (1939), 507–520[13] Nagata, M.: Local rings , Wiley Interscience, New York (1962)[14] Ribenboim, P. :
Th´eorie des valuations , Les Presses de l’Universit´e de Montr´eal, Montr´eal,1st ed. (1964), 2nd ed. (1968)[15] Warner, S. :
Nonuniqueness of immediate maximal extensions of a valuation , Math. Scand. (1985), 191–202[16] Warner, S. : Topological fields , Mathematics studies , North Holland, Amsterdam (1989)[17] Zariski, O. – Samuel, P.:
Commutative Algebra , Vol. II, New York–Heidelberg–Berlin (1960), Vol. II, New York–Heidelberg–Berlin (1960)