Fixed points, combs and generalized power series

Abh. Math. Sem. Univ. Hamburg 63 (1993), 227-244

Fixed Points, Combs and Generalized Power Series

By S. PRIESS-CRAMPE and P. RIBENBOIM

Introduction

The present paper has several sources.

1) Let (F, <) be a totally ordered abelian group, let (K, v) be a valued field with value group F. Consider the ordered set (F U {0}, <~ where <o denotes the oppposite order and 0 <~ 7 for every 7 ~ F.

Let d : K • K ~ F U {0} be defined by

d(x,x)=O a n d i f x @ y , d(x,y)=v(x-y).

Then d is an ultrametric distance on K. KRULL showed that (K, v) is a maximal valued field (i.e. it has no proper

valued extension with the same residue field and value group) if and only if every pseudo-convergent family of elements of K has a pseudo-limit, that is K is pseudo-complete in the ultrametric d. Equivalently, K is spherical complete, that is the intersection of any chain of balls is non-empty.

In many instances (see [6], [5]) a maximal valued field is isomorphic to a field of generalized power series, having exponents in F and coefficients in the residue field R. Special groups of generalized power series had been previously considered by HAHN [4].

2) In [8] a result analogous to Banach's fixed point theorem was proved. It holds for contracting maps over spherically complete ultrametric spaces (with totally ordered value sets). Among the applications, it was given a simple proof that maximal valued fields are henselian.

3) In a series of papers ([10], [11], [12], [13]) a systematic study of gemneralized power series was undertaken; the exponents are assumed to be a partially ordered (but not necessarily totally ordered) set, most often an abelian group.

Just like in 1), the series studied in 3) lead to the consideration of generalized ultrametric distances, with values in a partially ordered set with a smallest element.

In w we extend the fixed point theorem of [8] to these generalized ultrametric spaces.

We show in w that if F is artinian and narrow, then the ultrametric space (X, d, F) is spherically complete. In particular, this is true if F is finite. For such ultrametric spaces, we may therefore apply the fixed point theorem.

228 S. Priel3-Crampe, P. Ribenboim

w is devoted to a study of conditions on the partially ordered set of values of the distance, which will be needed in w for groups of generalized power series. The highlight is the notion of an infinite comb. There is an extended study on conditions for the existence of infinite combs in products of ordered sets and in ordered sets whose order is obtained by perforation from a total order (these notions are thoroughly explained in the text).

In w we consider specifically groups A of generalized power series with exponents in an ordered set (S, <) and coefficients in an additive abelian group R. We introduce an order on the set F(S) of finite antichains of (S, <) (according to the procedure indicated in [14]) and define an ultrametric distance on the group A, with values in F(S). We are able to show that this ultrametric space is spherically complete if and only if (S, <) does not have an infinite comb.

The final section contains applications of the generalized fixed point theorem, similar to the ones in [8] and concerning the inversion of generalized power series and Hensel's lemma.

1 The Fixed Point Theorem

We will generalize the definition of an ultrametric space, which was given in [8]: Let (F,<) be an ordered set with 0 q~ F and 0 < 7 for every 7 ~ F; we do not assume that the order is total. Let X ~ 0 be a set. A mapping d : X • X ~ F U {0} will be called an ultrametric distance (and (X, d, F) an ultrametric space), if the following properties hold for all x, y, z E X, 7 E F:

(D1) d(x, y) = 0 if and only if x = y.

(D2) d(x,y) = d(y,x).

(D3) I f d(x,y) < 7, d(y,z) < 7, then d(x,z) < 7.

Often we shall only write X or (X, F) instead of (X, d, F). Let X be an ultrametric space, let ~ E F and a E X. The set B~(a) = {x E

X I d(x,a) < ~} will be called a ball with centre a and radius ~. As an immediate consequence of the properties of an ultrametric distance

we obtain:

1.1. Let ~, fl E F and a, b E X. If ~ < fl and a E Bfl(b), then B~(a) ~_ Bfl(b).

Hence every point of a ball is also its centre. From (1.1) we deduce:

1.2. If a, b E X, ~, fl E F and B~ ~ Bfl(b), then fl ;gct.

Indeed, if fl _< c~, then B~(a) ~ Bfl(b) = Bfl(a) ~_ B~(a), which is absurd.

The ultrametric space X will be called spherically complete, when every non-empty set ~3 of balls of X, which is totally ordered by inclusion, has a non-empty intersection. If (F, <) is totally ordered, the relation between the

Fixed Points, Combs and Generalized Power Series 229

concepts of spherical completeness, of pseudo-convergence and completeness was examined in [8].

We shall give examples of spherically complete spaces in the next section. A map (p: X ~ X is said to be contracting, when x, y E X, x ~ y, implies d(q)(x),q)(y)) < d(x,y).

As a generalization of the Banach Fixed Point Theorem (and Satz 3 of [8]) we will prove:

Theorem 1. I f X is spherically complete and q~: X --~ X is contracting, then q9 has exactly one fixed point.

Proof Assume, lrx = d(x, cp(x)) 4= 0 for every x E X. Let Bx = B~x(X). The set ~ = {Bx I x E X} is ordered by inclusion. Let cg be a maximal chain in ~. Since X is sperically complete, there exists an element z E N{Bx I Bx E cg}. Then Bz c_ Bx for every Bx E c~. Indeed, this is obvious, if z = x. If z ~ x, then d(cp(z),cp(x)) <_ d(z,x) <_ rr~ = d(x, cp(x)); rtz = d(cp(z),z) <_ rr~. Hence it results from (1.1) that Bz m_ Bx. Since cg is a maximal chain in N', then Bz is the smallest element of cg. But 7r~tz) = d (r r < d (z, r = rcz and therefore Be(z ) ~ Bz, contradicting the maximality of cg. Hence there exists an element x E X with r = x. If also r = y for x :p y, then d(x,y) = d(tp(x),q~(y)) < d(x,y), which is absurd. Thus there exists exactly one fixed point for ~p. []

Remark. Analysing the proof of this theorem, we see that to prove the existence of a fixed point for the mapping q~:X ~ X, it suffices to assume the following property. For any x, y E X, d(q)(x),~o(y)) < d(x,y) and for x :p ~0(x), d < d (x,

2 First Examples of Spherically Complete Ultrametric Spaces

In this section, we shall give a condition on the ordered set (F, <) which guarantees that the ultrametric space (X, d, F) is spherically complete.

To begin, we recall some definitions and results which will be needed in the sequel (see, for example [10]).

Let (S,<) be an ordered set. Two elements x, y E S will be called incomparable, if neither x < y nor y _< x, and we will denote this by x I I Y. If all elements of S are pairwise incomparable, S is said to be trivially ordered. A trivially ordered subset is also called an antichain. A subset T of S will always be endowed with the restricted order. (S, <) is called narrow, when every trivially ordered subset of S is finite. (S, <) is artinian, when S does not contain any infinite strictly decreasing sequence sl > s2 > s3 .... S is noetherian, when S does not contain any infinite strictly increasing sequence

S1 < $2 ' ~ $3 < . . . .

We recall the well-known facts:

230 S. PrieB-Crampe, P. Ribenboim

2.1. An ordered set (S, <) is finite if and only if it is artinian, noetherian and narrow.

2.2. An ordered set (S, <) is artinian and narrow when for every subset {sn ] n > 1} of S, there exist nl < n2 < ... such that s,, < sn2 < . . . . Equivalently, there exist indices m < n such that Sm< sn.

We obtain the following result for the ultrametric space X = (X, d, F):

Proposition 2. I f (F, _<) is artinian and narrow, then X is spherically complete.

Proof We show that every chain ~ @ 0 of balls of X has a smallest ball, thus X is spherically complete.

Assume that the chain ~ does not have a smallest ball. Hence there exists an infinite strictly decreasing sequence B~ ~ B2 ~ ... of balls in ~'. Let an E X, an E F be such that Bn = B~, (an) (ball of centre an, radius ~n) and consider the set {~1, a2,...}, which is artinian and narrow by hypothesis.

By (2.2) there exist indices m < n such that a,~ < an. However B,, ~ Bn implies that am ~g an, as it was shown in (1.2). This is a contradiction. []

Corollary 3. I f F is finite, then X is spherically complete.

3 Infinite Combs in Ordered Sets

In view of indicating more examples of spherically complete ultrametric spaces in w we need first to study certain conditions on an ordered set (S, <).

Let F(S) be the set of all finite non-empty antichains of (S, _<). We define the relation < on F(S) as follows: if X, Y c F(S), then X < Y if for every y E Y, there exists x E X such that x < y. This is an order relation extending the order on S; moreover, if Y _ X then X < Y. (These considerations are found in [14]).

The subset cs = {zn [ n >_ 1} U {an [ n >_ 0} of (S, <) is called an infinite comb (~-comb) when the following relations are satisfied:

Zn+l

Zn

an+l

at/

an-I

a, < an+l, a, < z,+l, an+l [] z,+l and Z,+l [[ z,,+l for all distinct n, m > 0. If in the sequel, we will consider an oo-comb, its elements z,, a, shall always have this special order.

a0

If X __c_ S, let V(X) =. {s ~ S ] there exists x E X such that x < s}. The following order-theoretic result will be of importance in the sequel:


Proposition 4. Let (S, <) be an ordered set. Then the following two conditions are equivalent:

(i) S does not contain an ~-comb.

(ii) I f Z = {zn I n > 1} is an infinite trivially ordered subset o f S, i f X1,X2 . . . . E F(S) with X1 < X2 < X3 < . . . . then there exists an integer n >_ 1, such that V(Xn) A Z = V(Xn+I) A Z . . . . .

Proo f 1) The implication (ii) ~ (i) is very easy. Therefore we will prove it first. Let cg = Z U {a, ] n _> 0}, where Z = {z, ] n > 1}, be an ~ - c o m b of S. Take Xn = {an}. Hence Xn < Xn+t < . . . and z, q~ V ( X n ) n Z ~_ {Zn+l,Zn+2 . . . . }, which implies that V (Xn) n Z ~ V(Xn+I) A Z , contradict ing (ii).

2) For the p roof o f the implication (i) =, (ii) we assume, that (ii) is not valid. Therefore, there exists an infinite trivially ordered subset Z ~ S and X1, X2 . . . . 6 F(S) with X1 < X2 < ... and natural numbers no < nl < n2 < ... such that V ( X , o ) n Z ~ V ( X n ~ ) n Z ~ . . . . With a new labelling we may assume V(Xo) n Z ~ V(X1) n Z ~ . . . . Hence the sets En = (V (Xn-O n Z ) \ V ( X n ) are non-void for all n > 1.

For every n > 1, we choose an element zn E En. I f n < m and if an E Xn, we say that zm is connected to an, when there

exist elements a,,-i E Xm-1, am-2 E Xm-2 . . . . . an+l E Xn+l with zm > am-1 >

am-2 ~_~.,. ~ an+l ~ an. Every zn, n > 1, is connected to some element o f X0. Since X0 is finite,

there exists an a0 E X0 and a strictly increasing subsequence 1 < nl < n2 < ... such that each zn, is connected to a0. With a new labelling we may therefore assume that all zn (with n > 1) are connected to a0.

Consider X ' 1 = {al E X1 I al > a0}. Again there exists an al E X ' 1 and a strictly increasing subsequence 1 < nl < n2 < ... such that each zn, is connected to al. Since zl E Et = ( V ( X o ) N Z ) k V ( X 1 ) and each zn~ E V ( X 1 ) ,

we have 1 < ni for all i > 1. Therefore, we can choose a new labelling such that zl is the first element and zn, the (i § 1) th element for all i > 1. It follows that now all zn, n > 2, are connected to al.

Repeat ing this argument, we obtain a sequence (zn)n_>l and an increasing sequence (an)n>0, such that zn E En and Zm > an for all m > n. (For the induction, we take at the step (j + 1) the elements zx . . . . . zj from the step before, and our new labelling starts with j + 1.)

We show that rg = {zn I n > 1} u {an I n > 0} is an ~ - c o m b . As Z is trivially ordered, all the elements zn (n > 1) are pairwise incomparable. I f zn > an ~ Xn, then zn E V(X, ) , which is contrary to the choice o f zn. I f zn < an, then zn < an < zn+l and hence zn < Zn+l, which is impossible. Thus z, II an for all n >__ 1. We have zn > an-l. I f we assume zn = an-l, then Zn+l >_ an ~_ an-1 = Zn which is again impossible, because z, [I z,+l. Hence zn > an-z for all n _ 1. Finally an-1 < an implies an-1 < an for all n _> 1, because otherwise Zn > a,-I = an, which was just excluded.

is therefore an ~ - c o m b of S, which is contrary to our hypothesis. []


Now we investigate the existence of infinite combs in products of ordered sets.

Let I be a non-empty set, let (Si, <i) (for i E I) be a family of non-empty ordered sets, S = 11 Si and < the product order on S; for simplicity we shall

also write < instead of <i. For each s ~ S, we denote by s(i) ~ S~ the i th coordinate o f s.

Proposition 5. (S, <) has an infinite comb if and only i f one o f the following conditions holds:

(a) J = {i E 1 I there exist s, t ~ Si such that s < t} is infinite.

(b) There exists i c I such that (Si, <i) has an infinite comb.

(c) There exist distinct indices i, j E 1 such that Si is not noetherian and Sj has an upper class U(x) = {s E Sj I s > x} which is not narrow.

(d) There exist distinct indices i, j E I such that Si is not noetherian and Sj has an upper class U (x) which is not artinian.

Proof. 1) We assume (a) and show that S has an infinite comb. It is immediate that it suffices to show that the set of all infinite sequences of elements in {0, 1}, with the coordinatewise order (and 0 < 1), has an infinite comb.

Let ao = (0,0,0 . . . . ), al -- (1,0,0, . . . ) , a2 = (1,1,0,0 . . . . ) . . . . and more generally

1 for l _ < i < n an(i)= 0 f o r n < i .

Let zl = (0, 1, O, 1, 0 . . . . ), i.e.

( i ) = ~ O f o r i - 1 ( rood2) Zl 1 f o r i - O (mod2) . L

Let z2 = (1,0, 1,0,0, 1,0,0 . . . . ), i.e.

Similarly, let

z3(i)

z~(i)

1 for i = 1

z2( i )= 0 f o r i = 2 1 f o r i = 0 (mod3) 0 f o r i ~ 0 ( m o d 3 ) , i > 3 .

1 for i = 1,2 0 for i = 3 1 f o r i - - 0 (mod4) 0 f o r i ~ 0 ( m o d 4 ) , i > 4 . . . . 1 for i = 1 , 2 , . . . , m - 1 0 for i = m 1 f o r i - 0 ( m o d m + l ) 0 f o r i ~ 0 ( m o d m + l ) , i > m + l .


With these choices, a0 < al < a2 < . . . . the elements zl, z2 . . . . . Zm . . . . are pairwise incomparable, am-1 < Zm (for every m > 1) and z,,, am are incomparable (for every m > 1). The verification is left to the reader.

2) We assume (b) and show that S has an infinite comb. For example, assume that $1 has the infinite comb cgl. Let S r = H si and s r ~ S r (without

i:pl restriction {1} 7~ I). Then c~ = {(t,s') c S~ x S r = S [ t ~ cr is an infinite comb o f S.

3) We assume (c) and show that S has an infinite comb. Say, for example that a0 < al < a2 < .. . is an infinite ascending chain in S1 and that the upper class o f x E $2 is not narrow, so there exists an infinite subset {zl ,zz . . . . } o f SE consisting o f pairwise incomparable elements such that x < zn (for every n > 1). Then $1 x $2 contains an infinite comb {(an, x) I n > 0} U {(an, zn+l) I n > 0}, because (ao, x) < (abx) < (a2,x) < . . . . the elements (an, zn+l) (for n > 0) are pairwise incomparable, (an, x) < (an, z,+l) and (an+bx), (an, zn+l) are incomparable (for every n > 0).

It follows at once that S has also an infinite comb.

4) We assume (d) and show that S has an infinite comb. Say, for example, that a0 < al < a2 < .. . is an infinite ascending chain in S1 and that zl > z2 > .. . > x is an infinite descending chain in the upper class o f x ~ $2, in $2. Then S1 • S2 contains the infinite comb {(an, x) I n > 0} U {(an, zn+l) I n > 0}, because (ao, x) < (abx) < (a2,x) < . . . . the elements (a~,zn+l) (for n > 0) are pairwise incomparable, (an, x) < (an, zn+l) and (an+l,x), (an, zn+l) are pairwise incomparable (for every n > 0).

5) Suppose that (g = {an [ n > 0} W {zn [ n > 1} is an infinite comb of S, where a0 < al < a2 < . . . . {zn I n > 1} is a set o f pairwise incomparable elements, with an < zn+l, zn+l and a,+l incomparable (for n > 0). We may assume that J is finite, say J = {1,2 . . . . . h} and that h > 2. Let jr = I \ J and i E jr. Since Si is trivially ordered and a0 < an, a0 < Zn for all n > 1, it results that ao(i) = an(i) = zn(i) for all n > 1. Therefore we may replace I by J.

Since a0 < al < a2 < . . . . there is an index i, say i = 1, and an infinite sequence 0 < no < nl < n2 < .. . such that an0(1) < an,(1) < .. . ; considering only the subsequence and with a new labelling: a0(1) < al(1) < a2(1) < . . . .

First case. For every j > 2, the set {zn(j) I n > 1} is finite. Let zrn = (z~(2) . . . . . zn(h)) ~ $2 x . . . x Sh, hence {zr, [ n > i} is finite. Thus

there exists an infinite subsequence 1 < nl < n2 < such that z r = z r (for all . . . . nk n l

i k > 1). Consider only this subsequence and with a new labelling, we have z n = zrl for all n > 1. Since {zn I n > 1} consists o f pairwise incomparable elements, then the same is true for {zn(1) I n > 1}, which is an infinite set contained in the upper class o f a0(1). I f there exists j > 2 such that Aj = {an(j) [ n > 0} is infinite, then this set is not noetherian and condit ion (c) is satisfied. Otherwise, each set Aj (j >_ 2) is finite and taking an appropria te subsequence and a new

r ' = ' (for every n > 0), where a n = .. . labelling, a n a 0 _ (an(2) . . . . . an(h)) c $2 x x Sh. T h e n C ~ = {an(1) l n > 0 } W { z n ( 1 ) [ n_> 1} is a comb in $1. Indeed with

234 S. Priel3-Crampe, R Ribenboim

the newer labelling, we have still a0(1) < al(1) < . . . . {zn(1) ] n > 1} consists of pairwise incomparable elements, zn(1) > an-l(1) (for every n > 1) and so Zn(1) 7 ~ an-l(1), otherwise Zn+l(1) >_ an(l) > an-l(1) = Zn(1), which is absurd. Finally, if zn(1) and an(l) are comparable, by the above zn(1) > an(l); since Z n! = Z 1! > a 0' = a n' (with the chosen labelling), then zn > an, which is contrary to the hypothesis. Thus, cs would be an w-comb in S1, hence condit ion (b) is satisfied.

Second case. There exists j, say j = 2, such that (with appropriate labelling), Z2 = {zn(2) I n > 1} consists of distinct elements.

If there exists k > 2 such that Zk = {zn(k) I n > 1} is not narrow or not artinian, then the upper class of ao(k) in Sk is not narrow or not artinian, so condit ion (c) or (d) holds.

It remains to consider the case where each Zk (for k > 2) is narrow and artinian, with Z2 infinite, hence not noetherian (by (2.1)). Then Z ' = Z2 x . . . • Zh is artinian and narrow (see [10]) hence by (2.2), with new labelling, Z 1' ~ Z 2' . . . . . . < Since Zn = (Zn(1), fn) and the elements zn (n > 1) are pairwise incomparable, if n < m then zn(1) ;g Zm(1). By (2.2) again, {zn(1) I n > l} is not artinian, or not narrow. Since Z2 is not noetherian, then condit ion (d) or condit ion (c) holds. This concludes the proof. []

Corollary 6. i) Let I be a non-empty set. Let Z I be endowed with the product order. Then Z I has no a-comb if and only i f I is finite.

ii) Q2 (endowed with the product order) has an a-comb.

Now we shall investigate the existence of w-combs in groups endowed with perforated orders (see [13]). All the groups considered henceforth are assumed to be abelian. Let (S ,+ , <) be a totally ordered group ~ {0}. The set of principal convex subgroups o f S is totally ordered by inclusion. Let (T, <) be an ordered set such that (Ct)t~r is the family of all principal convex subgroups :~ {0} and t < t' if and only if Ct c. Ct,. Denote by C t the largest convex subgroup properly contained in Ct.

The mapping v: S --* T U {0}, v(0) = 0 and for s 4: O, v(s) = t if s E Ct\C T, is a group valuation (cp. [7], here V(Sl + s2) <_ Max{v(sO, v(s2)}). For every t E T, C t /C; is an archimedian totally ordered group, hence it is o-isomorphic to a subgroup of IR. Let q)t:Ct ~ IR be a fixed order-preserving group homomorph i sm with kernel C t . Let q)t(Ct)>_o denote the set of non-negative elements o f (pt(Ct).

If g = (gt)tcr E 1-I ~pt(Ct)>_o, define <g on S as follows: s <g s', when tET

s = s' or, if s ~ s' and s ' - s E Ct\C F, then gt < qgt(s I - s). One verifies immediately that _<g is a compatible order relation on S, which is said to be an order perforated from _<, or perforated order. Since IR is archimedian, _<g is subtotal (cp. w before (5.6)). We define the support o f g as being supp(g) = {t E T I gt ~ 0}. Let g E r I q~t(ct)>_o; we define a new element

tET

g* of the above set, as follows: gt = 0 if q)t(Ct)>o has the smallest positive


element r > 0 and gt = r; gt = gt, otherwise. Clearly, the perforated orders <g and <g. coincide. Hence we need only to consider orders <g such that if 0 < gt, there exists an element ct ~ Ct such that 0 < ~ot(ct) < gt. We will need this fact in the proof of the following proposition.

Proposition 7. Le t (S, +, <) be a totally ordered group and let <g be a perfo-

rated order f o r S. Then the fo l lowing two conditions are equivalent:

i) (S, <g) has an oo-comb.

ii) supp(g) is not artinian.

Proof. 1) ii) ~ i): Since supp(g) is not artinian, there exists a strictly decreasing sequence (ti)i>_l in supp(g). For every i E N, we choose si, ci E C t , \ C ~

such that ~ot,(si) = gt, and 0 < (ot,(ci) < gt,. As ~0t, is orderpreserving, we n

have in (S,<) that 0 < ci < si for all i ~ N. Choose a0 = 0, an = ~ si and i = 1

n

zn = ~ si + cn+l for n ~ N . i = 1

Then obviously, an <g an+l and an+l I[g zn+l for n > 0. Furthermore Zn - an-1 = Sn + Cn+l E Ctn because cn+l E Ct,+~ c_ C~ ~ Ct,. Then ~Ot,(Z n - -

an-l) = tpt~ = gt,, therefore an-i <g zn for all n > 1. Let n < m, then Zm - zn = Sn+l "[- . . . -'~ Sm At- Cm+l - - Cn+l , hence Zm - zn ~ Ct,+~ and ~ t , + l ( Z m - Zn) = g t n + l - ([)tn+a(Cn+l) �9 Thus Zm, zn are incomparable for all m, n > 1 with m r n. Consequently, the set {zn I n > 1}td{an I n > 0} is an oo-comb of (S, <g).

2) i) ~ ii) : Suppose on the contrary that supp(g) is artinian. By hypothesis, there exists an c~-comb cg = {zn I n > 1} U {an [ n > 0} in (S, <g).

We show: (a) v(zn+l - an) = V(an+l - an) for all n _> O. Assume, there exists n _> 0 such that t = v(zn+l - an) (: t' = v(an+l - an).

Then, since zn+~ - an+l = Zn+l -- an + an -- an+l, one obtains: If t < t' then zn+l - an+l ~ C t , \C t. and rpt,(zn+l - an+l) = qgt,(an - an+l) -< - g t ' ; similarly, if t' < t, qgt(Zn+l - an+l) = ~Ot(Zn+l -- an) >_ gt. Hence an+l and zn+l would be comparable which is not true.

We will prove now: (b) If n < m, then v(a,,+~ - am) <_ v(an+l - an) for all n, m >_ O. Assume that (b) is not valid. Then there exists n _> 0 and a minimal

m > 0 such that m > n and v(am+l - a m ) > v(an+l - a n ) . Put t r = V(am+l - a m ) ,

t = v(an+l - an). We have Zm+ 1 - - Zn+ 1 = Zm+ 1 - - am + a,, - am-I + " " -k an - Zn+l. From (a) we deduce v ( z m + l - am) = t' and v ( a n - z n + l ) = t. Since m was chosen minimal, we have v(am-am_l ) <_ t , . . . , v(an+ 1-an) <_ t. Hence v ( z , ,+ l -Z ,+l ) = t'

and ~ot,(Zm+l - Z n + l ) = ~Ot'(Zm+l - -am) > gt'. Thus Zm+l >g Zn+l, which is a contradiction.

Because of (b) we have the following two possibilities:

236

(1)

(2)

S. Prieg-Crampe, P. Ribenboim

q 3 V v ( a n + l - a n ) = t. tET no>l n>no

V 3 V(am+l - a m ) < V(an+l - a n ) . n>l m>n

Case (1): We have Zno+3--Zno+l = Zno+3--ano+2+ano+2--ano+l +a.o+l--zno+l. Fur thermore v(z.o+3-a.o+2 ) = t (by (a)), v(a.o+2-a.o+l ) = t and a.o+l I[g Z.o+l. From a.o+l - z . o+ l = a.o+~ - a . o + ano -Zno+l it follows that ano+l -Zno+l E Ct. Hence Z.o+3- Z.o+ 1 E Ct and qh(Z.o+3- z.o+�94 ) = tpt(Z.o+3- a.o+2 ) + tpt(ano+2- a.o+l) + ~ot(a.o+l - z . o+ l ) > St + S t - S t = St. This implies that Z.o+3 >g z.o+l, if gt ~ 0, If St = 0, then qh(ano+l -Z.o+l) = 0, tpt(z.o+3-a.o+2 ) > 0 and therefore ~ot(Z.o+3-z.o+l) > 0, which yields also z.o+3 >g z.o+l. But this is absurd. Thus, there remains only the other possibility (2).

Case (2): Now (v(a.+l -a.))._>�94 contains a strictly decreasing subsequence ( v ( a . j + l - a.j))j_>l. F rom this and the fact that supp(s) is artinian, one obtains:

(,) V V [v(an+l - an) >_ t >_ v(a,n+l - am) ~ t ~ supp(g)] . i>l ni<n<m tET

Let n >_ 1, n >_ hi. We have Zn+ 1 --an+l = Z n + l - a n - } - a n - an+l and v(z.+l - a.) = v(an - a.+l). Hence t' := v(z.+l - a.+l) < t := v(a. - a.+l). If t' q~ supp(g), then St' = 0, thus z.+l and a.+l would be comparable, which is not true. If t' E supp(g ), then t ~ ~ t, hence t t < t. We consider zn+l - an+2 = Zn+l--an+l +an+l--an+2. Because of (*), we have v(a.+l-a.+2) > t' and therefore !)(Zn+l--an+2) ~--" V(an+l--an+2) =: t. Hence (pT(an+z--Zn+l) = (pi (an+2--an+l) > S7

and thus an+2 ~>g Zn+l. This is a contradiction. []

Corollary 8. i) A product o f f initely many subgroups o f ~ , endowed with any order perforated f rom the lexicographic order, has no oo-comb.

o0 ii) A product I I si o f subgroups o f R , endowed with an order (lex)g perfo-

i= l

rated f rom the lexicographic order, has an oc-comb when supp(g) is infinite.

The next result involves ordered monoids: A monoid is a commutat ive semigroup with a neutral element. (S, +, _<) is

an ordered monoid when (S, _<) is an ordered set and the order is compatible with addition: if s, t, u E S and s _ t, then s + u ___ t + u.

A monoid (S, +) is said to be cancellative, if from s, s', t E S and s + t = s '+ t it follows that s = s'. Let (S, +, _<) be a cancellative ordered monoid. The order _< is a natural order if s _ t implies that there exists an u E S such that 0 < u and t = s + u. The order of an ordered group is a natural order.

An ordered set (S, _<) is called a tree, when for each s c S the lower class L(s) = {t E S [ t <_ s} is totally ordered. In particular, a trivially ordered set is a tree.

Proposition 9. Let (S, +, _<) be a cancellative ordered monoid, which is a tree, and let <_ be a natural order. Then S does not contain an oo-comb.


Proof By [1], (3.2) the subset P = {s E S I 0 < s} of (S, <) is totally ordered and the order < on S is given by s < s' if and only if there exists an element p E P such t h a t s + p = s ' .

Obviously, P is a totally ordered submonoid of S. Assume now, that {zn I n > 1} U {an I n > 0} is an w-comb of (S, <). We

have zn+l > an >_ ao for all n > 0. Hence there exist elements Pn, P'n E P such that zn+l = an+fin and an = ao+pn. Thus Zn+l = ao+qn with qn E P. I fn ~ m, then zn+l ~ zm+l and therefore qn ~ qm. But since P is totally ordered, qn and qm are comparable and hence also Zn+l and zm+l. This is a contradiction. []

4 Ultrametric Spaces of Generalized Power Series

Let now (R, +) be an abelian group ~ {0} and (S, <) an ordered set. For each f : S ~ R we define the support o f f by supp(f) = {s E S r f(s)

0}. The zero-mapping has empty support. Consider A = { f :S ~ R I supp(f) is artinian and narrow}. With pointwise addition, A is an abelian additive group. A is called the group of generalized power series with exponents in the ordered set (S, <) and coefficients in the group (R, +). We use the notation A = [[RS'<-]].

As in w let F(S) be the set of finite non-empty antichains of S, with the order X < Y whenever for every y E Y there exists x E X such that x < y. We adjoin to F(S) an element 0 (the empty set) with the convention that X < 0 for every X E F(S).

For each f E A let n(f) = 0 when f = 0, and n(f) = Minsupp(f) when f ~ 0, where Min supp(f) denotes the set of all minimal elements of supp(f). We note that since the support of f is artinian and narrow, then Min supp(f) ~ 0 and it is a finite antichain of S. As easily seen, n is a group-valuation (cp. [14], (3.4)).

Now we consider F(S) U {0} endowed with the order opposite to <, which we denote by <~ thus if X, Y E F(S) then 0 <~ X and X <~ Y when for every x E X there exists y E Y such that y < x.

Let d:A x A ~ (F(S) U {0},< ~ be defined by d(f ,g) = n(f - g ) . This defines an ultrametric distance on A (see [14]).

In the following result, we characterize groups of generalized power series which are spherically complete.

Theorem 10. Let (S, <) be an ordered set and (R, +) :fi {0} an abelian group. The following conditions are equivalent:

(i) A = I[RS'<-]] is spherically complete.

(ii) (S, <) does not contain an w-comb.

Proof 1) i) ~ ii): Assume that (S, <) contains the w-comb ff = {Zn I n > 1 } W {an I n > 0}. Choose 0 @ r E R. Let et E A for t E S be defined by

r i f s = t et(s)= 0 i fs~=t .

238 S. Priefl-Crampe, E Ribenboim

We consider the balls Bm = B{~}(fm) with f m = ~', ez, (for m > 1). Since i=1

d(fm+l,fm) = 7 ~ ( f m + l - fro) = 7r(ez,.+,) = {Zm+l} <o {am} we have fm+l E Bm n Bm+l. As {am+l} <o {am} this implies (by (1.1)) that B1 _~ B2 __P_ " " .

oo Hence by hypothesis ["1 B,, =/= 13. Thus there exists an element f C A such

rn=l ~x3

that f E ["1 B,, and therefore supp(f) is narrow. However, we show that m=l

Z = {z, I n _> 1} ~ supp(f). Indeed, f E Bm implies that n(f - fro) > {am}. If f(Zm) = 0, then Zm E supp(f --fro) and hence {Zm} > zc(f - fro) > {am}, which yields the contradiction z,, > am. Thus f(Zm) 4= 0 for every rn > 1, showing that Z _ supp(f).

2) ii) ~ i) : We assume that the equivalent condition (ii) of proposition 4 holds.

Let ~ be any non-empty chain of balls. If there exists a smallest ball in ~ , then obviously N B 4= 13. Therefore we assume now that ~ has no smallest

ball. This implies that there exists a limit ordinal p and a strictly decreasing sequence (B~)~<; of balls of ~ which is cofinal in ~ (relative to 2). It suffices to show that ['-1 Bz 4= 13.

6<p Let Bz = Bx~ (fa), where X~ c F(S). This implies (a) zr(f~ - f ~ , ) _> Xz for all 6 < 6' < p.

We show: (b) If 6 < 6' < p then X~ _< Xz,. Indeed, let g:S --. R be given by g(s) = fz,(s), if s q~ X~, and g(s) 4= f~,(s),

if s E Xa,. Then supp(g) ~ supp(fa,) U Xa, ; since this last set is artinian and narrow, because X~, is finite then g c A. By definition, =(g - f~ , ) = Xa, which implies that g c Bx~, (f~,) ~_ Bx~ (fz). From lr(g-f~) > Xz and 7r(fa - f a , ) > Xa it follows that zr(g - f z , ) = X~, > Xz.

We show: (c) If 6 < 6' < p and f,~ (s) 4= f,s, (s), then X,~ < {s}. Indeed, if f6(s) 4= f~,(s), then s c supp(f~ - fa,). Hence there exists an

x 6 ~ ( f a - f ~ , ) such that x < s. Because of (a) we find y ~ Xa such that y < x < s. Thus X6 _< {s}.

Let f : S ~ R be defined by

f(s) = ~ fz(s), if there exists 6 < p such that Xz ~ {s}

k 0, otherwise.

We have to confirm that f is well-defined. Assume that there exist 6 < 6' < p such that X~ ~ {s}, Xz, ~ {s} and fz(s) ~ fz,(s). By (c) this last inequality implies that X6 < {s}, which is absurd. Hence f is well-defined.

We show that supp(f) is artinian. Suppose that s~ E supp(f), s~ > s~+l for all v > 0. By definition of f , there

exists 60 < p such that Xa 0 ~ {so} and f(so) = f6o(SO). As s~ < so for all v > 0,


it results that X60 ~ {Sv} for all v > 0. Hence f(sv) = f6o(Sv) for all v > 0. But then {s~ I v > 0} _ supp(f60). Because supp(f60) is artinian, we find therefore a/~ > 0 such that s~ = su for all v > #. Thus supp(f) is artinian.

We show that supp(f) is narrow. Assume that {zn I n > 1} is an infinite trivially ordered subset of supp(f) .

For every i > 1, let 6i be the smallest ordinal such that {zi} ~ X6,. Since every set of ordinals is wellordered the sequence (6i)i>__~ contains a non-decreasing subsequence 6i 1 < (~i2 "( "" "" By considering an infinite subset of {zn ] n > 1} we may assume with a new labelling that 6~ _< b2 _< 63 _< " ". If there exists an ordinal 6 < p such that bi _< b for all i _> 1, then by (c) this implies f~(zi) = f~,(zi) = f(zi) for every i > 1, so {zn I n > 1} ~ supp(f~), with f~ E A. This is impossible. Thus lim 6i = p and since p is a limit ordinal, there is an

i

infinite subset {zil,zi2 . . . . } of {zn I n > 1} such that 6i, < 6i2 < " " . With a new labelling we may therefore assume that 61 < b2 < " " < p, lim fin = p. Thus

n

{zn} ~ X~. for n > 1, {zn} > X~,_, . . . . . {z,} > X~, for all n > 2. By taking Z = {zn In > 1}, we obtain V ( X J n Z = {Zn+l,Zn+2 . . . . }. This is contrary to the hypothesis o f condit ion (ii) o f proposi t ion 4. Hence f E A and obviously T E N B ~ . []

6 < p

If the ordered set (S,_<) is narrow or noetherian, then obviously it does not contain an ~ -comb . Therefore we obtain:

Corollary 11. Let (S, <) be an ordered set which is narrow or noetherian. Then A = [[RS,<-]] is spherically complete.

We apply the results of w to obtain other examples o f spherically complete power series groups.

Let (Si, <i) (i = 1,2 . . . . . n) be ordered monoids, let S = f i Si, endowed i = 1

with the product order <.

Corollary 12. Assume:

1) Each Si has no ~-comb.

2) I f i -J= j and Si is not noetherian, then in Sj every upper class U(x) = {s E Sj I s > x} is narrow and artinian.

Then A = [[RS,<-]] is spherically complete.

Proof This follows at once from proposi t ion 5 and theorem 10. []

In particular I[RZ",-<]] is spherically complete, but I[R~,-<]] is not spherically complete.

Now we apply proposi t ion 7, with the notat ions therein. Let (S, <) be a totally ordered group, let <g be the perforated order

defined by g E 11 q~t(Ct)>_o. From theorem 10, it follows: tET


Corollary 13. Ifsupp(g) is artinian, then Ag = [[Rs,<-g]] is spherically complete.

In particular, the result holds when the set of convex subgroups of S, ordered by inclusion, is artinian, say when it is finite.

Let (R, <) be an ordered abelian group, (S, <) an ordered set. We define on A = [[RS'<-]] the following relation:

I f f , g c A let f < g when f = g or f (s) < g(s) for every s E rt(f - g ) . The relation < is a compatible order. The following result was shown in

[21:

4.1. (A, <) is a lattice ordered group if and only if one of the following conditions holds:

1) (S, <) is trivially ordered and (R, <) is a lattice ordered group.

2) (S, _<) is a tree and (R, N) is a totally ordered group.

Corollary 14. Let (R, <) be a totally ordered group, let (S, +, <) be an ordered cancellative monoid, such that the order is natural and (S, <) is a tree. Then the lattice ordered group A = [[RS,<-]] is spherically complete.

Proof This follows at once from proposition 9. []

5 Applications

We shall now apply the Fixed Point Theorem to rings of generalized power series.

An ordered monoid (S, +, <) is strictly ordered if s, t, u ~ S, s < t imply that s + u < t + u.

Every cancellative monoid or every trivially ordered monoid is strictly ordered; in particular, every ordered group is strictly ordered.

We shall assume henceforth that (S, +, <) is strictly ordered.

Let S be a monoid and X ~_ S. We denote by < X > the submonoid of S generated by X. The following result is due to ERD6S and RAP6 (see for example [3]):

5.1. Let (S, _<) be a (strictly) ordered monoid, let X ~_ S and assume that 0 < x for every x E X. If X is an artinian and narrow subset, then so is <X>.

An extension of this result is the following ([3]):

5.2. Let (S, +, _<) be an ordered group, and let X be an artinian and narrow subset of S. Assume that for every x c X there exists an integer k > 1 such that 0 < kx. Then < X > is artinian and narrow.

Let R :~ {0} be a commutative ring. We will define a multiplication on .4 = [[RS,~-H; in the special case when (S, <) is totally ordered, this yields the usual multiplication of formal power series. For details, see [11], [12].


5.3. For every s ~ S and f , g E A, let

Xs( f ,g) = {(t,u) E S • S I t + u = s , f ( t ) =/= O,g(u) -'/= 0}.

Then X A f , g) is finite.

This allows to define a multiplication on A by

( f " g)(s) = Z f (t)g(u) . (t,u)EXs(f ,g)

With this multiplication (and the pointwise addition which was explained in the beginning of w A = [[RS'<-]] becomes a commutative ring with unit element e, where e(0) = 1, e(s) = 0 for every s 6 S, s :~ 0.

Let (S, +, <) be a (strictly) ordered monoid; we can define an addition on the set of all finite trivially ordered subsets of S. (This is done explicitly in [14]). Let X, Y E F(S) U {0}. Then define X + Y as the set of all minimal

a

elements of {x + y I x ~ X,y E Y }. Obviously X + Y E F(S) u {0}. Endowed a

with this addition and the order which was explained in w F(S) becomes an ordered monoid ([14]) with {0} as its neutral element.

If A = l[RS,<-]], the group valuation n : A --. F(S)U {0} has now the additional property

5.4. n ( f . g) >_ n( f ) + n(g) for all f , g E A. a

For the next proposition, we will need also the following result:

5.5. Let (S, +, _<) be a strictly ordered monoid. F(S) U {0}. Then {0} < X implies Z < Z + X.

a

Let Z E F(S) and X E

Proof If {0} < X, then Z = Z + {0} < Z + X, since F(S) is an ordered a a

monoid. Now, {0} < X implies that 0 < x for all x E X and that there exists an x0 E X such that 0 < x0. But all the elements of X are pairwise incomparable. Hence 0 < x for all x E X.

Assume Z = Z + X. Then an element z E Z may be written as z = z' + x g

for z' E Z and x 6 X. As 0 < x this implies z' < z which is impossible, since Z is trivially ordered. []

A proof of the following proposition may also be found in ([11], (2.2)). The proof, which will given here, uses the Fixed Point Theorem (Theorem 1).

Proposition 15. Let (S, +, <_) be a strictly ordered monoid and R --/= {0} a commutative ring. Let f E A -= [[RS'<-]. Then the following two conditions are equivalent:

i) f is a unit o f A.

242 S. Prieg-Crampe, R Ribenboim

ii) There exists h E A, such that supp(e - f . h) _ {s E S I 0 < s}.

Proof Set Sumd(s) = {t E S I There exists u E S such that t + u = s} and D = U{Sumd(s) 10 "it: s}.

1) i) ~ ii): Let f .g = e and define h:S ---, R by

g(t), when t E D h( t )= 0, w h e n t C D .

Then supp(h) __q D n supp(g) ~ supp(g), which implies that supp(h) is artinian and narrow. Hence h E A. Let s E S, 0 r s; then (f.g)(s) = ~, f(t)g(u). Since

t + U = S

u E Sumd(s), then u E D; therefore g(u) = h(u). Thus e(s) = ~ f(t)h(u), t + u = s

which yields supp(e - f . h) ~ {s E S t 0 < s}.

2) ii) =~ i): Let h E A be such that s u p p ( e - f - h) _ {s E S I 0 < s}. Let k = e - f . h . Consider the mapping qo:A --* A, r = e+l .k . For 11, 12 c A with 11 ~ 12 we have d(q)(lO, r = n(q)(ll)-q)(12)) = n((ll-12)'k) > n(ll-12)+n(k),

a

where for the last inequality we used (5.4). Since supp(k) _ {s E S [ 0 < s}, we have n(k) > {0}. Hence we obtain by (5.5), that rt(ll - / 2 ) + rt(k) > n(ll - / 2 ) =

a

d(ll, 12). Recalling that F(S) is endowed with the order <~ opposite to _% then q) is a contracting map of A, relative to the distance d : A x A --* (F(S),_<~ Let T be the submonoid of S, which is generated by supp(k). Then by (5.1) it follows that T is narrow. Corollary l l implies that B = [RT'<-] is spherically complete. But B is identified to a subring of A. For l E B we have supp(cp(/)) ___ supp(e) t3 supp(l - k) ~ {0} U {supp(I) + supp(k)} ___ T. Hence the restriction q~[n is a mapping from B to B. Theorem 1 therefore implies that ~0lB has a fixed point l E B ~ A. This yields l = e + l .k and hence l ( e - k ) = e, which implies f ( l . h) = e. Thus f is a unit of A. []

Let now (S, +,_<) be a torsion-free ordered group. For s, t E S we define: s _<' t if there exists an integer k > 0 such that ks <_ kt. Then _<' is a refinement of the order _< of S (i.e. s _< t implies s _<r t), and _<P is compatible with the addition of S. If this order _<' is total, the order _< is called subtotal ([12] and [3]). In [3], Theorem 1, the following result is proved:

5.6. The ring A = [[RS,<-ll of generalized power series is a field if and only if R is a field, S is a torsion-free group and the order of S is subtotal.

Let R be a field and let (S, +, <) be a subtotally ordered torsion-free group. A = [IRS,<-] consists of all the mappings from S to R with artinian and narrow support (in the order _< of S). If a subset of S is artinian and narrow with respect to the order _<, then it is wellordered with respect to the refinement _<' of _< ([10]). Thus A can be considered a subfield of A' = [[RS,<-'ll, and the natural valuation rt" A' --* (S, <'), 0 :fi f ~ Min_<, supp(f) may be restricted to A. With respect to this valuation, A has the value group (S, _<') and residue field R. Hence W is an immediate extension of A. In [12] it is proved by using


RAYNER'S results on field-families [9], that (A, n') is henselian. We will now give another proof which is based on the Fixed Point Theorem. This proof is close to that given in [8] for maximal valued fields.

Proposition 16. Le t (S, +, _<) be a subtotally ordered torsion-free group and let R be a field. Then A = I[RS,<-] is henselian with respect to the valuation n'.

P r o o f We denote by B~,, Mn, the valuation ring and valuation ideal of A. There are several equivalent conditions by which the property henselian

for a valued field is defined. We will prove that the following one (cp. [15], p.349) is fulfilled for (A, n'):

If f(x) = x n + c , _ l x "-1 + ' " + C l X + C o E B~,[x] with co --- 0 mod M~, and Cl ~ 0 mod M,,, then there exists an element z E M,, such that f ( z ) = O.

Consider the mapping q): M~, --* M~,, z ~ z - c l e f ( z ) . We show that q) is contracting with respect to the ultrametric d(z, y) = n ' ( z - y ) (and the value set (S, <') ordered dually). Let z , y E M~,. Then (p(z) -q~(y) = z - y - c~-I (f(z) - f (y)) = c-{ 1 ((y" - z") + c,-1 (y~-i _ zn-1) + . . . + c2(y2 _ z2)) = c71 (z - y )m with m E m,,. Hence n'(z - y) <' n'(~0(z) - q)(y)).

n-1 Denote Y = U supp(ci) u supp(ci-1). Since the supports of ci, c71 are

i=0 artinian and narrow (with respect to <), so Y is artinian and narrow. Further 0 <' y for all y 6 Y. Hence there exists for every y E Y an integer k > 1 such that 0 < ky. Therefore, we have by (5.2) that T = < Y > is artinian and narrow. This (with respect to <) implies that [[RT'<-]] = ~RT'<-']]. Thus F = I[RT'<-']] is a subring of A. By corollary 11 F is spherically complete with respect to the restricted ultrametric d ( z , y ) = n ' (z - y ) of A. Hence also G = M~, A F is spherically complete. Thus the contracting mapping q): G ~ G has (by Theorem 1) a fixed point z E G. The element z has all the properties which we were looking for, because z E M~, and f ( z ) = O. []

References

[1] A. BENHISSt and P. RIBENBOIM, Ordered Rings of Generalized Power Series, to appear.

[2] P. CONRAD, J. HARVEY and C. HOLLAND, The Hahn Embedding Theorem for Abelian Lattice-ordered Groups. Trans. Amer. Math. Soc. 108 (1963), 143-169.

[3] G.A. ELLIOWr and P. RIBENBOIM, Fields of Generalized Power Series. Arch. Math. 54 (1990), 365-371.

[4] H. HAHN, Uber die nichtarchimedischen Gr6gensysteme. Sitz.-Ber. Akad. Wiss. Wien, math.-naturw. Kl. Abt. IIa, 116 (1907), 601-655.

[5] I. KAPLANSKY, Maximal Fields with Valuations. Duke Math. J. 9 (1942), 303-321.

[6] W. KRULL, Allgemeine Bcwcrtungstheorie, J. rcinc angew. Math. 167 (1932), 160-196.

[7] S. PRIESS-CRAMPE, Angeordnete Strukturen: Gruppen, K/brper, projektive Ebenen. Springer, Berlin-Heidelberg-New York-Tokyo, 1983.

244 S. Prie6-Crampe, E Ribenboim

[8] S. PRIESS-CRAMPE, Der Banachsche Fixpunktsatz f'fir ultrametrische R~iume, Re- sults in Math. 18 (1990) 178-186.

[9] F. RAYNER, An Algebraically Closed Field. Glasgow Math. J. 9 (1968), 146-151.

[10] P. RIBENBOIM, Rings of Generalized Power Series: Nilpotent Elements. Abh. Math. Sere. Univ. Hamburg 61 (1991), 1-19.

[11] E RIBENBOIM, Rings of Generalized Power Series II: Units, Zerodivisors and Semi-simplicity, to appear in J. Algebra.

[12] E RIBENBOIM, Fields: Algebraically Closed and Others, to appear in Manu. Math.

[13] P. RIBENBOIM, Some Examples of Valued Fields. Compt. Rend. Acad. Sci. Paris (1991).

[14] P. RIBENBOIM, Ordering the Set of Antichains of an Ordered Set, preprint.

[15] S. WARNER, Topological Fields, North-Holland Amsterdam-New York-Oxford- Tokyo, 1989.

Eingegangen am: 08.07.1992

Authors' adresses: Sibylla Priess-Crampe, Mathematisches Institut der Universit~it Miinchen, Theresienstr. 39, 8000 MiJnchen 2, Germany, Paulo Ribenboim, Depart- ment of Mathematics and Statistics, Queen's University Kingston, Ontario, K7L3N6, Canada.

Documents

Fixed points, combs and generalized power series