198, 327]338 Ž1997.
JA977063
JOURNAL OF ALGEBRA
ARTICLE NO.
Semisimple Rings and Von Neumann Regular
Rings of Generalized Power Series
Paulo Ribenboim
Department of Mathematics and Statistics, Queen’s Uni¨ ersity, Jeffrey Hall,
Kingston, K7L 3N6, Canada
Communicated by D. A. Buchsbaum
Received August 15, 1996
DEDICATED TO MY FRIEND CHRISTIAN JENSEN, AT THE BEGINNING OF
HIS SECOND YOUTH
In this paper we continue our investigation of generalized power series.
The main theorem determines rings of generalized power series which are
Von Neumann regular rings and semisimple rings. In the final section we
give a new proof of Neumann’s theorem on skewfields of generalized
power series with totally ordered group of exponents. Using a result of
Erdos
¨ and Rado,
` we deduce a simple proof of a theorem in w1x, which is
proven here also for skewfields.
It is my pleasure to thank C. U. Jensen and K. R. Goodearl for useful
comments while this paper was in preparation.
1. ORDERED SETS
The ordered set Ž S, F. is said to be artinian Žresp. noetherian, resp.
narrow . if it does not contain any infinite strictly descending subsets Žresp.
any infinite strictly ascending subsets, resp. any infinite trivially ordered
subsets .. A set which is artinian and narrow is also called a quasi-wellordered set.
A totally ordered set is artinian if and only if it is well-ordered. The
following facts are well-known:
(1.1) An ordered set Ž S, F. is finite if and only if it is noetherian, artinian,
and narrow.
327
0021-8693r97 $25.00
Copyright Q 1997 by Academic Press
All rights of reproduction in any form reserved.
328
PAULO RIBENBOIM
(1.2) Let Ž S, F. be an ordered set. The following conditions are equi¨ alent:
(1) Ž S, F. is artinian and narrow.
(2) If s1 , s2 , . . . g S there exist natural numbers i1 - i 2 - ??? such that
si1 - si 2 F ??? .
(3) If s1 , s2 , . . . g S there exist natural numbers i - j such that si - s j .
k
(1.3) Let Ž S1 , F1 ., . . . , Ž Sk , Fk . be ordered sets, S s Ł is1
Si endowed
with the product order F . Then Ž S, F. is artinian and narrow if and only if
each Ž Si , Fi . is artinian and narrow.
We note that if F , F9 are orders on S such that s F t implies s F9t
for any s, t g S, and if T is a subset of S which is artinian and narrow with
respect to F , then it is also artinian and narrow with respect to F9.
2. ORDERED MONOIDS
Let Ž S, q. be a monoid with a neutral element 0. Let F be an order
relation on S. Ž S, q, F. is an ordered monoid if the following conditions
are satisfied: if s F s9 then s q t F s9 q t and t q s F t q s9 for all
s, s9, t g S. The order is strict when s - s9 implies s q t - s9 q t and
t q s - t q s9 for all s, s9, t g S. A monoid S endowed with a compatible
strict total order is cancellati¨ e if s q t s s9 q t or if t q s s t q s9 then
s s s9.
We shall require the following well-known fact:
(2.1) Let T1 , . . . , Tk Ž k G 1. be artinian and narrow subsets of Ž S, q, F..
Then T1 q ??? qTk is artinian and narrow.
Neumann proved:
(2.2) Let Ž S, q, F. be a totally ordered commutati¨ e monoid, T a wellordered subset of S such that 0 - t for e¨ ery t g T. Then the monoid
²T : s D `ns 1 nT generated by T is also a well-ordered subset of S Ž here
nT s t 1 q ??? qt n < t i g T for each i4..
This result was extended by Erdos
¨ and Rado
` Žsee Rosenstein w9x.:
(2.3) Let Ž S, q, F. be a commutati¨ e non-tri¨ ially ordered monoid and let
T be an artinian and narrow subset of S such that 0 - t for e¨ ery t g T. Then
²T : s D `ns 1 nT is artinian and narrow.
The order F on S is said to be subtotal when for any s, t g S there
exists an integer k G 1 such that ks F kt or kt F ks. If S is commutative
group, the order F is subtotal if and only if for every s g S there exists
k G 1 such that ks G 0 or ks F 0.
GENERALIZED POWER SERIES
329
Let F be a subtotal order on the commutative monoid S. We define
the binary relation F9 on S by letting s F9t whenever there exists an
integer k G 1 such that ks F kt. If S is a torsion-free commutative
monoid, then F9 is a compatible total order, finer then F , that is, if
s F t then s F9t.
3. REGULAR RINGS
Let R be a ring with unit element 1. R is said to be a ŽVon Neumann.
regular ring when for every x g R there exists y g R such that xyx s x.
A cartesian product of regular rings is a regular ring. Every semisimple
ring is regular.
(3.1) If R is a regular ring and x g R, x / 0, is not a zero di¨ isor, then x is
in¨ ertible.
Proof. Let x g R, x / 0 and let y g R be such that xyx s x. Then
x Ž yx y 1. s Ž xy y 1. x s 0. So xy s yx s 1.
(3.2) Let R be a ring. R is a skewfield if and only if it is regular and has no
zero di¨ isors.
Proof. One implication is obvious and the other follows at once from
Ž3.1..
We shall require the following theorem:
(3.3) Let R be a ring. Then R is semisimple if and only if R is regular and
e¨ ery set of non-zero mutually orthogonal idempotents of R is finite.
Proof. It is well known that if R is semisimple then it is regular and
each set of non-zero mutually orthogonal idempotents is finite. The
converse is due to Kaplansky Žsee Goodearl w3, Corollary 2.16x..
4. GENERALIZED POWER SERIES
Let Ž S, q, F. be a strictly ordered monoid and let Ž R, q, ? . be a ring. It
is not assumed that S is abelian, nor that R is commutative.
If f : S ª R let
supp Ž f . s s g S < f Ž s . / 0 4 .
Let A be the set of all f : S ª R such that suppŽ f . is artinian and narrow.
A is a subgroup of the additive group of all mappings from S to R with
pointwise addition, because suppŽ f q g . : suppŽ f . j suppŽ g . and the
union of two artinian and narrow subsets of S is again artinian and
330
PAULO RIBENBOIM
narrow. The mapping 0: S ª R with 0Ž s . s 0 for all s g S is the neutral
element of Ž A, q..
(4.1) Let s g S, f 1 , . . . , f k g A Žwith k G 1.. Then the set
X s X Ž s, f 1 , . . . , f k .
s Ž t 1 , . . . , t k . g S k < s s t 1 q ??? qt k , f i Ž t i . / 0 for i s 1, . . . , k 4
is finite.
Proof. It is trivial when k s 1, so we assume that k G 2. If X is
infinite, there exists j, 1 F j F k, such that the jth projection pr j Ž X . is
infinite. By renumbering, we assume that pr1Ž X . is infinite. Since pr1Ž X .
: suppŽ f 1 ., it is artinian and narrow and by Ž1.1., there exist t 11 , t 12 , . . . g
pr1Ž X . with t 11 - t 12 - ??? . For each j we choose Ž t 1 j , t 2 j , . . . , t k j . g X.
k
The set T s Ž t 2 j , . . . , t k j .< j s 1, 2, . . . 4 : Ł is2
suppŽ f i ., so by Ž1.3., T is
artinian and narrow. By Ž1.2., there exists j1 - j2 - ??? such that t i j1 F t i j 2
F ??? Žfor all i s 2, . . . , k .. Hence s s t 1 j1 q t 2 j1 q ??? qt k j1 - t 1 j 2 q t 2 j 2
q ??? qt k j 2 s s, which is absurd.
If f, g g A, we define the mapping fg: S ª R as
Ž fg . Ž s . s
Ý
f Ž t . g Ž u. .
Ž t , u .gX Ž s, f , g .
Note that there are only finitely many non-zero summands. If Ž fg .Ž s . / 0,
then there exists Ž t, u. g X Ž s, f, g . so s s t q u g suppŽ f . q suppŽ g .. So
suppŽ fg . : suppŽ f . q suppŽ g ., thus by Ž2.1., suppŽ fg . is artinian and
narrow, hence fg g A. This defines a binary operation of multiplication
on A.
It is routine to verify that the operation is associative, right and left
distributive with respect to the addition. Moreover, the unit element of A
is e: S ª R given by eŽ0. s 1, eŽ s . s 0 for all s g S, s / 0.
Thus Ž A, q, ? . is a ring, called the ring of generalized power series, with
coefficients in R and exponents in S. We use the notation A s ww R S, F xx.
The special case where R is a commutative ring and Ž S, q, F. is a strictly
ordered monoid has been introduced and studied in several papers Žsee w6,
7, 1x..
We shall require the following remark. Assume that F , F 9 are compatible orders on the monoid S such that if s F t then s F9t; then
A s ww R S, F xx is a subring of A9 s ww R S, F 9 xx.
For each f g A, f / 0, suppŽ f . is non-empty artinian and narrow. Let
p Ž f . denote the set of minimal elements of suppŽ f .. If Ž S, F. is totally
ordered, then p Ž f . consists of only one element, which is still denoted by
p Ž f ..
331
GENERALIZED POWER SERIES
(4.2) Let Ž S, F. be totally ordered. If f, g g A _ 04 , then:
(1) p Ž f q g . G minp Ž f ., p Ž g .4 . If p Ž f . - p Ž g . then p Ž f q g . s
p Ž f ..
(2) p Ž fg . G p Ž f . q p Ž g .. If R has no zero-di¨ isors, then p Ž fg . s
p Ž f . q p Ž g ..
Proof. Let p Ž f . s s, p Ž g . s t. Ž1. If s F t and s9 - s then f Ž s9. s
g Ž s9. s 0, so Ž f q g .Ž s9. s 0, hence p Ž f q g . G s. If s - t then Ž f q g .Ž s .
s f Ž s . / 0, so p Ž f q g . s s.
Ž2. Let u g suppŽ fg ., so
0 / Ž fg . Ž u . s
f Ž u9 . g Ž u0 . .
Ý
Ž u9, u0 .gX Ž u , f , g .
So there exist u9 g suppŽ f ., u0 g suppŽ g ., u9 q u0 s u. So u9 G s, u0 G t,
and u G s q t, showing that p Ž f q g . G s q t. Since the order on S is
strict and Ž u9, u0 . g X Ž s q t, f, g ., u9 q u0 s s q t, u9 G s, u0 G t, hence
u9 s s, u0 s t. Thus Ž fg .Ž s q t . s f Ž s . g Ž t . / 0, if R has no zero-divisors.
If n G 1 and R is a ring, let Mn= nŽ R . denote the ring of all n = n
matrices with entries in R. We have the following canonical isomorphism:
(4.3) ww Mn= nŽ R . S, F xx ( Mn=nŽww R S, F xx..
Proof. Let f g ww Mn= nŽ R . S, F xx with f Ž s . equal to the matrix with the
Ž i, j .-entry denoted by f Ž s .Ž i, j. . Let w Ž f . be the matrix with the Ž i, j .-entry
equal to f i, j : S ª R, defined by f i, j Ž s . s f Ž s .Ž i, j. . We have
n
supp Ž f . s
D
supp Ž f i , j . ,
i , js1
so each f i j g ww R S, F xx, so w Ž f . g Mn=nŽww R S, F xx..
It is clear that w is a bijection and also that w Ž f q g . s w Ž f . q w Ž g ..
Now we verify that w Ž fg . s w Ž f . w Ž g .: For every i, j s 1, . . . , n and s g S
we have
Ž w Ž fg . . i , j Ž s . s Ž Ž fg . Ž s . . Ž i , j .
s
ž
f Ž t . g Ž u.
Ý
Ž t , u .gX Ž s, f , g .
/
Ž i , j.
n
s
Ý
Ž t , u .gX Ž s, f , g .
ž
Ý f Ž t . Ž i , k . g Ž u. Ž k , j.
ks1
n
s
Ý Ý f Ž t . Ži , k . g Ž u. Ž k , j. ,
ks1 Y
/
332
PAULO RIBENBOIM
where Y s Ž t, u.< t q u s s, f Ž t .Ž i, k . / 0, g Ž u.Ž k, j. / 04 . From f Ž t .Ž i, k . s
w Ž f . i, k Ž t . and g Ž u.Ž k, j. s w Ž g . k, j Ž u., we have Y s X Ž s, w Ž f .Ž i, k . , w Ž g .Ž k, j. ..
On the other hand
n
Ž w Ž f . w Ž g . . i, j Ž s. s Ý w Ž f . i, k w Ž g . k, j Ž s.
ž
/
ks1
n
s
Ý
ks1
ž
Ý
w Ž f . i , k Ž t . w Ž g . k , j Ž u.
Ž t , u .gX Ž s, w Ž f . i, k w Ž g . k , j .
/
n
s
Ý
Ý
f Ž t . Ži , k . g Ž u. Ž k , j. .
ks1 Ž t , u .gY
We conclude that w Ž fg . s w Ž f . w Ž g . and w is a canonical ring isomorphism.
We indicate conditions for A to be a skewfield. Neumann showed, using
Ž2.2.:
(4.4) Assume that the order on S is total. Then A is a skewfield if and only
if R is a skewfield and S is a group.
We shall give in Section 6 another proof of this theorem, not involving
Ž2.2.. The above result was extended by Elliott and Ribenboim w1x for the
case when R is commutative. The proof involved Ž2.3. and holds also when
R is not commutative:
(4.5) Assume that S is a commutati¨ e monoid. Then A is a skewfield if and
only if R is a skewfield, S is a torsion-free group, and the order F on S is
subtotal.
5. REGULAR AND SEMISIMPLE RINGS OF
GENERALIZED POWER SERIES
We prove our main theorem:
Ž5.1. THEOREM. Let R be a ring with unit element 1 and let Ž S, q, F. be
a strictly ordered monoid, A s ww R S, F xx. Assume the following:
Ž18. There exists an artinian and narrow subset T of S which is infinite
Ž hence the order on S is not tri¨ ial ..
Ž28. If the order on S is not total, then S is commutati¨ e and torsion-free.
GENERALIZED POWER SERIES
333
Then the following conditions are equi¨ alent:
(1) A is regular.
(2) Ža. S is a group.
Žb. If S is commutati¨ e, the order on S is subtotal; if S is non-commutati¨ e, the order on S is total.
Žc. R is a regular ring.
Žd. E¨ ery set of non-zero mutually orthogonal idempotents of R is
finite.
(3) R is a semisimple ring and conditions Ža. and Žb. of Ž2. hold.
(4) A is a semisimple ring.
Proof. (1) « (2). Ža. S is a group: let s g S and let e s : S ª R be
defined by e s Ž s . s 1, e s Ž t . s 0 for all t g S, t / s. By hypothesis, there
exists g g A such that e s s e s ge s . Then 1 s e s Ž s . s Ž e s ge s .Ž s .. Hence
there exists t g S such that s q t q s s s. Since the order is strict, Ž S, q.
is cancellative, so s q t s t q s s 0, showing that S is a group.
Žb. By hypothesis, if S is not commutative, the order on S is total. Let S
be commutative. We show that the order on S is subtotal. Let s g S,
s / 0. Since A is regular, there exists g g A such that Ž e0 q e s . g Ž e0 q e s .
s e0 q e s . Computing both sides at 0, s, 2 s, 3s, . . . , ys, y2 s, y3s, . . . , we
have the following relations:
1 s g Ž 0 . q g Ž ys . q g Ž ys . q g Ž y2 s .
1 s g Ž 0 . q g Ž 0 . q g Ž 0 . q g Ž ys .
0 s g Ž 2 s . q g Ž s . q g Ž s . q g Ž 0.
0 s g Ž 3s . q g Ž 2 s . q g Ž 2 s . q g Ž s .
0 s g Ž 4 s . q g Ž 3s . q g Ž 3s . q g Ž 2 s .
0 s g Ž 5s . q g Ž 4 s . q g Ž 4 s . q g Ž 3s .
???
0 s g Ž ys . q g Ž y2 s . q g Ž y2 s . q g Ž y3s .
0 s g Ž y2 s . q g Ž y3s . q g Ž y3s . q g Ž y4 s .
0 s g Ž y3s . q g Ž y4 s . q g Ž y4 s . q g Ž y5s .
???
Then
Ži.
Žii.
g Ž0. q g Žys . / 0 or
g Žys . q g Žy2 s . / 0,
334
PAULO RIBENBOIM
and
Ži.
Žiii.
g Ž0. q g Žys . / 0 or
g Ž0. q g Ž s . / 0.
In case Žii., g Žys . q g Žy2 s . / 0, g Žy2 s . q g Žy3s . / 0, g Žy3s . q
g Žy4 s . / 0, . . . , etc. So there exist infinitely many integers 0 - n1 - n 2 n 3 - ??? such that g Žyn i s . / 0 so yn1 s, yn 2 s, . . . g suppŽ g .. By Ž1.2.
there exist n i - n j such that yn i s F yn j s, so Ž n j y n i . s F 0.
In case Žiii., g Ž0. q g Ž s . / 0, g Ž s . q g Ž2 s . / 0, g Ž2 s . q g Ž3s . / 0, . . . ,
so there exist integers 0 - n1 - n 2 - n 3 - ??? such that g Ž n i s . / 0 for all
i s 1, 2, . . . . So n1 s, n 2 s, . . . g suppŽ g ., hence by Ž1.2. there exist n i - n j
such that n i s F n j s, hence 0 F Ž n j y n i . s.
If Ži. holds but Žii. and Žiii. do not hold, then g Ž0. q g Ž s . s 0, g Ž s . q
Ž
g 2 s . s 0, g Ž2 s . q g Ž3s . s 0, . . . , and g Žys . q g Žy2 s . s 0, g Žy2 s . q
g Žy3s . s 0, g Žy3s . q g Žy4 s . s 0, . . . . If g Ž0. / 0, then 0, s, 2 s, . . . g
suppŽ g .. Hence, as before, there exist n i - n j such that Ž n j y n i . s G 0. If
g Žys . / 0 then ys, y2 s, y3s, . . . g suppŽ g . and as before there exist
n i - n j such that Ž n j y n i . s F 0. This shows the order on S is subtotal.
Žc. R is a regular ring: Let a g R, a / 0 and let f : S ª R be given by
Ž
f 0. s a, f Ž s . s 0 for every s g S, s / 0. By hypothesis, there exists g g A
such that fgf s f. Then a s f Ž0. s Ž fgf .Ž0. s ag Ž0. a, so R is a regular
ring.
Žd. ŽPart 1. We show that there exists an artinian and narrow subset
T : S such that if M is a set of non-zero mutually orthogonal idempotents
of R, then < M < - < T <. If this is not true, for every artinian and narrow
subset T of S, there exists a set M as indicated, with < T < F < M <. By
hypothesis we may choose T to be infinite. There exists an injective map
u : T ª M. Let f : S ª R be defined by f Ž t . s u Ž t . for every t g T and
f Ž s . s 0 for every s g S _ T. Thus suppŽ f . s T, hence f g A. Since A is
regular, there exists g g A such that f s fgf, so g / 0. For every t g T
f Ž t. s
Ý
f Ž t9 . g Ž u . f Ž t0 . .
Ž t 9, u , t 0 .gX Ž t , f , g , f .
Then f Ž t . s f Ž t . f Ž t . f Ž t . s f Ž t .wÝ f Ž t9. g Ž u. f Ž t0 .x f Ž t .. Since f Ž t . f Ž t9. s
f Ž t0 . f Ž t . s 0 when t9 / t, t0 / t, and since f Ž t . / 0, Ž t, yt, t . g
X Ž t, f, g, f . and f Ž t . s f Ž t . g Žyt . f Ž t .. Thus yt < t g suppŽ f . s T 4 :
suppŽ g ., so yt < t g T 4 is artinian and narrow. Hence T is noetherian; but
T is artinian and narrow, so T is finite by Ž1.1., and this is a contradiction.
Žd. ŽPart 2. Every set M of non-zero mutually orthogonal idempotents
of R is finite: Indeed, if this is false, let M0 be an infinite set of non-zero
mutually orthogonal idempotents of R. By Zorn’s Lemma, M0 is contained
in a maximal set M of non-zero mutually orthogonal idempotents and M
335
GENERALIZED POWER SERIES
is infinite. By Žd. ŽPart 1., there exists an artinian and narrow subset T of
S such that < M < - < T <. Let w : M ª T be an injective map and let
f : S ª R be defined by f Ž w Ž a.. s a for all a g M, f Ž t . s 0 for all
t g S _ w Ž M .. Then suppŽ f . : T and so f g A.
We show that f is not a zero-divisor: Let fg s 0 with g g A, g / 0. Let
a0 g M be arbitrary and t 0 g suppŽ g .. Then there exists r g R such that
g Ž t 0 . rg Ž t 0 . s g Ž t 0 ., so g Ž t 0 . r is a non-zero idempotent. Also
0 s Ž fg . Ž w Ž a0 . q t 0 . s
f Ž s. g Ž t . .
Ý
Ž s, t .gX Ž w Ž a 0 .qt 0 , f , g .
Then 0 s a0 0 s a0 ŽÝ f Ž s . g Ž t .. s a0 a0 g Ž t 0 . s a0 g Ž t 0 .. Thus 0 s a0 g Ž t 0 . r
for every a0 g M. Then g Ž a0 . r f M and M j g Ž t 0 . r 4 is still a set of
non-zero mutually orthogonal idempotents of R, which is absurd, due to
the maximality of M.
Since A is regular and f is not a zero-divisor, then f is invertible. Let
h g A be such that e0 s fh. So
1 s e0 Ž 0. s
f Ž s . hŽ t . .
Ý
Ž s, t .gX Ž0, f , h .
The set X Ž0, f, h. is finite. Since M is infinite, there exists a g M _
f Ž s .<Ž s, t . g X Ž0, f, h.4 . Then
asa?1sa
ž
f Ž s . h Ž t . s 0.
Ý
/
Ž s, t .gX Ž0, f , h .
This is an absurdity which completes the proof of (1) « (2).
(2) « (3). By Ž3.3., R is a semisimple ring.
k
(3) « (4). We have R s Ł is1
R i where each R i is a simple ring.
R i s Mn i=n iŽ K i ., the ring of all n i = n i matrices with entries in a skewfield
K i . Then
k
As
ž /
Ł Ri
is1
S, F
k
s
Ł
is1
R iS , F
.
But ww Mn i=n iŽ K i . S, F xx ( Mn i=n i ww K iS, F xx, by Ž4.3.. Since K i is a skewfield, if
the order on the group S is total, by the theorem of Neumann, L i s ww K iS, F xx
is a skewfield. If the order on S is not total, then S is an abelian group, the
order on S is subtotal. By Ž4.5., L i s ww K iS, F xx is a skewfield. So A i s
k
Mn i=n iŽ L i . is a simple ring and A s Ł is1
A i is a semisimple ring.
(4) « (1). This is trivial.
We mention explicitly the corollary concerning Laurent series A s
RŽŽ X .. s ww R Z, F xx ŽF is the usual order on Z..
336
PAULO RIBENBOIM
(5.2) The following conditions are equi¨ alent:
(1) A is a regular ring.
(2) R is regular and e¨ ery set of non-zero mutually orthogonal idempotents of R is finite.
(3) R is a semisimple ring.
(4) A is a semisimple ring.
Thus, if K is a field, R s K N Žthe ring of sequences of elements in K .,
then A s RŽŽ X .. is not a regular ring.
6. SKEWFIELDS OF GENERALIZED POWER SERIES
In this final section we give a new proof of Neumann’s theorem Ž4.4.,
without applying Ž2.2.. We obtain as a corollary the theorem Ž4.5., using
Ž2.3..
(6.1) If Ž S, q, F. is a totally ordered group and R is a skewfield, then
A s ww R S, F xx is a skewfield.
Proof. Let f g A, f / 0, with p Ž f . s 0, f Ž0. s 1. We shall prove that
f has inverse in A. This suffices to show that A is a skewfield. Indeed, if
g g A, g / 0. Let p Ž g . s s, let h g A be given by hŽys . s g Ž s .y1 ,
hŽ t . s 0 for all t g S, t / ys. Then f s gh s hg is such that p Ž f . s 0,
f Ž0. s 1, so f is invertible, hence so is g.
We show that if f is not invertible, if l is an ordinal such that < l < ) < S <,
then there exists a family Ž ka .a - l where each ka g A, such that, assuming
fa s e y fka / 0, if p Ž fa . s sa , then sa - sb for all a - b - l. This is,
however, impossible because < l < ) < S <. The family Ž ka .a - l is defined by
transfinite induction.
Let k 0 s 0 so s0 s p Ž e y fk 0 . s 0. Let m be an ordinal, m - l. We
assume that ka has been defined for all a - m , so that if sa s p Ž e y fka .
then sa - sb when a - b - m.
First Case. m s n q 1. Let fn s e y fkn and let gnq1: S ª R be defined by gnq1Ž sn . s fn Ž sn ., gnq1Ž t . s 0 for all t g S, t / sn . Let knq1 s kn
q gnq1 and fnq1 s e y fknq1 s fn y fgnq1. So fnq1Ž sn . s 0, hence snq1 s
p Ž fnq1 . ) sn .
Second Case. m is a limit ordinal. We define km : S ª R as follows:
kmŽ u. s 0 if u G sa for all a - m ; kmŽ u. s ka Ž u.Ž u. where a Ž u. - m , a Ž u.
is the smallest ordinal such that m - sa Ž u. . We note that if a Ž u. - b - m
then ka Ž u.Ž u. s kb Ž u.. Indeed, p Ž ka Ž u. y kb . s p Ž f Ž ka Ž u. y kb .. s p ŽŽ e
y fkb . y Ž e y kfa Ž u. .. s sa Ž u. since sa Ž u. - sb . Since u - sa Ž u. , ka Ž u.Ž u. s
kb Ž u..
GENERALIZED POWER SERIES
337
Next we show that suppŽ km . is well-ordered, so km g A. Let U :
suppŽ km ., U / B. If u g U, there exists a Ž u. - m such that kmŽ u. s
ka Ž u.Ž u.. We consider the non-empty set of ordinals a Ž u.< u g U 4 . Let a 1
be its smallest element. Then U l suppŽ ka 1 . / B, so it is a well-ordered
set and let u1 be its smallest element. Now we show that u1 F u for every
u g U. If u g U and a Ž u. s a 1 , then u g U l suppŽ ka 1 ., so u1 F u. If
u g U and a 1 - a Ž u., then u1 - sa 1 F u.
Let fm s e y fkm , sm s p Ž fm .. We show that sa F sm for all a - m ; then
sa - sm for all a - m. If not, there exists a - m such that sm - sa . Hence
p Ž ka y km . s p Ž f Ž ka y km .. s p Ž fm y ka . s sm . But kmŽ sm . s ka Ž sm .Ž sm .
s kb Ž sm . for all b G a Ž sm .. Then a - a Ž sm .. So sm - sa - sa Ž sm . . But by
definition a Ž sm . F a , hence sa Ž sm . F sa and this is a contradiction.
This concludes the transfinite induction leading to the family Ž ka .a - l
with sa - sb for all a - b - l, which is impossible, thereby proving the
theorem.
Now we have:
(6.2) Let Ž S, q, F. be a torsion-free abelian group with F subtotal and
let R be a skewfield. Then A is a skewfield.
Proof. Let F9 be the total order on S associated to the subtotal order
as indicated in Section 2. Let A9 s ww R S, F9 xx. By Ž6.1., A9 is a skewfield. As
noted in Section 4, A s ww R S, F xx is a subring of A9. We show that A is a
skewfield. Let f g A, f / 0 with suppŽ f . : s g S <0 F9s4 and f Ž0. s 1. So
there exists k g A9 such that fk s kf s e. Then suppŽ k . is well-ordered
Žwith respect to F9. and contained in s g S <0 F9s4 , and k Ž0. s 1. We
show that suppŽ k . is artinian and narrow Žwith respect to F., so k g A.
For this purpose we prove that suppŽ k . : ²suppŽ f . _ 04:. If this is not
true, there exists s g suppŽ k ., smallest Žwith respect to F9. such that
s f ²suppŽ f . _ 04:. We have
0 s e Ž s . s fk Ž s . s f Ž 0 . k Ž s . q
Ý f Ž t. kŽ s y t. .
0-tFs
So
0 / k Ž s. s y
Ý f Ž t. kŽ s y t.
0-tFs
and there exists t, 0 - t F s, such that f Ž t . / 0, k Ž s y t . / 0. So t g
suppŽ s . _ 04 , s y t - s, and s y t g suppŽ k ., thus s y t g ²suppŽ f . _
04:, hence s g ²suppŽ f . _ 04: which is a contradiction. By Ž2.3., ²suppŽ f .
_ 04: is artinian and narrow, hence suppŽ k . is artinian and narrow, so
k g A and f is invertible in A. It follows at once that every g g A, g / 0,
is invertible in A, so A is a skewfield.
338
PAULO RIBENBOIM
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