Table Of Content3
9
9 Explicitly Non-Standard Uniserial Modules
1
n
a
J P. C. Eklof S. Shelah
5 University of California, Irvine∗ Hebrew University and
1
Rutgers University†
]
O
L February 1, 2008
.
h
t
a
m
Abstract
[
1 A new construction is given of non-standard uniserial modules over certain
v valuation domains; the construction resembles that of a special Aronszajn tree in
7
set theory. A consequence is the proof of a sufficient condition for the existence of
0
2 non-standard uniserial modules; this is a theorem of ZFC which complements an
1
earlier independence result.
0
3
9
/ Introduction
h
t
a
m This paper is a sequel to [ESh]. Both papers deal with the existence of non-standard
: uniserial modules over valuation domains; we refer to [ESh] for history and motivation.
v
i While the main result of the previous paper was an independence result, the main results
X
of this one are theorems of ZFC, which complement and extend the results of [ESh].
r
a We are interested in necessary and sufficient conditions for a valuation domain R
to have the property that there is a non-standard uniserial R-module of a given type
J/R. (Precise definitions are given below.) The question is interesting only when R is
uncountable, and since additional complications arise for higher cardinals, we confine
ourselves to rings of cardinality ℵ . Associated to any type J/R is an invariant, denoted
1
Γ(J/R), which is a member of a Boolean algebra D(ω ) (equal to P(ω ) modulo the filter
1 1
of closed unbounded sets). For example, if R is an almost maximal valuation domain,
then Γ(J/R) = 0 for all types J/R; but there are natural and easily defined examples
where Γ(J/R) = 1.
∗ThankstoRutgersUniversityforitssupportofthisresearchthroughitsfundingofthefirstauthor’s
visits to Rutgers.
†Partially supported by Basic Research Fund, Israeli Academy of Sciences. Pub. No. 461
1
2
It is a fact that
if Γ(J/R) = 0, then there is no non-standard uniserial R-module of type J/R
(cf. [ESh, Lemma 5]). In [ESh] we showed that the converse is independent of ZFC
+ GCH; the consistency proof that the converse fails involved the construction of a
valuation domain R associated with a stationary and co-stationary subset of ω — that
1
is, 0 < Γ(J/R) < 1. The existence of such sets requires a use of the Axiom of Choice;
no such set can be explicitly given. Thus — without attempting to give a mathematical
definition of “natural” — we could say that for natural valuation domains, R, it is the
case that for every type J/R, Γ(J/R) is either 0 or 1. For natural valuation domains,
it turns out that the converse is true: if there is no non-standard uniserial R-module of
type J/R, then Γ(J/R) = 0. This is a consequence of the following result which is proved
below (for all valuation domains of cardinality ℵ ):
1
if Γ(J/R) = 1, then there is a non-standard uniserial R-module of type J/R.
(Theorem 12.) This vindicates a conjecture made by Barbara Osofsky in [O1, (9), p.
164]. (See also the Remark following Theorem 12.)
The proof of Theorem 12 divides into several cases; the key new result which is
used is a construction of a non-standard uniserial module in the essentially countable
case; this construction is done in ZFC and is motivated by the construction of a special
Aronszajn tree. (See Theorem 7.) Moreover, the uniserial constructed is “explicitly
non-standard” in that there is an associated “special function” which demonstrates that
it is non-standard. This special function continues to serve the same purpose in any
extension of the universe, V, of set theory, so the module is “absolutely” non-standard.
In contrast, this may not be the case with non-standard uniserials constructed using a
prediction (diamond) principle. (See the last section.)
The first author would like to thank L. Salce and S. Bazzoni for their critical reading
of a draft of this paper, and A. Kechris for a helpful conversation on absoluteness.
Preliminaries
For any ring R, we will use R∗ to denote the group of units of R. If r ∈ R we will write
x ≡ y (mod r) to mean x−y ∈ rR.
A module is called uniserial if its submodules are linearly ordered by inclusion. An
integral domain R is called a valuation domain if it is a uniserial R-module. If R is a
valuation domain, let Q denote its quotient field; we assume Q 6= R. The residue field
3
of R is R/P, where P is the maximal ideal of R. [FS] is a general reference for modules
over valuation domains.
If J and A are R-submodules of Q with A ⊆ J, then J/A is a uniserial R-module,
which is said to be standard. A uniserial R-module U is said to be non-standard if it is
not isomorphic to a standard uniserial.
Given a uniserial module U, and a non-zero element, a, of U, let Ann(a) = {r ∈
R:ra = 0} and let D(a) = ∪{r−1R:r divides a in U}. We say U is of type J/A if
∼
J/A = D(a)/Ann(a). This is well-defined in that if b is another non-zero element of U,
∼
then D(a)/Ann(a) = D(b)/Ann(b). For example, U has type Q/R if and only if U is
divisible torsion and the annihilator ideal of every non-zero element of U is principal.
(But notice that there is no a ∈ U with Ann(a) = R.) It is not hard to see that if U has
type J/A, then U is standard if and only if it is isomorphic to J/A. We will only consider
types of the form J/R; it is a consequence of results of [BFS] that the question of the
existence of a non-standard uniserial R-module of type J /A can always be reduced to
1
the question of the existence of a non-standard uniserial of type J/R for an appropriate
J.
¿From now on we will assume that R has cardinality ℵ . We always have
1
(∗) J = ∪ r−1R
σ<ω1 σ
forsomesequence ofelements {r :σ < ω }such thatforallτ < σ, r |r . IfJ iscountably
σ 1 τ σ
generated, then U is standard, so generally we will be assuming that J is not countably
generated; then it has a set of generators as in (∗), where, furthermore, r does not divide
σ
r if τ < σ.
τ
If δ ∈ lim(ω ), let
1
(∗∗) J d=ef∪ r−1R.
δ σ<δ σ
By results in [BS] every uniserial module U, of type J/R, is described up to isomor-
phism by a family of units, {eτ:σ < τ < ω } such that
σ 1
(†) eδeτ ≡ eδ (mod r )
τ σ σ σ
for all σ < τ < δ < ω . Indeed, U is a direct limit of submodules a R where Ann(a ) =
1 σ σ
r R; then a R ∼= r−1R/R and U is isomorphic to a direct limit of the r−1R/R, where
σ σ σ σ
the morphism from r−1R/R to r−1R/R takes r−1 to eτr−1 if a = eτr−1r a .
σ τ σ σ σ σ σ σ τ τ
If U is given by (†), then U is standard if and only if there exists a family {c :σ < ω }
σ 1
of units of R such that
(††) c ≡ eτc (mod r )
τ σ σ σ
4
for all σ < τ < ω . Indeed, if the family {c :σ < ω } satisfying (††) exists, then mul-
1 σ 1
tiplication by the c gives rise to isomorphisms from r−1R/R to a R, which induce an
σ σ σ
isomorphism of J/R with U.
Essentially Countable Types
Definition. SupposeJ = ∪ r−1Rasin(*). CallthetypeJ/Ressentiallyuncountable
σ<ω1 σ
if for every σ < ω there exists τ > σ such that r R/r R is uncountable. Otherwise,
1 σ τ
J/R is essentially countable; this is equivalent to saying that there is a γ < ω such that
1
for all γ < σ < ω , r R/r R is countable. Say that J/R is strongly countable if for all
1 γ σ
σ < ω , R/r R is countable; clearly, a strongly countable type is essentially countable.
1 σ
It is easily seen that the notions of being essentially or strongly countable are well-
defined, that is, independent of the choice of the representation (*). If the residue field
of R is uncountable, then, except in trivial cases, the types J/R have to be essentially
uncountable; but if the residue field is countable, the question is more delicate.
Proposition 1 If the residue field of R is uncountable, then every type J/R such that
J is not countably generated is essentially uncountable.
Proof. Let J = ∪ r−1R as in (*). It suffices to prove that if σ < τ, then r R/r R
σ<ω1 σ σ τ
is uncountable. But r R/r R ∼= R/tR where t = r r−1 ∈ P. So we have tR ⊆ P ⊆ R,
σ τ τ σ
∼
and hence (R/tR)/(P/tR) = R/P, the residue field of R. Since R/P is uncountable, so
is R/tR. 2
Theorem 2 For any countable field K there are valuation domains R and R , both of
1 2
cardinality ℵ with the same residue field K and the same value group, whose quotient
1
fields, Q and Q , respectively, are generated by ℵ but not countably many elements, and
1 2 1
such that Q /R is essentially uncountable and Q /R is strongly countable.
1 1 2 2
Proof. Let G be the ordered abelian group which is the direct sum ⊕ Zα ordered
α<ω1
anti-lexicographically; that is, Σ n α > 0 if and only if n > 0, where β is maximal such
α α β
that n 6= 0. In particular, the basis elements have their natural order and if α < β, then
β
kα < β in G for any k ∈ Z. Let G+ = {g ∈ G:g ≥ 0}.
Let Rˆ = K[[G]], that is, Rˆ = { k Xg:k ∈ K,∆ a well-ordered subset of G+},
Pg∈∆ g g
with the obvious addition and multiplication (cf. [O1, p. 156]). Given an element y =
k Xg of Rˆ, let supp(y) = {g ∈ ∆:k 6= 0}; let p-supp(y) = {α ∈ ω :∃g ∈ supp(y)
Pg∈∆ g g 1
whose projection on Zα is non-zero}. Define v(y) = the least element of supp(y). If
X ⊆ G, then y|X is defined to be k Xg. Let y|ν = y|{g ∈ G+:g < ν}.
Pg∈X∩∆ g
5
Let R = {y ∈ Rˆ:p-supp(y) is finite}. Then R is a valuation domain since p-
1 1
supp(xy−1) ⊆ p-supp(x) ∪ p-supp(y). Let R be the valuation subring of R generated
2 1
by {Xg : g ∈ G}. We have Q = ∪ X−αR for j = 1, 2. Now Q /R is essentially
j α<ω1 j 1 1
uncountable since for all β > α, XαR /XβR contains the 2ℵ0 elements of the form
1 1
Xζ(n)X(n+1)α
n∈ω
(with p-supp = {α}) where ζ is any function: ω → 2. R has cardinality 2ℵ0; if 2ℵ0 > ℵ ,
1 1
to get an example of cardinality ℵ , choose a valuation subring of R which contains all
1 1
the monomials Xg (g ∈ G) and ℵ of the elements ζ(n)X(n+1)α for each α.
1 Pn∈ω
We claimthatQ /R isessentially countable. LetK[G] bethesubring of Rˆ generated
2 2
by {Xg:g ∈ G+}; thus K[G] consists of the elements of Rˆ with finite support; we shall
refer to them as polynomials. R consists of all elements of the form xy−1 where x
2
and y are polynomials and v(x) ≥ v(y). We claim that R /XβR is countable for any
2 2
β < ω . There are uncountably many polynomials, but we have to show that there are
1
only countably many truncations xy−1|β.
Given polynomials x and y with v(x) ≥ v(y), there is a finitely generated subgroup
⊕ Zσ of G (with σ < σ < ... < σ ) such that x and y are linear combinations of
1≤i≤d i 1 2 d
monomialsXg withg ∈ ⊕ Zσ . Moreprecisely, thereexistk,r ∈ ω anda(k+r)-tuple
1≤i≤d i
(a , ...,a ) of elements of K and k +r linear terms t of the form
1 k+r j
d
tj = Xnijvi
i=1
(n ∈ Z, v variables) such that if we let t (σ) denote d n σ , then
ij i j Pi=1 ij i
(♮) x = k a Xtj(σ) and y = r a Xtj(σ).
Pj=1 j Pj=k+1 j
Finally, there is q ≤ d such that σ is maximal with σ < β.
q q
Now,considerationofthealgorithmforcomputingxy−1 showsthat,forfixed(a ,...,a )
1 k+r
and t , there are linear terms
j
d
sℓ = Xmiℓvi
i=1
(m ∈ Z, ℓ ∈ ω) and elements c ∈ K such that for any strictly increasing sequence
iℓ ℓ
σ = hσ ,...,σ i, if x and y are as in (♮), then
1 d
xy−1 = XcℓXsℓ(σ).
ℓ∈ω
6
For any q ≤ d, only certain of the s involve only variables v with i ≤ q (i.e. m = 0 if
ℓ i iℓ
i > q); say these are the s with ℓ ∈ T (T ⊆ ω). If σ is such that σ < β iff i ≤ q, then
ℓ i
xy−1|β = c Xsℓ(σ).
Pℓ∈T ℓ
There are only countably many choices for q, d, k and r in ω, (a , ..., a ) ∈ Kk+r,
1 k+r
and for σ < ... < σ < β. Therefore, there are only countably many possibilities for the
1 q
truncations xy−1|β. 2
By the first part of the following, the type Q /R of the previous theorem must be
2 2
strongly countable; on the other hand, there are types which are essentially countable
but not strongly countable.
Proposition 3 (i) If Q/R is essentially countable, then it is strongly countable.
(ii) For any countable field K, there is a valuation domain R with residue field K
which has a type J/R which is essentially countable but not strongly countable.
Proof. (i) Since Q/R is essentially countable, we can write Q = ∪ r−1R such that
σ<ω1 σ
for all σ < τ, r R/r R is countable. We claim that R/r R is countable, which clearly
σ τ 0
is equivalent to R/r R countable for all τ < ω . Suppose not. There is a σ < ω such
τ 1 1
that r = r2t for some t ∈ R (since r−2 ∈ Q). But then r R/r R ∼= R/tr R, which is
σ 0 0 0 σ 0
uncountable since R/r R is uncountable, and this contradicts the choice of the r .
0 σ
(ii) Let G = ⊕ Zα, ordered anti-lexicographically. Let Rˆ = K[[G]] (cf. proof
α≤ω1
ˆ
of Theorem 2), and let R be the smallest valuation subring of R containing all the
monomials Xg (g ∈ G). Let J = ∪ r−1R where r = Xα+ω1. Then the proof that
α<ω1 α α
r R/r R ∼= R/Xβ−αR is countable for all α < β is the same as in Theorem 2. But
α β
R/r R = R/Xω1R is clearly uncountable. 2
0
Remark. More generally, referring to a dichotomy in [O1, Prop. 7, p. 155], if the type
J/R is essentially countable and falls into Case (A), then J/R is strongly countable; if it
falls into case (B), then it is not strongly countable.
Gamma Invariants
A subset C of ω is called a cub — short for closed unbounded set — if supC = ω
1 1
and for all Y ⊆ C, supY ∈ ω implies supY ∈ C. Call two subsets, S and S , of ω
1 1 2 1
equivalent iff there is a cub C such that S ∩C = S ∩C. Let S˜ denote the equivalence
1 2
class of S. The inclusion relation induces a partial order on the set, D(ω ), of equivalence
1
˜ ˜
classes, i.e., S ≤ S if and only if there is a cub C such that S ∩C ⊆ S ∩C. In fact,
1 2 1 2
this induces a Boolean algebra structure on D(ω ), with least element, 0, the equivalence
1
class of sets disjoint from a cub; and greatest element, 1, the equivalence class of sets
7
containing a cub. We say S is stationary if S˜ 6= 0, i.e., for every cub C, C ∩S 6= ∅. We
say S is co-stationary if ω \S is stationary.
1
˜
Given R and a type J/R, where J is as in (*), define Γ(J/R) to be S, where
S = {δ ∈ lim(ω ):R/∩ r R is not complete}
1 σ<δ σ
where the topology on R/ ∩ r R is the metrizable linear topology with a basis of
σ<δ σ
neighborhoods of 0 given by the submodules r R (σ < δ). This definition is independent
σ
of the choice of the representation of J as in (*) — see [ESh].
For any limit ordinal δ < ω , let
1
Tδ = {hu :σ < δi:∀σ < τ < δ(u ∈ R∗, and u −u ∈ r R)};
J/R σ σ τ σ σ
that is, T δ consists of sequences of units which are Cauchy in the metrizable topology
J/R
on R/∩ r R. Let Lδ consist of those members of T δ which have limits in R, i.e.
σ<δ σ J/R J/R
Lδ = {hu :σ < δi ∈ T δ :∃u ∈ R∗ s.t. ∀σ < δ(u −u ∈ r R)}.
J/R σ J/R δ δ σ σ
Note that Γ(J/R) = S˜ where
S = {δ ∈ lim(ω ):T δ 6= Lδ }.
1 J/R J/R
If J is not countably generated, then ¬CH implies that Γ(J/R) = 1, since the com-
pletion of R/∩ r R has cardinality 2ℵ0 > ℵ . An ω -filtration of R by subrings is an
σ<δ σ 1 1
increasing chain {N :α ∈ ω } of countable subrings of R such that R = ∪ N , and
α 1 α∈ω1 α
for limit α, N = ∪ N .
α β<α β
Define Γ′(J/R) = E˜′ where
E′ = {δ ∈ lim(ω ):∃hu :σ < δi ∈ T δ s.t. ∀f ∈ R∗∃σ < δ s.t.
1 σ J/R
u f ∈/ N (mod r )}.
σ δ σ
Again, it can be shown that the definition does not depend on the choice of {r :ν <
ν
ω } or of {N :α < ω }. Notice that Γ′(J/R) ≤ Γ(J/R) since if T δ = Lδ , then we
1 α 1 J/R J/R
can let f be a limit of hu−1:σ < δi.
σ
In [ESh, Theorem 7] it is proved that if Γ′(J/R) 6= 0, then there is a non-standard
uniserial R-module.
Theorem 4 Suppose J/R is essentially countable. Then
(i) Γ′(J/R) = 0;
(ii) Γ(J/R) = 1.
8
Proof. Without loss of generality we can assume that J = ∪ r−1R where r R/r R
σ<ω1 σ 0 σ
is countable for all σ < ω .
1
(i) We can also assume that the ω -filtration of R by subrings, R = ∪ N , has the
1 α<ω1 α
property that for all α, N contains a complete set of representatives of r R/r R for each
α 0 σ
σ < α. For any δ ∈ lim(ω ), and any hu : σ < δi in T δ , let f = u−1. To show that
1 σ J/R 0
δ ∈/ E′, it suffices to show that u f ∈ N (mod r ) for all σ < δ. Now u f = u u−1 ≡ 1
σ δ σ σ σ 0
(mod r ), since u ≡ u (mod r ), by definition of Tδ . Say u u−1−1 = y ∈ r R. By the
0 σ 0 0 J/R σ 0 0
assumption on N , there exists a ∈ N such that y ≡ a (mod r ). Then u f = u u−1 =
δ δ σ σ σ 0
1+y ≡ 1+a (mod r ), and 1+a ∈ N since N is a subring of R.
σ δ δ
(ii) To show that Γ(J/R) = 1, it suffices to show that for all limit ordinals δ < ω ,
1
R/∩ r Risnotcomplete. Assuming thatitiscomplete, weshallobtainacontradiction
ν<δ ν
by showing that r R/r R is uncountable. Fix a ladder on δ, i.e., a strictly increasing
0 δ
sequence hν : n ∈ ωi whose sup is δ. For each function ζ:ω → 2, define uζ = huζ : σ <
n σ
δi ∈ T δ as follows: if ν < σ ≤ ν , then
J/R m m+1
uσ = Xζ(i)rνi.
i≤m
Clearly u ∈ r R, and if τ > σ, where ν < τ ≤ ν , then m ≤ k and
σ 0 k k+1
k
uτ −uσ = X ζ(i)rνi ∈ rνm+1R ⊆ rσR.
i=m+1
Since R/∩ r R is assumed to be complete, for each ζ there is an element uζ ∈ R which
ν<δ ν ∗
represents the limit of huζ : σ < δi in R/∩ r R. To obtain a contradiction, we need
σ ν<δ ν
only show that if η 6= ζ, then uζ −uη ∈/ r R. Without loss of generality there exists m
∗ ∗ δ
such that ζ|m = η|m and ζ(m) = 0, η(m) = 1, then
uη −uη ∈ r R; and
∗ νm+1 νm+1
uζ −uζ ∈ r R;
∗ νm+1 νm+1
but uη −uζ = r ∈/ r R, so uζ −uη ∈/ r R ⊇ r R. 2
νm+1 νm+1 νm νm+1 ∗ ∗ νm+1 δ
Special Aronszajn trees
This section contains standard material from set theory. (See, for example, [J, §22] or
[Dr, Ch. 7, §3].) It is included simply to provide motivation for the notation and proof
in the next section.
9
A tree is a partially ordered set (T, <) such that the predecessors of any element
are well ordered. An element x of T is said to have height α, denoted ht(x) = α, if the
order-type of {y ∈ T:y < x} is α. The height of T is defined to be sup{ht(x)+1:x ∈ T}.
If T is a tree, a branch of T is a maximal linearly ordered initial subset of T; the length
of a branch is its order type. If T is a tree, let T = {y ∈ T : ht(y) = α}. We say that a
α
tree T is a κ-Aronszajn tree if T is of height κ, |T | < κ for every α < κ, and T has no
α
branch of length κ.
A tree T of height ω is a special Aronszajn tree if T is countable for all α < ω and
1 α 1
for each α < ω there is a function f :T → Q such that
1 α α
whenever x ∈ T and y ∈ T and x < y, then
α β
($)
f (x) < f (y).
α β
Notice thata special Aronszajn treeis anω -Aronszajntree, since anuncountable branch
1
would give rise to an uncountable increasing sequence of rationals.
K¨onig’s Lemma implies that there is no ω-Aronszajn tree. However, there is an
ω -Aronszajn tree:
1
Theorem 5 There is a special Aronszajn tree.
Proof. Let <αω denote the set of all functions from {β ∈ ω :β < α} to ω. We shall
1
construct T and f by induction on α < ω such that T is a countable subset of <αω
α α 1 α
and the partial ordering is inclusion, i.e., if x ∈ T and y ∈ T then x < y if and only if
α β
α < β and y|α = x. Finally, T will be defined to be T .
Sα<ω1 α
Let T = {∅}, f (∅) = 0, T = {0}ω, and f : T → Q be onto (0,∞). Suppose now
0 0 1 1 1
that T and f have been defined for all α < δ such that for all σ < ρ < δ:
α α
for any ǫ > 0, and x ∈ T there is y ∈ T such that x < y
σ ρ
(⋆)
and f (y) < f (x)+ǫ.
ρ σ
There are two cases. In the first case, if δ is a successor ordinal, δ = τ +1, let
T = {x∪{(τ, n)} : n ∈ ω, x ∈ T }.
δ τ
Define f so that for every x ∈ T ,
δ τ
{f (x∪{(τ, n)}) : n ∈ ω} = {r ∈ Q : r > f (x)}.
δ τ
Clearly (⋆) continues to hold.
In the second case, δ is a limit ordinal. Choose a ladder hν : n ∈ ωi on δ. For each
n
σ < δ, x ∈ T and k > 0, by inductive hypothesis (⋆) there exists a sequence hy : n ∈ ω
σ n
s.t. ν > σisuchthaty ∈ T , x < y < y forn < mandf (y ) < f (x)+(1/k−1/n).
n n νn n m νn n σ
Let y[σ, x, k] = y ∈ <δω . Let T consist of one such y[σ, x, k] for each σ, x, k.
Sn∈ω n δ
Define f (y[σ, x, k]) = f (x)+1/k. Then it is clear that (⋆) still holds. 2
δ σ
10
Special Uniserial Modules
Definition. Suppose U is a uniserial module of type J/R where J = ∪ r−1R as
σ<ω1 σ
in (*). For each σ > ω , fix an element a of U such that Ann(a ) = r R (so that
1 σ σ σ
the submodule a R of U is isomorphic to R/r R ∼= r−1R/R). Let I be the set of all
σ σ σ σ
R-module isomorphisms ϕ:a R → r−1R/R. We say that {f :σ ∈ ω } is a special family
σ σ σ 1
of functions for U if for each σ < ω , f : I → Q such that whenever σ < ρ and ϕ ∈ I
1 σ σ ρ
extends ψ ∈ I , then f (ψ) < f (ϕ).
σ σ ρ
Lemma 6 If U has a special family of functions, then U is non-standard.
Proof. Supposethereisanisomorphismθ:U → J/R. Thenforevery σ < ω , θ restricts
1
to an isomorphism ϕ of a R onto r−1R/R. But then hf (ϕ ):σ < ω i is an uncountable
σ σ σ σ σ 1
2
strictly increasing sequence of rationals, a contradiction.
With this lemma as justification, we will say that U is explicitly non-standard if U
has a special family of functions.
If the uniserial module uniserial module U, of type J/R, is described up to isomor-
phism by a family of units, {eρ:σ < ρ < ω } as in (†), then it is clear that U is explicitly
σ 1
non-standard if and only if for every σ < ω , there is a function
1
f :(R/r R)∗ −→ Q
σ σ
such that
whenever σ < ρ and c ,c ∈ R∗ satisfy c ≡ c eρ
($$) σ ρ ρ σ σ
(mod r ), then f (c ) < f (c ).
σ σ σ ρ ρ
(Here, and hereafter, we abuse notation and regard f and f as functions on R∗.)
ρ σ
Note that we have a tree, T, such that T = (R/r R)∗ and the partial ordering is
σ σ
given by:
c +r R < c +r R ⇐⇒ σ < ρ and c ≡ c eρ (mod r ).
σ σ ρ ρ ρ σ σ σ
Assume σ < ρ. Each c has at least one successor of height ρ, namely c eρ, and if
σ σ σ
r R/r R is countable, then c has only countably many successors of height ρ. For each
σ ρ σ
c ∈ T , its unique predecessor in T is c (eρ)−1. (Here again we abuse notation and
ρ ρ σ ρ σ
write, for example, c for an element of T instead of c +r R.)
ρ ρ ρ ρ
Without loss of generality we can assume that the r are such that for all σ < ω ,
σ 1
r R/r R is infinite. (Just choose a subsequence of the original r ’s if necessary.) Thus
σ σ+1 σ
for all σ < ω , there is an infinite subset W of R∗ such that for all u 6= v ∈ W , u ≡ 1
1 σ σ
(mod r ) and u 6≡ v (mod r ).
σ σ+1
