FACTORIAL PRESERVATION
YOUSSEF FARES
Abstract. Let A be a Dedekind domain with finite residue fields
and with a finite unit group. Let S be a subset of A and f be a polynomial with coefficients in the quotient field of A. We show that
if the subsets S and f (S) have the same factorials (in Bhargava’s
sense), then f is of degree 1. In particular, we answer Gilmer and
Smith’s question [10] (when A=Z): if S and f (S) are polynomially
equivalent (in McQuillan’s sense), then f is of degree 1.
1. Introduction
Let S be a subset of Z and f be a un polynomial with rational
coefficients. About the behavior of f with regard to S, let us consider
the following three equalities:
(1) f (S) = S
(2) Int(S, Z) = Int(f (S), Z)
(3) (n!)S = (n!)f (S) for all n ∈ N
These three equalities are connected; more precisely, the first implies
the second which implies the third. Concerning the first equality, one
knows that if f (S) = S then f is of degree 1 [13, chap. IX]. For the
second equality, let us recall that Int(S, Z) denotes the set of integervalued polynomials over S, that is
Int(S, Z) = {g ∈ Q[X] | g(S) ⊆ Z}.
Gilmer and Smith [10, Thm. 4.2] have proved that, when S is dense in
Z for almost all p-adic topologies, then f is of degree 1. They ask the
question to know, whether the polynomial f shoud not be of degree 1
whatever the infinite subset S of Z. About the third equality, recall
that (n!)S denotes the n-th factorial of S defined as the GCD of the
leading coefficients of polynomials of Int(S, Z) of degree ≤ n. Let us
also remind that Bhargava [1], [2] and [3] introduced the notion of
factorials of a subset S and
Q that one has:
GCD{ 0≤i<j≤n (si − sj ) | s0 , · · · , sn ∈ S}
Q
.
(n!)S =
GCD{ 0≤i<j<n (si − sj ) | s0 , · · · , sn−1 ∈ S}
The aim of this paper is to show that, without any hypothesis on S,
the equality of the factorials of S and f (S) implies that f is of degree 1. As corollaries, we shall answer by the affirmative to Gilmer and
1991 Mathematics Subject Classification. 11B65, 13F05, 13F20.
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2
YOUSSEF FARES
Smith’s question [10]. In fact, we are going to treat this problem in a
more general way by working in Dedekind domains. But let us begin
by reminding some general definitions.
Notation. Let A be a domain with quotient field K and let S be
a subset of A. We denote by Int(S, A) the A-module formed by the
integer-valued polynomials over S, that is,
Int(S, A) = {P ∈ K[X] | P (S) ⊂ A}.
We denote by Intn (S, A) the sub-A-module of Int(S, A) formed by
polynomials with degree ≤ n. We call n-th factorial of S relatively to
A and note (n!)A
S the ideal of A defined by:
(n!)A
S = {y ∈ A | yInt(S, A) ⊆ A[X]}.
When A is a Dedekind domain, the previous notions behave well by
localization: for every maximal ideal M of A, one has [6, Prop I.2.7]:
Int(S, A)M = Int(S, AM ),
and
M
(n!)A
= (n!)A
S AM
S
so,
(n!)A
S =
\
M
(n!)A
S .
M∈M ax(A)
We are thus going to begin with a local study in a valuation domain
and recall in that case the notion of v-ordering sequence introduced by
Bhargava [1].
2. v-ORDERINGS
Although the localizations of Dedekind domains are discrete valuation domains, we consider here valuations of rank one, that is, valuations whose value groups are subgroups of R. In this section, V denotes
the ring of a rank-one valuation v and S denotes some subset of V .
Definition. Let N ∈ N ∪ {∞}. A sequence {an }0≤n≤N of elements of
S is a v-ordering of lengh N if, for every 0 ≤ n ≤ N , one has:
v(
n−1
Y
k=0
(an − ak )) ≤ v(
n−1
Y
(x − ak )) for all x ∈ S.
k=0
If v is discrete or if S is a finite subset, there always exist v-ordering
of S. The following propositions are due to Bhargava [1] in the case of
a discrete valuation. They also hold for rank-one valuations [9].
Proposition 1. The sequence {an }0≤n≤N is a v-ordering of S if and
Q
X−ak
only if the polynomials fn (X) = n−1
k=0 an −ak (0 ≤ n ≤ N ) form a basis
of the V -module IntN (S, V ).
FACTORIAL PRESERVATION
3
Both following propositions are straightforward consequences of the
previous one.
Proposition 2. Let {an }0≤n≤N be a v-ordering of S. Then, for every
0 ≤ n ≤ N,
n−1
Y
wS (n) = v( (an − ak ))
k=0
does not depend on the v-ordering {an }0≤n≤N of S.
Proposition 3. Let {an }0≤n≤N be a v-ordering of S. Then,
(n!)VS
=
n−1
Y
(an − ak )V
k=0
for all 0 ≤ n ≤ N .
Remarks.
(1) If S is finite of cardinality s, a v-ordering a0 , a1 , · · · , as−1 of
lengh s − 1 is formed by all elements of S and, as soon as n ≥ s,
wS (n) = +∞ and (n!)VS = V.
(2) If S is an infinite precompact subset of V , then there are vorderings of S of any lengh [7, Lemma 4.5].
(3) If S ⊂ T , then wT (n) ≤ wS (n) for every n because Int(T, V ) ⊂
Int(S, V ).
(4) If T = {as + b | s ∈ S} where a et b are two elements of A, then
(n!)VT = an (n!)VS for every n ∈ N. In particular, if a is inversible
in A, then S and T have the same sequence of factorials, that
is : (n!)VT = (n!)VS for all n ∈ N.
Here is another characterization of v-orderings given by Bhargava [2]
in the case of subsets of Z.
Proposition 4. Let N ∈ N. The sequence {an }0≤n≤N of elements
of S is v-ordering if and only if for every 0 ≤ n ≤ N and every
x1 , · · · , x n ∈ S
Y
Y
v(
(ai − aj )) ≤ v(
(xi − xj )).
0≤i<j≤n
0≤i<j≤n
Proof. Let us suppose that for every 0 ≤ n ≤ N and for every x1 , · · · , xn ∈
S one has:
Y
Y
v(
(ai − aj )) ≤ v(
(xi − xj )).
0≤i<j≤n
0≤i<j≤n
We are going to prove by induction that the sequence {an }0≤n≤N is a
v-ordering of S. For n = 1, this is obvious. Let 1 ≤ n ≤ N −1. Assume
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YOUSSEF FARES
that a0 , · · · , an is a v-ordering of S. According to the hypothesis, one
has, for x = xn+1 ∈ S and x0 = a0 · · · , xn = an :
!
!
à n
à n
Y
Y
Y
Y
(x − ak )
(ai − aj ) .
(an+1 − ak )
(ai − aj ) ≤ v
v
k=0
0≤i<j≤n
k=0
0≤i<j≤n
Qn
Qn
We deduce that v( k=0 (an+1 −ak ) ≤ v( k=0 (x−ak )). Thus, a0 , . . . , an+1
is a v-ordering of S.
Conversely, let us suppose that {an }0≤n≤N is a v-ordering of S. Let
0 ≤ n ≤ N and x0 , · · · , xn ∈ S. Let T = {x0 , · · · , xn }. By reordering
the elements of T , one may asssume that x0 , · · · , xn is a v-ordering of
T . By Proposition 2:
n
n
Y
X
Y
X
v(
(xi − xj )) =
wT (k) and v(
(ai − aj )) =
wS (k).
0≤i<j≤n
k=1
0≤i<j≤n
k=1
Now T ⊂ S, thus wS (k) ≤ wT (k) for every 0 ≤ k ≤ n. Consequently,
Y
Y
v(
(ai − aj )) ≤ v(
(xi − xj )).
0≤i<j≤n
0≤i<j≤n
¤
Recall that a subset S of V is said precompact if the topological
closure of S in the completion of V is compact. One knows that S
is precompact if and only if, for any ideal I of V, S meets at most a
finite number of classes modulo I. In such a case the set {v(x − a); x ∈
S} ∩ [0; n] is finite for every n and for every a ∈ S.
Proposition 5. If S is any infinite precompact subset of V then, every
v-ordering sequence of S is dense in S.
Remark. When V is a discrete valuation domain with finite residue
field, Proposition 5 is a consequence of the p-adic Stone-Weierstrass
theorem given by Bhargava and Kedlaya [4]: Int(S, V ) is dense in the
ring C(Ŝ, V̂ ) of continuous functions from Ŝ to V̂ for the uniform convergence topology. But, it seems that the proof of this result implicitly
needs Proposition 5. However one finds in [8] a proof which does not depend on Proposition 5 and which holds for rank-one valuation domain
with any residue field provided that S is precompact. Nevertheless it
appears to us that a direct proof deserves to be given, specially because
we are going to use it.
Proof. Let {an }n∈N be a v-ordering of S and let x ∈ S. Let M =
supn∈N v(x − an ) and suppose that M is finite. The subset S being
precompact, one can write {v(x − an ), n ∈ N} = {v0 , · · · vr } where
r ∈ N and v0 < v1 < · · · < vr = M . For 0 ≤ j ≤ r, let C(vj ) = {ak |
v(x − ak ) ≥ vj }. As S is precompact, the subsets Cj are precompact
as well. By means of a finite induction, we are going to show that the
FACTORIAL PRESERVATION
5
cardinality cj of Cj is finite for all 0 ≤ j ≤ r. Let j be an integer
such that 0 ≤ j ≤ r and cj+1 , cj+2 , · · · , cr are finite. Assume that cj
is infinite and write Cj = {akn } as the subsequence of {an } formed by
the elements belonging to Cj . One knows that the sequence {akn } is a
v-ordering of Cj [5, Prop. 2]. Consequently, for every m ∈ N, one has:
v(
m−1
Y
(akm − aki )) ≤ v(
m−1
Y
(x − aki )).
i=0
i=0
From the precompactness of Cj , it follows that, for every M1 > M ,
there exist m, l ∈ N such that l < m and v(akm − akl ) > M1 .
By construction, on the one hand one has:
v(
m−1
Y
(x − aki )) ≤ cr vr + · · · + cj+1 vj+1 + (m −
i=0
r
X
ck )vj
k=j+1
≤ mvj +
r
X
ck (vk − vj )
k=j+1
Q
and, on the other hand, v( m−1
i=0 (x−aki )) ≥ (m−1)vj +M1 . By choosing
M1 large enough, one obtains a contradiction. Thus, the hypothesis M
is finite leads to c0 is finite. But, C0 = {an } is infinite, and then, M is
infinite.
¤
3. Isometries and factorial preservation
In this section, V still denotes the ring of a rank-one valuation v. A
map ϕ from S to V will be said contracting if v(ϕ(x)−ϕ(y)) ≥ v(x−y)
for every x, y ∈ S.
Proposition 6. Let S be a precompact subset of V for the v-adic topology and let ϕ : S −→ V be a contracting map. The following assertions
are equivalent:
(1) (k!)VS = (k!)Vϕ(S) fo all k ∈ N.
(2) ϕ is an isometry from S onto ϕ(S).
Proof. As the set S is precompact, ϕ(S) is also precompact. In particular, S and ϕ(S) admit infinite v-orderings. Let {ak }k≥0 and {ϕ(bk )}k≥0
be two v-orderings of S and ϕ(S) respectively. Let n ∈ N. Assume
that (k!)VS = (k!)Vϕ(S) for all 0 ≤ k ≤ n. According to Proposition 4,
one has
Y
Y
Y
v(
(ϕ(bi ) − ϕ(bj )) = v(
(ai − aj )) ≤ v(
(bi − bj )).
0≤i<j≤n
0≤i<j≤n
0≤i<j≤n
As ϕ is contracting, one has:
Y
Y
v(
(bi − bj )) ≤ v(
(ϕ(bi ) − ϕ(bj )),
0≤i<j≤n
0≤i<j≤n
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YOUSSEF FARES
and, as a consequence,
Y
Y
Y
v(
(ϕ(bi ) − ϕ(bj )) = v(
(ai − aj )) = v(
(bi − bj )).
0≤i<j≤n
0≤i<j≤n
0≤i<j≤n
Thus, for every (i, j) such that 0 ≤ i, j ≤ n, one has:
v(ϕ(bi ) − ϕ(bj )) = v(bi − bj ).
If this holds for all n, the sequence {bk }k≥0 is a v-ordering of S.
Let x, y be in S with x 6= y. Let r = v(x − y). The subset S being
precompact, according to Proposition 5, there exist n, m ∈ N such
that v(x − bn ) > r and v(y − bm ) > r. Consequently, v(bm − bn ) = r.
Since the map ϕ is contracting , v(ϕ(bn ) − ϕ(x)) ≥ v(bn − x) > r and
v(ϕ(bm ) − ϕ(y) ≥ v(bm − y) > r. The equality v(ϕ(bm ) − ϕ(bn )) =
v(bm − bn ) implies v(ϕ(x) − ϕ(y)) = v(x − y). In other words, ϕ is an
isometry from S onto ϕ(S) and (1) implies (2).
Conversely, if ϕ is an isometry from S onto ϕ(S), then ϕ transforms
any v-ordering of S into a v-ordering of ϕ(S) and thus preserves the
factorials.
¤
Corollary 7. Let f ∈ V [X]. The subsets S and f (S) have the same
sequence of factorials if and only if f is an isometry from S onto f (S).
Indeed, every polynomial f ∈ V [X] defines a contracting map from
V into V .
Remarks.
Let f (X) =
for all k ≥ 0, then
(1) f is injective,
(2) inf 1≤k≤m v(ak ) = 0.
Pm
k=0
ak X k ∈ V [X] . If (k!)VS = (k!)Vf(S)
4. Factorial preservation in Dedekind domains
Notation. In this paragraph, A denotes a Dedekind domain with
quotient field K. For every maximal ideal M ∈ M ax(A), denote by
vM the valuation on K associated to M . Let U (A) denotes the group
of units of A. Let S be a subset of A and let f be a polynomial in
Int(S, A) = {P ∈ K[X] | P (S) ⊂ A}.
M
M
When f (X) = uX + a, then (n!)A
= (n!)A
S
f (S) for all n ∈ N and
all maximal ideal M of A not containing u. The following proposition
gives the converse under finiteness hypotheses.
Proposition 8. Assume that the group U (A) is finite and that the
M
M
= (n!)A
subset S is precompact for all M-adic topologies. If (n!)A
S
f (S)
for all n ∈ N and all maximal ideals M but a finite number M1 , . . . , Mr ,
then f is of degree 1.
FACTORIAL PRESERVATION
7
Proof. Let d ∈ A\{0} be such that df ∈ A[X]. Let Mr+1 , . . . , Ms be the
maximal ideals of A containing d. Since U (A) is finite, A is not a semilocal domain and there exists at least one maximal ideal N of A different
AN
N
from M1 , . . . , Ms . For such an ideal N, one has (n!)A
S = (n!)f (S) for all
n ∈ N. According to Proposition 6 and thanks to the precompactness of
S, f is an isometry from S onto f (S) relatively to vN and, in particular,
f is injective on S.
Let us consider the formal identity d(f (X) − f (Y )) = (X − Y )h(X, Y )
where h ∈ A[X, Y ]. To prove that f is of degree 1, it is enough to
prove that there exists a ∈ S such that the polynomial h(a, X) is
constant. Assuming that there is no such a ∈ S, we are going to obtain
a contradiction.
Let a1 , · · · , as+1 be distinct elements of S. For every maximal ideal M
different from M1 , · · · , Ms , one has:
vM (h(a1 , x)) = vM (f (a1 ) − f (x)) − vM (a1 − x) = 0 for all x ∈ S
because f is an isometry on S relatively to vM . According to the
lemma below, there exists i ∈ {1, · · · , s} such that vMi (h(a1 , x)) is not
bounded for x ∈ S. Thus, there exists a sequence {xn }n∈N of elements
of S such that
lim vMi (h(a1 , xn )) = +∞.
n→+∞
Because of the compactness of S with respect to the Mi -adic topol(1)
ogy, there exists an extracted subsequence {xn }n∈N of {xn } which
converges toward x(1) in the Mi -adic completion of A. By continuity,
h(a1 , x(1) ) = 0; in particular, f (a1 ) = f (x(1) ).
(1)
The same arguments applied to S1 = {xn | n ∈ N} and to the
polynomial h(a2 , X) shows the existence of j ∈ {1, · · · , s} such that
vMj (h(a2 , x) is not bounded for x ∈ S1 . Consequently, there exists
(2)
a subsequence {xn }n∈N which converges toward x(2) for the Mj -adic
topology, and f (a2 ) = f (x(2) ). If i = j, then x(1) = x(2) , and hence,
f (a1 ) = f (a2 ) in contradiction with the fact that f is injective on S.
Thus, i 6= j.
(2)
The same arguments applied to S2 = {xn | n ∈ N} and to the polynomial h(a3 , X) shows the existence of k ∈ {1, · · · , s} different of i and
j. So, by iteration, one would prove the existence of s + 1 distinct elements in {1, · · · , s}. Thus, there exists a ∈ S such that the polynomial
h(a, X) is constant.
¤
Lemma 9. Let A be a Dedekind domain such that U (A) is finite. Let
E be an infinite subset of A and let g ∈ A[X]. If there are a finite
number of maximal ideals M1 , . . . , Ms and an integer M such that, for
all x ∈ E, one has:
(1) vM (g(x)) = 0 for every M ∈ M ax(A) \ {M1 , . . . , Ms },
(2) vM (g(x)) ≤ M for every M ∈ {M1 , . . . , Ms },
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YOUSSEF FARES
then the polynomial g is constant.
Proof. With these hypotheses, one can write, for all x ∈ E,
g(x)A =
s
Y
MvMi (g(x))
i=1
where 0 ≤ vMi (g(x)) ≤ M . As a consequence, the number of different
ideals g(x)A is bounded by M s . Since U (A) is finite, there are only a
finite number of values g(x) with x ∈ E. Finally, the polynomial g is
constant.
¤
Proposition 8 has for consequence:
Theorem 10. Let A be Dedekind domain with finite residue fields and
finite group of units. Let S be an infinite subset of A and let f ∈
Int(S, A). Then, S and f (S) have the same factorial sequence if and
only if f (X) = uX + a where u is a unit of A and a ∈ A.
Proof. As the residue fields are finite, S is precompact. If S and f (S)
have the same factorial sequence then, according to the previous propon
A
sition, f (X) = uX + a where a, u ∈ K. Then, (n!)A
f (S) = u (n!)S , and
hence, u ∈ U (A). Finally, for x ∈ S, ux ∈ A and ux + a ∈ A imply
a ∈ A.
¤
Remark. If A is the ring of integers of a number field K, the hypotheses of the previous theorem hold if and only if A = Z or K is a
quadratic imaginary field. This hypotheses also hold for A = OP , the
ring of regular functions outside P of a function field K defined on a
finite field Fq where P denotes any place of K.
Corollary 11. Under the hypotheses of the previous theorem, one has
Int(S, A) = Int(f (S), A) if and only if f (X) = uX+a where u ∈ U (A)
and a ∈ A.
In other words, if S and f (S) are two polynomially equivalent subsets, then f is of degree 1. In the particular case when A = Z, this is
the answer to Gilmer and Smith’s question [10].
Let us give now a result analogous to Proposition 8 where, on the
one hand, S = A and, on the other hand, the hypotheses on A are
weakened. Recall that a domain R is d-ring if, for every non-constant
polynomial g ∈ R[X], there are a ∈ R and M ∈ M ax(R) such that
g(a) ∈ M. Recall also:
Lemma 12. [11, Prop. 1] [6, Prop. VII. 2.3] Let R be a d-ring and let
g ∈ R[X] \ R. Then there are infinitely many maximal ideals M of A
for which the equation g(x) ≡ 0 (M) has a solution in A.
FACTORIAL PRESERVATION
9
Proposition 13. Let A be a Dedekind with finite residue fields which
is a d-ring and let f ∈ Int(A) = {g(X) ∈ K[X] | g(A) ⊆ A}. If one
M
M
has (n!)A
= (n!)A
A
f (A) for all n ∈ N and for all maximal ideals M of A
but a finite number, then f is of degree 1.
M
M
Proof. Let us suppose (n!)A
= (n!)A
A
f (A) for all n ∈ N and all M ∈
M ax(A) \ {M1 , . . . , Mr }. Let d ∈ A \ {0} be such that df ∈ A[X]. Let
Mr+1 , . . . , Ms be the maximal ideals of A containing d. Let d(f (X) −
f (Y )) = (X − Y )h(X, Y ). Let a ∈ A and let M ∈ M ax(A) \
{M1 , · · · , Ms }. Analogously to the proof of Proposition 8, we have
vM h(a, x) = 0 for all x ∈ A. Lemma 12 shows that h(a, X) is a constant polynomial, and hence, f is of degree 1.
¤
Corollary 14. Let A be the ring of integers of a number field or of a
function field and let f ∈ Int(A). Then (n!)A = (n!)f (A) if and only if
f (X) = uX + a where u ∈ U (A) and a ∈ A.
Indeed, the ring of integers of a number field or of a function field is
always a d-ring [6, Examples VII.2.12]. The previous corollary gives, in
the particular case where A = Z, Theorem 4.2 of [10]: if A is the ring
of integers of a number field or of a function field, and if f ∈ Int(A),
is such that Int(A) = Int(f (A), A), then f is of degree 1.
Remark. In the case where the subset S is finite, the results are
completely different. Indeed, let R be a domain with quotient field F
and let S and T be two finite subsets of R. Then
(1) Int(S, R) = Int(T, R) if and only if S = T [6, Exercise IV.1].
(2) Let f ∈ F [X]. The Lagrangian interpolation shows that f (S) =
S if and only if
X
Y (X − t)
Y
f (X) =
σ(s)
+ g(X) (X − s)
(s − t)
s∈S
t∈S
s∈S
t6=s
where g ∈ F [X] and σ is a permutation of S.
(3) If n!S = n!T for all n ≥ 0, then S and T have the same cardinality, but there does not always exist elements u ∈ U (R) and
a ∈ R such that T = uS + a as shown by the following example.
Example. Let R = Z, S = {0, 1, 6} and T = {0, 2, 5}. One has:
1!S = 1!T = 1 , 2!S = 2!T = 30 and n!S = n!T = 0 for all n ≥ 3. The
subsets S et T have the same factorial sequence, but T is not the image
of S by a polynomial of degree 1 .
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YOUSSEF FARES
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Youssef Fares, lamfa cnrs umr 6140, Université de Picardie, 80039
Amiens, France
E-mail address: [email protected]