P-orderings: a metric viewpoint and the non-existence of simultaneous orderings

http://www.elsevier.com/locate/jnt

Journal of Number Theory 99 (2003) 36–56

P-orderings: a metric viewpoint and thenon-existence of simultaneous orderings

Melanie Wood

Department of Mathematics, Duke University, Durham NC 27708, USA

Received 10 May 2001; revised 30 June 2002

Abstract

For a prime ideal Y and a subset S of a Dedekind ring R; a Y-ordering of S is a sequence of

elements of S with a certain minimizing property. These Y-orderings were introduced in

Bhargava (J. Reine Angew. Math., 490 (1997) 101) to generalize the usual factorial function

and many classical results were thereby extended, including results about integer-valued

polynomials. We consider Y-orderings from the viewpoint of the Y-adic metric on R: We find

that the Y-sequences of S depend only on the closure of S in RY: When R is a Dedekind

domain and R0 is the integral closure of R in a finite extension of the fraction field of R; we

relate theY-sequences of R and R0: Lastly, we investigate orderings that are simultaneously Y-

orderings for all prime ideals YCR; and show that such simultaneous orderings do not exist

for imaginary quadratic number rings.

r 2002 Elsevier Science (USA). All rights reserved.

Keywords: p-orderings; p-sequences; Generalized factorials; Simultaneous orderings; Fixed divisors;

Integer-valued polynomials

1. Introduction

The factorial function arises in many classical number theoretical problems, and in[1], Bhargava introduced the notion of Y-ordering to generalize factorials toarbitrary subsets of Dedekind rings. The generalized factorials of a subset S of aDedekind ring R are certain functions from the positive integers to the set of ideals ofR; and they lead to generalizations of these classical number theoretical results thatinvolve the factorial function. Some such problems for a Dedekind domain R are

E-mail address: [email protected].

0022-314X/03/$ - see front matter r 2002 Elsevier Science (USA). All rights reserved.

PII: S 0 0 2 2 - 3 1 4 X ( 0 2 ) 0 0 0 5 6 - 2

finding the fixed divisor of a given primitive polynomial over S; finding what idealsof R can be fixed divisors of a primitive polynomial over S; and finding thepolynomials with coefficients in the fraction field of R that map S to R: (The fixeddivisor of a polynomial FðxÞ over S is the ideal of R generated by FðSÞ:) Generalizedfactorials depend on the Y-sequence of S for each non-zero prime ideal Y of R;which in turn depends on Y-orderings of S: If we can find the Y-sequence of S foreach Y; then we can find its factorials and thus solve the problems listed above.

Let us recall from [1] the definitions of Dedekind ring, Y-ordering, Y-sequence,and generalized factorial.

Definition 1.1. A Dedekind ring is any Noetherian, locally principal ring in which allnonzero prime ideals are maximal. This includes Dedekind domains and theirquotients.

Fix a Dedekind ring R and a set SCR:

Definition 1.2. For a prime ideal Y of R; a Y-ordering is any sequence a0; a1; a2;yconstructed inductively as follows. First, let a0 be any element of S: Then, oncea0; a1;y; ak�1 have been selected, let ak be any element x of S that minimizes thepower of Y that divides

Yk�1

i¼0

ðx � aiÞ

over all xAS:

As any a0; a1; a2;y chosen in this way is a Y-ordering of S; it follows that thereare many different Y-orderings.

Definition 1.3. For a given Y-ordering, the corresponding Y-sequence of S is thesequence n1ðS;YÞ; n2ðS;YÞ;y; where nkðS;YÞ is the power of Y that dividesQk�1

i¼0 ðak � aiÞ:

Surprisingly, we have the following result.

Theorem 1.1 (Bhargava [1, Theorem 1]). The Y-sequence of S depends only on S and

Y; and is independent of the Y-ordering.

Lemma 1.1 (Bhargava [1, Lemma 3]). Let S be a subset of a Dedekind ring R: Then

for any integer 0pkojSj; nkðS;YÞaR for only finitely many primes Y:

Of course for kXjSj; nkðS;YÞ ¼ ð0Þ since two elements of a0; a1;y; ak must be thesame. Lemma 1.1 allows us to define generalized factorials.

M. Wood / Journal of Number Theory 99 (2003) 36–56 37

Definition 1.4. For each non-negative integer k; the generalized factorial is thefollowing product over all prime ideals of R:

k!S;R ¼YY

nkðS;YÞ:

Since non-zero ideals in a Dedekind ring have unique factorization into prime-power ideals, the factorials simply combine all of the Y-sequences into a singleobject, with no loss of information.

In Section 2, we recall from [1–3] some of the above-mentioned applications ofgeneralized factorials. In Section 3, we use the metric structure of S in the completion

RY of R at Y to help find Y-sequences of S: We find that the Y-sequences of S

depend only on the isometry class of %S; and that if %T is a proper subset of %S; then T

and S have different Y-sequences. From this it follows that many properties of fixed

divisors and integer-valued polynomials depend only on the isometry class of %S in thelocalization at each prime ideal. In particular, we can compute Y-sequences when S

is dense in RY:

In Section 4, we relate factorials for a Dedekind domain A; and an extension B;where B is the integral closure of A in a finite extension of the fraction field of A: Wefind that for SCA; the factorials of S in A are the same as the factorials of S in B:We also relate k!A;A to k!B;B in terms of the behavior of prime ideals when lifted from

A to B: We see that when a prime ideal Y of A is inert or ramifies in B; it has a largercontribution to A’s factorial than it does to B’s factorial.

In Section 5, we discuss simultaneous orderings, which are sequences that are Y-orderings for all prime ideals of R simultaneously. Whether a Dedekind domain R

has a simultaneous ordering has implications for it having a certain type of regularbasis for integer-valued polynomials with coefficients in its fraction field. Also, it issuggested that natural gamma functions interpolating the generalized factorialsmight only exist for sets S that have simultaneous orderings. We show thatimaginary quadratic number rings cannot have simultaneous orderings. Also, weoutline further possible investigations related to simultaneous orderings, as well aspossible further investigations to find rings with regular bases for integer-valuedpolynomials.

2. Applications of P-orderings and generalized factorials

To see why these functions are called generalized factorials, consider the followingexample.

Example 2.1. It is easy to see that for S ¼ R ¼ Z; ak ¼ k is a Y-orderingfor all primes in Z; as is ak ¼ �k and ak ¼ k þ 5: All of these Y-orderingsgive the same Y-sequence: nkðZ;YÞ is the power of Y that divides k!: Thusk!Z;Z ¼ ðk!Þ:

M. Wood / Journal of Number Theory 99 (2003) 36–5638

In general, it is difficult to find Y-orderings because one has to take a minimumover a set that is usually infinite. This is why it is useful to have theorems thatdescribe factorials in terms of properties such as the metric structure of S; or therelation of R to other Dedekind domains where the Y-orderings might be easier tofind. Once we know the factorials of S; though, we can apply many results.

Definition 2.1. The fixed divisor of a polynomial FðxÞAR½x over S is the ideal of R

generated by fFðaÞ j aASg and is denoted dðS;FÞ:

Theorem 2.1 (Bhargava [2, Theorems 2 and 4]). Let S be any subset of a Dedekind

domain R; and let FAR½x be any primitive polynomial of degree k: Then dðS;FÞdivides k!S;R:

Conversely, if an ideal I divides k!S;R; there is a primitive polynomial FAR½x of

degree k such that dðS;FÞ ¼ I :

A method for finding dðS;FÞ; the fixed divisor, given S and the coefficients of F isgiven in [2, Theorem 3], along with the following useful lemma.

Lemma 2.1 (Bhargava [2, Lemma 1]). If S is a subset of a discrete valuation ring R;with maximal ideal Y; and a0; a1;y is a Y-ordering of S; then for any FAR½x of

degree k;

dðS;FÞ ¼ ðFða0Þ;Fða1Þ;y;FðakÞÞ:

So, for a Dedekind domain R; given a polynomial F of degree k; if we knowthe finitely many primes that divide k!S;R then we can find dðS;FÞ by looking

at the localization of dðS;FÞ at each of these primes Y and using Y-orderings of S

up to ak:Another interesting application of generalized factorials is the following.

Theorem 2.2 (Bhargava [3, Theorem 10]). Let S be a subset of a Dedekind domain R;and b0; b1;y; bn be any elements of S: Then

Qioj ðbi � bjÞ is a multiple of

0!S;R1!S;R?n!S;R:This is sharp, i.e. the products

Qioj ðbi � bjÞ over all ðn þ 1Þ-tuples of S elements

generate the ideal LnðS;RÞ ¼ 0!S;R1!S;R?n!S;R:

Given SCR; let fai;kgk�1i¼0 be a sequence in R that, for each prime ideal Y*k!S;R;

is termwise congruent modulo nkðS;YÞ to some Y-ordering of S: We can alwaysfind such a sequence since we may apply the Chinese Remainder Theorem byLemma 1.1.

Theorem 2.3 (Bhargava [1, Theorem 13]). Let R be a Dedekind domain, and

K its fraction field. Let SCR be any subset, and IntðS;RÞ be the ring of all

polynomials in K½x that map S into R: Then any FAIntðS;RÞ can be represented in


the form Xn

k¼0

bkðx � a0;kÞðx � a1;kÞ?ðx � ak�1;kÞ;

where n is a finite integer, and the bkAk!�1S;R are uniquely determined by F : Conversely,

any polynomial of the above form with each bkAk!�1S;R is an element of IntðS;RÞ: Thus

IntðS;RÞ and the direct sum

"N

k¼0k!�1

S;R

are isomorphic as R-modules.

Corollary 2.1 (Bhargava [1, Theorem 14 and Corollary 3]). Let SCR as in Theorem

2.3. Then, IntðS;RÞ has a regular basis (i.e. IntðS;RÞ is a free R-module generated by

exactly one polynomial of each degree) if and only if all the k!S;R are non-zero principal

ideals. (For example, this is always the case when R is a PID and S is infinite.)

Notation 2.4. In this paper, our context will be that R is a Dedekind domain. Also, Ywill be a prime ideal of whatever Dedekind domain we are working in. When Y is

obvious from context, we will use nkðSÞ in place of nkðS;YÞ: We use n�ðS;YÞ to signify

fnkðS;YÞgNk¼1; the entire Y-sequence. Lastly, we sometimes say ‘‘the factorials of S’’

or ‘‘k!S’’ as shorthand for k!S;R when the ring R is clear from context.

3. A metric viewpoint

One important perspective on Y-orderings is a metric one. We can use the usual

Y-adic metric, and imagine R (as well as S) sitting inside RY; the Y-adic completion

of R: The exponent of Y in nkðS;YÞ is the Y-adic valuation ofQk�1

i¼0 ðak � aiÞ: The

definition of Y-ordering can be seen as a purely metric one, with each choice of ak

maximizing the product of the distances jak � aij:Some immediate consequences of this viewpoint are the following.

Proposition 3.1. Let S and T be subsets of R: If there is a Y-adic isometry of R (such

as translation, or multiplication by an element not in Y) that takes S to T ; then

n�ðS;YÞ ¼ n�ðT ;YÞ:

Proposition 3.2. Let S and T be subsets of R: If there is a Y-adic homothety of R of

valuation e (such as multiplication by an element in Ye\Yeþ1) that takes S to T ; then

YeknkðS;YÞ ¼ nkðT ;YÞ:

Propositions 3.1 and 3.2 show why k!aZþb;Z ¼ akk! [3, Example 17] and why in

general k!aRþb;R ¼ akk!R;R:


From the fact that the Y-adic metric satisfies the ultrametric inequality, we candeduce a useful fact about Y-orderings.

Proposition 3.3. Let SCR; Y a prime ideal of R: If a0; a1;y; ak is the start of a Y-ordering of S; then there exists a d > 0 such that for all xAS with jak � xjod; the

sequence a0; a1;y; ak�1; x is the start of a Y-ordering of S.

Proof. Let d ¼ minfjak � aij j i ¼ 0; 1;y; k � 1g: If jak � xjod; then for all i ¼0; 1;y; k � 1; jak � aij > jak � xj; which implies that jx � aij ¼ jak � aij: So theproposition follows from the definition of Y-ordering. &

So at any point while constructing a Y-ordering, when a choice of ak will suffice,then any element sufficiently close to ak will also suffice.

A fundamental question about Y-orderings is, given R and Y; for which subsets S

and T of R one has n�ðS;YÞ ¼ n�ðT ;YÞ: For the rest of the section, we fix R and Y toinvestigate this question. By definition fk!Sgk ¼ fk!Tgk exactly when n�ðS;YÞ ¼n�ðT ;YÞ for all prime ideals of R: If we know that fk!Sgk ¼ fk!Tgk; we can apply

Theorems 2.1–2.3 to get results comparing S and T : Theorem 2.1 would tell us thatprimitive polynomials of a given degree to R; from S and T ; respectively, have thesame set of fixed divisors. Theorem 2.2 would say that the ideals generated by allproducts of pairwise differences from n-tuples, in S and T ; respectively, are the same.Theorem 2.3 would tell us that IntðS;RÞEIntðT ;RÞ as R-modules. To approach thequestion of when n�ðTÞ ¼ n�ðSÞ; we can first investigate the case where TCS:

Proposition 3.4. Let TCS: Then, the following are equivalent.

(1) For all k; nkðT ;YÞ ¼ nkðS;YÞ:(2) All Y-orderings of T are also Y-orderings of S:(3) S has a Y-ordering all of whose elements are in T :

Proof. For (1) ) (2), suppose n�ðTÞ ¼ n�ðSÞ: Then, take a Y-ordering fakg of T : Wecan always have a0 start a Y-ordering of S: Suppose that a0;y; ak�1 is the start of aY-ordering of S: Then, the power of Y dividing

Qðak � aiÞ is nkðTÞ ¼ nkðSÞ; and

thus a0;y; ak is the start of a Y-ordering of S: So, inductively, any Y-ordering of T

is also a Y-ordering of S:For (2) ) (3), take a Y-ordering of T ; and it will be a Y-ordering of S; all of

whose elements are in T :For (3) ) (1) note that any Y-ordering of S with elements in T is also a Y-

ordering of T : This is because at each step, if akAT gives a minimum over allelements of S; then it certainly gives a minimum over all elements of TCS: Soinductively, the Y-ordering of S is also one for T ; and thus they have equal Y-sequences. &

Combining this proposition with Theorem 2.1 and Lemma 2.1 has the followingimmediate consequence.


Corollary 3.1. If TCSCR; and T and S have the same sets of fixed divisors for each

degree, then for any polynomial FðxÞAR½x; dðS;FÞ ¼ dðT ;FÞ:

Proof. Since S and T have the same sets of fixed divisors for each degree, they musthave the same factorials by Theorem 2.1, and thus the same Y-sequences for allprimes. So by Proposition 3.4, for each prime, T and S have a common Y-ordering.This means that at any localization, dðT ;FÞ ¼ dðS;FÞ by Lemma 2.1, and thecorollary follows. &

Even if S does not have a Y-ordering with all of its elements in T ; we can still starta Y-ordering fakg of S with a0AT : Let k be minimal such that in constructing a Y-ordering of S; we can choose a0; a1;y; ak�1 in T ; but we cannot choose akAT : Thenwe have that nkðS;YÞ strictly divides nkðT ;YÞ because the minimum power ofQðak � aiÞ is smaller over S than T : So, for the first term in which n�ðSÞ and n�ðTÞ

differ, niðSÞ is smaller than niðTÞ: In fact, in general nkðSÞ will be no larger thannkðTÞ:

Proposition 3.5 (Bhargava [3, Lemma 13]). If TCS; then k!S j k!T :

Since Y-orderings of a set S often do not use all the elements of S (e.g. our Y-ordering of Z that uses only non-negative integers), we have an example of a propersubset of S; the elements of a Y-ordering fa0; a1;yg ¼ TCS; such that n�ðTÞ ¼n�ðSÞ: This leads us to the important question: What elements of S are actuallyneeded in a Y-ordering of S?

Proposition 3.6. If jR=YjoN; or the image of S-R=Yn is finite for all n; then for

every xAS; every open set U containing x; and every Y-ordering of S; there is an

element of the Y-ordering in U :

We note that the finiteness condition is necessary by considering the followingexample: R ¼ Q½t; Y ¼ ðtÞ; S ¼ Q; U ¼ fx j x � 1AðtÞg: Clearly, f2; 3; 4; 5;yg is aðtÞ-ordering for S because it gives the ðtÞ-sequence f1; 1; 1;yg: However,f2; 3; 4; 5;yg contains no elements of U : With the finiteness condition, though, ifwe do not take elements of U ; we have to take infinitely many elements that are nearU ; which we will see leads to a contradiction.

Proof (Proof of Proposition 3.6). Suppose that there is a Y-ordering fakg of S with

no elements in U : Let an element of valuation r have distance 1Nr from 0: Choose a

ball C; centered at x; contained in U ; such that C ¼ fy j jy � xjp 1Neg for some e: Let

f be maximal so that infinitely many elements of fakg are in E ¼ fy j jy � xjp 1Nfg:

Such a maximum exists since fakg has infinitely many elements, and f is bounded

above by e: So, finitely many elements of fakg are in D ¼ fy j jy � xjp 1Nfþ1g: Since

the image of S in R=Yfþ1 is finite, we can choose a ball D0CE of radius 1Nfþ1 with

infinitely many elements of fakg in it.


Fix some j with ajAD0: We will compare jaj � aij to jx � aij:

1. If aiAD0; then jaj � aijp 1Nfþ1; jx � aij ¼ 1

Nf ;

2. if aiAD; then jaj � aij ¼ 1Nf ; jx � aijp 1

Nfþ1;

3. otherwise, jaj � aij ¼ jx � aij:

Therefore if we consider the ratio

rj ¼ðQj�1

i¼0 jx � aijÞðQj�1

i¼0 jaj � aijÞ;

we see that for j larger than j0 ¼ maxfi j ai�1ADg; we have rjXrj0Nej ; where ej is the

cardinality of fijj0pioj and ajAD0g: Thus, rj-N; which implies that eventually the

power of Y dividingQj�1

i¼0 ðx � aiÞ is less than the power dividingQj�1

i¼0 ðaj � aiÞ;which contradicts the definition of aj : This shows that any Y-ordering must

eventually intersect U : &

This proposition shows that if a Y-ordering does not contain a certain element x

of S; it must contain elements arbitrarily close to x: Also, we note from Proposition3.3 that at any step of the construction of a Y-ordering, we could always take a pointsufficiently close to x (that is not x), if there exists such a point, instead of taking x:This suggests that when Y-ordering S for a given Y; it does not matter whether or

not xAS; but that it only matters whether or not x belongs to the closure %S of S in

RY: In other words, the Y-sequence might depend more specifically on %S than on S:

Again, consider the case that TCS: In Proposition 3.4 we determined whethern�ðTÞ ¼ n�ðSÞ by considering elements in the Y-orderings of S: However, in general,it may be difficult to construct a Y-ordering. We can give an alternate condition thatmay be more useful depending on what we know about T and S:

Theorem 3.1. Let TCS: If %T ¼ %S in RY; then n�ðT ;YÞ ¼ n�ðS;YÞ: If jR=YjoN; or

the image of S-R=Yn is finite for all n; then the converse is true.

Proof. First, assume %T ¼ %S: Then all points of S\T are limit points of T since

S\TCSC %S ¼ %T: Thus, we can construct a Y-ordering of S with only elements of T :At any point where an element x of S would suffice for ak; by Proposition 3.3, allelements sufficiently close to x will also suffice, and T has elements arbitrarily closeto x: So, in a Y-ordering of S; we can always make a choice for ak that is in T : ByProposition 3.4 we have that n�ðTÞ ¼ n�ðSÞ:

Now, for the other direction, suppose %Ta %S: Let W be the complement of %T in RY:

Then W is open, and W does not intersect T ; since TC %T: However, if S does not

intersect W ; then we have SC %T: This implies %SC %T; a contradiction since TCS and%Ta %S: So, S intersects W ; which means that by Proposition 3.6, every Y-ordering of

S contains an element of W ; and thus an element not in T : Thus Proposition 3.4implies that n�ðTÞan�ðSÞ:


Theorem 3.1 is related to a result of McQuillan [5, Theorem 2]. McQuillanconsiders an S that is closed in every localization. He shows that (for R a Dedekind

domain with finite residue fields) %T ¼ S in every localization if and only ifIntðT ;RÞ ¼ IntðS;RÞ: The ‘‘only if’’ direction follows from Theorem 3.1, which saysT and S have the same Y-sequences. Then from Proposition 3.4, T and S have

common Y-orderings, and so we can pick the same fai;kgNi¼0 (in terms of Theorem

2.3) to show that IntðT ;RÞ ¼ IntðS;RÞ: The ‘‘if’’ direction follows from the conversepart of Theorem 3.1. If IntðT ;RÞ ¼ IntðS;RÞ and U ¼ T,S; then IntðT ;RÞ ¼IntðU ;RÞ; which means fk!Tgk ¼ fk!Ugk: The converse then gives that %T ¼ %U in

every localization, which means that %SC %T: By symmetry we have %T ¼ %S ¼ S in everylocalization. The language of Y-orderings allows us to understand the result moregenerally (for all Dedekind domains R and for S not closed at every localization) andto see how analogous statements can be made for other applications of Y-orderings.

If we view RY as a ring, we see that for SCR; n�ðS;YÞ ¼ n�ðS;YRYÞ; since the Y-

adic valuation on R extends to the YRY-adic valuation on RY: So for a fixed prime

ideal YCR; we can talk about n�ðS;YRYÞ for SCRY: This allows us to strengthen

one direction of Theorem 3.1.

Corollary 3.2. If %T ¼ %S in RY; then n�ðS;YÞ ¼ n�ðT ;YÞ:

Proof. Apply Theorem 3.1 with RY as the ring, and T and %T as the two subsets.

Since TC %T and T and %T have the same closure, n�ðT ;YRYÞ ¼ n�ð %T;YRYÞ: Thus, if

%T ¼ %S; n�ðT ;YÞ ¼ n�ðT ;YRYÞ ¼ n�ð %T;YRYÞ ¼ n�ð %S;YRYÞ ¼ n�ðS;YRYÞ ¼ n�ðS;YÞ;as desired. &

Note that Corollary 3.2 does not have a converse, because %S could be the image of%T under an isometry, and S and T would have the same Y-sequences but notnecessarily have the same closure. Moreover, consider the example R ¼ Zð3Þ; Y ¼ð3Þ; S ¼ f0; 1; 4; 7g; and T ¼ f0; 1; 4; 9g: Now, fk!Sgk ¼ fk!Tgk ¼f1; ð3Þ; ð3Þ2; 0; 0;yg; but S ¼ %S is not the image of T ¼ %T under any isometry.So, S and T can have equal Y-sequences (and even equal factorials) without theirclosures being isometric.

When the set S is dense in RY; we can find the Y-sequences as follows.

Proposition 3.7 (cf. Bhargava [1, Example 3]). Let R be any Dedekind domain. Let

jR=Yj ¼ N: If %S ¼ RY; then nkðS;YÞ ¼ YP

N

i¼1I k

Nim:

Also, Propositions 3.2 and 3.7 combine to give the Y-sequences when %S is a whole

residue class modYiRY: This immediately gives the fixed divisors, LnðS;RÞ; and

IntðS;RÞ of any subset S such that SCr þYiR for some r and S has an element in all

the residue classes modYk (for every k > i) that reduce to r ðmodYiÞ:


4. Extensions of Dedekind domains

Notation 4.1. For this section, we always use A and B to indicate Dedekind domains

such that B is the integral closure of A in a finite extension of A’s fraction field. We also

use p to indicate a prime ideal of A and Yi; indexed by i; to be the prime ideals of B

lying over p with ei and fi the degrees of ramification and inertia. Finally, for an ideal a

of a ring R; NðaÞ ¼ jR=aj:

A natural question to ask about this situation is for SCA; how k!S;A relates to

k!S;B:

Proposition 4.1. For SCA; k!S;AB ¼ k!S;B:

Proof. This follows immediately from Theorem 2.2.

So we see that in a Dedekind domain, if we want to work with the factorialsof S; we can do so allowing the possibility of S being placed in a larger ring.This result applies to many of our above examples and theorems. We canfind the factorials of S in a ring A that is simple enough to either allowdirect computation, as in our examples, or to allow use of our metric theorems

because %S has a nice description in the Ap: Then these factorials are also the

factorials of S in B: For example, ðk!Þ are still the factorials of Z as a subset of any

number ring, and ð2kk!Þ are still the factorials of the even rational integers in anynumber ring.

Another natural question to ask in this situation is how k!A;A compares to k!B;B:We have expressions for these factorials from Proposition 3.7. Let VkðB; pÞ ¼Q

Yi jpB nkðB;YiÞ: This is the contribution to k!B;B from the prime ideals of B that lie

above pCA; so k!B;B ¼Q

pCA VkðB; pÞ: Using that NðYiÞ ¼ NðpÞfi ; we see that if

there is no inertia at Yi; the powers of the Yi-sequence of B will be the same as thepowers in the p-sequence for A: However, inertia will cause the powers in the Yi-sequence to be smaller, making VkðB; pÞ smaller than nkðA; pÞB:

If p has any ramification in B; then the primes of B that divide pB will have aproduct smaller than pB; making VkðB; pÞ smaller than nkðA; pÞB: For example, if p

totally ramifies to degree e; then VkðB; pÞe= nkðA; pÞB; since the lack of inertia willkeep the powers in the sequences the same, but the only prime divisor of pB in B isY1; where Ye

1 ¼ pB:Finally, we see that if a prime p of A totally splits, then the powers in all

the sequences (of p and Yi) will be the same since there is no inertia, and thenpB ¼

QYi implies that VkðB; pÞ ¼ nkðA; pÞB: Since k!B;B j k!A;B ¼ k!A;AB; all

prime ideals of B that appear in k!B;B; must lie above a prime ideal of A that

divides k!A;A: So, we can find k!B;B if we know how all the prime divisors of k!A;A lift

to B: In particular we have the following generalization of a result of Bhargava[1, Theorem 20].


Corollary 4.1. All primes of A split completely in B if and only if fk!A;ABgk ¼fk!B;Bgk:

Corollary 4.1 has several applications with our theorems from Section 2 when wetake S ¼ A and R ¼ B:

Proposition 4.2. Let A and B be as above. The following are equivalent:

(1) All primes of A split completely in B:(2) A and B have the same factorials in the ring B for all k:(3) For any FAB½x; dðA;FÞ ¼ dðB;FÞ:(4) For all n; LnðA;BÞ ¼ LnðB;BÞ:(5) IntðA;BÞ ¼ IntðB;BÞ:

Proof. Note that with B as the ring, (2) is equivalent to A and B having the same setof fixed divisors of degree k for each k; and (3) is equivalent to A and B havingcommon Y-orderings for every prime ideal Y of B: From Corollary 4.1 andProposition 4.1, (1) is equivalent to (2). Clearly (3) implies (2), and by Corollary 3.1,since ACB; (2) implies (3). By Theorem 2.2, (2) is equivalent to (4). Theorem 2.3

shows that (5) implies (2). Also, given (3) we can pick the same fai;kgNi¼0 (in the terms

of Theorem 2.3) for A and B; and thus (5) follows from (3). &

Note that we can always find A and B that satisfy the conditions of Proposition 4.2by taking any A and B and inverting all the primes of A that do not spilt completelyin B:

As another application of our analysis, we have the following proposition, whichwe will use several times in Section 5. Recall that nkðZ; pÞ is just the power of p that

divides k!; and that in a quadratic number ring Nðk!Þ ¼ k!2:

Proposition 4.3. For a quadratic number ring R; Nðk!R;RÞ ¼Q

p okðpÞ; where the

product is over all positive rational primes ppk; and

okðpÞ ¼

nkðZ; pÞ p ramifies in R;

nkðZ; pÞ2 p splits in R;

p2P

N

i¼1I k

p2im p is inert in R:

8>><>>:

5. Simultaneous Y-orderings

For each non-zero prime ideal Y and subset S of a Dedekind ring R; we havedefined a sequence fakg that is a Y-ordering. However, in the case of S ¼ R ¼ Z we

have a sequence fakg ¼ fkgNk¼0 that is a Y-ordering for all primes YCZ:


Definition 5.1. A simultaneous ordering (as in [1]) of SCR is a sequence fakg ofelements of S that is a Y-ordering for all primes YCR: A simultaneous ordering ofR is a simultaneous ordering for S ¼ R:

An important open question about Y-orderings posed in [1, Question 3; 3,Question 30] and is when simultaneous orderings exist. It is suggested in [3] thatnatural gamma functions interpolating the factorials might exist exactly for S thathave such simultaneous orderings. As noted in [3], when S has a simultaneous

ordering fakg; k!S;R ¼ ðQk�1

i¼0 ðak � aiÞÞ; which implies that all factorials are

principal, and if S is infinite, then we can use Corollary 2.1 to find that IntðS;RÞhas a regular basis. In fact, if we look for Dedekind domains R with regular basesanalogous to the classical regular bases for IntðZ;ZÞ we see the following.

Theorem 5.1 (Bhargava [1, Theorem 19]). If faigNi¼0 is a simultaneous ordering of R;then IntðR;RÞ has the regular basis

ðx � a0Þðx � a1Þ?ðx � ak�1Þðak � a0Þðak � a1Þ?ðak � ak�1Þ

; ð1Þ

k ¼ 0; 1; 2;y:Conversely, the sequence of polynomials (1), for k ¼ 0; 1; 2;y; form a regular basis

for R only if faigNi¼0 is a simultaneous ordering of R:

In the case of S ¼ R; we have a simultaneous ordering for R ¼ Z; and one forR ¼ Fq½t is given in [3]. Also, if R has only finitely many prime ideals, we can use the

Chinese Remainder Theorem to construct a simultaneous ordering. CombiningCorollary 4.1 and Proposition 3.4, we have the following result similar to Theorem20 of Bhargava [1].

Proposition 5.1. If A is a Dedekind domain with a simultaneous ordering for S ¼ R ¼A; and B is an integrally closed finite extension of A as in Section 4, then all primes of A

split completely in B if and only if A’s simultaneous ordering is also a simultaneous

ordering for S ¼ R ¼ B:

For a number ring R; not all rational primes will split completely in R (unlessR ¼ Z), and thus 0; 1; 2; 3;y will not be a simultaneous ordering for R: This doesnot rule out the possibility of R having some simultaneous ordering, but no numberrings other than the rational integers are known to have simultaneous orderings.

One path of investigation to determine which rings have simultaneous orderings isto first find in which rings all factorials are principal for S ¼ R: Since by Corollary2.1 this is equivalent to having a regular basis for integer-valued polynomials, thisquestion is of interest of its own right. Though the condition of principal factorials isclearly necessary for a simultaneous ordering, it is not sufficient as we will see thatthe PID Z½i does not have a simultaneous ordering. The following proposition maybe of use in this investigation.


Proposition 5.2. Let R be a Dedekind domain. Then k!R;R is principal for all k if and

only if for all N YY primejR=Yj¼N

Y ð*Þ

is principal.

Proof. Since the powers of the Y-sequence depend only on jR=Yj; if (*) is principalfor all N then the k!R;R will be principal. For the converse, assume all the k!R;R are

principal, and we induct on N to show that (*) is principal. For N ¼ 2 note that2!R;R is just (*) for N ¼ 2: Suppose that (*) is principal for all NoM: We know that

ðM � 1Þ!R;R and M!R;R are principal, and from Proposition 3.7 we can write M!R;R ¼aðM � 1Þ!R;R for some ideal a which must be principal. We note that a can be written

as the product of several terms, all of which are just (*) for some N that divides M:All the terms where NoM are principal by induction and there is only one term forwhich N ¼ M; and thus that term must be principal and we are done. &

When R is a number ring, we can use Notation 4.1 with A ¼ Z and B ¼ R andrephrase Proposition 5.2 as follows.

Corollary 5.1. For a number ring B; k!B;B is principal for all k if and only if for all

primes p of A ¼ Z and for all s Yi such that

fi¼s

Yi

is principal.

In addition to number rings, the case of rings of integers in function fields is also ofinterest. Specifically in Notation 4.1 let A ¼ Fq½t and the question is which possible

B have simultaneous orderings. However, here we investigate the case S ¼ R ¼RD ¼ the ring of integers in Qð

ffiffiffiffiffiffiffiffi�D

pÞ; where D is a square-free positive integer.

Corollary 5.1 is of immediate use.

Lemma 5.1. If RD has k!RD;RDprincipal for all k (such as in the case that RD has a

simultaneous ordering), then either D ¼ 1 or D is prime.

Proof. Suppose RD has all principal factorials and D is composite. Then let p be aprime that divides D and thus divides the discriminant of RD: So p ramifies in RD and

we can write p ¼ Y2; and by Corollary 5.1 Y ¼ ðpÞ: Since D is not 1 or 3; the only

units in RD are 71 and thus p2 ¼ 7p: However, neitherffiffiffiffiffiffiffiffi7p

pis in Qð


pÞ so this

is impossible. Thus if RD has all principal factorials, D is 1 or prime. &

We will in fact prove the following stronger result.


Theorem 5.2. Let D be a positive integer, and R be the ring of integers in Qðffiffiffiffiffiffiffiffi�D

pÞ:

Then, R does not have a simultaneous ordering.

Notation 5.3. In the following, we let N denote the norm NK=Q (K ¼ Qðffiffiffiffiffiffiffiffi�D

pÞ) and its

generalization to ideals as it was used in Section 4 (NðaÞ ¼ jRD=aj). We note that an

element of K can be written as x þ yffiffiffiffiffiffiffiffi�D

p; where x and y are rational, and

Nðx þ yffiffiffiffiffiffiffiffi�D

pÞ ¼ x2 þ Dy2 ¼ jR=ðx þ y


pÞj:

Also, N is multiplicative. We will often denote RD as R when D is general or clear from

context.

Lemma 5.2. Z½i has no simultaneous ordering.

Proof. Suppose by way of contradiction that Z½i has a simultaneous ordering fakg:Since Nða1 � a0Þ ¼ Nð1!Z½i;Z½iÞ ¼ 1; and translation and multiplication by units are

isometries for all primes, we can assume without loss of generality that fa0; a1g ¼f0; 1g: By Proposition 4.3, Nða2ða2 � 1ÞÞ ¼ 2 and Nða3ða3 � 1Þða3 � a2ÞÞ ¼ 2; so bya translation and multiplication by a unit we can assume fa0; a1; a2; a3g ¼ f0; 1; i; 1þig: Proposition 4.3 also says that

Nða4ða4 � 1Þða4 � iÞða4 � i � 1ÞÞ ¼ 8:

However, this is impossible, because for all elements xAZ½i outside of f0; 1; i; 1þ ig;

Nðxðx � 1Þðx � iÞðx � i � 1ÞÞ > Nð2ð2þ iÞÞ ¼ 20: &

Lemma 5.3. Z½1þffiffiffiffiffi�3

p

2 has no simultaneous ordering.

Proof. Suppose by way of contradiction that R ¼ R3 has a simultaneous orderingfakg: Proposition 4.3 tells us that Nð1!RÞ ¼ Nð2!RÞ ¼ 1 which implies we can assume

by translation and multiplication by a unit that fa0; a1; a2g ¼ f0; 1; 1þffiffiffiffiffi�3

p

2g: Since

Nð3!RÞ ¼ 3 implies

a3ða3 � 1Þ a3 �1þ

ffiffiffiffiffiffiffi�3

p

2

! !¼ ð


pÞ;

which is a prime ideal, a3 must be a unit distance from 2 of f0; 1; 1þffiffiffiffiffi�3

p

2g; and we can

similarly assume

fa0; a1; a2; a3g ¼ 0; 1;1þ


p

2;3þ


p

2

( ):


Also,

Y3i¼0

ða4 � aiÞ !

¼ ð2Þðffiffiffiffiffiffiffi�3

pÞ;

which are prime ideals, so a4 must be a unit distance from 2 of the elements offa0;y; a3g: Thus we can assume

fa0;y; a4g ¼ 0; 1;1þ


p

2;3þ


p

2;�1þ


p

2

( ):

However,

Y4i¼0

ða5 � aiÞ !

¼ ð2Þðffiffiffiffiffiffiffi�3

pÞ;

so a5 must be a unit distance from 3 of fa0;y; a4g; but this is impossible. &

These cases suggest a general principle that the first few elements of a simultaneousordering would have to be very ‘‘close’’ to one another, and there do not existelements that close to one another. We have this requirement of ‘‘closeness’’ because

NYk�1

i¼0

ðak � aiÞ !

¼ Nðk!RÞpNðk!Þ

since ZCR implies that k!Rjk!: In fact, from Proposition 4.3 we see that Nðk!RÞ willbe increasingly less than Nðk!Þ the more rational primes less than k that ramify or areinert in R: For large D; this bound on how close elements must be, in combinationwith the fact that elements must be further away from each other for larger D (e.g.the only units for D > 3 are 71) and that some small primes must ramify or be inert,will establish Theorem 5.2. We deal with one more small case before embarking intothe lemmas used for proving the theorem.

Lemma 5.4. Z½1þffiffiffiffiffi�7

p

2 has no simultaneous ordering.

Proof. Suppose by way of contradiction that R ¼ R7 has a simultaneous orderingfakg: Since 1!R ¼ ð1Þ; we can assume fa0; a1g ¼ f0; 1g: Then,

2!R ¼ ð2Þ ¼ 1þffiffiffiffiffiffiffi�7

p

2

� �1�


p

2

� �

and we can assume that either

fa0; a1; a2g ¼ f0; 1; 2g


or

fa0; a1; a2g ¼ 0; 1;1þ


p

2

� �:

In either case, since 3!R ¼ ð2Þ; which is the product of two prime ideals, as above, in asimultaneous ordering, a3 would have to be a unit distance from one of a0; a1; or a2:Since the only units in R are 71; we can easily check that none of the possibilitiesgives

ða3 � a2Þða3 � a1Þða3 � a0Þ ¼ ð2Þ

and thus there is no simultaneous ordering for R: &

Lemma 5.5. If RD has a simultaneous ordering, then D � 7 ðmod 8Þ:

Proof. Suppose R ¼ RD has a simultaneous ordering. We know from Lemmas 5.2and 5.3 that the only units in R are 71: Since 1!R ¼ ð1Þ; we can assume fa0; a1g ¼f0; 1g: We consider three cases based on the behavior of (2) in R:

(1) If (2) is inert in R; then Proposition 3.7 implies that 2!R ¼ ð1Þ: So, a2 would haveto be a unit distance from both 0 and 1; which is impossible.

(2) If ð2Þ ramifies in R; then ð2Þ ¼ Y2 for some prime ideal Y of R: Proposition 3.7implies that 2!R ¼ Y; so

ðða2 � 1Þða2 � 0ÞÞ ¼ 2!R ¼ Y;

which implies Y is principal. Let ðpÞ ¼ Y; since 71 are the units of R; p is analgebraic number whose square is 72; which implies that D ¼ 2: However, in R2

it is easy to check that there is no a2 such that ðða2 � 1Þða2 � 0ÞÞ ¼ ðffiffiffiffiffiffiffi�2

pÞ; since

ðffiffiffiffiffiffiffi�2

pÞ is prime in R2; so such an a2 would have to be a unit distance from either

0 or 1:(3) If ð2Þ splits in R; then D � 7 ðmod 8Þ; as desired. &

Now, we begin to work towards the proof of Theorem 5.2, knowing that the onlycases left to eliminate have D � 7 ðmod 8Þ: Supposing that R has a simultaneousordering, we see that since ða0 � a1Þ ¼ 1!R will always be (1), we always know thatfa0; a1g is a translation of f0; 1g since the only units in R are 71: The motivatingquestion is, what is the largest k such that fa0; a1;y; akg is a translation off0; 1;y; kg?

Definition 5.2. Given that the set fa0; a1;y; ak�1g of points in the complex plane is atranslation of the set f0; 1;y; k � 1g; we define SðkÞ to be the minimum of

NðQk�1

i¼0 ðx � aiÞÞ over all xAR such that ðx � a0ÞeZ (i.e., those x off the line

containing a0; a1;y; ak�1). Clearly SðkÞ does not depend on fa0; a1;y; ak�1g:


When SðkÞ is larger than Nðk!RÞ; we will be able to say that ak is on that line. Sofirst we determine how large SðkÞ is.

Lemma 5.6. If kX1 and ðk þ 2Þ2p2D; then SðkÞ > ðk!Þ2:

Proof. We note that

SðkÞ ¼ minYk�1

i¼0

½ðx � iÞ2 þ Dy2 j x; yA1

2Z; x þ yAZ; ya0

( ):

If jyjX1; then, using the assumption that DXðkþ2Þ2

2; we have

Yk�1

i¼0

½ðx � iÞ2 þ Dy2XDkX

ðk þ 2Þ2

2

!k

>ðk þ 1Þ2

4

!k

Xðk!Þ2

as desired.The discriminant of

2j þ 1

2

� �2

þD

4� ðj þ 1Þðk � jÞ;

considered as a quadratic in j; is ð�k þ 2Þ2 � 2ðD þ 1� 4kÞ: Because ðk þ 2Þ2p2D;this discriminant is strictly negative, so

2j þ 1

2

� �2

þD

4> ðj þ 1Þðk � jÞ

for all j: Thus, if jyj ¼ 12; then xA1

2Z but xeZ and so we have

(1) If k is even, then

Yk�1

i¼0

½ðx � iÞ2 þ Dy2XYk�1

i¼0

k � 1

2� i

� �2

þDy2

" #

¼Yk�22

j¼0

2j þ 1

2

� �2

þD

4

" #0B@

1CA

2

>Yk�22

j¼0

ðj þ 1Þðk � jÞ

0B@

1CA

2

¼ðk!Þ2:


(2) If k is odd, then

Yk�1

i¼0

½ðx � iÞ2 þ Dy2XYk�1

i¼0

k

2� i

� �2

þDy2

" #

¼ k2 þ D

4

Yk�32

j¼0

2j þ 1

2

� �2

þD

4

" #0B@

1CA

2

>k þ 1

2

Yk�32

j¼0

ðj þ 1Þðk � jÞ

0B@

1CA

2

¼ðk!Þ2;

where the change in the left factor of the last inequality comes from the fact that,

because DXðkþ2Þ2

2X2k þ 2; we know k2þD

4Xðkþ1

2Þ2: &

Using this lower bound for SðkÞ; we can answer our above question about thelargest k such that fa0; a1;y; akg could be a translation of f0; 1;y; kg:

Lemma 5.7. Let m be the largest integer such that ðm þ 2Þ2p2D: If fakg is a

simultaneous ordering of R; then fa0; a1;y; amg is a translation of f0; 1;y;mg by

some element of R: Also, Nðl!RÞ ¼ Nðl!ZÞ for 1plpm:

Proof. Clearly the lemma is true for m ¼ 0: Let us suppose that fa0; a1;y; al�1g is a

translation of f0; 1;y; l � 1g and ðl þ 2Þ2p2D: Then SðlÞ > Nðl!Þ by our previous

lemma, and Nðl!ÞXNðl!RÞ: So, for any element xAR whoseffiffiffiffiffiffiffiffi�D

pcoordinate is

different from those of fa0; a1;y; al�1g; we have NðQl�1

i¼0 ðx � aiÞÞ > Nðl!RÞ: Thus,

al � ai for 0pipl � 1 must be l consecutive integers, and NðQl�1

i¼0 ðal � aiÞÞXNðl!Þwith equality exactly when fa0; a1;y; alg is a translation of f0; 1;y; lg: We must

have equality since NðQl�1

i¼0 ðal � aiÞÞ ¼ Nðl!RÞpNðl!Þ: So we have the lemma by

induction. &

Next, we will see the implication of the fact that Nðl!RÞ ¼ Nðl!Þ for the behavior ofsmall rational primes.

Lemma 5.8. Let m be the largest integer such that ðm þ 2Þ2p2D: Then if R has a

simultaneous ordering, all primes in Z less than or equal to m split in R:

Proof. From the previous lemma, Nðl!RÞ ¼ Nðl!Þ for 1plpm: From Proposition4.3, this can only happen when all primes less than or equal to m split. &


Finally, we can prove Theorem 5.2 because this phenomenon of all the smallprimes splitting does not occur in the RD:

Proof of Theorem 5.2. Suppose for the sake of contradiction that RD has asimultaneous ordering. Then D is not 1 by Lemma 5.2 and so D is prime by Lemma5.1. Since D � 7 ðmod 8Þ from Lemma 5.5, 2 is a quadratic residue mod D: We knowthat an odd rational prime, paD; splits in R exactly when �D is a quadratic residuein Z=pZ: So, if p is a rational prime less than or equal to m (defined as in Lemma 5.8),then p splits in R by Lemma 5.8. By quadratic reciprocity,

1 ¼ �D

p

� �¼ ð�1Þ

p�12

D

p

� �¼ð�1Þ

p�12 ð�1Þ

p�12

D�12

p

D

� �

¼ð�1Þp�12

Dþ12

p

D

� �

¼ p

D

� �:

So all primes less than or equal to m are quadratic residues mod D: But allnumbers less than or equal to m can be written as a product of such primes, so1; 2;y;m are all quadratic residues mod D: Godwin [4] proved that if n is the

smallest non-residue mod D; either n2 þ 3n þ 1pD or np6:If mX6; then nXm þ 1X7 and

ðm þ 1Þ2 þ 3ðm þ 1Þ þ 1pD:

However,

2D � Iffiffiffiffiffiffiffi2D

pm� 2 ¼ 2D � m � 4o ðm þ 3Þ2 � m � 4

¼ðm þ 1Þ2 þ 3ðm þ 1Þ þ 1

pD

but

2D � Iffiffiffiffiffiffiffi2D

pm� 2oD

implies Do6: This contradicts the fact that D � 7 ðmod 8Þ:So, mp5 and thus 2Do64: This with D � 7 ðmod 8Þ; D prime, and Lemma 5.4

gives that D ¼ 23 or 31: But, 3 is not a square modulo 31; so if R has a simult-aneous ordering, D must be 23: However, (5) is inert in R23; so there are no primes inR23 of norm 5; and Nð5!R23

Þ ¼ Nð4!R23Þ: From Lemma 5.6, Sð4Þ > Nð4!Þ: This gives

that

Sð5ÞXSð4Þ > Nð4!Þ ¼ Nð4!R23Þ ¼ Nð5!R23

Þ;


which implies that in the simultaneous ordering, a5 lies on the line with a0;y; a4 andthis tells us that ð5Þ splits in R23 by the same reasoning as in Lemma 5.8. This is acontradiction, and thus there does not exist a D for which RD has a simultaneousordering. &

So we have taken care of the question of simultaneous orderings for the ring ofintegers in an imaginary quadratic number field. In rings where there are more units,there is not the prohibition of ‘‘closeness’’ that our proof took advantage of. We dostill hope though that this result is a step towards understanding the answers to someof the following questions.

Question 5.4. Which number rings (and which rings of integers in function fields) have

simultaneous orderings?

Question 5.5. What are necessary and sufficient conditions for a ring R (those

mentioned above or others) to have all principal factorials k!R;R; equivalently to have a

regular basis for integer-valued polynomials?

Question 5.6. Could an imaginary quadratic number ring (or a number ring in general

or the ring of integers in a function field) have a sequence fakg that is a Y-ordering for

all but finitely many primes Y?

Note that having a sequence as in Questions 5.6 is related to having a

simultaneous ordering of S�1R if all the excluded primes are principal, where S is

generated by the generators of the excluded prime ideals. Since R is dense in S�1R;an fakg that was a Y-ordering for all primes but those that generate S would be a

simultaneous ordering for S�1R: However, though for all Y we can Y-order S�1R

with elements of R; there might be a simultaneous ordering of S�1R but none thatonly contains elements of R:

Acknowledgments

This work was done during the 2000 Research Experience for Undergraduates atthe University of Minnesota, Duluth, directed by Dr. Joseph A. Gallian andsponsored by the National Science Foundation (Grant DMS-92820179) and theNational Security Agency (Grant 904-00-1-0026). The author thanks ManjulBhargava, Mike Develin, Joseph Gallian, Stephen Hartke, Manish Patnaik, andDavid Witte for their encouragement and valuable suggestions.

References

[1] M. Bhargava, P-orderings and polynomial functions on arbitrary subsets of Dedekind rings, J. Reine

Angew. Math. 490 (1997) 101–127.


[2] M. Bhargava, Generalized factorials and fixed divisors over subsets of a Dedekind domain, J. Number

Theory 72 (1998) 67–75.

[3] M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly 107 (2000)

783–799.

[4] H.J. Godwin, On the least quadratic non-residue, Proc. Cambridge Philos. Soc. 61 (1965)

671–672.

[5] D.L. McQuillan, On a Theorem of R. Gilmer, J. Number Theory 39 (1991) 245–250.


P-orderings: a metric viewpoint and the non-existence of simultaneous orderings

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Transcript of P-orderings: a metric viewpoint and the non-existence of simultaneous orderings