VARGA S THEOREM IN NUMBER FIELDS. Pete L. Clark Department of Mathematics, University of Georgia, Athens, Georgia. Lori D.

Transcription

1 #A74 INTEGERS 18 (2018) VARGA S THEOREM IN NUMBER FIELDS Pete L. Clark Department of Mathematics, University of Georgia, Athens, Georgia Lori D. Watson 1 Department of Mathematics, University of Georgia, Athens, Georgia Received: 11/9/17, Accepted: 8/21/18, Published: 8/31/18 Abstract We give a number field version of a recent result of Varga on solutions of polynomial equations with binary input variables and relaxed output variables. 1. Introduction This note gives a contribution to the study of solution sets of systems of polynomial equations over finite local principal rings in the restricted input / relaxed output setting. The following recent result should help to explain the setting and scope. Let n, a 1,..., a n 2 Z + and 1 apple N apple P n i=1 a i. Put ( 1 if N < n m(a 1,..., a n ; N) = min Q n i=1 y i if n apple N apple P n i=1 a ; i the minimum is over (y 1,..., y n ) 2 Z n with 1 apple y i apple a i for all i and P n i=1 y i = N. Theorem 1. ([6, Thm. 1.7]) Let R be a Dedekind domain, and let p be a maximal ideal in R with finite residue field R/p = F q. Let n, r, v 1,..., v r 2 Z +. Let A 1,..., A n, B 1,..., B r R be nonempty subsets each having the property that no two distinct elements are congruent modulo p. Let r, v 1,..., v r 2 Z +. Let P 1,..., P r 2 R[t 1,..., t n ] be nonzero polynomials, and put z B A := #{x 2 ny A i 81 apple j apple m P j (x) 2 B j (mod p vj )}. i=1 1 Partial support from National Science Foundation grant DMS

2 INTEGERS: 18 (2018) 2 Then za B = 0 or 0 1 nx 1,..., #A n ; #A i rx (q vj #B j ) deg(p j ) A. z B A i=1 Remark 1. In the notation of Theorem 1, put v := max 1applejappler v j. Then Theorem 1 may be viewed as a result on polynomials with coe cients in the residue ring R/p v (cf. [6, Thm. 1.6]), a finite, local principal ring. For every finite local principal ring r, there is a number field K, a prime ideal p of the ring of integers Z K of K, and v 2 Z + such that r = Z K /p v ([7, Thm. 2], [2, Thm. 1.12]). Henceforth we will work in the setting of residue rings of Z K. If in Theorem 1 we take v 1 = = v r = 1, A i = F q for all i and B j = {0} for all j, then we recover a result of E. Warning. Theorem 2. (Warning s Second Theorem [9]) Let P 1,..., P r 2 F q [t 1,..., t n ] be nonzero polynomials, and let z = #{x = (x 1,..., x n ) 2 F n q P 1 (x) = = P r (x) = 0}. Then z = 0 or z q n P n deg(pj). By Remark 1, we may write F q as Z K /p for some maximal ideal p in the ring of integers Z K of a suitable number field K. Having done so, Theorem 2 can be interpreted in terms of solutions to a congruence modulo p, whereas Theorem 1 concerns congruences modulo powers of p. At the same time, we are restricting the input variables x 1,..., x n to lie in certain subsets A 1,..., A n and also relaxing the output variables: we do not require that P j (x) = 0 but only that P j (x) lies in a certain subset B j modulo p vj. There is however a tradeo : Theorem 1 contains the hypothesis that no two elements of any A i (resp. B j ) are congruent modulo p. Thus, whereas when v j = 1 for all j we are restricting variables by choice e.g. we could take each A i to be a complete set of coset representatives for p in Z K as done above when v j > 1 we are restricting variables by necessity we cannot take A i to be a complete set of coset representatives for p vj in Z K. We would like to have a version of Theorem 1 in which the A i s can be any nonempty finite subsets of Z K, and the B j can be any nonempty finite subsets of Z K containing {0}. However, to do so the degree conditions need to be modified in order to take care of the arithmetic of the rings Z K /p dj. In general this seems like a di cult and worthy problem. An interesting special case was resolved in recent work of L. Varga [8]. His degree bound comes in terms of a new invariant of a subset B Z/p d Z \ {0} called the price of B and denoted pr(b) that makes connections to the theory of integer-valued polynomials.

3 INTEGERS: 18 (2018) 3 Theorem 3. (Varga [8, Thm. 6]) Let P 1,..., P r 2 Z[t 1,..., t n ]\{0} be polynomials without constant terms. For 1 apple j apple r, let d j 2 Z +, and let B j Z/p dj Z be a subset containing 0. If rx deg(p j ) pr(z/p dj Z \ B j ) < n, then #{x 2 {0, 1} n 81 apple j apple r, P j (x) 2 B j (mod p dj )} 2. In this note we will revisit and extend Varga s work. Here is our main result. Theorem 4. Let K be a number field of degree N, and let e 1,..., e N be a Z- basis for Z K. Let p be a nonzero prime ideal of Z K, and let d 1,..., d r 2 Z +. Let P 1,..., P r 2 Z K [t 1,..., t n ] be nonzero polynomials without constant terms. For each 1 apple j apple r, there are unique {' j,k } 1applekappleN 2 Z[t 1,..., t n ] such that P j (t) = NX ' j,k e j. (1) k=1 For 1 apple j apple r, let B j be a subset of Z K /p dj S := rx that contains 0 (mod p dj ). Let! NX deg(' j,k ) pr(z K /p dj \ B j ). k=1 Then #{x 2 {0, 1} n 81 apple j apple r, P j (x) (mod p dj ) 2 B j } 2 n S. Thus we extend Varga s Theorem 3 from Z to Z K number of solutions. and refine the bound on the In Section 2 we discuss the price of a subset of Z K /p d. It seems to us that Varga s definition of the price has minor technical flaws: as we understand it, he tacitly assumes that for an integer-valued polynomial f 2 Q[t] and m, n 2 Z, the output f(n) modulo m depends only on the input modulo m. This is not true: for instance t(t 1) if f(t) = 2, then f(n) modulo 2 depends on n modulo 4, not just modulo 2. So we take up the discussion from scratch, in the context of residue rings of Z K. The proof of Theorem 4 occupies Section 3. After setting notation in Section 3.1 and developing some preliminaries on multivariate Gregory-Newton expansions in Section 3.2, the proof proper occurs in Section 3.3.

4 INTEGERS: 18 (2018) 4 2. The Price Consider the ring of integer-valued polynomials We have inclusions of rings Let Int(Z K, Z K ) = {f 2 K[t] f(z K ) Z K }. Z K [t] Int(Z K, Z K ) K[t]. m(p, 0) := {f 2 Int(Z K, Z K ) f(0) 0 (mod p)}. Observe that m(p, 0) is the kernel of a ring homomorphism Int(Z K, Z K )! Z K /p: first evaluate f at 0 and then reduce modulo p. So m(p, 0) is a maximal ideal of Int(Z K, Z K ). We put U(p, 0) := Int(Z K, Z K ) \ m(p, 0) = {f 2 Int(Z K, Z K ) f(0) /2 p}. Let d 2 Z +, and let B be a subset of Z K /p d. We say that h 2 U(p, 0) covers B if: for all b 2 Z K such that b (mod p d ) 2 B, we have h(b) 2 p. The price of B, denoted pr(b), is the least degree of a polynomial h 2 U(p, 0) that covers B, or 1 if there is no such polynomial. Remark 2. a) If B 1, B 2 are subsets of Z K /p d \ {0}, then pr(b 1 [ B 2 ) apple pr(b 1 ) + pr(b 2 ) : If for i = 1, 2 the polynomial h i 2 U(p, 0) covers B i and has degree d i, then h 1 h 2 2 U(p, 0) covers B 1 [ B 2 and has degree d 1 + d 2. b) If 0 (mod p d ) 2 B, then pr(b) = 1: Since 0 2 B we need h(0) 2 p, contradicting h 2 U(p, 0). c) If d = 1, then for any subset B Z K /p \ {0}, we have pr(b) apple #B: Let B be any lift of B to Z K. Then h = Y x2 B(t x) 2 Z K [t] Int(Z K, Z K ) covers B and has degree #B. Note that here we use polynoimals with Z K -coe cients. It is clear that #B is the minimal degree of a covering polynomial h with Z K - coe cients: we can then reduce modulo p to get a polynomial in F q [t] that we want to be 0 at the points of B and nonzero at 0, so of course it must have degree at least #B. d) If we assume no element of B is 0 modulo p, let B be the image of B under the natural map Z K /p d! Z K /p = F q ; then our assumption gives 0 /2 B. Above we constructed a polynomial h 2 Z K [t] of degree #B such that h(0) /2 p and for all x 2 Z K such that x (mod p) 2 B, we have h(x) 2 p. This same polynomial h covers B and shows that pr(b) apple pr(b) apple #B.

5 INTEGERS: 18 (2018) 5 For B Z K /p d \ {0} we define apple(b) 2 Z +, as follows. For 1 apple i apple d we will recursively define B i Z K /p i \ {0} and k i 1 2 N. Put B d = B, and let k d 1 be the number of elements of B d that lie in p d 1. Having defined B i and k i 1, we let B i 1 be the set of x 2 Z K /p i 1 such that there are more than k i 1 elements of B i mapping to x under reduction modulo p i 1. We let k i 2 be the number of elements of B i 1 that lie in p i 2. Notice that 0 /2 B i for all i: indeed, B i is defined as the set of elements x such that the fiber under the map Z K /p i+1! Z K /p i has more elements of B i+1 than does the fiber over 0. We put Lemma 3. We have apple(b) apple q d 1. Xd 1 apple(b) := k i q i. Proof. Each k i is a set of elements in a fiber of a q-to-1 map, so certainly k i apple q. In order to have k i = q, then B i+1 would need to contain the entire fiber over 0 2 Z K /p i, but this fiber includes 0 2 Z K /p i+1, which as above does not lie in B i+1. So Xd 1 Xd 1 apple(b) = k i q i apple (q 1)q i = q d 1. i=0 Theorem 5. For any subset B Z K /p d \ {0}, we have pr(b) apple apple(b). i=0 Proof. Step 1: For r 1, let A = {a 1,..., a q d 1} Z K /p d be a complete residue system modulo p d 1 none of whose elements lie in p d. We will show how to cover A with f 2 U(p, 0) of degree q d 1. We denote by v p the p-adic valuation on K. Let 2 Z K be an element with v p ( ) = P d 2 j=0 qj, and let 2 Z K be an element such that v p ( ) = 0 and, for all nonzero prime ideals q 6= p of Z K, we have v q ( ) v q ( ). (Such elements exist by the Chinese Remainder Theorem.) Put g A (t) := q d 1 i=0 Y (t a j ) 2 Z K [t], h A (t) := g A (t) 2 K[t]. For all x 2 Z K, {x a 1,..., x a q d 1} is a complete residue system modulo p d 1, so in Q q d 1 (x a j), for all 0 apple j apple d 1 there are q d 1 j factors in p j, so P d 2 v p (g A (x)) j=0 qj and thus v p (h A (x)) 0. For any prime ideal q 6= p of Z K, both v q (g A (x)) and v q ( ) are non-negative, so v q (h A (x)) 0. Thus h A 2 Int Z K. Moreover, the condition that no a j lies in p d ensures that v p (g A (0)) = P d 2 j=0 qj, so h A 2 U(p, 0). If x 2 Z K is such that x a j (mod p d ) for some j, then v p (x a j ) d. Since in the above lower bounds of v p (g A (x)) we obtained a lower bound of at

6 INTEGERS: 18 (2018) 6 most d 1 on the p-adic valuation of each factor, this gives an extra divisibility and shows that v p (h A (x)) 0. Thus h A covers A with price at most q d 1. Step 2: Now let B Z K /p d \ {0}. The number of elements of B that lie in p d 1 is k d 1. For each of these elements x i we choose a complete residue system A i modulo p d 1 containing it; since no x i lies in p d this system satisfies the hypothesis of Step 1, so we can cover each A i with price at most q d 1 and thus (using Remark (2a)) all of the A i s with price at most k d 1 q d 1. However, by suitably choosing the A i s we can cover many other elements as well. Indeed, because we are choosing k d 1 complete residue systems modulo p d 1, we can cover every element x that is congruent modulo p d 1 to at most k d 1 elements of B. By definition of B d 1, this means that we can cover all elements of B that do not map modulo p d 1 into B d 1. Now suppose that we can cover B d 1 by h 2 U(p, 0) of degree apple 0. This means that for every x 2 Z K such that x (mod p d 1 ) lies in B d 1, h(x) 2 p. But then every element of B whose image in p d 1 lies in B d 1 is covered by h, so altogether we get pr(b) apple k d 1 q d 1 + pr(b d 1 ). Now applying the same argument successively to B d 1,..., B 1 gives pr(b i ) apple k i 1 q i 1 + pr(b i 1 ), and thus Xd 1 pr(b) apple k i q i = apple(b). i=0 3. Proof of the Main Theorem 3.1. Notation Let K be a number field of degree N, and let e 1,..., e N be a Z-basis for Z K. A Z-basis for Z K [t 1,..., t n ] is given by e j t I as j ranges over elements of {1,..., N} and I ranges over elements of N n. So for any f 2 Z K [t 1,..., t n ], we may write Then we have f = ' 1 (t 1,..., t n )e ' N (t 1,..., t n )e N, ' i 2 Z[t 1,..., t n ]. (2) For a subset B Z K /p d, we put deg f = max deg ' i. i B = Z K /p d \ B.

7 INTEGERS: 18 (2018) Multivariable Newton Expansions Lemma 4. If f 2 Q[t] is a polynomial and f(n) Z, then f(z) Z. Proof. See e.g. [3, p. 2]. Theorem 6. Let f 2 K[t]. a) There is a unique function (f) : N N! K, r 7! r (f) such that (i) we have r (f) = 0 for all but finitely many r 2 N N, and (ii) for all x = x 1 e x N e N 2 Z K, we have f(x) = X r2n N r (f) x1 r 1 xn b) The following are equivalent: (i) We have f 2 Int(Z K, Z K ). (ii) For all r 2 N N, r (f) 2 Z K. We call the r (f) the Gregory-Newton coe cients of f. r N. (3) Proof. Step 1: Let f 2 K[t]. Let e 1,..., e N be a Z-basis for Z K. We introduce new independent indeterminates t 1,..., t N and make the substitution to get a polynomial NX t = e k t k k=1 f 2 K[t]. This polynomial induces a map K N! K hence, by restriction, a map Z N! K. For x = (x 1,..., x N ) 2 Z N, write x = x 1 e x N e N 2 Z K. Then we have f(x) = f(x). Let M = Maps(Z N, K) be the set of all such functions, and let P be the K-subspace of M consisting of functions obtained by evaluating a polynomial in K[t] on Z N, as above. By the CATS Lemma [5, Thm. 12], the map K[t]! P is an isomorphism of K-vector spaces. Henceforth we will identify K[t] with P inside M. Step 2: For all 1 apple k apple N, we define a K-linear endomorphism k of M, the kth partial di erence operator: k(g) : x 2 Z N 7! g(x + e k ) g(x). These endomorphisms all commute with each other: ( i j)(g) = g(x + e i + e j ) g(x + e j ) g(x + e i ) + g(x) = ( j i)(g).

8 INTEGERS: 18 (2018) 8 Let 0 k be the identity operator on M, and for i 2 Z +, let i k be the i-fold composition of k. For I = (i 1,..., i N ) 2 N N, put I = i1... i N 2 End K (M). When we apply k to a monomial t I, we get another polynomial. More precisely, if deg tk (t I ) = 0 then kt I is the zero polynomial; otherwise deg tk ( kt I ) = (deg tk t I ) 1; 8l 6= k, deg tl ( kt I ) = deg tl t I. Thus for each f 2 P, for all but finitely many I 2 N N, we have that I (f) = 0. For the one variable di erence operator, we have x x + 1 x x = =. r r r r 1 From this it follows that for I, r 2 N N we have I x1 xn 0 (0) = r 1 r N r 1 i 1 r N 0 i N = r,i. (4) So if : N N! K is any finitely nonzero function then for all I 2 N N we have I ( X x1 xn )(0) = I, (5) r2n N and thus there is at most one such function satisfying (3), namely r r 1 r N (f) : r 7! r (f)(0). So for any f 2 M and r 2 N N, we define the Gregory-Newton coe cient r (f) := r (f)(0) 2 K. of Gregory-Newton coef- We may view the assignment of the package { r (f)} r2n N ficients to f 2 M as a K-linear mapping M! K NN. If we put M + = Maps(N N, K), then we get a factorization M! M +! K N N, where the first map restricts from Z N to N N, and the factorization occurs because the Gregory-Newton coe cients depend only on the values of f on N N. We make several observations.

9 INTEGERS: 18 (2018) 9 First Observation: The map is an isomorphism. Indeed, knowing all the successive di erences at 0 is equivalent to knowing all the values on N N, and all possible packages of Gregory-Newton coe cients arise. Namely, let S n be the assertion that for all x 2 N N with P k x k = n and all f 2 M, then f(x) is a Z-linear combination of its Gregory-Newton coe cients. The case n = 0 is clear: f(0) = 0 (f). Suppose S n holds for n, let x 2 N N be such that P k x k = n + 1, and choose k such that x = y + e k ; thus P k y k = n. Then f(x) = f(y) + k f(y). By induction, f(y) is a Z-linear combination of the Gregory-Newton coe cients of f and kf(y) is a Z-linear combination of the Gregory-Newton coe cients of kf. But every Gregory-Newton coe cient of kf is also a Gregory-Newton coe cient of f, completing the induction. Second Observation: The composite map K[t]! M! M +! K N N is an injection. Indeed, the kernel of M! K NN is the set of functions that vanish on Z N \ N N. In particular, any element of the kernel vanishes on the infinite Cartesian subset (Z <0 ) N and thus by the CATS Lemma is the zero polynomial. Third Observation: For a subring R K and f 2 M, we have f(n N ) R i all of the Gregory-Newton coe cients of f lie in R. This is a consequence of the First Observation: the Gregory-Newton coe cients are Z-linear combinations of the values of f on N N and conversely. Step 3: For F 2 K[t], we define the Newton expansion T (F ) = X t1 tn r (F ) 2 K[t]. r 1 r N r2n N This is a finite sum. Moreover, by definition of r (F ) and by (5) we get that for all r 2 N N, r (T (F )) = r (F ). It now follows from Step 2 that T (F ) = F 2 K[t]. Applying this to the f associated to f 2 K[t] in Step 1 completes the proof of part a). Step 4: If we assume that f 2 Int(Z K, Z K ) then f(z N ) Z K, so all the Gregory- Newton coe cents lie in Z K. Conversely, if all the Gregory-Newton coe cients of f lie in Z K, then for x = x 1 e x N e N 2 Z K, by Lemma 4 and (3) we have f(x) = f(x 1,..., x N ) 2 Z K, so f 2 Int(Z K, Z K ) Proof of Theorem 4 We begin by recalling the following result.

10 INTEGERS: 18 (2018) 10 Theorem 7. Let F be a field, and let P 2 F [t 1,..., t n ] be a polynomial. Let Then either #U = 0 or #U 2 n deg(p ). U := {x 2 {0, 1} n P (x) 6= 0}. Proof. This is a special case of a result of Alon-Füredi [1, Thm. 5]. We now turn to the proof of Theorem 4. Put Z := {x 2 {0, 1} n 81 apple j apple r, P j (x) (mod p dj ) 2 B j }. Step 0: If q = #Z K /p is a power of p, then we have p d 2 p d. Therefore in (1) if we modify any coe cient of ' j,k (t) by a multiple of p d, it does not change P j modulo p d and thus does not change the set Z. We may thus assume that every coe cient of every ' j,k is non-negative. Step 1: For w = P k i=1 tii a sum of monomials and 0 apple r apple k, we put X r(w) := t Ii 1 t I ir. 1applei 1<i 2<...<i rapplek For x 2 {0, 1} n, we have w(x) = #{1 apple i apple k x Ii = 1}, so w(x) r(w)(x) =. r For f 2 Z K [t 1,..., t n ], write f = P N k=1 ' k(t)e k and suppose that all the coe cients of each ' k are non-negative equivalently, each ' k (t) is a sum of monomials. For r 2 N N, we put r(f) := r 1 (' 1 ) r N (' N ) 2 Z[t]. For h 2 Int(Z K, Z K ) with Gregory-Newton coe h (f) := X r2n N r r(f) 2 Z K [t]. cients r, we put For x 2 {0, 1} n, we have so using (2) we get h (f)(x) = X r r(x) = NY 'k (x), k=1 r r(f)(x) = X r r k '1 (x) 'N (x) r r 1 r N = h(' 1 (x)e ' N (x)e N ) = h(f(x)).

11 INTEGERS: 18 (2018) 11 Step 2: For 1 apple j apple r, let h j 2 Int(Z K, Z K ) have degree pr(b j ) and cover B j. Put ry F := Note that rx deg(f ) apple deg h j (P j ) (mod p) 2 Z K /p[t] = F q [t].! rx nx h j (P j ) apple deg(h j ) deg(' j,k ) = S. k=1 Here is the key observation: for x 2 {0, 1} n, if F (x) 6= 0, then for all 1 apple j apple r we have p - h j (P j )(x) = h j (P j (x)), so P j (x) (mod p dj ) /2 B j, and thus x 2 Z. Step 3: For all 1 apple j apple r we have P j (0) = 0 and h j 2 U(p, 0), so h j (0) /2 p, so F (0) = ry h j (P j (0)) = ry h j (P j (0)) (mod p) = Applying Alon-Füredi to F, we get ry h j (0) (mod p) 6= 0. #Z #{x 2 {0, 1} n F (x) 6= 0} 2 n deg F 2 n S, completing the proof of Theorem 4. References [1] N. Alon and Z. Füredi, Covering the cube by a ne hyperplanes, European J. Combin. 14 (1993), [2] A. Brunyate and P.L. Clark, Extending the Zolotarev-Frobenius approach to quadratic reciprocity, Ramanujan J. 37 (2015), [3] P.-J. Cahen and J.-L. Chabert, Integer-Valued Polynomials, Mathematical Surveys and Monographs, 48, American Mathematical Society, Providence, RI, [4] P.L. Clark, A. Forrow and J.R. Schmitt, Warning s Second Theorem with restricted variables, Combinatorica 37 (2017), [5] P.L. Clark, The Combinatorial Nullstellensätze revisited,electron. J. Combin. 21, no. 4 (2014). Paper P4.15. [6] P.L. Clark, Warning s second theorem with relaxed outputs, pete/main Theorem.pdf [7] A.A. Nečaev, The structure of finite commutative rings with unity, Mat. Zametki 10 (1971), [8] L. Varga, Combinatorial Nullstellensatz modulo prime powers and the parity argument, Electron. J. Combin. 21, no. 4 (2014), Paper [9] E. Warning, Bemerkung zur vorstehenden Arbeit von Herrn Chevalley, Abh. Math. Semin. Univ. Hambg. 11 (1935),