Decomposition of Pascal s Kernels Mod p s

Transcription

1 San Jose State University SJSU ScholarWorks Faculty Publications Mathematics January 2002 Decomposition of Pascal s Kernels Mod p s Richard P. Kubelka San Jose State University, richard.kubelka@ssu.edu Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Richard P. Kubelka. "Decomposition of Pascal s Kernels Mod p s " Integers: Electronic Journal of Combinatorial Number Theory 2002: 1-6. This Article is brought to you for free and open access by the Mathematics at SJSU ScholarWorks. It has been accepted for inclusion in Faculty Publications by an authorized administrator of SJSU ScholarWorks. For more information, please contact scholarworks@ssu.edu.

2 DECOMPOSITION OF PASCAL S KERNELS MOD p s Richard P. Kubelka Department of Mathematics, San Jose State University, San Jose, CA , USA kubelka@math.ssu.edu Received: 12/11/01, Revised: 11/18/02, Accepted: 12/18/02, Published: 12/18/02 Abstract For a prime p we define Pascal s Kernel Kp, s =[kp, s i ] i,=0 as the infinite matrix satisfying kp, s i = 1 p s mod p if is divisible by p s and kp, s i = 0 otherwise. While the individual entries of Pascal s Kernel can be computed using a formula of Kazandzidis that has been known for some time, our purpose here will be to use that formula to explain the global geometric patterns that occur in Kp, s. Indeed, if we consider the finite truncated versions of Kp, s, we find that they can be decomposed into superpositions of tensor products of certain primitive p p matrices. Most of us have seen the beautiful geometric patterns that result when Pascal s Triangle is viewed modulo a prime p. In fact, if we consider Pascal s Rectangle, the infinite matrix R =[r i ] i,=0, where r i =, and its truncated non-infinite versions R n =[r i ] pn 1 i,=0, then we can describe these patterns explicitly in a concise manner by noting that R n R 1 R 1 mod p, 1 where there are n factors on the right. Here signifies the Kronecker tensor product of two matrices [Ser]: if A is an m n matrix and B is an r s matrix, then A B is the mr ns matrix whose ith r s block is a i B. Moreover, if M 0,...,M n are p p matrices and a 0 + a 1 p a n p n and b 0 + b 1 p b n p n are the p-adic expansions of a and b, then M 0 M n ab =M 0 anb n M n a0 b 0 2 The proof of 1 follows readily from 2 and from the result of Lucas [Luc] that a n ai mod p 3 b i=0 b i

3 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY , #A13 2 It should be noted that by the multinomial version of Lucas Theorem [Dick], results similar to 1 hold for Pascal s Simplices, higher dimensional arrays of multinomial coefficients. Modulo powers of a prime p, we still have beautiful geometric patterns, but they are subtler and less easily described. In this paper, as a first step toward describing and explaining these patterns, we will examine Pascal s Kernels, rectangular arrays obtained from those binomial coefficients which are divisible by p s. To be more precise, let us recall the notation employed in [Sing]: if a b = de p e + + d n p n and d e 0, let E a b = e and F a b = de. Now define the Pascal s Kernel Kp, s =[kp, s i ] i,=0 as the infinite matrix satisfying kp, s i = 1 p s mod p if E s, and kp, si = 0 otherwise. Note: kp, s i = d s in the p-adic expansion of a if s e; in particular, if E = s, then kp, s i = F b. As above, define truncated matrices Kp, s, n bykp, s, n = [kp, s i ] pn 1 i,=0. Remark 1. The name Pascal s Kernel is suggested by the fact that the entries of Kp, s, n live in Z p as it is mapped isomorphically onto the kernel of the mod p s reduction map in the exact sequence 0 Z p p s Z p s+1 Z p s Figure 1: K3, 1, 3 Figure 2: K3, 2, 3 Figures 1 and 2 show K3, 1, 3 and K3, 2, 3, for example, with zeros suppressed. Note that our definition implies that for s t, Kp, s, n and Kp, t, n have no overlap. That is, for all i,, at most one of kp, s, n i and kp, t, n i is nonzero. While a formula for F a b has been known for some time [Kaz], it is the purpose of this paper to interpret that formula in a way that explains the geometric patterns that occur

4 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY , #A13 3 in Pascal s Kernels in terms of a superposition of tensor products of certain primitive matrices. Specifically, we have the following Theorem. Let p be a prime. For s>n, Kp, s, n =0. For s n, there exist p p matrices M 1,0, M 0,1, M 0,0, and M 1,1, which depend only on p, such that Kp, s, n is the sum of all tensor products M tn 1,t n 2 M tn 2,t n 3 M t1,t 0 M t0,0 with t k {0, 1} and precisely s of the t 0,t 1,...t n 1 equal 1. Remark 2. There are evidently n s summands in Kp, s, n. The p p matrix M r,s is defined as having i-th entry Φ r,s i, mod p if 0 i + + s pr < p and 0 otherwise, where r i + + s pr! Φ r,s i, = 1. i!! Remark 3. i M 1,0 i 1 p mod p if i + p and M1,0 i = 0 otherwise. In fact, M 1,0 = Kp, 1, 1. ii M 0,1 i = +1! i!! =i + +1 mod p. iii M 0,0 i =! i!! = mod p. In fact, M0,0 = Kp, 0, 1. iv M 1,1 i +1 p mod p if i + p 1 and M1,1 i = 0 otherwise. v M 1,1 i M 0,0 1 p 1 p 1 i = 2p 2 i 1 p 1 i if i + p 1 and M1,1 i =0 otherwise. Therefore M 1,1 is ust M 0,0 reflected across its anti-diagonal with its nonzero entries inverted; here inverses are taken in the field Z/p. vi M 1,0 i M 0,1 1 p 1 p 1 i = 2p 1 i 1 2p 2 i 1 p 1 i if i + p and M 1,0 i = 0 otherwise. Therefore M 1,0 is ust M 0,1 reflected across its antidiagonal with its nonzero entries inverted and negated; inverses and negatives are taken in the field Z/p. Remark 4. As we will see in the course of the proof below, although Kp, s, n is expressed as a sum of matrices, these matrices don t overlap. That is, for each entry of Kp, s, n, at most one of the summands is nonzero in that position. In terms of the geometric picture, Kp, s, n is a mosaic, with little pieces set in the holes between the bigger pieces. Examples. Kp, 0, 1 = M 0,0 and Kp, 1, 1 = M 1,0, as noted above. Kp, 1, 3 = M 1,0 M 0,0 M 0,0 + M 0,1 M 1,0 M 0,0 + M 0,0 M 0,1 M 1,0 4

5 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY , #A13 4 Kp, 2, 3 = M 1,0 M 0,1 M 1,0 + M 0,1 M 1,1 M 1,0 + M 1,1 M 1,0 M 0,0 5 In the case p =3,wehave: M 1,0 = 0 0 1, M 0,1 = , M 0,0 = and the reader may verify 4 and 5 by examining Figures 1 and 2., M 1,1 = Proof of the Theorem. Consider kp, s, n i. Our definition implies that this equals F if E = s and zero otherwise. We will show that if E s, then the i-th entry of every one of the tensor product summands called for in the Theorem must be 0. On the other hand, we will show that if E = s, the i-th entries of all such summands will be 0 except for precisely one, which will equal F. First express i,, and i + p-adically: i = b 0 + b 1 p b n 1 p n 1, = c 0 + c 1 p c n 1 p n 1, and i + = a 0 + a 1 p a n p n, where 0 a k,b k,c k <pas usual and b n and c n are both 0. We take our main tools from [Sing]: i + n e = E = b k + c k a k /p 1 6 i + n F 1 e a k! mod p 7 b k!c k! Remark 5. It is crucial to note as was pointed out in [Sing] that the number e obtained in 6 is exactly the number of carries that we get when we add i and p-adically. For example, if there are no carries, then e =0,a k = b k + c k for all k, and 7 reduces to 3. Considering the p-adic addition of i and more closely, let r k denote the carry digit from the k-th place to the k+1-st place. So a k = b k +c k +r k 1 pr k, with r k,r k 1 {0, 1} and r 1 and r n both 0. By Remark 5, n r k = e, and so 7 becomes i + n F 1 e b k + c k + r k 1 pr k! b k!c k! n 1 r b 8 k k + c k + r k 1 pr k! n = Φ rk,r b k!c k! k 1 b k,c k mod p., That 6 gives the number of carries implies immediately that E n. So if s>n, we must have kp, s, n i =0.

6 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY , #A13 5 Now suppose s n and consider a typical M tn 1,t n 2 M tn 2,t n 3 M t1,t 0 M t0,0 summand of Kp, s, n as called for by the Theorem. By 2 its contribution to kp, s, n i should be M tn 1,t n 2 bn 1 c n 1 M t0,0 b0 c 0 9 with n t k = s. If E = n r k >s, then for some smallest k, r k = 1 and t k = 0. But r k = 1 implies that b k + c k p 1 if k > 0 and r k 1 = 1 also, or b k + c k p if r k 1 = 0. In the former case our assumption that k is minimal implies that M tk,t k 1 = M 0,1. Hence b k + c k p 1 here implies that M tk,t k 1 bk c k = 0. In the latter case, b k + c k p implies that M tk,t k 1 bk c k = 0 regardless of whether t k 1 equals 0 or 1. Both cases therefore yield zero summands, which is exactly what we need. The case where E <sis dispatched by the same argument, with the roles of 0 and 1 for the r k s and t k s reversed. If E = s, we noted earlier that kp, s, ni = F. We will show that a typical summand of kp, s, n i as called for by the Theorem and as given by 9 will be nonzero precisely when t k = r k for k =0,...n 1 and that in that case the value of the summand will be given by the last expression in 8. We start our inductive proof of this by considering the case where b 0 + c 0 p 1, and thus r 0 =0. Ift 0 = 1 in our summand, then M t0,0 b0 c 0 =M 1,0 b0 c 0 = 0, by the definition of M 1,0. On the other hand, if t 0 =0, then M t0,0 b0 c 0 =M 0,0 b0 c 0 =Φ 0,0 b 0,c 0 =Φ r0,r 1 b 0,c 0. Thus, working from right to left, the first factor in 9 will match the k = 0 factor in 8. Moreover, in the case where b 0 + c 0 p, a similar argument will also give us the match of those factors. Now proceeding inductively, we assume that t k = r k for all k<mand so M tk,t k 1 bk c k =M rk,r k 1 bk c k =Φ rk,r k 1 b k,c k 10 for all k<m. Thus the first m factors in 8 match the first m factors in 9 read right-to-left. To advance the induction, we need to show that t m = r m. This and the inductive hypothesis will then guarantee that M tm,tm 1 bmcm =Φ rm,rm 1 b m,c m, and then the first m + 1 factors will match à la 10, and our induction will be complete. The proof that t m = r m proceeds by contradiction, using the same argument with the minimal k replaced by m that we employed above in considering the cases where E >sand E <s. Proof of Remaining Items from Remark 3. For item i, we must show 1 p p! i!! mod p when i + p. This amounts to showing that i +!/p i + p! mod p. 11

7 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY , #A13 6 But i +!/i + p! is a product of p consecutive integers, one from each residue class mod p. Since 0 i, < p, one of those factors will be exactly p and no other will be divisible by p. Therefore i +! p 1! 1 mod p, p i + p! this last congruence being Wilson s Theorem [Edg]. This completes the proof of 11, and hence of item i. A nearly identical argument gives us item iv, that +1 p +1 p! mod p i!! when i + p 1. For items v and vi, note that if H is a p p matrix with rows and columns numbered from 0 to p 1, the matrix H with entries h i = h p 1 p 1 i is H reflected across its anti-diagonal. The proofs of items v and vi combine this fact with straightforward applications of the following Lemma. For 0 m p 1, [p 1 m!] 1 1 p m m! mod p. The proof of the Lemma once again follows directly from Wilson s Theorem. Remark 6. As a final aside, let us note that the matrix M 0,0 has the further property that M 2 0,0 is ust M 0,0 reflected across its anti-diagonal. Furthermore, M 4 0,0 = M 0,0 and, since M 0,0 is invertible, M 3 0,0 = I. See [Stra] for details. Acknowledgement. I would like to thank the referee for suggestions that markedly simplified the statement and proof of the results in this paper. References [Dick] L. E. Dickson, Theorems on the residues of multinomial coefficients with respect to a prime modulus, Quart. J. Pure Appl. Math., , [Edg] H. M. Edgar, A First Course in Number Theory, Wadsworth, Belmont, 1988, 41. [Kaz] G. S. Kazandzidis, Congruences on the binomial coefficients, Bull. Soc. Math. Grèce NS, , [Luc] E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris 1891, [Ser] J.-P. Serre, Linear Representations of Finite Groups, Springer, New York 1977, 8. [Sing] D. Singmaster, Notes on binomial coefficients I a generalization of Lucas congruence, J. London Math. Soc. 2, , [Stra] N. Strauss and I. Gessel, Solution to Advanced Problem #6527, Amer. Math. Monthly, ,