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BINOMIAL COEFFICIENTS AND p-ADIC LIMITS KEITH CONRAD √ √ √ Here are the power series for 1 + x, 3 1 + x, and 6 1 + x: √ 1 1 1 5 4 7 5 21 6 1 + x = 1 + x − x2 + x3 − x + x − x + ··· , 2 8 16 128 256 1024 √ 1 1 5 10 4 22 5 154 6 3 1 + x = 1 + x − x2 + x3 − x + x − x + ··· , 3 9 81 243 729 6561 √ 1 5 55 3 935 4 4301 5 124729 6 6 1 + x = 1 + x − x2 + x − x + x − x + ··· . 6 72 1296 31104 186624 6718464 The coefficients are all rational and √ are written in reduced form. It is not hard to get a formula for the coefficient of xk in n 1 + x = (1 + x)1/n : dk 1 1 1 1 1/n (1 + x) = −1 − 2 ··· − (k − 1) (1 + x)1/n−k n n n n dxk so √ ( n 1 + x)(k) (1/n)(1/n − 1) · · · (1/n − (k − 1)) 1/n = = . k! k! k x=0 These numbers are binomial coefficients evaluated at 1/n. √ Look at the primes in the denominators of the coefficients. In 1 + x each coefficient √ has denominator that is a power of 2, in 3 1√ + x each coefficient has denominator that is a power of 3 (243 = 35 and 729 = 36 ), and in 6 1 + x each coefficient is a power of 2 times a power of 3 (1296 = 24 34 , 31104 = 27 35 , 186624 = 28 36 , and 6718464 = 210 38 ). For r ∈ Q the power series for (1 + x)r at x = 0 has coefficients that are binomial coefficients evaluated at r: X r X r(r − 1)(r − 2) · · · (r − (k − 1)) r k x = xk . (1 + x) = k! k k≥0 k≥0 r What primes occur in the denominators of k ? Theorem 1. For a rational number r, a prime dividing the denominator of kr must divide the denominator of r. Proof. We want to show if a prime p divides the denominator of kr then it divides the denominator of r. Writing r = a/b in reduced form, a(a − b)(a − 2b) · · · (a − kb) a/b (a/b)(a/b − 1)(a/b − 2) · · · (a/b − (k − 1)) = = k! k bk k! and it is obvious that the only possible primes in the denominator are prime factors of b or a prime factor of k!. It is not obvious why every prime factor of k! that is not a factor of b gets completely canceled out when the ratio in kr is simplified. We will explain this purely algebraic phenomenon by using p-adic limits. 1 2 KEITH CONRAD To prove the theorem we will prove its contrapositive: each prime p that does not divide r the denominator of r also does not divide the denominator of any k . Expressed in terms r of p-adic absolute values, this says: if |r|p ≤ 1 then | k |p ≤ 1 for k ≥ 0. To prove this, observe that the expression x x(x − 1)(x − 2) · · · (x − (k − 1)) = k! k is a polynomial with rational coefficients, and therefore it is a continuous function Qp → Qp just as much as it is a continuous function R → R (addition, multiplication, and division in a field are all continuous for any absolute value on the field). When |r|p ≤ 1, r lies in Zp so r is a p-adic limit of nonnegative integers (use the truncations of the p-adic expansion of r): write r = limi→∞ ai where ai ∈ N. By p-adic continuity of polynomials, ai r . = lim i→∞ k k Each aki isan integer since binomialcoefficients with nonnegative integers upstairs are ai combinatorics). Thus integers ( aki = 0 if 0 ≤ ai < k and aki ∈ Z+ if ai ≥ k by k ∈ Zp ai r for each i, and Zp is closed in Qp , so a p-adic limit of k is in Zp . Hence | k |p ≤ 1. 33/20 is expanded out and simplified, the Example 2. When the binomial coefficient 7 denominator can only have prime factors 2 and 5. Explicitly, 352590381 33/20 352590381 =− =− . 102400000000 218 58 7 The theorem we proved admits a converse. Theorem 3. Each prime p that divides the denominator of r also divides the denominator of every kr for k ≥ 1. Obviously this doesn’t hold at k = 0, since 0r = 1. Proof. terms of p-adic absolute values: if |r|p > 1 then Let’s reformulate the theorem in | kr |p > 1 for all k ≥ 1. The top of kr is r(r − 1) · · · (r − (k − 1)), and for each positive integer j we have |r − j|p = |r|p since |r|p > 1 ≥ |j|p so r = r(r − 1)(r − 2) · · · (r − (k − 1)) k k! p = |r|p |r − 1|p |r − 2|p · · · |r − (k − 1)|p |k!|p = |r|kp |k!|p = |r|kp pordp (k!) ≥ |r|kp > 1. √ 6 Example 4. Every Taylor coefficient of 1 + x besides the constant term must have its reduced form denominator divisible by 2 and by 3; it can never be a power of just one of those primes.