Download Binomial coefficients and p-adic limits

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.