Solutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010.

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1 Solutions, Project on p-adic Numbers, Real Analysis I, Fall, (24) The p-adic numbers. Let p {2, 3, 5, 7, 11,... } be a prime number. (a) For x Q, define { 0 for x = 0, x p = p n for x = p n (a/b), where a, b are not multiples of p. Show that x y p is a metric on Q. Solution: It is obvious that x y p is nonnegative, symmetric in x, y, and is 0 if and only if x = y. The remaining thing to check is the Triangle Inequality. So for x, y, z Q, we want to check that x z p x y p + y z p. So if a = x y, b = y z, we want to prove a + b p a p + b p. If either a or b is 0, this is obvious. So we may assume a = s t pn, b = u v pm for s, t, u, v nonzero integers which are not multiples of p, and n, m are integers. In this case, a p = p n and b p = p m. Now compute a + b in two cases: if n m, we may assume without loss of generality that n > m, and then a + b = s t pn + u v pm = svpn m + ut p m tv Thus in this case, since the numerator and denominator of the fraction on the right cannot be multiples of p, a + b p = p m = max( a p, b p ) a p + b p. In the other case, if n = m, then a + b = s t pn + u sv + tu v pn = p n. tv In this case, the denominator tv cannot be a multiple of p, but the numerator sv + tu is of the form p k q for k 0 and q not a multiple of p. Then a + b p = p n k p n = max( a p, b p ) a p + b p. 1

2 (b) Define the p-adic numbers Q p to be the completion of Q with respect to the metric x y p. Let k be an integer. Assume a j {0, 1,..., p 1} for j k. Then show that a j p j converges in Q p. (In other words, show the partial sums of the series are a Cauchy sequence in Q with respect to the p-adic metric.) Solution: Consider the sequence of partial sums n s n = a j p j. If n, m N, then assume without loss of generality that n m. In this case, n s n s m = a j p j. j=m+1 Then, to compute s n s m p, there are a few cases. If m = n or a j = 0 for all j = m + 1,..., n, then s n s m p = 0. If not, there is a smallest l {m + 1,..., n} so that a l 0, and thus in this case s n s m p = a l p l + + a n p n p = p l (a l + + a n p n l ) p = p l. Therefore, in all cases, since l m + 1, s n s m p p (m+1) p (N+1). To recap, for n, m N, s n s m p p (N+1), which goes to 0 as N. Thus the sequence of partial sums {s n } is a Cauchy sequence with respect to the p-adic metric, and so its limit in the completion Q p can be regarded as the infinite sum a jp j. (c) Let { l } T = a j p j k, l Z, k l, aj {0, 1,..., p 1}. Show that T = {sp n s, n are nonnegative integers}. Solution: If sp n T for s, n nonnegative integers, then s can be written as a finite sum (the base-p representation) s = a 0 + a 1 p + + a l p l, where 0 a i p 1. Therefore, 2

3 sp n = a 0 p n + a 1 p n+1 + a l p n+l, which is clearly in T. On the other hand, for v = l a jp j T, we may assume that k 0 by setting the coefficients a 0,..., a k 1 = 0 if k > 0. Then v = p k (a k + a k p l k ) T since k = n for n a nonnegative integer. (d) Show that the completion of T with respect to the p-adic metric x y p may be identified with the set { } X = a j p j k Z, aj {0,..., p 1}. What is the p-adic metric on X? Solution: It will be useful to regard elements of T (finite sums) as elements of X by filling out the remaining coefficients with 0 s. For s, t T, we may similarly assume that that s = a j p j, t = b j p j for the same k. s t p = 0 if and only if a j = b j for all j k. Similarly, s t p = p n if and only if n is the smallest integer for which a n b n, since in this case, we have for some l > n s t = p n [(a n b n ) + (a n+1 b n+1 )p + (a l b l )p l n ]. (Note that a n b n for a n, b n {0,..., p 1} implies a n b n is not divisible by p.) It follows that {t n = j a n,jp j } is a Cauchy sequence in T if and only if for all l, there is an N = N(l) such that a n,j = a m,j for all j l and m, n N. (Proof: Such a {t n } is a Cauchy sequence since for every ɛ > 0, there is an l so that p l 1 < ɛ and then for N = N(l), if m, n N, we have t n t m p p l 1 < ɛ. On the other hand, if {t n } is a Cauchy sequence, then for all l, there is an ɛ > 0 so that ɛ < p l, and thus there is an N = N(ɛ) so that if n, m N, then t n t m p < ɛ < p l. This in turn implies that for all j l we have a n,j = a m,j.) We also remark that by the same consideration, if {t n } is a Cauchy sequence, there is a uniform k, independent 3

4 of n, so that t n = a n,jp j. To prove this, write each t n = n a n,j p j. Then for l = 0, there is an N so that a m,j = a N,j for all j 0 and m N. This implies that for all j min(0, k N ) and m N, that a m,j = 0. Then if we let k = min(0, k 1, k 2,..., k N ), j k implies a m,j = 0 for all m. Now we are in a position to show X is the completion of T. In order to do this, we need to find a metric on X and an isometric injection of T into X whose image is dense and so that every Cauchy sequence in T converges in X. So for x = a j p j, y = b j p j, define x y p = p n, where n is the smallest number so that a n b n. (If all a j = b j, define x y p = 0.) Then by the above computations, the natural injection of T into X (by realizing extending elements of T to power series by introducing zero coefficients of p j for large j) is an injective isometry. Moreover, by the above paragraph, for every Cauchy sequence {t n = j a n,jp j } T, the coefficients all satisfy lim n a n,j = a j (since the coefficient sequences are all eventually constant). Then the previous paragraph shows that for t = a jp j, we have t n t p 0. Thus every Cauchy sequence in T converges to a limit in X. To show T is dense in X, for x = a jp j, the sequence of partial sums of the series (which are elements of T ) converge to x. Thus X is naturally the completion of T with respect to the p-adic metric. (e) Show that if x, y X that x + y X and xy X. Define x+y and xy by addition and multiplication of power series, with the usual carrying operation from decimal arithmetic to handle the case of coefficients p. Solution: This process of multiplication and addition of power series in p obviously produces elements in X. It is also easy to check that multiplication and addition are each commutative and associative, and that multiplication distributes over addition. (f) Show that 1 = (p 1)[1 + p + p 2 + ] X. Solution: Compute 1+[(p 1)+(p 1)p+(p 1)p 2 + ] via the carrying process to be p + (p 1)p + (p 1)p 2 +, 4

5 and then carry to the p place to find 0+p p+(p 1)p 2 +. Carry again to get p p 2 +, and so forth. By indefinitely repeating this process, we find the answer to be = 0. Thus it makes sense to set 1 = (p 1) + (p 1)p + (p 1)p 2 + in X. (g) Define the set Z p to consist of all elements of X for which a j = 0 for all j < 0. (Elements of Z p are called the p-adic integers.) Show that Z p = {x X x p 1}. Solution: This follows by the description of p in part(d). (h) If x j Z p for all j k, show that the sum x j p j converges in X. Solution: By part (g), if x j Z p, it is of the form x j = a m p m m=0 for a j {0,..., p 1}. Therefore, p j x j = a m p m+j. m=0 This implies x j p j p p j. The Triangle Inequality, together with the convergence of the geometric series shows that the partial sums of the series form a Cauchy sequence. (We have checked the Triangle Inequality for rational numbers; to check it is true in X is largely the same.) Since X is complete (by (d)), the series converges in X. (i) If q {1,, p 1}, show there is a unique r {1,..., p 1} so that qr = 1 + ap for some a {0,, p 1}. (This is just the usual fact that Z/pZ is a field.) Solution: Consider the set {0, q, 2q,..., (p 1)q}, and the set of remainders upon dividing each of these by p {0 = s 0,..., s p 1 } {0,..., p 1}. We claim there are no repetitions in the s i : for if s i = s j, for i, j {0,..., p 1} (assume j > i), then jq iq = (j i)q is divisible by the prime p. Since q has no p factors, and j i {0,..., p 1}, this forces j i = 0 and so j = i. Therefore, the {s 0,..., s p 1 } are p distinct elements of {0,..., p 1}. Since these two finite sets have the same cardinality, the 5 j=0 p j,

6 inclusion is a one-to-one correspondence. Therefore, there is a unique r {0, 1,..., p 1} so that s r = 1. (Note r = 0 is impossible.) Therefore, by the definition of the remainder, we have qr = 1 + ap for an integer a. It is easy to see that a {0,..., p 1}. (j) Show that the multiplicative inverse of 1 + ap exists in X and is given by 1 ap + a 2 p 2 a 3 p 3 +. Solution: 1 Z p by (f) and (g). The series 1 ap + a 2 p 2 is then of the form considered in (h), and thus converges in X. Our definition of multiplication in (e) shows, by a standard computation, that the multiplicative inverse of 1 + ap is the given series. (k) Show that any rational number can be represented by a unique element of X. (Hints: Write any rational number as a product of ±1, an element of T, and a reciprocal q 1 of a positive integer q which is not divisible by p. It will also be useful to know that b Z p implies b Z p. To show uniqueness, analyze the power series term by term to ensure the relevant equations are satisfied: For example, we must have for each rational number t that rational number t satisfies t + ( t) = 0.) Solution: We ve already shown the elements of T, of the form sp n for s, n 0 by (c), are in X. All rational numbers can be written in the form ± sp n q 1 for q a positive integer which is not divisible by p. We know that 1 X and (e) shows that if x X, then x = ( 1)x X. It remains to show q 1 X. Since q is not divisible by p, we may write q as an element of T by ( ) q = a 0 + a 1 p + a l p l with a 0 0. Use (i) and (j) to find an inverse for a 0 by finding an r so that a 0 r = 1+ap and thus a 1 0 = r(1+ap) 1. This is in X by (e). Now multiply (*) by a 1 0 to find qa 1 0 = 1 + b 1 p + = 1 + bp for b Z p by (g). Now we may write q 1 = a 0 (1 + bp) 1 = a 0 (1 bp + b 2 p 2 ). The coefficients of the series on the right ( b) k are elements of Z p, since one can easily check that c, d Z p implies cd Z p. Then a 0 (1 bp + b 2 p 2 ) X by (e). Thus 6

7 for all q which are not multiples of p, q 1 is an element of X. Thus by (e) again, ± sp n q 1 X. To show uniqueness, first it is clear that elements of T have unique representatives in X. Then we build in taking a minus sign and multiplying by q 1. The basic idea is that to solve relevant equations involving power series term by term with coefficients in {0,..., p 1}, by using properties such as (i). For example, if t is a rational number defined by a k p k + a k+1 p k+1 +, for a k 0, a j {0,..., p 1}, define t = b k p k + b k+1 p k+1 + so that t + ( t) = 0. Adding the series together, we find for the p k terms, that we must have a k + b k = p, and then a 1 is carried to the p k+1 terms. then we may solve a k+1 + b k = p to determine the next term b k+1, etc. Clearly, this process can be repeated indefinitely to determine all the coefficients of t. On the other hand, we have shown above that t = ( 1)t as in (e) and (f). Thus this is the unique solution. A similar analysis works for multiplying by q 1 to show uniqueness of any rational number. (l) Show that X is the completion of Q with respect to the p-adic metric (and so we may identify X with Q p ). (Hint: Show that for each rational number, the two definitions of p, on Q and on X, agree.) Solution: We have seen X is the completion of T with respect to the p-adic metric, and T is a subset of Q. But then by (k), each rational number can be represented in X as well. To show that X is the completion in Q, we need to check that the p-adic metric for rational numbers is the same as the metric we ve defined on X. So if α, β Q, let the corresponding elements of X be x, y. Then we must check that α β p = x y p. By the algebra of elements of X we have developed (addition of series and multiplying by 1), it suffices to check that γ p = z p for γ = α β Q and z = x y X. For this, write γ = p k a b for a, b not divisible by p, and then γ p = p k. On the other hand, by the techniques in (k), we have p k a b = a pk (±b 1 ) for b a positive integer not divisible by p. In this case, we may check that a (±b 1 ) is a product of elements of Z p 7

8 with nonzero initial coefficient. a = a 0 + a 1 p + + a l p l, 1 = (p 1) +, b 1 = b 0 + b 1 p + Thus their product, by multiplication of power series, is an element of Z p with nonzero initial coefficient (equal to a 0 b 0 modulo p if the element is positive, and a 0 b 0 (p 1) if it s negative). This implies z = p k a (±b 1 ) = p k (c 0 + c 1 p + ) for c 0 0. Thus, for z as an element of X, z p = p k = γ p. Therefore, the two definitions of p agree on Q and X, and thus we may embed Q isometrically into X. Therefore, we have isometric injections (which we write simply using the subset notation) T Q X, and we know that the completion of T is X. Now X, with the p-adic metric, is a complete metric space (since it is the completion of T ), and Q X provides an isometric injection of Q into X. Thus, X satisfies the definition of a completion of Q, and so by uniqueness of the completions, we may say X is the completion of Q, the p-adic numbers Q p. Q T with respect to the p-adic metric. (m) Show that Q p is a field. (So in addition, you must show that nonzero elements have multiplicative inverses.) Solution: If x = a jp j is nonzero, we may assume a k 0. Then as in (l) above, we may write x 1 = p k a 1 k (1 + bp) 1 for b Z p. Then as in (l) again, p k a 1 k (1 + bp) 1 is an element of X. (n) Show that every real number can be represented as a (not quite unique) sum k ± a j p j, k Z, a j {0,..., p 1}, j= which is convergent with respect to the usual metric x y on R. Solution: This is just the base-p representation of a real number. 8