# Chapter 8. P-adic numbers. 8.1 Absolute values

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1 Chapter 8 P-adic numbers Literature: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap. I,III 8.1 Absolute values The p-adic absolute value p on Q is defined as follows: if a Q, a 0 then write a = p m b/c where b, c are integers not divisible by p and put a p = p m ; further, put 0 p = 0. Example. Let a = Then a 2 = 2 7, a 3 = 3 8, a 5 = 5 3, a p = 1 for p 7. We give some properties: ab p = a p b p for a, b Q ; a + b p max( a p, b p ) for a, b Q (ultrametric inequality). Notice that the last property implies that a + b p = max( a p, b p ) if a p b p. It is common to write the ordinary absolute value a = max(a, a) on Q as a, to call the infinite prime and to define M Q := { } {primes}. Then we 123

2 have the important product formula: p M Q a p = 1 for a Q, a 0. We define more generally absolute values on fields. Let K be any field. An absolute value on K is a function : K R 0 with the following properties: ab = a b for a, b K; a + b a + b for a, b K (triangle inequality); a = 0 a = 0. Notice that these properties imply that 1 = 1. The absolute value is called nonarchimedean if the triangle inequality can be replaced by the stronger ultrametric inequality or strong triangle inequality a + b max( a, b ) for a, b K. An absolute value not satisfying the ultrametric inequality is called archimedean. If K is a field with absolute value and L an extension of K, then an extension or continuation of to L is an absolute value on L whose restriction to K is. Examples. 1) Every field K can be endowed with the trivial absolute value, given by a = 0 if a = 0 and a = 1 if a 0. It is not hard to show that if K is a finite field then there are no non-trivial absolute values on K. 2) The ordinary absolute value on Q is archimedean, while the p-adic absolute values are all non-archimedean. 3) Let K be any field, and K(t) the field of rational functions of K. For a polynomial f K[t] define f = 0 if f = 0 and f = e deg f if f 0. Further, for a rational function f/g with f, g K[t] define f/g = f / g. Verify that this defines a non-archimedean absolute value on K(t). Let K be a field. Two absolute values 1, 2 on K are called equivalent if there is α > 0 such that x 2 = x α 1 for all x K. We state without proof the following result: Theorem 8.1. (Ostrowski) Every non-trivial absolute value on Q is equivalent to either the ordinary absolute value or a p-adic absolute value for some prime number p. 124

3 8.2 Completions Let K be a field, a non-trivial absolute value on K, and {a k } k=0 a sequence in K. We say that {a k } k=0 converges to α with respect to if lim k a k α = 0. Further, {a k } k=0 is called a Cauchy sequence with respect to if lim m,n a m a n = 0. Notice that any convergent sequence is a Cauchy sequence. We say that K is complete with respect to if every Cauchy sequence w.r.t. in K converges to a limit in K. For instance, R and C are complete w.r.t. the ordinary absolute value. Ostrowski proved that any field complete with respect to an archimedean absolute value is isomorphic to R or C. Every field K with an absolute value can be extended to an up to isomorphism complete field, the completion of K. Theorem 8.2. Let K be a field with non-trivial absolute value. There is an up to absolute value preserving isomorphism unique extension field K of K, called the completion of K, having the following properties: (i) can be continued to an absolute value on K, also denoted, such that K is complete w.r.t. ; (ii) K is dense in K, i.e., every element of K is the limit of a sequence from K. Proof. Basically one has to mimic the construction of R from Q or the construction of a completion of a metric space in topology. We give a sketch. Cauchy sequences, limits, etc. are all with respect to. The set of Cauchy sequences in K with respect to is closed under termwise addition and multiplication {a n } + {b n } := {a n + b n }, {a n } {b n } := {a n b n }. With these operations they form a ring, which we denote by R. It is not difficult to verify that the sequences {a n } such that a n 0 with respect to form a maximal ideal in R, which we denote by M. Thus, the quotient R/M is a field, which is our completion K. We define the absolute value α of α K by choosing a representative {a n } of α, 125

4 and putting α := lim n a n, where now the limit is with respect to the ordinary absolute value on R. It is not difficult to verify that this is well-defined, that is, the limit exists and is independent of the choice of the representative {a n }. We may view K as a subfield of K by identifying a K with the element of K represented by the constant Cauchy sequence {a}. In this manner, the absolute value on K constructed above extends that of K, and moreover, every element of K is a limit of a sequence from K. So K is dense in K. One shows that K is complete, that is, any Cauchy sequence {a n } in K has a limit in K, by taking very good approximations b n K of a n and then taking the limit of the b n. Finally, if K is another complete field with absolute value extending that on K such that K is dense in K one obtains an isomorphism from K to K as follows: Take α K. Choose a sequence {a k } in K converging to α; this is necessarily a Cauchy sequence. Then map α to the limit of {a k } in K. Corollary 8.3. Assume that is a non-trivial, non-archimedean absolute value on K. Then the extension of to K is also non-archimedean. Proof. Let a, b K. Choose sequences {a k }, {b k } in K that converge to a, b, respectively. Then a + b = lim k a k + b k lim k max( a k, b k ) = max( a, b ). 8.3 p-adic Numbers and p-adic integers In everything that follows, p is a prime number. The completion of Q with respect to p is called the field of p-adic numbers, notation Q p. The continuation of p to Q p is also denoted by p. This is a non-archimedean absolute value on Q p. Convergence, limits, Cauchy sequences and the like will all be with respect to p. As mentioned before, by identifying a Q with the class of the constant Cauchy sequence {a}, we may view Q as a subfield of Q p. Lemma 8.4. The value set of p on Q p is {0} {p m : m Z}. 126

5 Proof. Let x Q p, x 0. Choose again a sequence {x k } in Q converging to x. Then x p = lim k x k p. For k sufficiently large we have x k p = p m k for some m k Z. Since the sequence of numbers p m k converges we must have mk = m Z for k sufficiently large. Hence x p = p m. The set Z p := {x Q p : x p 1} is called the ring of p-adic integers. Notice that if x, y Z p then x y p max( x p, y p ) 1. Hence x y Z p. Further, if x, y Z p then xy p 1 which implies xy Z p. So Z p is indeed a ring. Viewing Q as a subfield of Q p, we have Z p Q = { a b : a, b Z, p b}. It is not hard to show that the group of units of Z p, these are the elements x Z p with x 1 Z p, is equal to Z p = {x Q p : x p = 1}. Further, M p := {x Q p : x p < 1} is an ideal of Z p. In fact, M p is the only maximal ideal of Z p since any ideal of Z p not contained in M p contains an element of Z p, hence generates the whole ring Z p. Noting x p < 1 x p p 1 x/p p 1 x/p Z p for x Q p, we see that M p = pz p. For α, β Q p we write α β (mod p m ) if (α β)/p m Z p. This is equivalent to α β p p m. Notice that if α = a 1 b, β = a 2 1 b with a 1, b 1, a 2, b 2 Z and p b 1 b 2, 2 then a 1 a 2 (mod p m ), b 1 b 2 (mod p m ) = α β (mod p m ). For p-adic numbers, very small means divisible by a high power of p, and two p-adic numbers α and β are p-adically close if and only if α β is divisible by a high power of p. Lemma 8.5. For every α Z p and every positive integer m there is a unique a m Z such that α a m p p m and 0 a m < p m. Hence Z is dense in Z p. Proof. There is a rational number a/b (with coprime a, b Z) such that α (a/b) p p m since Q is dense in Q p. At most one of a, b is divisible by p and 127

6 it cannot be b since a/b p 1. Hence there is an integer a m with ba m a (mod p m ) and 0 a m < p m. Thus, α a m p max( α (a/b) p, (a/b) a m p ) p m. This shows the existence of a m. As for the unicity, if a m is another integer with the properties specified in the lemma, we have a m a m p p m, hence a m a m (mod p m ), implying a m = a m. Theorem 8.6. The non-zero ideals of Z p are p m Z p (m = 0, 1, 2,...) and Z p /p m Z p = Z/p m Z. In particular, Z p /pz p = Fp. Proof. Let I be a non-zero ideal of Z p and choose α I for which α p is maximal. Then α p = p m with m Z 0. We have p m α Z p, hence p m I. Further, for β I we have βp m p 1, hence β p m Z p. Hence I p m Z p. This implies I = p m Z p. The homomorphism Z/p m Z Z p /p m Z p : a (mod p m ) a (mod p m ) is clearly injective. and also surjective in view of Lemma 8.5. Hence Z/p m Z = Z p /p m Z p. Lemma 8.7. Let {a k } k=0 be a sequence in Q p. Then k=0 a k converges in Q p if and only if lim k a k = 0. Further, every convergent series in Q p is unconditionally convergent, i.e., neither the convergence, nor the value of the series, are affected if the terms a k are rearranged. Proof. Suppose that α := k=0 a k converges. Then a n = n n 1 a k a k α α = 0. k=0 k=0 Conversely, suppose that a k 0 as k. Let α n := n k=0 a k. Then for any integers m, n with 0 < m < n we have α n α m p = n k=m+1 a k p max( a m+1 p,..., a n p ) 0 as m, n. So the partial sums α n form a Cauchy sequence, hence must converge to a limit in Q p. 128

7 To prove the second part of the lemma, let σ be a bijection from Z 0 to Z 0. We have to prove that k=0 a σ(k) = k=0 a k. Equivalently, we have to prove that M k=0 a k M k=0 a σ(k) 0 as M, i.e., for every ε > 0 there is N such that M a k k=0 M a σ(k) p < ε for every M > N. k=0 Let ε > 0. There is N such that a k p < ε for all k N. Choose N 1 > N such that {σ(0),..., σ(n 1 )} contains {0,..., N} and let M > N 1. Then in the sum S := M k=0 a k M k=0 a σ(k), only terms a k with k > N and a σ(k) with σ(k) > N occur. Hence each term in S has p-adic absolute value < ε and therefore, by the ultrametric inequality, S p < ε. We now show that every element of Z p has a Taylor series expansion, and every element of Q p a Laurent series expansion where instead of powers of a variable X one takes powers of p. Theorem 8.8. (i) Every element of Z p can be expressed uniquely as k=0 b kp k with b k {0,..., p 1} for k 0 and conversely, every such series belongs to Z p. (ii) Every element of Q p can be expressed uniquely as k= k 0 b k p k with k 0 Z, b k {0,..., p 1} for k k 0 and b k0 0 and conversely, every such series belongs to Q p. Proof. We first prove part (i). First observe that by Lemma 8.7, a series k=0 b kp k with b k {0,..., p 1} converges in Q p. Further, it belongs to Z p, since k=0 b kp k p max k 0 b k p k p 1. Let α Z p. Define sequences {α k } k=0 in Z p, {b k } k=0 in {0,..., p 1} inductively as follows: α 0 := α; (8.1) For k = 0, 1,..., let b k {0,..., p 1} be the integer with α k b k (mod p) and put α k+1 := (α k b k )/p. By induction on k, one easily deduces that for k 0, α k Z p, α = k b j p j + p k+1 α k. j=0 129

8 Hence α k j=0 b jp j p p k 1 for k 0. It follows that α = lim k k b j p j = j=0 b j p j. j=0 Notice that the integer a m from Lemma 8.5 is precisely m 1 k=0 b kp k. Since a m is uniquely determined, so must be the integers b k. We prove part (ii). As above, any series k= k 0 b k p k with b k {0,..., p 1} converges in Q p. Let α Q p with α 0. Suppose that α p = p k 0. Then β := p k 0 α has β p = 1, so it belongs to Z p. Applying (i) to β we get α = p k 0 β = p k 0 with c k {0,..., p 1} which implies (ii). Corollary 8.9. Z p is uncountable. Proof. Apply Cantor s diagonal method. We use the following notation: α = 0. b 0 b 1... (p) α = b k0 b 1. b 0 b 1... (p) c k p k k=0 if α = k=0 b kp k, if α = k= k 0 b k p k with k 0 < 0. We can describe various of the definitions given above in terms of p-adic expansions. For instance, for α Q p we have α p = p m where α = k=m b kp k with b k {0,..., p 1} for k m and b m 0. next, if α = k=0 a kp k, β = k=0 b kp k Z p with a k, b k {0,..., p 1}, then α β (mod p m ) a k = b k for k < m. For p-adic numbers given in their p-adic expansions, one has the same addition with carry algorithm as for real numbers given in their decimal expansions, except that for p-adic numbers one has to work from left to right instead of right to left. Likewise, one has subtraction and multiplication algorithms for p-adic numbers which are precisely the same as for real numbers apart from that one has to work from left to right instead of right to left. 130

9 Theorem Let α = k= k 0 b k p k with b k {0,..., p 1} for k k 0. Then α Q {b k } k= k 0 is ultimately periodic. Proof. = Exercise. = Without loss of generality, we assume that α Z p (if α Q p with α p = p k 0, say, then we proceed further with β := p k 0 α which is in Z p ). Suppose that α = A/B with A, B Z, gcd(a, B) = 1. Then p does not divide B (otherwise α p > 1). Let C := max( A, B ). Let {α k } k=0 be the sequence defined by (8.1). Notice that α k determines uniquely the numbers b k, b k+1,.... Claim. α k = A k /B with A k Z, A k C. This is proved by induction on k. For k = 0 the claim is obviously true. Suppose the claim is true for k = m where m 0. Then α m+1 = α m b m p = (A m b m B)/p. B Since α m b m (mod p) we have that A m b m B is divisible by p. So A m+1 := (A m b m B)/p Z. Further, A m+1 C + (p 1)B p C. This proves our claim. Now since the integers A k all belong to { C,..., C}, there must be indices l < m with A l = A m, that is, α l = α m. But then, b k+m l = b k for all k l, proving that {b k } k=0 is ultimately periodic. Examples. (i) We determine the 3-adic expansion of 2. We compute the numbers 5 α k, b k according to (8.1). Notice that 1 2 (mod 3). 5 k α k b k

10 It follows that the sequence of 3-adic digits {b k } k=0 of 2 is periodic with period 5 2, 1, 0, 1 and that 2 5 = = (2) = (2). 1 1 (ii) We determine the 2-adic expansion of. Notice that = We start 7 with the 2-adic expansion of 1. 7 k α k b k So 1 7 = (2), 1 56 = (2). 8.4 The p-adic topology The ball with center a Q p and radius r in the value set {0} {p m : m Z} of p is defined by B(a, r) := {x Q p : x a p r}. Notice that if b B(a, r) then b a p r. So by the ultrametric inequality, for x B(a, r) we have x b p max( x a p, a b p ) r, i.e. x B(b, r). So B(a, r) B(b, r). Similarly one proves B(b, r) B(a, r). Hence B(a, r) = B(b, r). In other words, any point in a ball can be taken as center of the ball. We define the p-adic topology on Q p as follows. A subset U of Q p is called open if for every a U there is m > 0 such that B(a, p m ) U. It is easy to see that this topology is Hausdorff: if a, b are distinct elements of Q p, and m is an integer with p m < a b p, then the balls B(a, p m ) and B(b, p m ) are disjoint. But apart from this, the p-adic topology has some strange properties. Theorem Let a Q p, m Z. Then B(a, p m ) is both open and compact in the p-adic topology. Proof. The ball B(a, p m ) is open since for every b B(a, p m ) we have B(b, p m ) = B(a, p m ). 132

11 To prove the compactness we modify the proof of the Heine-Borel theorem stating that every closed bounded set in R is compact. Assume that B 0 := B(a, p m ) is not compact. Then there is an infinite open cover {U α } α A of B 0 no finite subcollection of which covers B 0. Take x B(a, p m ). Then (x a)/p m p 1. Hence there is b {0,..., p 1} such that x a b (mod p). But then, x B(a + bp m, p m 1 ). So p m B(a, p m ) = p 1 b=0 B(a + bpm, p m 1 ) is the union of p balls of radius p m 1. It follows that there is a ball B 1 B(a, p m ) of radius p m 1 which can not be covered by finitely many sets from {U α } α A. By continuing this argument we find an infinite sequence of balls B 0 B 1 B 2, where B i has radius p m i, such that B i can not be covered by finitely many sets from {U α } α A. We show that the intersection of the balls B i is non-empty. For i 0, choose x i B i. Thus, B i = B(x i, p m i ). Then {x i } i 0 is a Cauchy sequence since x i x j p p m min(i,j) 0 as i, j. Hence this sequence has a limit x in Q p. Now we have x i x p = lim j x i x j p p m i, hence x B i, and so B i = B(x, p m i ) for i 0. The point x belongs to one of the sets, U, say, of {U α } α A. Since U is open, for i sufficiently large the ball B i must be contained in U. This gives a contradiction. Corollary Every non-empty open subset of Q p is disconnected. Proof. Let U be an open non-empty subset of Q p. Take a U. Then B := B(a, p m ) U for some m Z. By increasing m we can arrange that B is strictly smaller than U. Now B is open and also U \ B is open since B is compact. Hence U is the union of two non-empty disjoint open sets. 8.5 Algebraic extensions of Q p We fix an algebraic closure Q p of Q p. We construct an extension of p to Q p. For polynomials f, g Z p [X] we write f g (mod p m ) if p m (f g) Z p [X]. Given f Z p [X] and a sequence of polynomials f m Z p [X] (m = 1, 2,...), we write lim m f m = f if for every k 0, the sequence of coefficients of X k in f m converges to the coefficient of X k in f. Clearly, lim m f m = f if and only if there is a sequence of non-negative integers a m with lim m a m (in R) such that f m f (mod p am ). 133

12 An important tool is the so-called Hensel s Lemma, which gives a method to derive, from a factorization of a polynomial f Z p [X] modulo p, a factorization of f in Z p [X]. Theorem Let f, g 1, h 1 be polynomials in Z p [X] such that f 0, f g 1 h 1 (mod p), gcd(g 1, h 1 ) 1 (mod p), g 1 is monic, 0 < deg g 1 < deg f, deg g 1 h 1 deg f. Then there exist polynomials g, h Z p [X] such that f = gh, g g 1 (mod p), h h 1 (mod p), g is monic, deg g = deg g 1. Proof. By induction on m, we prove that there are polynomials g m, h m Z p [X] such that { f gm h (8.2) m (mod p m ), g m g 1 (mod p), h m h 1 (mod p), g m is monic, deg g m = deg g 1, deg g m h m deg f. For m = 1 this follows from our assumption. Let m 2, and suppose that there are polynomials g m 1, h m 1 satisfying (8.2) with m 1 instead of m. We try to find u, v Z p [X] such that g m = g m 1 + p m 1 u, h m = h m 1 + p m 1 v satisfy (8.2). By assumption, A := p 1 m (f g m 1 h m 1 ) Z p [X]. Notice that f g m h m (mod p m ) if and only if f (g m 1 + p m u)(h m 1 + p m v) 0 (mod p m ) A vg m 1 + uh m 1 (mod p) A vg 1 + uh 1 (mod p). Thanks to our assumption gcd(g 1, h 1 ) 1 (mod p) such u, v exist, and in fact, we can choose u with deg u < deg g 1. Then clearly, g m = g m 1 + p m 1 u, h m = h m 1 + p m 1 v satisfy (8.2). Now for each term X k, the coefficients of X k in the g m form a Cauchy sequence, hence have a limit, so we can take g := lim m g m. Then g is monic, and 0 < deg g < deg f. Likewise, we can define h := lim m h m. Then This completes our proof. f gh = lim m (f g mh m ) =

13 Corollary Let f = a 0 X n + a 1 X n a n Q p [X] be irreducible. Put M := max( a 0 p,..., a n p ). Let k be the smallest index i such that a i p = M. Then k = 0 or k = n. Proof. Assume that 0 < k < n. So a i p < a k p for i < k and a i p a k p for i k. Put f := b 1 k f. Then f = b 0 X n + + b k 1 X n k+1 + X n k + b k+1 X n k b n with b i p < 1 for i < k and b i p 1 for i > k. Now f Z p [X], b 0,..., b k 1 are divisible by p, and thus, f (X n k + b k+1 X n k b n ) 1 (mod p). By applying Hensel s Lemma, we infer that there are polynomials g, h Z p [X] such that f = gh and deg g = n k. Then f, hence f, is reducible, contrary to our assumption. We are now ready to define an extension of p to Q p. Given α Q p, let f = X n + a 1 X n a n Q p [X] be the monic minimal polynomial of α over Q p, that is the monic polynomial in Q p [X] of smallest degree having α as a root. Then we put α p := a n 1/n p. Let α (1) = α,..., α (n) be the conjugates of α, i.e., the roots of f in Q p. Let L be any finite extension of Q p containing α, and suppose that [L : Q p ] = m. Then L has precisely m embeddings in Q p that leave the elements of Q p unchanged, say σ 1,..., σ m. Now in the sequence σ 1 (α),..., σ m (α), each of the conjugates α (1),..., α (n) occurs precisely m/n times. Define the norm N L/Qp (α) := σ 1 (α) σ m (α). Then α p = a n 1/n p = α (1) α (n) 1/n p = N L/Qp (α) p 1/[L:Qp]. In case that α Q p, the minimal polynomial of α is X α, and thus we get back our already defined α p. Theorem p defines a non-archimedean absolute value on Q p. 135

14 Proof. Let α, β Q p, and take L = Q p (α, β). Then αβ p = N L/Qp (αβ) 1/[L:Qp] p = N L/Qp (α) 1/[L:Qp] p N L/Qp (β) 1/[L:Qp] p = α p β p. To prove that α + β p max( α p, β p ), assume without loss of generality that α p β p and put γ := α/β. Then γ p 1, and we have to prove that 1+γ p 1. Let f = X n + a 1 X n a n be the minimal polynomial of γ over Q p. Then a n p = γ n p 1, and by Corollary 8.14, also a i p 1 for i = 1,..., n 1. Now the minimal polynomial of γ + 1 is f(x 1) = X n + + f( 1) and so as required. γ + 1 p = f( 1) 1/n p = ( 1) n + a 1 ( 1) n a 0 1/n p max(1, a 1 p,..., a n p ) 1/n 1, We recall Eisenstein s irreducibility criterion for polynomials in Z p. Lemma Let f(x) = X n + a 1 X n a n 1 X + a n Z p [X] be such that a i 0 (mod p) for i = 1,..., n, and a n 0 (mod p 2 ). Then f is irreducible in Q p [X]. Proof. Completely similar as the Eisenstein criterion for polynomials in Z[X]. Example. Let α be a zero of X 3 8X + 10 in Q 2. The polynomial X 3 8X + 10 is irreducible in Q 2 [X], hence it is the minimal polynomial of α. It follows that α 2 = 10 1/3 2 = 2 1/3. We finish with some facts which we state without proof. Theorem (i) Let K be a finite extension of Q p. Then there is precisely one absolute value on K whose restriction to Q p is p, and this is given by N K/Qp ( ) 1/[K:Qp] p. Further, K is complete with respect to this absolute value. (ii) Q p is not complete with respect to p. (iii) The completion C p of Q p with respect to p is algebraically closed. 136

15 8.6 Exercises In the exercises below, p always denotes a prime number and convergence is with respect to p. Exercise 8.1. (a) Determine the p-adic expansion of 1. (b) Let α = k=0 b kp k with b k {0,..., p 1} for k 0. Determine the p-adic expansion of α. Exercise 8.2. Let α Q p, α 0. Prove that α has a finite p-adic expansion if and only if α = a/p r where a is a positive integer and r a non-negative integer. Exercise 8.3. Let α = k= k 0 b k p k Q p where b k {0,..., p 1} for k k 0 and b k0 0. Suppose that the sequence {b k } k= k 0 is ultimately periodic, i.e., there exist r, s with r k 0, s > 0 such that a k+s = a k for all k r. Prove that α Q. Exercise 8.4. Let α Z p with α 1 p p 1. In this exercise you are asked to define α x for x Z p and to show that this exponentiation has the expected properties. You may use without proof that the limit of the sum, product etc. of two sequences in Z p is the sum, product etc. of the limits. (a) Prove that α p 1 α 1 p p 1. (b) Let u be a positive integer. Prove that α u 1 p u p α 1 p. Hint. Write u = p m b where b is not divisible by p and use induction on m. (c) Let u, v be positive integers. Prove that α u α v p u v p α 1 p. (d) We now define α x for x Z p as follows. Take a sequence of positive integers {a k } k=0 such that lim k a k = x and define α x := lim k α a k. Prove that this is well-defined, i.e., the limit exists and is independent of the choice of the sequence {a k } k=0. (e) Prove that for x, y Z p we have α x α y p x y p α 1 p. (Hint. Take sequences of positive integers converging to x, y.) Then show that if {x k } k=0 is a sequence in Z p such that lim k x k = x then lim k α x k = α x (so the function x α x is continuous). 137

16 (f) Prove the following properties of the above defined exponentiation: (i) (αβ) x = α x β x for α, β Z p, x Z p with α 1 p p 1, β 1 p p 1 ; (ii) α x+y = α x α y, (α x ) y = α xy for α Z p with α 1 p p 1, x, y Z p. Remark. In 1935, Mahler proved the following p-adic analogue of the Gel fond- Schneider Theorem: let α, β be elements of Z p, both algebraic over Q, such that α 1 p p 1 and β Q. Then α β is transcendental over Q. Exercise 8.5. Denote by C((t)) the field of formal Laurent series k=k 0 b k t k with k 0 Z, b k C for k k 0. We define an absolute value 0 on C((t)) by 0 0 := 0 and α 0 := c k 0 (c > 1 some constant) where α = k=k 0 b k t k with b k0 0. This absolute value is clearly non-archimedean. (a) Prove that C((t)) is complete w.r.t. 0. (b) Define 0 on the field of rational functions C(t) by 0 0 := 0 and α 0 := c k 0 if α 0, where k 0 is the integer such that α = t k 0 f/g with f, g polynomials not divisible by t. Prove that C((t)) is the completion of C(t) w.r.t. 0. Exercise 8.6. In this exercise you are asked to work out a p-adic analogue of Newton s method to approximate the roots of a polynomial (which is in fact a special case of Hensel s Lemma). Let f = a 0 X n + + a n Z p [X]. The derivative of f is f = na 0 X n a n 1. (a) Let a, x Z p and suppose that x 0 (mod p m ) for some positive integer m. Prove that f(a + x) f(a) (mod p m ) and f(a + x) f(a) + f (a)x (mod p 2m ). Hint. Use that f(a + X) Z p [X]. 138

17 (b) Let x 0 Z such that f(x 0 ) 0 (mod p), f (x 0 ) 0 (mod p). Define the sequence {x n } n=0 recursively by x n+1 := x n f(x n) f (x n ) (n 0). Prove that x n Z p, f(x n ) 0 (mod p 2n ), f (x n ) 0 (mod p) for n 0. (c) Prove that x n converges to a zero of f in Z p. (d) Prove that f has precisely one zero ξ Z p such that ξ x 0 (mod p). Exercise 8.7. In this exercise, p is a prime > 2. (a) Let d be a positive integer such that d 0 (mod p) and x 2 solvable. Show that x 2 = d is solvable in Z p. d (mod p) is (b) Let a, b be two positive integers such that none of the congruence equations x 2 a (mod p), x 2 b (mod p) is solvable in x Z. Prove that ax 2 b (mod p) is solvable in x Z. Hint. Use that the multiplicative group (Z/pZ) is cyclic of order p 1. This implies that there is an integer g such that (Z/pZ) = {g m mod p : m = 0,..., p 2}. (c) Let K be a quadratic extension of Q p. Prove that K = Q p ( d) for some d Z p. Next, prove that Q p ( d 1 ) = Q p ( d 2 ) if and only if d 1 /d 2 is a square in Q p. (d) Determine all quadratic extensions of Q 5. (e) Prove that for any prime p > 2, Q p has up to isomorphism only three distinct quadratic extensions. Exercise 8.8. (a) Prove that x p 1 = 1 has precisely p 1 solutions in Z p, and that these solutions are different modulo p. (b) Let S consist of 0 and of the solutions in Z p of x p 1 = 1. Let α Z p. Prove that for any positive integer m, there are ξ 0,..., ξ m 1 S such that α m 1 k=0 ξ kp k (mod p m ). Then prove that there is a sequence {ξ k } k=0 in S such that α = k=0 ξ kp k. (This is called the Teichmüller representation of α). 139

18 Exercise 8.9. In this exercise you may use the following facts on p-adic power series (the coefficients are always in Q p, and m, m Z). 1) Suppose f(x) = n=0 a n(x x 0 ) n, g(x) = n=0 b n(x x 0 ) n converge and are equal on B(x 0, p m ). Then a n = b n for all n 0. 2) Suppose that for x B(x 0, p m ), f(x) = n=0 a n(x x 0 ) n converges and f(x) f(x 0 ) p p m. Further, suppose that g(x) = n=0 b n(x f(x 0 )) n converges on B(f(x 0 ), p m ). Then the composition g(f(x)) can be expanded as a power series n=0 c n(x x 0 ) n which converges on B(x 0, p m ). 3) We define the derivative of f(x) = n=0 a n(x x 0 ) n by f (x) := na n (x x 0 ) n 1. n=1 If f converges on B(x 0, p m ) then so does f. The derivative satisfies the same sum rules, product rule, quotient rule and chain rule as the derivative of a function on R, e.g., g(f(x)) = g (f(x))f (x). Now define the p-adic exponential function and p-adic logarithm by exp p x := n=0 x n n!, log p x := ( 1) n 1 n=1 n (x 1) n. Further, let r = 1 if p > 2, r = 2 if p = 2. Prove the following properties. (a) Prove that exp p (x) converges and exp p (x) 1 p = x p for x B(0, p r ). Hint. Prove that x n /n! p 0 as n, and x n /n! p < x p for n 2. (b) Prove that log p (x) converges and log p x p = x 1 p for x B(1, p r ). (c) Prove that exp p (x + y) = exp p (x) exp p (y) for x, y B(0, p r ). Hint. Fix y and consider the function in x, f(x) := exp p (y) 1 exp p (x + y). Then f(x) can be expanded as a power series n=0 a nx n. Its derivative f (x) can be computed in the same way as one should do it for real or complex functions. This leads to conditions on the coefficients a n. (d) Prove that log p (xy) = log p (x) + log p (y) for x, y B(1, p r ). 140

19 (e) Prove that log p (exp p x) = x for x B(0, p r ). (f) Prove that exp p (log p x) = x for x B(1, p r ). 141

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