Chapter 8. Padic numbers. 8.1 Absolute values


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1 Chapter 8 Padic numbers Literature: N. Koblitz, padic Numbers, padic Analysis, and ZetaFunctions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap. I,III 8.1 Absolute values The padic 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 nontrivial absolute values on K. 2) The ordinary absolute value on Q is archimedean, while the padic absolute values are all nonarchimedean. 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 nonarchimedean 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 nontrivial absolute value on Q is equivalent to either the ordinary absolute value or a padic absolute value for some prime number p. 124
3 8.2 Completions Let K be a field, a nontrivial 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 nontrivial 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 welldefined, 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 nontrivial, nonarchimedean absolute value on K. Then the extension of to K is also nonarchimedean. 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 padic Numbers and padic integers In everything that follows, p is a prime number. The completion of Q with respect to p is called the field of padic numbers, notation Q p. The continuation of p to Q p is also denoted by p. This is a nonarchimedean 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 padic 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 padic numbers, very small means divisible by a high power of p, and two padic numbers α and β are padically 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 nonzero 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 nonzero 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 padic 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 padic 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 padic numbers given in their padic expansions, one has the same addition with carry algorithm as for real numbers given in their decimal expansions, except that for padic numbers one has to work from left to right instead of right to left. Likewise, one has subtraction and multiplication algorithms for padic 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 3adic 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 3adic 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 2adic expansion of. Notice that = We start 7 with the 2adic expansion of 1. 7 k α k b k So 1 7 = (2), 1 56 = (2). 8.4 The padic 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 padic 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 padic topology has some strange properties. Theorem Let a Q p, m Z. Then B(a, p m ) is both open and compact in the padic 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 HeineBorel 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 nonempty. 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 nonempty open subset of Q p is disconnected. Proof. Let U be an open nonempty 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 nonempty 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 nonnegative 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 socalled 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 nonarchimedean 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 padic expansion of 1. (b) Let α = k=0 b kp k with b k {0,..., p 1} for k 0. Determine the padic expansion of α. Exercise 8.2. Let α Q p, α 0. Prove that α has a finite padic expansion if and only if α = a/p r where a is a positive integer and r a nonnegative 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 welldefined, 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 padic 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 nonarchimedean. (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 padic 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 padic 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 padic exponential function and padic 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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