PMATH 940: p-ADIC ANALYSIS NOTES HEE-SUNG YANG Abstract. Electronic version of class notes for PMATH 940: p-adic Analysis. 1. Introduction 1.1. Introduction to valuation. Definition 1. A map |·| from field K to R is said to be a valuation if (i) (Positive-semidefinite) For all a ∈ K, |a| ≥ 0, and |a| = 0 iff a = 0. (ii) (Multiplicativity) For all a,b ∈ K, |ab| = |a|·|b|. (iii) There exists C > 0 such that for all a ∈ K with |a| ≤ 1, then |1+a| ≤ C. Remark 2. (iii) is often replaced by the triangle inequality. Example 3. The ordinary absolute value |·| on C. In this case, we can take C = 2. Example 4. The p-adic valuation |·| on C is defined in the following manner: Let p be a p prime number. Let ord a (for a ∈ Z) be the largest power of p dividing a. Extend this idea p to Q by putting ord (a/b) = ord a−ord b. Now define |·| on Q by letting |0| = 0 and p p p p p |a/b| = p−ordp(a/b). In this case, we can take C = 1. Thus p-adic valuation is an example of p a non-Archimedean valuation, which will be discussed in Section 1.3 in greater detail. Example 5. Let k be a field and consider K = k(T), where T is transcendental over k. Let λ ∈ R with 0 < λ < 1. Let p(T) ∈ k[T] be irreducible. Observe that every non- zero element of K has a representation of the form h(T) = p(T)qf(T) where q ∈ Z and g(T) (f,p) = (g,p) = 1. Note that q is uniquely determined. Define | · | on K by |0| = 0 and |h(T)| = λq. Axioms 1(i) and (ii) are immediate. For (iii), note that if |h| ≤ 1, then q ≥ 0 (cid:12) (cid:12) so |1+h| = (cid:12)1+p(T)qf(T)(cid:12) ≤ 1. We can take C = 1. (cid:12) g(T)(cid:12) Example 6. Let K be any field, and put a valuation |·| known as the trivial valuation, i.e., 0 |0| = 0 and |a| = 1 for all nonzero a ∈ K. We can (trivially) take C = 1. 0 0 Example 7. Let K = k(T) as in Example 5. Let γ ∈ R with γ > 1. We define |·| first on k[T]. If f(T) = a +a T +···+a Tn (a (cid:54)= 0) 0 1 n n and a ∈ K for all i, we put |f| = γn. i (cid:12) (cid:12) Extend this to elements f(T)/g(T) in K with g (cid:54)= 0 by putting (cid:12)f(cid:12) = |f|. As always, (cid:12)g(cid:12) |g| |0| = 0. One can check that |·| satisfies (iii) with C = 1. Date: 25 November 2014. 1 1.2. Properties of valuation. (1) |1| = 1 (Note that |1| = |1·1| = |1|·|1|.) (2) If |an| = 1, then |a| = 1. Note that |−1| = 1 and |−a| = a for all a ∈ K. (3) If K = F then the only valuation on K is the trivial valuation, by (2). p (4) If |·| is a valuation on K and λ ∈ R+, then |·| defined by |a| := |a|λ for all a ∈ K 1 1 is also a valuation of K: take C = Cλ. 1 Definition 8. If |·| and |·| are valuations on a field K, then we say they are equivalent if 1 there exists λ ∈ R+ such that |a|λ = |a| for all a ∈ K. This gives us an equivalence class 1 on a field K of valuations, and such an equivalence class of valuations is known as a place of K. Lemma 9. A valuation |·| on K satisfies the triangle inequality if and only if for all a ∈ K with |a| ≤ 1, we have |1+a| ≤ 2. Proof. (⇐) Suppose a ,a ∈ K. If a = 0 or a = 0 then clearly |a +a | ≤ 2max(|a |,|a |). 1 2 1 2 1 2 1 2 Suppose neither is zero, and without loss of generality, assume |a | ≥ |a |. Then |a +a | = 1 2 1 2 |a | · |1 + a /a | ≤ 2|a | = 2max(|a |,|a |). Thus, we have |a + a | ≤ 2max(|a |,|a |) for 1 2 1 1 1 2 1 2 1 2 any a ,a ∈ K. Now apply induction to a ,a ,...,a to derive 1 2 1 2 2n |a +a +···+a | ≤ 2nmax(|a |,|a |,...,|a |). 1 2 2n 1 2 2n Givena ,a ,...,a ∈ K whereN issufficientlylarge,wecanchoosensothat2n−1 < N ≤ 2n 1 2 N and define a = a = ··· = a = 0. Then for any a ,a ,...,a ∈ K, N+1 N+2 2n 1 2 N |a +a +···+a | ≤ 2nmax(|a |,|a |,...,|a |) ≤ 2N max (|a |), 1 2 N 1 2 N j 1≤j≤N from which we can take a = a = ··· = a to derive |N| ≤ 2N. Let b,c inK and n ∈ Z . 1 2 N + Then (cid:12) (cid:12) (cid:12)(cid:88)n (cid:18)n(cid:19) (cid:12) (cid:12)(cid:12)(cid:18)n(cid:19) (cid:12)(cid:12) |b+c|n = |(b+c)n| = (cid:12) brcn−r(cid:12) ≤ 2(n+1)max(cid:12) brcn−r(cid:12) (cid:12)(cid:12) r (cid:12)(cid:12) r (cid:12) r (cid:12) r=0 (cid:12)(cid:18) (cid:19)(cid:12) (cid:12) n (cid:12) ≤ 4(n+1)max(cid:12) (cid:12)|b|r|c|n−r (since |N| ≤ 2N) r (cid:12) r (cid:12) n (cid:18) (cid:19) (cid:88) n ≤ 4(n+1) |b|r|c|n−r r r=0 ≤ 4(n+1)(|b|+|c|)n. Thus, it follows that |b+c| ≤ (4(n+1))1/n(|b|+|c|), and letting n → ∞ gives us the triangle inequality. (⇒) This is immediate, since |1+a| ≤ |1|+|a| ≤ 2. (cid:3) Corollary 10. Every valuation on K is equivalent to a valuation satisfying the triangle inequality. 1.3. Non-Archimedean valuation. Definition 11. A valuation |·| on a field K is said to be non-Archimedean if we can choose C = 1 in Definition 1(iii). Otherwise, it is called Archimedean. 2 Remark 12. Observe that any valuation equivalent to a non-Archimedean valuation is also non-Archimedean. Lemma 13. A valuation on K is non-Archimedean if and only if | · | satisfies the strong triangle inequality for all a,b ∈ K. Proof. (⇒) Without loss of generality, suppose a,b ∈ K with |a| ≤ |b|. Note that we have |a+b| = |b(1+ a)| = |b||1+ a| ≤ |b|·|1|, according to Definition 11. Thus |a+b| ≤ |b| = b b max(|a|,|b|), as required. (⇐)Again, weassumethata,b ∈ K with|a| ≤ |b|. Thus, bythestrongtriangleinequality, we have |a+b| = |b||1+ a| ≤ |b|. This implies that |1+ a| ≤ 1 for any a,b ∈ K, so we can b b choose C = 1. Therefore |·| is non-Archimedean. (cid:3) Lemma 14. Let |·| be a valuation on a field K. Then |·| is non-Archimedean if and only if |e| ≤ 1 for all e ∈ R , where R denotes the ring generated by 1 in K. K K Remark 15. We cannot assume that the ring generated by 1 in K is Z, since that is no longer the case if K has a positive characteristic. Proof. (⇐) Any non-Archimedean valuation is equivalent to a non-Archimedean valuation, and since any non-Archimedean valuation satisfies the triangle inequality, one can replace the original valuation to the one satisfying the triangle inequality. Suppose e ∈ R and K |e| ≤ 1. Apply the triangle inequality: n (cid:12)(cid:18) (cid:19)(cid:12) n (cid:88)(cid:12) n (cid:12) (cid:88) |1+e|n = |(1+e)n| ≤ (cid:12) (cid:12)|e| ≤ |a| ≤ n+1. (cid:12) j (cid:12) j=0 j=0 Take the n-th root on both sides and let n → ∞ to get |1+e| ≤ 1 for any e ∈ R . Thus we K can take C = 1, as required. (⇒) This is immediate, since |1+1| ≤ |1| by the triangle inequality. Apply induction to derive |e| ≤ 1 for all multiples of 1. (cid:3) Corollary 16. If K and k are fields with k ⊆ K, and | · | a valuation on K, then | · | is non-Archimedean on K if and only if its restriction to k is non-Archimedean also. Proof. Apply the previous lemma for the ⇐ direction. The ⇒ direction is immediate. (cid:3) Corollary 17. If |·| is a valuation on a field K with charK > 0, then |·| is non-Archimedean. Proof. This follows from the fact that the only valuation on the finite field F is the trivial p valuation, and the trivial valuation is (trivially) non-Archimedean. (cid:3) 2. Ostrowski’s Theorem Theorem 18 (Ostrowski’s Theorem). All non-trivial valuations on Q is equivalent to either the ordinary absolute value or the p-adic valuation, where p is a prime. Proof. Let | · | be a valuation on Q. By Corollary 10 we may assume that | · | satisfies the triangle inequality. Let b > 1 and c > 0 with b,c ∈ Z. Write c in terms of b: c = c bm + c bm−1 + ··· + c , where c ,...,c are taken from {0,1,...,b − 1} and where m m−1 0 0 m c (cid:54)= 0. Note that m ≤ logc/logb. By the triangle inequality, m |c| ≤ (m+1) max |c |max(1,|b|m) ≤ (m+1)M max(1,|b|m) i 0≤i≤m 3 where M −max(|1|,...,|b−1|). Let a ∈ Z and put c = an for n ∈ Z . Then + + (cid:18) (cid:19) nloga |a|n = |an| = +1 M max(1,|b|nlologgba). logb Take n-th roots and let n → ∞: loga |a| ≤ max(1,|b|logb). (1) Suppose first that there is some positive a with |a| > 1. Then from (1), we see that |b| > 1 for all b > 1 with b ∈ Z. Interchanging the roles of a and b in (1), we see that |a|1/loga = |b|1/logb. Thus, there exists a real number λ with λ > 1 such that for all positive integers a, we have |a| = aλ and hence |·| is equivalent to the ordinary absolute value on Q (i.e., |a/b| = |a/b|λ , ∞ where |a/b| denotes the ordinary absolute value). ∞ Now suppose that |a| ≤ 1 for all positive integers a. If |a| = 1 for all positive integers a then |·| is the trivial valuation. Thus there is a smallest positive integer a for which |a| < 1. Notice that a is a prime by the multiplicative property of valuations. Let c be an integer such that p (cid:45) c. We can write c = up+v with v ∈ {1,2,...,p−1}. Then |v| = 1 since p is the smallest positive integer with valuation less than 1. Furthermore, |up| = |u| · |p| < 1. Since |a| ≤ 1 for all a ∈ Z , we see that | · | is non-Archimedean. + Therefore |c| = |up+v| = |v|. This means that the valuation on Q is determined once we know that its value on p. Thus it is equivalent to the p-adic valuation, as desired. (cid:3) We will give a criterion for the two valuations on a field K to be equivalent. Proposition 19. Let K be a field and let |·| and let |·| be valuations in K. If |·| is not 1 2 1 the trivial valuation and for a ∈ K |a| < 1 ⇒ |a| < 1, 1 2 then |·| and |·| are equivalent. 1 2 Proof. By taking inverse we see that |a| > 1 implies |a| > 1. Next suppose that |a| = 1 1 2 1 and |a| > 1. Since our valuation |·| is not a trivial valuation, these exists c ∈ K\{0} with 2 1 c (cid:54)= 0 and |c| < 1. Then for each n ∈ Z we have 1 + |can| = |c| |an| > 1 for n sufficiently large. 2 2 2 But |can| = |c| |a|n < 1 for all n ∈ Z and this contradicts our assumption. We thus 1 1 1 + conclude that |a| < 1 ⇒ |a| < 1 and |a| > 1 ⇒ |a| > 1. Since |·| is non-trivial there 1 2 1 2 1 exists a c ∈ K with c (cid:54)= 0,|c| > 1. Then |c| > 1. Now let a ∈ K,a (cid:54)= 0 and define γ by 1 2 |a| = |c|γ. Let m,n ∈ Z with m/n > γ. Then |a| < |c|m/n. Hence 1 1 + 1 1 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)an(cid:12) (cid:12)an(cid:12) (cid:12) (cid:12) < 1 ⇒ (cid:12) (cid:12) < 1. (cid:12)cm(cid:12) (cid:12)cm(cid:12) 1 2 Therefore |an| < |cm| so |a| < |c|m/n. 2 2 2 2 Similarly if m/n < γ we have |a| > |c|m/n. 2 2 4 Therefore |a| = |c|γ. Thus γ = log|a| /log|c| = log|a| /log|c| and so 2 2 1 1 2 2 log|a| log|c| 1 1 = log|a| log|c| 2 2 is equal to λ for some λ ∈ R,|λ| > 0. Accordingly, |a| = |a|λ, completing the proof. (cid:3) 1 2 Given any function |·| on a field K satisfying axioms (i) and (ii) in Definition 1 we can use it to define a topology on K. For each ε > 0 and x ∈ K we defined the fundamental basis 0 of neighbourhoods of x by the inequalities |x−x | < ε. Axiom (iii) in Definition 1 ensures 0 0 that our space is Hausdorff. We can define a metric d on K by putting d(a,b) = |a−b|. Remark 20. The induced topology is the discrete topology whenever |·| is the trivial valua- tion. Proposition 21. Let |·| and |·| be valuations on a field K. |·| and |·| induce he same 1 2 1 2 topology on K if and only if they are equivalent. Proof. (⇒) We may suppose that both |·| and |·| are non-trivial valuations on K. Suppose 1 2 thataininK with|a| < 1. Thus|a| < 1 ⇒ |a|n| → 0asn → ∞. Since|·| and|·| induce 1 1 1 1 2 the same topology , |an| → 0. But then |a| < 1. The result now follows by Proposition 2 2 19. (cid:3) Inequivalent valuations on a field K are independent in the following sense: Proposition 22. Let | · | ,...,| · | are non-trivial valuations on a field K with pairwise 1 H non-equivalence. Then there exists a ∈ K such that |a| > 1 while |a| < 1 1 j for all 2 ≤ j ≤ H. Proof. The proof is via induction on H. Let H = 2 (base case). Since |·| is not the trivial 1 valuation and |·| and |·| are not equivalent, there exists some b ∈ K with |b| < 1 and 1 2 1 |b| ≥ 1. Similarly, since |·| is non-trivial, there exists c ∈ K such that |c| < 1 and |c| ≥ 1. 2 2 2 1 Then we can take a = cb−1. Notice that |a| > 1 and |a| < 1. 1 2 Suppose that H > 2 and the result holds for j = 2,...,H −1. Then there exists, by the inductive hypothesis, b ∈ K with |b| > 1 but |b| < 1 for all j = 2,...,H − 1. We can 1 j also consider |·| and |·| , and by the inductive hypothesis one can find d ∈ K such that 1 H |d| > 1 while |d| < 1. 1 H Notice that if |b| < 1 we are done, since we can take a = b. If |b| = 1 then we can take H H a = bnd for n sufficiently large. If |b| > 1 we can take a = bn d. a works for sufficiently H 1+bn large n since (cid:40) bn 1 1 (|·| and |·| ) 1 H = → . 1+bn 1+b−n 0 all other valuations This completes the proof. (cid:3) 5 3. September 25: Approximation theorem and completion 3.1. Approximation theorem. Theorem 23 (Approximation theorem). Let | · | ,...,| · | be pairwise inequivalent, non- 1 H trivial valuations. Let b ,...,b ∈ K. Let ε > 0 be a positive real number. Then there exists 1 H an element a ∈ K so that |a−b | < ε for i = 1,2,...,H. i i Proof. By Proposition 22, there exist c ∈ K(1 ≤ j ≤ H) with |c | > 1 and |c | < 1 for all j j j j i i (cid:54)= j. Notice that for j = 1,...,H, (cid:40) (cid:12)(cid:12) cnj (cid:12)(cid:12) 1 if i = j (cid:12) (cid:12) → (cid:12)1+cn(cid:12) 0 if i (cid:54)= j. j i as n → ∞. The result now follows from the triangle inequality by taking (cid:88)H cn a = j b . (cid:3) 1+cn j j=1 j Remark 24. This can be viewed as an analogue of the Chinese Remainder Theorem. Definition 25. Let K be a field and | · | be the valuation on K. We say that a sequence (a ,a ,...) of elements of K converges to a limit b in K with respect to |·| if given ε > 0 1 2 there exists n (ε) such that for n > n (ε) we have |a −b| < ε. 0 0 n Remark 26. By axiom (i) of Definition 1, if the limit exists it is unique. Definition 27. We define a Cauchy sequence (z ,z ,...) to be a sequence such that for each 1 2 ε > 0 there exists n (ε) such that whenever m,n > n (ε) we have |a −a | < ε. Note that 1 1 m n if (a ,a ,...) has a limit then it is a Cauchy sequence. 1 2 Definition 28. A sequence (a ,a ,...) is said to be a null sequence with respect to |·| if it 1 2 has limit zero. Definition 29. A field K is complete with respect to |·| if every Cauchy sequence in K has a limit in K. Remark 30. Q is not complete with respect to | · | (the ordinary absolute value on Q), ∞ √ since, for example, one can choose a Cauchy sequence which converges to 2. So, what about Q and |·| ? Let p = 5. We now will construct a sequence with respect to p the 5-adic valuation |·| which does not have a limit in Q. Here, |·| is normalized so that 5 5 |5| = 5−1. To show that Q is not complete with respect to | · | , we construct a sequence 5 5 of integers (a ,a ,...) satisfying a2 − 6 ≡ 0 (mod 5n+1) and a ≡ a (mod 5n) for all 1 2 n n+1 n n ∈ N. Define b ,b ,... from {0,1,2,3,4} inductively. Let b = 1 and choose b so that 0 1 0 1 (1+b 5)2 ≡ 1+2b 5 ≡ 6 (mod 52). Thus 2 1 ≡ 1 (mod 5) so we can choose b = 3. More 1 1 b 1 generally suppose that b ,b ,...,b have been chosen hence determining a ,...,a . Then 0 1 n 1 n a2 = (b +b 5+···+b 5n)2 ≡ 6 (mod 5n+1). Thus (a +b 5n+1)2 ≡ a2 +2b 5n+1 ≡ n 0 1 n n n+1 n n+1 6+c5n+1+2b 5n+1 ≡ 6+(c+2b )5n+1 (mod 5n+2) for some integer c. Thus it suffices n+1 n+1 to choose b so that c+2b ≡ 0 (mod 5). Observe that such b can be found. Thus n+1 n+1 n+1 a2 ≡ 6 (mod 5n+2). Further,wehavea ≡ a (mod 5n+1). Thiscompletestheinductive n+1 n+1 n construction. 6 Notice that (a ,a ,...) is a Cauchy sequence with respect to | · | since a ≡ a 1 2 5 n+1 n (mod 5n+1) for n = 1,2,.... If (a ,a ,...) converged to a limit d in Q, then since |a2−6| ≤ 1 2 √ n 5 5−n+1 we see that |d2−6| = 0. In other words, d = 6 ∈ Q and this is a contradiction. One 5 can show in a smilier manner that Q is not complete with respect to any p-adic valuation |·| . Plainly, the sequence (5n)∞ is a null sequence with respect to |·| but is not a Cauchy p n=1 5 sequence with respect to |·| or any p-adic valuation |·| where p (cid:54)= 5. ∞ p Definition 31. Let K be a field with a valuation | · |. Let L be a field containing K. A valuation |·| on L extends |·| on K if |α| = α for all α ∈ K. 1 1 Definition 32. Let K be a field with a valuation on | · |. We say that a field L together with a valuation |·| with extends on K is a completion of K if L is complete and L is the 1 closure of K in the topology induced by |·| . 1 Given a field K with valuation |·|, how do we construct a completion of K with respect to | · |? First suppose that | · | satisfies the triangle inequality. The Cauchy sequences of elements of K form a ring (call this ring R) under term-wise addition and multiplication, i.e. (a ) + (b ) = (a + b ) and (a ) · (b ) = (a b ). Then the set of all null sequences of n n n n n n n n elements in K with respect to | · | (call this set N) forms a maximal ideal in the ring R. Thus L := R/N formulates a field. We now define a valuation | · | on L in the following 1 way. If α ∈ L, then α = (a ,a ,...)+N where (a ,a ,...) is a Cauchy sequence in K with 1 2 1 2 respect to |·|. Define |α| = lim |a |. However, we need to ensure that |·| is well-defined. 1 n 1 n→∞ We first check that the limit exists, and then that the limit does not depend on our choice of representative of the equivalence class we choose. We first check that (|a |) is a Cauchy n n sequence. Notice by the triangle inequality that: • |a | ≤ |a −a |+|a | ⇒ |a |−|a | ≤ |a −a | and |a |−|a | ≤ |a −a |. Hence, n n m m n m n m m n n m ||a |−|a || ≤ |a − a |. Thus we see that the sequence (|a |) is Cauchy and so n m ∞ n m n converges to a limit. • Now we need to show that the choice of a representative does not matter (well- defined). Note that if (a ,a ,...) ∼ (b ,b ,...), then we must show that lim|a | = 1 2 1 2 n lim|b |. But then n ||a |−|b || ≤ |a −b |. (2) n n ∞ n n Since (a −b ) is a null sequence we see from (2) that lim|a | = lim|b |. n n n n n Finally, we must check that | · | is a valuation on L but this is routine. First, we can 1 (naturally) embed K in L with the map ϕ : K → L by ϕ(a) = (a,a,a,...) + N. Then ϕ is an injective field homomorphism satisfying |a| = |ϕ(a)| . Let K(cid:48) = ϕ(K). Then K(cid:48) is 1 everywhere dense in L. For let α = (a ,a ,...) ∈ L, then given ε > 0 there exists n (ε) such 1 2 0 that for all n > n (ε), we have |α−ϕ(a )| < ε since (a ,a ,...) is a Cauchy sequence. 0 n 1 1 2 We now verify if L is complete. Let (a ) be a Cauchy sequence in L. Since K(cid:48) is n n everywhere dense in L, there exists a sequence (ϕ(a )) such that |ϕ(a )−a | < 1. Thus, n n n n 1 n the sequence (ϕ(a )−a )) is a null sequence in L. Hence (ϕ(a )) is a Cauchy sequence n n n n n in L. Thus (a ) is a Cauchy sequence in K. In particular, it determines an element β in L n n with lim |ϕ(a )−β| = 0. n 1 n→∞ Thus lim |a −β| = 0 so (a ) converges to an element in L. n→∞ n 1 n n 7 4. September 30 Let |·| be the p-adic valuation on Q normalized so that |p| = p−1. We will denote by p p Q “the” completion of Q with respect to | · | . We say “the” to refer to the construction p p described previously. Further, we will denote the valuation |·| on Q by |·| . 1p p p Theorem 33. Every α ∈ Q with α ≤ 1 has a unique representative Cauchy sequence (a ) p p n n such that: • a ∈ Z for all n = 1,2,3,... n • 0 ≤ a < pi for i = 1,2,... i • a ≡ a (mod pi) for i = 1,2,3,.... i+1 i Proof. We first prove uniqueness. Suppose that (a ) and (b ) are two such sequences n n n n representingsomeα ∈ Q . If(a ) (cid:54)= (b ) thenfor someintegeri, we havea (cid:54)= b . Butthen p n n n n i i for n > i, it is known that a ≡ a (mod pi) and b ≡ b (mod pi). Hence a (cid:54)≡ b (mod pi) n i n i n n therefore |a −b | ≥ p−i. Thus (a −b ) is not a null sequence, hence a contradiction. It n n p n n remains to show that each element in Q has such a representation. To do so, we will need p the following result: Lemma 34. If x ∈ Q and |x| ≤ 1 then for any positive integer i there exists an integer c p with 0 ≤ c < pi such that |x−c| ≤ p−i. p Proof of Lemma. Theresult isimmediatewhenx = 0. So supposex (cid:54)= 0. Now writex = a/b with (a,b) = 1. Since |a| ≤ 1, we see that p (cid:45) b. Thus there exist integers m and n such b p that mb+npi = 1. Put c = am. Then 1 (cid:12) (cid:12) (cid:12)(cid:12)a −c (cid:12)(cid:12) = (cid:12)(cid:12)a(cid:12)(cid:12) (cid:12)(cid:12)1− c1b(cid:12)(cid:12) = |1−mb| ≤ p−i. (cid:12)b 1(cid:12)p (cid:12)b(cid:12)p(cid:12) a (cid:12) p p We now choose c with 0 ≤ c < pi so that c ≡ c (mod pi). (cid:3) 1 Now we are ready to prove existence. Suppose that (b ) a Cauchy sequence which rep- n n resents α. Then for each positive integer j there exists an N(j) such that |b −b | ≤ p−j i k p whenever i,k > N(j). We may suppose, without loss of generality, that N(1) < N(2) < ··· so N(j) ≥ j. Further, |b | ≤ 1 for i ≥ N(1) since for all k > N(1) we have (by the strong i p triangle inequality) (cid:18) (cid:19) 1 |b | = |(b )+(b −b )| ≤ max(|b | ,|b −b | ) ≤ max |b | , . i p k i k p k p i k p k p p But |α| ≤ 1 and |b | → |α| as k → ∞ hence |b | ≤ 1. p k p p i p WenowuseLemma34tofindasequenceofintegers(a ) with0 ≤ a < pj forj = 1,2,..., n n j and |a −b | ≤ p−j. j N(j) p We now show that a ≡ a (mod p ). We have j+1 j j |a −a | = |(a −b )+(b −b )+(b −a )| j+1 j p j+1 N(j+1) N(j+1) N(j) N(j) j p ≤ max(|a −b | ,|b −b | ,|b −a | ) j+1 N(j+1) p N(j+1) N(j) p N(j) j p ≤ max(p−j+1,p−j,p−j) ≤ p−j. 8 Therefore a ≡ a (mod pj). Further, (a −b ) is a null sequence since for each positive j+1 j i i i integer j we have, for i > N(j), |a −b | = |(a −a )+(a −b )+(b −b )| i i p i j j N(j) N(j) i p ≤ max(|a −a | ,|a −b | ,|b −b | ) i j p j N(j) p N(j) i p ≤ max(p−j,p−j,p−j) = p−j. Thus |a −b | → 0 as i → ∞. i i p If a ∈ Q and |a| > 1 then for some integer k, |pka| ≤ 1. We can find an appropriate p p p representative for pka, say (a ) . We then represent a by (p−ka ) . Now suppose that α ∈ Q i i i i p with|α| ≤ 1. Let(a ) beasequenceasinTheorem33whichrepresentsα. Wecanrepresent p i i a in base p, i.e., a = b +b p+···+b pi−1 where b ,b ,...,b ∈ {0,1,...,p−1}. Since i i 0 1 i−1 0 1 i−1 a ≡ a (mod pi) we have a = b +b p+···+b pi. Thus we can view α has having the i+1 i i+1 0 1 i unique power series expansion α = b +b p+b p2+···. Further, if a ∈ Q and |α| = pk for 0 1 2 p p k ∈ Z then α has the power series expansion α = b p−k +b p−k+1 +···+b +b p+ + −k −k+1 0 1 b p2 +···. (cid:3) 2 Remark 35. This is a natural representation in terms of digits {0,1,2,...,p − 1}. There are other “natural” representations, for instance the Teichmu¨ller representation. It is worth noting that each element α ∈ Q has a unique base p expansion. Contrast this with the base p 10 expansion of elements R. Then of course 1.0000... and 0.999... are both equal to 1. Definition 36. Z is the set of p-adic integers and is defined by p Z := {x ∈ Q ,|x| ≤ 1}. p p p Addition, subtraction, multiplication, and division can be performed as in R but with car- rying taken from right to left instead of left to right. Let k be a field and P an indeterminate over k. Let k(T) be the field of rational functions in T with coefficients in k. Let γ be a real number with 0 < γ < 1. For any α ∈ k(T), we can write f(T) α = Ta g(T) where f,g ∈ k[T] with f(0),g(0) (cid:54)= 0 and a ∈ Z. We define a valuation | · | on k(T) by |α| = γa. Let k((T)) be the completion of k(T) with respect to | · |. Denote the expansion of | · | from k(T) to k((T)) by |·|. Let N ∈ Z and let (f ,f ,...) be a sequence of elements + N N+1 of k. Then put m (cid:88) f(m) = f Tn n n=N for m = N,N+1,.... Note that f(m) ∈ k[T] for m = N,N+1,.... The sequence (f(m))∞ m=N is a Cauchy sequence of elements of k(T) with respect to |·|, since |f(i) −f(j)| ≤ γmin(i,j)+1. Denote the element of k((T)) of which f(m) for which (f(m))∞ is a representative by m=N ∞ (cid:88) f(T) = f Tn. n m=N 9 5. October 2 ∞ Recallthatk((T))isthecompletionofk(T)withrespectto|·|. Letf(T) = (cid:80) f Tn,f ∈ n n n=N k. f is the limit of (f(m))∞ where f(m)(t) is the partial sum from N to m. Conside the m=N set S of all expressions (cid:88) f tn, n n(cid:29)−∞ where f ∈ k. When we say n (cid:29) −∞, we mean that there exists some integer N ∈ Z such n that f = 0 for n < N. S forms a commutative ring with identity under the usual rules for n multiplying and adding the power series. ∞ Let f ∈ S \{0}. Then f(T) = Tab(1+ (cid:80) g Tn) for a ∈ Z and b ∈ k \{0}. Put n n=1 (cid:32) (cid:32) (cid:33)(cid:33) ∞ ∞ (cid:88) (cid:88) (cid:88) h(T) = T−ab−1 1+ − g Tn = h Tn. n n m=1 n=1 n(cid:29)−∞ Further, f(T)h(T) = 1. Thus S is a field. Note that k(T) ⊆ S and the closure of k(T) with respect to |·| is contained in S Thus S is isomorphic to k((T)). Definition 37. We define k[[T]] to be the element of k((T)) with |f| ≤ 1. Then f ∈ k[[T]] ⇔ f = (cid:80)f Tn. k[[T]] is a subring of k((T)). It is the ring of formal power series with n coefficients in k. ∞ Definition 38. f ∈ Q[[T]] with f = (cid:80) f Tn is said to satisfy Eisenstein’s condition if there n n=0 is a non-zero integer l such that lnf is an integer for all n ≥ 0. n Theorem 39 (Eisenstein’s theorem). Let f ∈ Q[[T]]. If f satisfies a non-trivial polynomial equation with coefficients in Q[T] then f satisfies Eisenstein’s condition. Proof. We may suppose that f satisfies g (T)+g (T)f(T)+...g (T)f(T)J = 0. Not all of 0 1 J the g ’s are the zero polynomial and g ∈ Q[T] for j = 0,1,...,J. Let us suppose that J is j j minimal. For indeterminates X and Y, put J (cid:88) H(X) = g (T)Xj ∈ Q[T,X], j j=0 and H(X +Y) = H(X)+H (X)Y +···+H (X)YJ where H (X) ∈ Q[T,X]. Since J is 1 J j minimal, we have H (f) (cid:54)= 0. Of course H(f) = 0. Now define the integer m as follows: 1 |H (f)| = γm. Obviously we have m ≥ 0. Put 1 f(T) = u(T)+Tm+1v(T), whereu(T) = f +f T+···+f Tm+1 ∈ Q[T]andv(T) = 0+f T+f T2+··· ∈ Q[[T]]. 0 1 m+1 m+2 m+3 Notice that it suffices to show that v(T) satisfies Eisenstein’s condition. Since H(f) = 0, we have (cid:88) 0 = H(u+Tm+1v) = H(u)+Tm+1H (u)v + T(m+1)jH (u)vj. (3) 1 j j≥2 10