p-adic Measures and Bernoulli Numbers

Transcription

1 -Adic Measures and Bernoulli Numbers Adam Bowers Introduction The constants B k in the Taylor series exansion t e t = t k B k k! k=0 are known as the Bernoulli numbers. The first few are,, 6, 0, 30, 0, 4,. These numbers lay an imortant role in number theory. For examle, the Reimann zeta-function essentially equals a Bernoulli number at negative integers: ζ k = B k k for k. A few of the interesting classical results about Bernoulli numbers are given in the following theorem. Theorem. Let k Z + and be a rime. Then. If a Z, then aa k B k Z.. If a Z, then a k a k B k k Z. This aer was written as an assignment in a course taught by Keith Conrad during Fall Semester 004 at the University of Connecticut. Anything correct is only so due to Professor Conrad s diligent revision.

2 3. If k 0 mod, then B k k Z. 4. For k and k even, if k k 0 mod, then B k k B k k mod. 5. For k even or k =,. B k + k Z This aer will develo the ideas needed to rove these claims bt -adic methods and in the rocess show the intimate relationshi between Bernoulli numbers and -adic measures. -Adic distributions Let X be any comact-oen subset of Q, such as Z or Z. A -adic distribution µ on X is defined to be an additive ma from the collection of comact-oen sets in X to Q. Additivity means n µ k= U k = n µu k, where n and {U,..., U n } is any collection of air-wise disjoint comactoen sets in X. Just from the definition, it is not clear how to construct a -adic distribution. Fortunately, Q has some forgiving toological roerties. The set Q has a toological basis of comact-oen sets of the form a + n Z, where a Q and n Z these are the closed balls in Q. Consequently, if U is any comact-oen subset of Q, it can be written as a finite disjoint union of sets k U = a j + N Z j= k=

3 3 for some N Z and a,..., a k Q. There are many such reresentations. In fact, each -adic ball a + n Z can be reresented as the union of smaller balls, a + n Z = b=0 a + b n + n+ Z. This follows from the fact that these balls are closed under intersection, so that they intersect if and only if one is contained in the other. The ability to decomose comact-oen subsets of Q into sets of the form a + N Z allows one to reformulate the additivity condition into a more ractical condition. Proosition. Every ma µ from the collection of comact-oen sets in X to Q for which µ a + n Z = µ a + b n + n+ Z b=0 holds whenever a + n Z X, extends to a -adic distribution on X. Equation is called a distribution relation. The roof follows from the additivity of µ and the toological roerties described above. For more details, see [], age 3. The roosition is quite useful. Any comact-oen set in X can be written as a disjoint union of balls a + n Z, so one can check whether or not a set function on these balls satisfies Equation and then extend it to a distribution on X. As a simle examle, consider the Haar distribution µ Haar defined by µ Haar a + n Z = n for a Z and n N. This extends to a distribution on Z since it satisfies the distribution relation: b=0 µ Haar a + b n + n+ Z = b=0 = n+ = µ n Haar a + n Z.

4 4 How does one construct distributions on Z? Can they be constructed from, say, olynomials? Suose f k is a olynomial of degree k. Define a ma µ k on the balls in Z as follows: µ k a + n Z = nk f k {a}n n where {a} n is the unique number in the set {0,,..., n } such that {a} n a mod n. What could f k look like? If µ k is a distribution, it must satisfy the distribution relation in, so taking a to be from the set {0,,..., n }, a a + b nk f k = µ n k a + b n + n+ Z = n+k n f k. b=0 Thus, in order to be a distribution, a a + b f k = k f n n k. b=0 b=0 b=0, n+ This is to hold for all n and every a {0,,..., n }, so µ k is a distribution on Z if and only if f k satisfies the identity x + b f k x = k f k. As it turns out, for each k N the only monic olynomial of degree k that satisfies the above equation is B k x, the k th Bernoulli olynomial see below. In fact, the k th Bernoulli olynomial satisfies the stronger condition: for any M N, M B k x = M k b=0 x + b B k. M What are the Bernoulli olynomials? Recall the definition of the Bernoulli numbers given in the introduction. They were defined by the following Taylor series exansion: t e t = t k B k k!. k=0

5 5 In the above series, B k is the k th Bernoulli number. The first few are B 0 =, B =, B = 6, B 3 = 0, B 4 = 30, B 5 = 0, B 6 = 4,. It can be shown that B k = 0 for all odd k >. For each k N, define the k th Bernoulli olynomial, denoted B k x, by the following formula: So for each k 0, B k x = k j=0 k j B k j x j. B k x = B 0 x k + kb x k + = x k k xk +. The first few Bernoulli olynomials are: B 0 x = B x = x B x = x x + 6 B 3 x = x 3 3 x + x B 4 x = x 4 x 3 + x 30 B 5 x = x 5 5 x x3 6 x B 6 x = x 6 3x x4 x + 4 B 7 x = x 7 7 x6 + 7 x5 7 6 x3 + 6 x Note that for each k N, and B k 0 = B k B kx = kb k x

6 6 for all x Q. The Bernoulli olynomials extend the notion of Bernoulli numbers in that they satisfy the formal identity te xt e t = k=0 B k x tk k!. Returning to the toic of -adic distributions, it can now be seen that µ k as defined by µ k a + n Z = nk {a}n B k n for a Z is in fact a distribution on Z, because the Bernoulli olynomials satisfy Equation. What are some of these Bernoulli distributions? Consider the cases k = 0,,. To simlify the following calculations, a will be taken from the set {0,,..., n }. When k = 0, the distribution is a familiar one, µ 0 a + n Z = n0 B 0 a n = n. This is just the Haar distribution, which was introduced above. When k =, µ a + n Z = n B a n = a n. This is known as the Mazur distribution and is denoted µ Mazur. A similar calculation reveals that a µ a + n Z = n B = a n a + n n 6. In general, µ k a + n Z = ak n + term in d k Z, where d k is the least common multile of the denominators of B 0, B,..., B k. [Note that d k is indeendent of a and n.]

7 7 Regardless of the choice of k, when a Z e.g., a = µ k a + n Z = n, which tends to infinity as n. Thus the small balls + n Z have large µ k -values. -Adic measures Again, let X be any comact-oen subset of Q, such as Z or Z. A -adic distribution µ on X is called a -adic measure on X if its values on comact-oen sets U X are bounded; i.e., there exists some B R such that µu B for all comact-oen sets U X. The least bound is denoted µ. The Dirac distribution µ α on Z, concentrated at α Z, is defined by { if α U, µ α U = 0 otherwise. This is a measure. The roof is simle and, as is all-too common with simle roofs, uninteresting. The Bernoulli distributions are much more interesting, but not bounded as was shown above, so they are not -adic measures. Is there any way to fix this roblem? For all α Z and each k N, define the k th regularized Bernoulli distributions on Z by µ k,α U = µ k U α k µ k α U. This adjustment to the k th Bernoulli distribution removes the divergent term, making the k th regularized Bernoulli distribution bounded, so actually a measure. The roof of this claim is involved and given below.

8 8 The case when k = 0 is quite simle: for any n N and 0 a n, µ 0,α a + n Z = µ 0 a + n Z α 0 µ 0 α a + n Z = n n = 0. Thus µ 0,α is identically zero, so is a measure, albeit not a very interesting one. What about the case when k =? It will be shown that µ,α a + n Z. Again let n N and 0 a n. Then µ,α a + n Z = µ a + n Z αµ α a + n Z = a {α n α a} n n = a α{α a} n + α. n In the above calculation, {α a} n is defined to be the coset reresentative of α a + n Z which is contained in {0,,,..., n }; that is and 0 α a n. {α a} n α a mod n So why is this bounded by? Consider the second term first: α = α. If then Z and α Z. If =, then since α Z, it follows that α 0 mod, i.e. α is a multile of. Either way, α Z, so α. Now consider the first term: a α{α a} n n.

9 9 By definition so {α a} n α a mod n, {α a} n = α a + A n, for some A Z. Thus solving for A or Hence αa = αα a n A = α a n α{α a} n n {α a} n n = a α{α a} n n. a α{α a} n = αa, n where this inequality follows from both α and A sitting inside Z. Therefore µ,α a + n Z, as required. It remains to show that the µ k,α are bounded for k >. This will follow from the next theorem. Theorem. Let d k be the least common denominator of the coefficients of B k x as before. Then d k µ k,α a + n Z d k ka k µ,α a + n Z mod n, 3 where both sides lie in Z. Before beginning the roof, a remark: Note d k is indeendent of a and n. So, as n gets large, 3 says that µ k,α a+ n Z is very close to ka k µ,α a+ n Z. Proof. By definition, d k µ k,α a + n Z = d k [ µ k a + n Z α k µ k α a + n Z ]

10 0 [ a {α = d k nk B k α k nk a} ] n B n k. n When these olynomials are exanded out by a few terms, the result is an unfriendly looking exression: k a + ] d k nk [ a n k k n [ {α α k k a} n k n ] {α k a} n +. The terms which are so conveniently concealed by have as denominators { nk, nk,... n, } resectively, since the olynomial d k B k x has integral coefficients by the construction of d k. So when these terms are multilied by the leading nk, the denominators will cancel, leaving a factor of n or greater; hence the hidden terms are congruent to zero mod n. This means the unfriendly exression above differs by an element of n Z from the slightly more friendly exression below: [ k a d k nk k k a {α α k k a} n + α k k ] {α k a} n, n n n n n which, if nothing else, at least fits on one line. Simlifying this a bit gives something even better: d k µ k,α a + n Z + term in n Z = d k [ a k k n ak αk {α a} k n n + k ] αk {α a} n k [ a k α k {α a} k n = d k k ] a k α k {α a} k n n.

11 By definition of {α a} n, the second term is congruent mod n to k a k α k α a k = k ak α, keeing in mind that the second term in fact has no denominator, desite the resence of the, because d k contains a factor of, by construction. The first term may have a denominator, namely n. Recall that {α a} n = α a A n, for some A Z. Then {α a} k n = α a A n k = k j=0 k j α a k j A n j by the binomial theorem. Thus {α a} k n n = k j=0 k j α a k j A j n j = α a k n + kα a k A + α a k n kα a k A mod n. Then, since α Z, α k {α a} k n αk α a k n n = ak n kαak A. kα k α a k A mod n Therefore, a k α k {α a} k n = ak n αk {α a} k n n n

12 ak a k n n kak αa mod n = ka k αa α = ka k a {α a} n α n = ka k a α{α a} n n. Putting all of these calculations together, d k µ k,α a + n Z + term in n Z [ a α{α = d k ka k a} n k ] n ak α ] [ a α{α = d k ka k a} n + α n = d k ka k µ,α a + n Z. This roves the theorem. Note that it is now known that everything is in Z, so the congruence in the statement of the theorem makes sense. Corollary. For each k > and α Z, the Bernoulli distribution µ k,α is bounded, hence a -adic measure. Proof. Let U be a comact-oen set in X. Without loss of generality, assume U = a + n Z for some n N and 0 a n. By the theorem, d k µ k,α U = d k µ k,α a + n Z = d k ka k µ,α a + n Z + A n, for some A Z. Thus { ka } µ k,α U max k µ,α a + n Z, A n d k

13 3 For n sufficiently large, { µ,α max a + n Z, n d k }. n d k <, so µ k,α U by the non-archimedean inequality. It also follows from the non-archimedean inequality that µ k,α U for all comact-oen U because any such set can be written as the disjoint union of -adic balls for large enough n. -Adic integration The main urose for constructing -adic measures is to integrate functions. The develoment arallels the classic treatment, beginning with the idea of Riemann sums. Suose µ is a -adic measure on some comact-oen subset X of Q, say Z or Z. Suose also that f is a continuous K-valued ma, where K is a finite extension of Q or ossibly Q itself. Let N N be fixed. Choose from each a + N Z contained in X a reresentative, say x a,n. Then the N th Riemann sum over {x a,n } is defined to be S N,{xa,N } = a+ N Z X fx a,n µa + N Z. Theorem. Let µ, X and f be as above. If f : X K is a continuous function, then the Riemann sums as defined above converge to a limit in K which does not deend on the choice of {x a,n }. This unique limit will be denoted fxdµx or simly X fdµ. X Proof. By definition, the -adic measure µ on X is bounded, so there is some B > 0 such that µu B for every comact-oen set U X.

14 4 First, convergence will be shown for a articular Riemann sum. For any given n N, Z = a + n Z 0 a n where the balls are airwise disjoint, so for each a {0,,..., n }, select the reresentative in a + n Z to be x a,n = a. It suffices to show the sequence {S N,{a} } is a Cauchy sequence. Let ɛ > 0 be given. The set X is comact, so f is not only continuous, but uniformly continuous. Thus, there exists some N 0 N such that fx fy < ɛ B whenever x y mod N, for all N N 0. Now choose N N 0 large enough that every ball a + N Z is either contained in X or disjoint from it. Let M > N. Two -adic balls intersect if and only if one is contained in the other, so for each a, a + N Z = 0 â M â a mod N â + M Z. Then by finite additivity of µ, µ a + N Z = 0 â M â a mod N µ â + M Z. Thus S N,{a} = a+ N Z X faµ a + N Z = â+ M Z X faµ â + M Z, where a is in the set {0,,..., N } and a â mod N.

15 5 By construction, for each â {0,,..., M }, the reresentative in the corresonding Riemann sum was xâ,m = â. Thus S M,{â} = â+ M Z X fâµ â + M Z. Consequently, S N,{a} S M,{â} = â+ M Z X fa fâ µ â + M Z max { fa fâ } µ â + M Z. But a â mod N, hence fa fâ < ɛ. Therefore SN,{a} S B M,{â} < ɛ, as required. It remains to show that the choice of reresentative does not matter. Suose for each a + n Z, the number x N,a is some other reresentative. Then S N,{a} S M,{x N,a } = a+ N Z X fa fx N,a µ a + N Z { fa max fx N,a } µ a + N Z.

16 6 As before, a x N,a mod N, so for an aroriately large N, hence the result. fa fx N,a < ɛ B, The following corollary follows directly from the theorem, so the roof is omitted. Corollary. If f : X K is a continuous function such that fx A for all x X, and if µu B for all comact-oen U X, then fx dµx A B. In articular, if f A, then fx dµ k,α x A. X X In -adic integration, the measure µ,α has a secial role. In some sense it can be thought of as the -adic analog to Lebesgue measure in classical analysis see [], age 37. This role is made more recise in the following theorem. Theorem. If f : Z K is a continuous function, then fxdµ k,α x = Z fx kx k dµ,α x. Z 4 Proof. The characteristic functions on -adic balls are dense in in CZ, K see [], so without loss of generality suose f = χ c+ n Z for some n N and 0 c n. Then Z fxdµ k,α x = lim N N a=0 faµ k,α a + N Z.

17 7 By Equation 3, for each N and a {0,,..., N }, d k µ k,α a + N Z = d k ka k µ,α a + N Z + N A N,a, for some A N,a Z. Thus N d k a=0 faµ k,α a + N Z = N a=0 ] fa [d k ka k µ,α a + N Z + N A N,a N = d k k a=0 faa k µ,α a + N Z N + N a=0 faa N,a. Observe, for any N N, the second term is in N Z A N,a Z. Thus since f and N lim N N faa N,a = 0. a=0 Therefore, N lim d k N a=0 N faµ k,α a + N Z = lim d kk N a=0 faa k µ,α a + N Z, hence d k fxdµ k,α x = d k k fx x Z k dµ,α x, Z as required.

18 8 Corollary. For each k N and each α Z not a root of unity, B k = k x k dµ α k,α x. Z Before beginning the roof, a remark: Note the right hand side in the above exression is indeendent of α, simly because the left hand side does not deend on α. Proof. Begin by observing that for any k and any α, µ k,α Z = µ k Z α k µz = α k B k 0 = α k B k. This follows from the definitions of µ k and µ k,α with a = 0 and n = 0. By the revious theorem, dµ k,α = kx k dµ,α x, Z Z but Z dµ k,α = lim N N a=0 µ k,α a + N Z = µ k,α Z. Thus kx k dµ,α x = α k B k, Z from which the result follows. Why boundedness is imortant

19 9 In the theory of -adic integration, does it matter that the distributions be bounded? To show that it does, consider a very simle case: let µ = µ Haar and let fx = x for all x Z. For each N N, Z = N a=0 so consider the sequence of artial sums N a=0 a + N Z, fx a,n µ a N + N Z = a=0 x a,n N, where x a,n is a reresentative of the ball a + N Z. For the integral to exist as defined above, the limit of this sum should not deend on the choice of x a,n a + N Z. First suose each x a,n = a. This makes sense since a a + N Z for each a {0,,..., N }. Then In Q, N a=0 x a,n N = N N a=0 a = N N N N lim N =. = N. Now, for each N, choose exactly one x a,n to be a + a 0 N for a fixed a 0 Z, leaving all the others as a. Then N x a,n = N a + a N N 0 N = N + a 0. a=0 a=0 In Q, N lim + a 0 = a 0 N. This limit is not indeendent of the choice of reresentative icked in each ball, so one cannot integrate fx = x against µ as defined here. While one

20 0 might be willing to accet a measure that does not allow certain functions to be integrable, fx = x is not usually one of those functions. As was shown, this roblem does not arise for a continuous function when using a bounded -adic distribution. This objection could be dealt with by insisting on taking the reresentative x a,n from the set {0,,..., N }. For further information on this develoment, see Volkenborn integral in [3]. Theorems about Bernoulli numbers In the introduction, several claims were made about Bernoulli numbers. The theory has been develoed enough that these can be roved. The roofs rely mainly on one of the corollaries given above: Corollary. For each k N and each α Z not a root of unity, B k = k x k dµ α k,α x. Z An imortant observation is that the value of this exression does not deend on α in any way, nor does it deend on choice of k. Keeing this in mind, recall the theorems from the introduction. Theorem. Let k Z + and be a rime. Then. If a Z, then aa k B k Z.. Sylvester-Lischitz If a Z, then a k a k B k k Z. 3. Adams If k 0 mod, then B k k Z. 4. Kummer For k and k even, if k k 0 mod, then B k k B k k mod.

21 5. Clausen - von Staudt For k even or k =, B k + k Z. Proof of Claim. If k =, then one of a or a must be even. Thus aa B = aa Z. Suose k >. Whenever k > is odd, B k = 0 and the result is trivial, so assume k is even. Consider the following lemma, given without roof see []. Lemma. If A Z for every rime, then A Z. Given this lemma, it suffices to show that aa k B k Z for every rime. Fix a rime. Then aa k B k = aak k x k dµ α k,α x, where α Z is not a root of unity. For any k and α, { } x k dµ,α x max x k µ, x Z Z so it only remains to show Z aa k k α k Z. 5 But the choice of α is arbitrary, so it suffices to find some α such that 5 is satisfied. If does not divide a, then a Z, so choose α = a. Then aa k k α k = aak k a k = ak,

22 which is in Z, as required. If does divide a, then a, so a Z. It follows that α cannot be a, so some other choice must be made. In this case, let α = +. The roof that this choice of α works relys on the following lemma, again given without roof see []. Lemma. Suose m, n Z. If and x + Z, then x m x n = x m n. If = and x + 4Z, then x m x n = x m n. First suose. Let α = + Z. Then by the lemma, α k = α k = k. Thus ak α k = a k k But a Z, so a. It follows that = a. ak α k Z. Since a k Z by the non-archimedean inequality, 5 follows. Now suose =. Let α = + = 3. When =, the lemma only alies if α mod 4Z, but 3 mod 4Z. However 3 k = 9 k/, and 9 is congruent to mod 4. Since k was assumed to be even, α k = 3 k = 9 k/ = 9 k/ = 4 k. Therefore ak α k = a k k = 4 a. 4 As before, a k Z and a Z, so aka k α k,

23 3 but this bound is insufficient to show 5. Recall the goal is to show aa k B k = akak α k Z x k dµ,α x Z. Given the bound on the first term in the roduct, it will suffice to show x k dµ,α x. Z For x Z, x k x mod Z, so x k dµ,3 x x dµ,3 x mod Z. Z Z By 4, with k = and α = 3, Z x dµ,3 x = Z dµ,3 x = µ,3z. To calculate this last exression, return to the definition of the Bernoulli distributions: µ k a + n Z = nk {a}n B k n for a Z. Thus Therefore µ k Z = B k 0 = B k. µ k,α Z = µ k Z α k µ k Z = α k B k. In this articular case, k = and α = 3, so µ,3z = 3 B = 8 = 6 3. However, 0 mod Z. Thus 3 x k dµ,3 x 0 mod Z, Z

24 4 so as required. It follows that x k dµ,3 x Z, aa k B k =, which roves the first claim of the theorem. Proof of Claim. By the lemma given in the roof of, it suffices to show that a k a k B k k = ak a k x k dµ α k,α x Z Z for every rime. As before, x k dµ,α x, Z so it only remains to show a k a k α k. 6 If does not divide a, then a =, so a Z. In this case, let α = a. Then a k a k α k = a k a k a k. If does divide a, then let α = +, as before. Since a, there exists some e N and a such that, a = and a = e a. Then a k = e a k = ek a k = ek.

25 5 Furthermore, since α Z, α k = α k = k = k. Observe that k = f k for some k Z and f Z + ossibly zero. Thus a k a k α k = ak a k k = +f ek. ek f If +f ek, then 6 will be verified. By construction, k = f k, so k > f. Because e, it follows that ek k > f. Therefore, ek k f. It follows that + f ek 0, so that +f ek, as required. Having verified 6 for all values of, the second claim in the theorem is roven. Proof of Claim 3. Begin by observing that >. The multilicative grou of nonzero residue classes of Z mod is cyclic of order see [], so choose α in {, 3,..., } so that α is the lowest ositive ower of α which is congruent to mod. Since k 0 mod, k is not a multile of. Thus α k mod. Therefore α Z. Thus k B k k = α k This roves the third claim. Z x k dµ,α x. Proof of Claim 4. To show B k B k mod, it must be shown that k k x k dµ α k,α x x k dµ α k,α x mod. Z Z

26 6 As before, α may be chosen from {, 3,..., } so that α is the lowest ositive ower of α which is congruent to mod. It will be shown that and Z α k α k mod x k dµ,α x x k dµ,α x mod. Z From this it follows that the roducts are congruent mod as well. To show the first of these two, observe that k = k + c for some c Z +, by assumtion. Thus α k = α k +c = α k α c α k mod. Now observe α k α = αk α k k α k α k = α k α k α k α k. By choice of α, since k, k 0 mod, the same argument as above shows α, k α k Z, hence so is the roduct of the two. Thus α k α k = α k α k α k α k = α k α k. This gives the first congruence. Now let x Z and {x} be the element from {0,,..., } that is congruent to x mod. Then x k {x} k = {x} k +c = {x} k {x} c {x} k x k mod. From this it follows that x k x k mod. Finally,

27 7 Z x k dµ,α x Z x k dµ,α x = x k x k dµ,α x su x k x k Z x Z. This rovides the second congruence, hence comletes the roof of the fourth claim. Proof of Claim 5. This claim, which is attributed to Clausen and von Staudt, states: For k even or k =, For each k it will be shown that B k + k B k + k Z. Z q for every rime q. This will give the required result, by the lemma above. First consider k =. The rime = is the only rime such that, so the sum is simly B + = + = 0. Clearly this is in Z q for every rime q. Now let k be an even integer and fix a rime q. If q does not divide k, then the sum will be over rimes not q; but for any q, Z q. Thus k Z q.

28 8 Furthermore, if q does not divide k, then k 0 mod q, so by the theorem of Adams see above B k k Z q. Since k Z, it follows that hence B k = B k k k Z q, B k + k Z q. Now suose that q does divide k. Then B k + = B k + q + k k q. As before, so it only remains to show that k q Z q, B k + q Z q. It will suffice to show qb k mod qz q, for then qb k = + qa for some A Z q and B k + q = qb k + q = + qa + q = A Z q. The roof of this congruence, which is called the Clausen - von Staudt congruence, is given below. Having shown B k + q Z q

29 9 if q k, it follows that for every rime q. Thus as required. B k + k B k + k Z q Z, This aer concludes by roving the congruence which is at the heart of the theorem of Clausen and von Staudt. Lemma. The Clausen - von Staudt congruence. If k for some rime, or = and k is even or k =, then B k mod Z, 7 Proof. Begin by recalling that B k = k α k Z x k dµ,α x, where α is any element of Z other than a root of unity. The roof is in two arts; when = and when. First suose. The choice of α is arbitrary, so let α = +. Then α k = + k k mod ordk+, whence Thus k mod. αk B k x k dµ,α x mod. Z Since Z = Z Z is a disjoint union, and k >,

30 30 x k dµ,α x = Z Z x k dµ,α x + Z x k dµ,α x Z x k dµ,α x mod, By assumtion k, so x k mod as has been seen before. Thus x k x mod for all x Z. Therefore, x k dµ,α x Z Z x dµ,α x mod. Define gx = {x} for x Z. Then x gx mod for all x Z. Thus x dµ,α x {x} dµ,α x mod. Z Z The function g is a ste function, and this last integral can be evaluated directly: Z {x} dµ,α x = = = a= a= a= a µ,αa + Z a a µ a + Z αµ α a + Z a {α a} α. Recall that α was chosen to be +. Thus

31 3 Z {x} dµ,α x = = = = a= a= a= a= a a a { + a} + a a a + + a + a + a a +. Now observe a= a a + = a= a= = + a a + mod Z mod Z mod Z Therefore B k mod Z, as required. The congruence has now been roven for all.

32 3 Now suose =. As before, B k = k α k Z x k dµ,α x. Choose α = 5 Z. Claim : k α k =. Notice that =. Thus 5 5 mod 4Z, so, by the lemma on age, Therefore, 5 k = 5 k = 4 5 k = k 4. k α k = k / k /4 =. Claim : Z x k dµ,α x =. By the same argument used for, x k dµ,α x x dµ,αx mod. Z Z Define gx = {x} for x Z, where {x} is the element in {0,,, 3} congruent to x mod 4. Then Z x dµ,αx Z gx dµ,α x mod 4.

33 33 This integral can be evaluated directly: Z {x} dµ,α x = = = 3 a= 3 a= 3 a= a µ,αa + 4Z a a µ a + 4Z 5 µ 5a + 4Z a 4 {5a} 5 4. Noting that {5} =, {0} =, and {5} = 3, this can be simlified: 3 a= a a 4 {5a} 5 4 = = 0 5. Observe that =. Thus 5 Z x dµ,αx mod 4 Z. This term is congruent to 0 mod, but not 0 mod 4, so it follows that x dµ,αx =. Z It has been shown that x k dµ,α x = Z Z x dµ,α x + C,

34 34 for some term C Z. Therefore, Z x k dµ,α x { = max Z } x dµ,α x, C =. This roves the second claim. Finally, B k = k α k x Z k dµ,α x Therefore B k Z, hence since =, This concludes the roof. B k mod Z. = =.

35 35 References [] K. Conrad, Introduction to local fields, in The Lectures of Keith Conrad as recorded by A. Bowers, Fall Semester 004. [] N. Koblitz, -adic Numbers, -adic Analysis, and Zeta-Functions, Sringer-Verlag New York, second ed., 984. [3] A. Robert, A Course in -adic Analysis, Sringer-Verlag, 000.