MATH 678: (MOTIVIC) L-FUNCTIONS

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1 MATH 678: (MOTIVIC) L-FUNCTIONS LECTURES BY PROF. KARTIK PRASANNA; NOTES BY ALEKSANDER HORAWA These are notes from Math 678 taught by Professor Kartik Prasanna in Fall 208, L A TEX ed by Aleksander Horawa (who is the only person responsible for any mistakes that may be found in them). This version is from January 5, 209. Check for the latest version of these notes at If you find any typos or mistakes, please let me know at ahorawa@umich.edu. Thanks to Peter Dillery for helping me catch up on anything I missed. The class will involve 0 homeworks and a final project/talk. The homework, together with some supporting papers, will be posted on the shared course file. Contents. Overview 2.. First example: the ζ-function 3.2. Brief overview of general conjectures 4.3. p-adic L-functions 6 2. L-functions of Dirichlet characters Dirichlet series Dirichlet characters Analytic continuation Functional equation Special values of L(s, χ) Zeta functions of number fields Evaluation of L(, χ) 8 3. Kubota Leopoldt p-adic L-function Power series over non-archimedean fields Kubota Leopoldt p-adic L-function 23 Date: January 5, 209.

2 2 KARTIK PRASANNA 3.3. p-adic logarithms, exponential functions, and power functions Leopoldt s formula for L p (, χ) p-adic class number formula p-adic measures and power series Power series Integration Alternative construction of L p (s, χ) Applications to class numbers in cyclotomic towers Main conjecture of Iwasawa theory Serre s construction of the p-adic L-function Classical modular forms Reduction modulo p Serre s p-adic modular forms Hecke operators Moduli-theoretic interpretation of modular forms Modular forms over C Modular forms over an arbitrary ring p-adic L-functions associated to Hecke characters Real analytic Eisenstein series Algebraicity of L-values p-adic interpolation p-adic L-function for Rankin Selberg L-function Algebraicity of critical values p-adic interpolation 20 Appendix A. Project topics 22 References 22. Overview One can associate L-functions to varieties (or even pure motives) over number fields. The question is: how do L-functions capture arithmetic of algebraic varieties (motives)?

3 MATH 678: (MOTIVIC) L-FUNCTIONS 3 In this class, we will use the language of motives very loosely. The reader interested in the precise definitions and more details should consult the papers [Mil3, Kim0] One can also associate p-adic L-functions to varieties (motives), which should conjecturally capture similar arithmetic information. {L-functions L(M, s)} {pure motives M} This class will focus on the following cases: {p-adic L-functions L p (M, s)} () Dirichlet characters, (2) Hecke characters of imaginary quadratic extensions k/q (3) elliptic curves over Q. General L-functions are unfortunately very hard to understand. In fact, the only cases where some of the conjectures are settled are those where one can associate an automorphic form to the geometric object. For example, it is known any elliptic curve E over Q is modular, i.e. L(E, s) = L(f, s) for some modular form f. In this case, both the Hasse Weil conjecture (about the analytic continuation and functional equation) and significant parts of the Birch Swinnerton-Dyer conjecture (and the order of vanishing and special values of the L-function) are known. In the above three cases, the associated automorphic forms are on the following groups: () GL() Q, (2) GL() k, (3) GL(2) Q. Time permitting, we might consider other reductive groups at the end of the class... First example: the ζ-function. We start with the simplest example of an L-function, the ζ-function. It is defined for Re(s) > by the following Dirichlet series ζ(s) = n. s n= It admits an Euler product ζ(s) = p. s p In fact, it is the L-function of the trivial motive. Analytic continuation. The function ζ(s) admits an analytic continuation to C with the exception of a pole of order one at s =.

4 4 KARTIK PRASANNA Functional equation. Let ξ(s) = π s/2 Γ(s/2)ζ(s). Then ξ(s) = ξ( s). The surprising factor π s/2 Γ(s/2) should be thought of as the Euler factor at. It is actually, in some sense, the most important Euler factor. Some basic properties of the Γ function are recalled for the reader s convenience: We recall some useful properties of the Γ function: () Γ(z + ) = zγ(z), (2) Γ(z)Γ( z) = π, sin πz (3) Γ(z)Γ(z + /2) = 2 2x πγ(2z), (4) Γ(z) has poles at 0,, 2,.... We are interested in the values of ζ(s) at integers. We know that ζ(2) = π2 π4, ζ(4) =, and 6 90 in general ζ(2k) π 2k Q for k. Using the functional equation, we see that so This shows that ξ( 2k) = ξ(2k), ( ) π 2k 2k 2 Γ ζ( 2k) = π k Γ(k) ζ(2k). 2 }{{}}{{} Q πq ζ( 2k) π 2k ζ(2k)q = Q. What about the odd positive numbers and even negative numbers? Let us first think where are the zeros of ζ are. Note, first, that there are no zeros of ζ(s) for Re(s) >, because of the Euler product expansion. By the functional equation and knowing the poles of Γ, we can hence conclude that ζ(s) does not have zeroes for Re(s) < 0 but it does have zeroes at 2, 4,.... What about the values of ζ(s) at positive odd integers? Because of the poles of the Γ function and zeros of the zeta function, we see here that ζ(2k+) is related by the functional equation to ζ ( 2k) for k 0. This is the truly interesting case that is hard to tackle. The remaining strip is the critical strip 0 Re(s). The zeros there are conjectured to be only at Re(s) = by the Riemann Hypothesis. 2 We finally summarize the situation in the following figure..2. Brief overview of general conjectures. Recall that we can attach L-function to pure motives. Conjecturally, the category of pure motives sits in a bigger category of mixed motives. The zeros of the L-function of the pure motive are then related to Ext in this larger category. In fact, the order r of vanishing of the L-function should be related to the rank of Ext. The arithmetic invariants of Ext are related to the values of L (r) (s).

5 MATH 678: (MOTIVIC) L-FUNCTIONS Figure. The green dots are the critical points, the red dots are values in the critical strip and were omitted from the discussion (s = is a pole of ζ and ζ(0) Q ), the blue dots are zeros of the zeta function, the gray dots are the mysterious points where not much is known. Definition. (Deligne, [Del79]). A critical point of a (motivic) L-function is an integer s = n such that the Γ-factors on either side of the functional equation have no poles. For example, the critical points for ζ(s) are the even positive numbers and the odd negative numbers. In these cases, the values of the ζ-function are known and can be written as a product of an explicit (transcendental) number and a rational number. Sporadic results on special values. Historically, people studied sums such as m+ni Z+Zi CM elliptic curves and we will discuss them later in the course.. These are related to period of (m+ni) 4 Later, Shimura [Shi76a] proved rationality of critical values of L-functions of modular forms. Then, Deligne [Del79] formulated the definition of critical points and stated a conjecture of rationality. For critical values n for M, he defined numbers Ω(M, n) and conjectured that L(M, n) Ω(M, n) Q. For example, for the ζ-function, Ω = (2πi) 2k, as we have seen above.

6 6 KARTIK PRASANNA The L-function of an elliptic curve is L(E, s) = ( α p p p s )( β p p s ) = p ( a p p s + p 2s, where a p = #E(F p ). The factor at infinity is Γ(s) and the functional equation then also involves Γ(2 s). Therefore, the only critical point is. This is the point relevant to the Birch Swinnerton-Dyer conjecture. No other points are critical for this L-function, but one could still study its values at noncritical integer points. The first person to do that was Bloch; for example, we studied L(E, 2) in the CM case. Later, Beilinson formulated a general conjecture for non-critical points. Beilinson s conjectures. Case. Let n be a point which is not the center of the center plus. Then Beilinson defines 2 Reg(M, n) and conjectures that L(M, n) Ω(M, n) Reg(M, n) Q. Case 2. The point is center plus. Beilinson states a similar conjecture in this case. Here, 2 the point could be a pole and it is related to the Tate conjecture. Case 3. The point is the center. Then the conjecture involves a height pairing. (See, for example, BSD.) The order in which one would study these conjectures is: () critical points away from the center, (2) non-critical points when the order of vanishing is and the center when the order of vanishing is, (3) the case where the order of vanishing is 2 (very little is known here)..3. p-adic L-functions. Let us first go back to the ζ-function. Recall that ζ( n) for n even is rational. In fact, it is ζ( n) = B n n where B n is the nth Bernoulli numbers, defined as te t e t = t n B n n!. n= Kummer congruences. When m n 0 mod p, then m B n mod p n (both sides are p-adic integers). When m n 0 mod p r (p ), then B m ( p m ) B m m ( pn ) B n n mod p n.

7 MATH 678: (MOTIVIC) L-FUNCTIONS 7 Kubota and Leopoldt constructed a p-adic analytic function ζ p (s) whose values interpolate the values of ζ at negative integers. In general, we already mentioned we can associate an L-function to a motive. We can also associate to it a p-adic L-function. In general, there can be more than one p-adic L-function and the theory seems richer than the archimedian theory. 2. L-functions of Dirichlet characters We will start by talking about L-functions of Dirichlet characters. The goals are the following: analytic continuation, functional equation, formula for values at critical points, L(, χ) for χ. 2.. Dirichlet series. Since the L-function is defined as a Dirichlet series, let us start by reviewing the general theory. A Dirichlet series is a series of the form a n n. s n= We want to determine a convergence criterion for this. Lemma 2. (Partial summation). Consider two sequence (b n ), (c n ) and let B n = b + b b n be the partial sum. Then l l b n c n = B l c l + B n (c n c n+ ) B k c k. Proof. We have that n=k n=k l l b n c n = (B n B n )c n n=k n=k = B l c l + l B n (c n c n+ ) B k c k. n=k This completes the proof. a Proposition 2.2. If n n s converges at s 0, then it converges for all s such that Re(s) > n= Re(s 0 ). Moreover, it converges uniformly on compact subsets, and hence the limit is an analytic function in s for Re(s) > Re(s 0 ). Proof. Write n= a n n = a n s n s 0 n= n s s 0

8 8 KARTIK PRASANNA and let P l (s 0 ) = Using Partial summation 2., we have that l a n l n = a n s n s 0 n s s 0 n=k n=k = P l (s 0 ) l s s 0 + l n=k l n= a n n s 0. ( ) P n (s 0 ) n s s 0 n s s 0 Note that n n+ s s 0 (n + ) = (s s 0) s s 0 n x dx. s s 0+ If Re(s) Re(s 0 ) + δ, this shows that n s s 0 (n + ) s s 0 s s 0. n δ+ Applying this to the above sum, we have that l a n n s P l(s 0 ) + c s s l δ 0 Since the series n= n=k l n=k P k (s 0 ) k. s s 0 n + P k (s 0 ). δ+ l s converges, this completes the proof. n δ+ Proposition 2.3. Consider a n n s and let n= s n = a + + a n. Let σ 0 and suppose A n C n σ for some C. Then and defines an analytic function. Proof. The proof is similar to the proof above. Consider the partial sum l P l (s) = a n n s. Then P l (s) P k (s) = l n=k = A l l s a n n s + l n= n= ( ) a n n A k. s (n + ) s k n=k }{{} s n+ xs+ dx n a n n s converges for Re(s) > σ

9 MATH 678: (MOTIVIC) L-FUNCTIONS 9 As in the previous proof, for Re(s) > σ, we take δ > 0 such that Re(s) σ + δ and complete the proof similarly Dirichlet characters. A Dirichlet character is a homomorphism ( ) Z χ: C. fz ( Z The smallest f such that χ factors through fz) is the conductor of χ. We may then define The L-function of χ is defined as χ(n) = To consider its convergence, note that l χ(n) n= { χ([n]) if (n, f) = 0 otherwise. L(s, χ) = { C l 0 C l Then Proposition 2.3 yields the following result. n= χ(n) n s. if χ is non-trivial, if χ is trivial. Proposition 2.4. If χ, L(s, χ) converges to an analytic function for Re(s) > 0. If χ =, L(s, χ) converges to an analytic function for Re(s) >. Exercise. Check that for Re(s) >, we have that ζ(s) = p L(s, χ) = p p s, χ(p)p s Analytic continuation. Fix a Dirichlet character χ of conductor f. Define F (z) = f a= χ(a) ze az e fz, G(z) = f a= χ(a) e az e fz. Note that F ( z) = z G(z). The poles of F are at 2πin/f.

10 0 KARTIK PRASANNA Note also that for t R 0 : G(t) = = f a= χ(a) e az e fz f χ(a)e at ( + e ft + e 2ft + ) a= = n χ(n)e nt which will provide a connection with the above L-function. Let C be the contour consisting of the negative real numbers with a small circle C ɛ around 0: Define H(s) = C s dz F (z)z z where z s = exp((s ) log(z)) and we take the principal branch of log. The key observation is that for s Z we have that H(s) = C ɛ F (z)z s dz z. Check that this integral converges absolutely and gives an analytic function of s. Substituting s s, we see that H(s) = C F ( z)( z) s dz z

11 MATH 678: (MOTIVIC) L-FUNCTIONS where ( z) s = exp(s log( z)) = z s e πis because log( z) = log(z) πi. Then H(s) = F ( z)z s e πis dz z C z = e πis G(z)z s dz C ) = e ( C πis G(z)z s dz + (e 2πis ) G(t)t s dt. ɛ Now assume σ = Re(s) >. Note that on C ɛ, G(z) < C ɛ on C ɛ. Also, z s < C 2 (s) ɛ σ. Then we see that G(z)z s dz C C ɛ ɛ C 2(s)ɛ σ 2πɛ 0 as ɛ 0. Therefore, letting ɛ 0, we see that Altogether, we have shown that H(s) = e πis (e 2πis ) = (e πis e πis ) = (e πis e πis ) = (e πis e πis ) and we know that H(s) is analytic. Finally, 0 0 G(t)t s dt χ(n)e nt t s dt n= χ(n) n= 0 ɛ e nt t s dt χ(n)n s e t t s dt. n= H(s) = (e πis e πis )Γ(s)L(s, χ) (e πis e πis )Γ(s) = 2i sin(πs)γ(s) = 0 2πi Γ( s). This shows that L(s, χ) = Γ( s)h(s) for Re(s) >. 2πi This will provide the analytic continuation. Indeed, the right hand side is defined for Re(s), except for a possible pole at s =. To determine whether or not L(s, χ) has a pole at s =, we need to check if H(s) has a zero at s =. We have that Recall that H() = C ɛ F (z) dz z = 2πi Res z=0 F (z) = f a= χ(a) ze az e fz. F (z). z

12 2 KARTIK PRASANNA Taking the power series expansions, we see that Res z=0 F (z) z = f a= χ(a) f = { 0 if χ, otherwise. This completes the analytic continuation Functional equation. We will now establish the functional equation. Recall that s dz H(s) = F (z)z z. Let Re(s) < 0. Consider the keyhole contour C R of radius R: C Then H(s) = lim R C R s dz F (z)z z. (Check that this is actually true.) Considering the residues, we have that H(s) = 2πi Res z=2πin/f (F (z)z s 2 ). We have that n=, n 0 R n := Res z=2πin/f F (z)z s 2 = = f f a= χ(a)e 2πina/f e (s )(log(2πn/f)+π/2) f a= f ( ) s 2πn χ(a)e 2πina/f e (s ) π 2 i. f

13 Recall that the Gauss sum is defined as Then a= MATH 678: (MOTIVIC) L-FUNCTIONS 3 g χ = f χ(a)e 2πia/f. a= R n = χ(n) f g χ ( 2πn f ) s e (s ) π 2 i for n. Similarly, R n = f χ(a)e 2πian/f e (s )(log(2πn/f) πi/2 = χ( n) ( ) s 2πn g χ e (s ) π 2 i. f f f Then H(s) = 2πi = n= n=, n 0 Res z=2πin/f (F (z)z s 2 ) ( 2πi 2πn g χ f f ) s ( e (s ) π = ig χ ( 2π f ) s ( e (s ) π 2 i + χ( )e (s ) π 2 i) χ(n) 2 i + χ( )e (s ) π i) 2 χ(n)n s n= ( ) s 2π = g χ (e s π 2 i χ( )e s π 2 i )L( s, χ) f } {{ } L( s,χ) since Re(s) < 0. Then L(s, χ) = ( ) s 2π Γ( s) (e s π 2 i χ( )e s π 2 i )L( s, χ). 2πi f Using Γ( s)γ(s) = Finally, this shows that π, we see that sin πs L(s, χ) = π 2πi sin(πs)γ(s) which is the functional equation. Finally, we have that L(s, χ) = ( ) s 2π (e s π 2 i χ( )e s π 2 i )L( s, χ). f ( ) s 2π g χ f L( s, χ) (e π 2 is + χ( )e π 2 is )Γ(s), { e π 2 is + χ( )e π 2 is 2 cos sπ if χ is even = 2 2i sin sπ if χ is odd. 2

14 4 KARTIK PRASANNA Letting δ be 0 or according to whether χ is even or odd, we can write e π 2 is + χ( )e π 2 is = 2i δ cos π (s δ). 2 We can finally state a cleaned up version of the functional equation. Theorem 2.5 (Functional equation). Suppose χ is a primitive character of conductor f. Let ( ) ( s+δ π 2 ) ( ) s + δ Λ(s, χ) = Γ L(s, χ). f 2 Then Λ(s, χ) = ( ) gχ Λ( s, χ). fi δ It is an exercise to show that g χ = f, which will show that the term value. ( ) gχ fi δ has absolute Proof. We have that Λ(s, χ) Λ( s, χ) = ( π f ( π = f ( π = f ( π f ) ( s+δ ) ( s+δ ) ) = g χ fi δ = g χ fi δ = g χ, fi δ 2 ) Γ ( s+δ 2 2 ) Γ ( s+δ 2 2 s Γ ( ) s+δ 2 Γ ( s+δ 2 ) 2 Γ ( s+δ 2 Γ ( s+δ 2 Γ ( s+δ 2 ) ) ) L(s, χ) ) L( s, χ) ( ) s 2π g χ f Γ(s)2i δ cos π (s δ) 2 Γ ( s 2 g χ π2 s 2 ) s 2iδ cos π(s δ) by Γ(z)Γ 2 ) Γ ( s+ 2 π ) ( Γ s+ δ ) cos π (s δ) 2 2 ) ( z + ) = 2 2z πγ(2z) 2 π sin π ( s+ δ 2 π cos π(s δ) by Γ( s)γ(s) = π 2 sin πs completing the proof Special values of L(s, χ). Recall that we showed that L(s, χ) = Γ( s)h(s). 2πi

15 MATH 678: (MOTIVIC) L-FUNCTIONS 5 Therefore: L( n, χ) = Γ(n)H( n) 2πi = 2πi Γ(n) F (z)z n dz C ɛ = Γ(n) Res z=0 (F (z)z n ). Let us take the equation f a= χ(a)te at e ft = n=0 B n,χ t n n! as the definition of B n,χ, the generalized Bernoulli numbers. Note that B n,χ Q(χ), the field of definition of χ. We then see that L( n, χ) = Γ(n) B n,χ n! = B n,χ n. By the functional equation: except when χ =, n =. B n,χ = 0 if n δ (2) Looking at the power series expansion of the terms in the definition of F, we see that Therefore, F (t) = = f a= f a= B 0,χ = χ(a)t( + at + (at)2 2! + ) ft + (ft)2 2! + χ(a) f f a= ft ( + at + )( 2 + ). χ(a) f = { 0 if χ, if χ =. Similarly, B,χ = f a= χ(a)a f 2 f f χ(a)a if χ, χ(a) = f a= if χ =. 2 a= What about L(, χ)? We will prove the following theorem in the next section (Corollary 2.9). Theorem 2.6 (Dirichlet). For χ, we have that L(, χ) 0. In particular, Dirichlet used this to show that there are infinitely many primes in arithmetic progressions. See, for example, [Ser73a].

16 6 KARTIK PRASANNA 2.6. Zeta functions of number fields. The reference for this is [Lan94]. Let K be any number field. Then we may define ζ K (s) = Na = s a O K To show convergence, we note that we may write p prime Np s. ζ K (s) = n s #{a Na = n}, n which is a Dirichlet series. We have reduced the convergence problem to bounding We define #{a Na = n}. ρ K = 2r (2π) r 2 h K R K w K dk where r, r 2 are the numbers of real and complex embeddings of K, h K is the class number, R K is the regulator, w K is the number of roots of unity, and d K = disc K/Q. Then one can show that #{a Na = n} = ρ K n + O(n [K:Q] ). This implies that ζ K (s) defines an analytic function for Re(s) > [K:Q] with a simple pole at s = of residue ρ K. This is the analytic class number formula. Similarly to the classical ζ-function, we have an analytic continuation and functional equation. Define Γ R (s) = π s/2 Γ(s/2), Then setting we have that Γ C (s) = (2π) s Γ(s). Λ(s) d K s Γ R (s) r Γ C (s) r 2 ζ K (s), Γ(s) = Γ( s). Example 2.7. Consider a primitive nth root of unity ζ n and K = Q(ζ n ). Then ( ) Gal(K/Q) = Z nz canonically. Given a primitive nth root of unity ζ, we have that σζ = ζ i(σ) for some i(σ) and the map is σ i(σ). The injectivity of the map is clear. We prove surjectivity. Suppose (p, n) =. We need to show that ζ ζ p is an automorphism of K/Q. Let f(x) be the irreducible polynomial of ζ. Then x n = f(x)h(x). If ζ p is not a root of f, then it is a root of h(x), so ζ is a root of h(x p ). Writing h(x p ) = f(x)g(x) and reducing modulo p, we see that h(x) p = f(x)g(x). This shows that f(x) has multiple roots. Since (n, p) =, x n is separable modulo p, which is a contradiction. Thus ζ p K.

17 Therefore, we have that MATH 678: (MOTIVIC) L-FUNCTIONS 7 (ζ ζ p ) [p] The element ζ ζ p is the Frobenius at p, Frob p. ( ) Z. nz Therefore, any character of conductor f dividing n is a character of the Galois group Gal(K/Q). Theorem 2.8. For K = Q(ζ n ), we have that ζ K (s) = χ primitive conductor dividing n L(s, χ). Proof. Write n = m p r for (p, m) =. Then K = Q(ζ n ) = Q(ζ m )Q(ζ p r) and we have the following diagram K = Q(ζ n ) K = Q(ζ m ) Q(ζ p r) Then K is the maximal extension of Q unramified at p. Letting Q F K be the maximal extension in which p splits completely, we know that K is the fixed field of inertia I p and F is the fixed field of the decomposition group G p. In particular, if p factors as in K, then the extensions have degrees i= Q (p) = (p... p g ) e K e K Ip = K f K Gp = F We consider the euler factor at p on both sides. On the left hand side, this is g ( ) g Np s = ( i p ) = ( fs tf ) g, residue fields have size equal to f. g Q. i=

18 8 KARTIK PRASANNA We now consider the right hand side. We first note that any Dirichlet character of conductor divisible by p will satisfy χ(p) = 0, so we may only consider characters that factor through (Z/mZ) = Gal(K /Q). Note that [p] corresponds to Frob p under this isomorphism, as mentioned above. In fact Frob p Gal(K /F ) = G p /I p and # Gal(K /F ) = f. In particular, χ(frob p ) is an fth root of unity. Moreover, any choice of fth root of unity determines the character χ of Gal(K /Q) and extends to exactly g characters of the whole Galois group Gal(K/Q). This shows that ( χ(p)t) = χ χ ( χ(frob p )t) = ( µt) g = ( t f ) g. µ f = Since the Euler factors at p agree on both sides, this proves the theorem. Corollary 2.9. For χ, we have that L(, χ) 0. Finally, we have the following generalization. Consider Q K Q(ζ n ). Then ζ K (s) = χ L(s, χ), where the product is over characters of ( Z nz) = Gal(Q(ζn )/Q) which factor through Gal(K/Q) Evaluation of L(, χ). Suppose χ. For χ odd, we have that L(, χ) = g χ ( 2π f 2i ) L(0, χ) = g χπ if ( B,χ) = iπg χ f f Corollary 2.0. If χ is an odd character, f χ(a)a 0. a= f χ(a)a. While this fact is very elementary, it would be really hard to prove it directly. Note that when χ is odd, is a critical point of L(s, χ), so we get this simple formula. Application. Suppose K/Q is imaginary quadratic. Then K Q(ζ n ) for some n by the Kronecker Weber theorem. Therefore, ζ K (s) = ζ(s)l(s, χ) for some χ. What is χ? It is the unique non-trivial quadratic character of conductor equal to d K. The residue at s = gives 2πh K w K dk = πig d K χ χ(a)a. d K d K a= a=

19 Therefore, taking absolute values MATH 678: (MOTIVIC) L-FUNCTIONS 9 h K = w K 2 d K d K χ(a)a. This allows to compute the class number in some cases. For example, one can use this to compute h Q( 23). This gives h K = 3. More examples are given in the homework. For real quadratic thing, we would obtain a factor corresponding to the regulator which could be harder to compute. Assume now χ is even and non-trivial. In this case, is not a critical point of L(s, χ). Nonetheless, we have that: χ(n) L(, χ) = n = n= n n= a= g χ f a= a= χ(a)e 2πian/f = f χ(a) g χ n e2πian/f n= = f χ(a) log( e 2πia/f ). g χ a= This is valid when χ is odd as well, but we will simply recover the previous formula Corollary Kubota Leopoldt p-adic L-function We will use the formulas for L( n, χ) to p-adically interpolate these values. This will give the Kubota Leopoldt p-adic L-function, which is our first example of a p-adic L-function. For most of this section, the reference is [Iwa72]. 3.. Power series over non-archimedean fields. We fix notation as follows: C p = completion of Q p Q p = algebraic closure of Q p K = finite extension of Q p Q p

20 20 KARTIK PRASANNA We will assume that the absolute value on Q p is normalized so that p = p. Consider K x, the ring of power series with coefficients in K. If A(x) K x. Then for any ξ K, A(ξ) = a n ξ n converges if and only if a n ξ n 0. Observe that if this power series converge for some ξ 0, then it converges for all ξ with ξ ξ 0. Lemma 3.. Let A(x), B(x) K x, both convergent in some neighborhood of 0. Suppose there is a sequence (ξ n ) C p of non-zero elements such that ξ n 0 as n, and A(ξ n ) = B(ξ n ) for all n. Then A = B. Proof. By looking at A B, we may assume that A(ξ n ) = 0 for all n and show that A = 0. Suppose A(x) = a m x m + a m+ x m+ + and a m 0. Then we have that 0 = a m ξ m k + a m+ ξ m+ k +, a m ξ m k = a m+ ξ m+ k +, a m = ξ k (a m+ + a m+2 ξ k + ). Since A(x) = a n x n converges in neighborhood of 0, there is an R such that a i R i 0, so a i C R i. For k large enough, ξ k < R l where we choose l such that a m+n ξ n k C R (m+n) (R l ) n. By choosing l appropriately, the sum a m+ + a m+2 ξ k + is bounded as k. Hence letting k in the expression above, ξ k 0, so a m = 0, which is a contradiction. For A = a n x n, we let A = sup a n. Let n P K = {A K x A < }. Note that: K[x] P K K x, A 0 and A = 0 if and only if A = 0, A + B max{ A, B }, ca = c A, AB A B. To check that P K is a Banach algebra, we just need to show it is complete. Proposition 3.2. The algebra P K is complete for. Proof. Suppose (A k ) is a Cauchy sequence in P K. Write A k (x) = a n,k x n. For l k, we have that n= A l (x) A k (x) = n (a n,l a n,k )x n.

21 MATH 678: (MOTIVIC) L-FUNCTIONS 2 Therefore, for each n, (a n,k ) is a Cauchy sequence. Let a n = lim a n,k and k A(x) = a n x n. Then a n,k A k. sup k ( A k ) <. n= To see that A(x) P K, note that a n = lim k a n,k and so a n We now want to show that lim k A k = A. Given ɛ > 0, there is an N such that for any l k N, A l A k < ɛ. Then so and hence This completes the proof. a n,l a n,k < ɛ for all n and l k N, a n a n,k ɛ for all n and k N, A A k ɛ for all k N. We want an estimate for ( X, n) where ( ) X X(X )... (X n + ) =. n n! For that, we need to estimate the p-adic absolute value of n!. Lemma 3.3. We have that p n p p n p n! np p n p. Proof. Write n = a 0 + a p + + a m p m where p m n < p m+ and 0 a i p. Let r = a 0 + a + + a m. The power of p dividing n! is [ ] [ ] [ ] n n n = (a p p 2 p m + + a m p m ) + (a 2 + a 3 p + ) + + a m = a + a 2 (p + ) + a 3 (p 2 + p + ) + + a m (p m + + ) = ( a (p ) + a (p 2 ) + + a m (p m ) ) p = p (n a 0 a a m ) = n r p. Finally, we obtain that n! = p n r n n p p p p p. For the other bound on n!, we have that r (m + )(p ), so and hence n! = p n r p n r p n (m + ), p p n p (m+) = p m+ p n p np p n p,

22 22 KARTIK PRASANNA proving the other inequality. The following theorem will be crucial in constructing p-adic L-functions by interpolation. Theorem 3.4. Let (b n ) n 0 be a sequence in K and (c n ) n 0 be defined by e t t n b n n! = t n c n n!. n=0 Suppose there is a real number R, 0 < R < p p a power series A(x) P K such that n=0 () A(x) coverges for all ξ such that ξ < R p p, (2) A(n) = b n. Proof. Let A k (x) = k i=0 < such that cn C R n. Then there is c i ( X i ). Note that deg Ak k. We claim that A k (x) is a Cauchy sequence in P K. We have that l ( ) X A l (x) A k (x) = c i i i=k+ ( ) max X k+ i l c i i max k+ i l CRi p i p. Let R = Rp p. Note that R <. Then by the above estimates showing that A k (x) is Cauchy. A l (x) A k (x) CR k+, By Proposition 3.2, there A k (x) converges, so let We may write A(x) = lim k A k (x) = A k (x) = a n x n. n=0 a n,k x n n=0 where we note that a n,k = 0 for n k. Then a n = lim a n,k. Since a n = lim a n,k, the k k estimate shows that a n C R n. a n,k = a n,k a n,n as a n,n = 0 A k A n C R n

23 MATH 678: (MOTIVIC) L-FUNCTIONS 23 Note that A(x) converges for all ξ such that ξ < R which shows (). We now need to show that (2) A(n) = b n. We claim that A(ξ) = lim k A k (ξ). We have that For n > k, a n,k = 0, so A(ξ) A k (ξ) = (a n a n,k )ξ n. n=0 (a n a n,k )ξ n = a n ξ n C R n ξ n C (R ξ ) k. For n k, (a n a n,k )ξ n A A k ξ n CR k+ ξ n, as A A k C R k+. Finally, this shows that { (a n a n,k )ξ n C R k+ if ξ <, C R (R ξ ) k if ξ. As k, these bounds show that lim (A(ξ) A k (ξ)) = 0. Finally, recall that A k (x) = k k ( c X ) i i, so i=0 A k (n) = k ( ) n c i = b n. i This shows existence and uniqueness follows from Lemma 3.. This is the main tool that we will use to construct p-adic L-functions. i= Kubota Leopoldt p-adic L-function. Let p be an odd prime. The short exact sequence + pz p Z p ( Z pz) splits and the splitting is given by the map ( ) Z ω : {roots of unity in Z p } pz defined by the relation ω(a) a (p). This is the Teichmuller character. We fix throughout embeddings Q C, Q Q p. Then ω defines a Dirichlet character via the first embedding. The split short exact sequence shows that there is an isomorphism ( ) Z Z p ( + pz p ) pz a (ω(a), a ). When p = 2, the situation is similar with p replaced by 4 where necessary:

24 24 KARTIK PRASANNA + 4Z 2 Z 2 ( Z 4Z). We hence define to deal with both cases simultaneously. { p if p odd q = 4 if p = 2 Theorem 3.5 (Kubota Leopoldt). Let χ be a Dirichlet character and K = Q p (χ). Then there is a unique power series A χ (x) P K such that A χ converges for ξ < q p p and A χ (n) = ( χω n (p)p n )B n,χω n. Note that for primitive characters χ, χ 2, by χ χ 2 we mean the primitive character associated to the product. We immediately note that the interpolation seems related to L( n, χω n ) and investigate this further. First, { A χ (0) = ( χ(p)p 0 if χ, )B 0,χ = ( p ) if χ =., so we let Then L p (s, χ) = s A χ( s). L p ( n, χ) = n ( χω n (p)p n )B n,χω n = ( χω n (p)p n )L( n, χω n ). Therefore, this p-adic function interpolates the L-values of L(s, χω n ). The intuitive reason the factor ( χω n (p)p n ) comes up is that the sum would n k behave badly p-adically if we allow p in the denominator. The reason for the Teichmuller character is a little harder to describe, but it is related to congruence between special values of L-functions. We will next work towards proving this theorem. Recall that we define Bernoulli numbers via te t e t = t n B n n!. n=0 The Bernoulli polynomials are similarly defined by e tx te t e t = B n (x) tn n!. Moreover, for a character χ, we defined the generalized Bernoulli numbers by f χ(a)te at e ft = t n B n,χ n!. a= n=0 n=0

25 MATH 678: (MOTIVIC) L-FUNCTIONS 25 Again, analogously, we define B n,χ (x) the generalized Bernoulli polynomials by e tx Proposition 3.6. We have that () B n, (x) = B n (x), (2) B n,χ (0) = B n,χ, (3) B n,χ (x) = n ) Bi,χ x n i, i=0 ( n i f χ(a)te at e ft = B n,χ (x) tn n!. a= (4) if χ, B n,χ ( x) = ( ) n χ( )B n,χ (x) and if χ =, B n, ( x) = ( ) n B n, (x ), (5) B n,χ = 0 if χ and n δ (2), (6) B n,χ = f n f ( a χ(a)b n ). f a=0 Proof. All of these properties are straightforward, so we just prove (6). We have that n=0 Then B n,χ = B n,χ (0) = f n B n,χ (x) tn n! = etx f a= = = f = f f a= f a= n=0 te at χ(a) e ft χ(a)te t(a+x) e ft f a= a= We may use Taylor expansions in the equality f a+x χ(a) (ft)eft f e ft f ( a + X χ(a) B n f n=0 ) (ft) n. n! χ(a)b n ( a f ). a= χ(a)te at = (e ft ) n=0 B n,χ t n n! to compute the Bernoulli numbers. In general, on the left side, we will get expressions of the form k S n,χ (k) = χ(a)a n. Lemma 3.7. We have that () S n,χ (kf) = (B n+ n+,χ(kf) B n+,χ (0)), (2) B n,χ = lim S h p h f n,χ(p h f) for any prime p (the limit is a p-adic limit). a=

26 26 KARTIK PRASANNA Proof. For (), we let Then F χ (t, x) = e tx F χ (t, x) F χ (t, x f) = e tx = = f a= f a= f a= χ(a)te at e ft. χ(a)te at e ft et(x f) χ(a)te at e ft etx ( e ft ) f χ(a)te t(x+a f). Taking coefficients of t n on both sides, we see that B n,χ (x) B n,χ (x f) f (x + a f)n = χ(a), n! (n )! a= B n,χ (x) B n,χ (x f) = n Plugging in x = f, 2f,..., kf, we see that a= B n,χ (f) B n,χ (0) = n B n,χ (2f) B n,χ (f) = n B n,χ (kf) B n,χ ((k + )f) = n and summing these together, we get This shows ().. f a= f χ(a)(x + a f) n. a= f χ(a)(a) n a= f χ(a)(a + f) n a= χ(a)te at e ft f χ(a)(a + (k )f) n a= B n,χ (kf) B n,χ (0) = n kf a= χ(a)a n. We will use () to show (2). We have that n+ ( ) n + B n+,χ (x) = B i,χ x n+ i. i Thus i=0 B n+,χ (kf) B n+,χ (0) = n ( ) n + B i,χ (kf) n+ i. i i=0

27 MATH 678: (MOTIVIC) L-FUNCTIONS 27 Let k = p h. Then B n+,χ (p h f) B n+,χ (0) p h f = n ( ) n + B i,χ (p h f) n i. i Now, take the p-adic limit as h : ( ) Bn+,χ (p h f) B n+,χ (0) lim = (n + )B h p h n,χ. f By (), we have that This shows (2). i=0 S n,χ (p h f) lim h p h f = B n,χ. We are finally ready to prove the existence of the p-adic L-function. Proof of Theorem 3.5. We want to use Theorem 3.4. Let χ n = χω n and b n = ( χ n (p)p n )B n,χn. Observe that B n,χn = lim S h p h f h,χ n (p h f). This is true by Lemma 3.7 after noting that the conductor of χ n is f p ɛ where ɛ {0,, }. Then b n = lim h p h f S n,χ n (p h f) χ n (p)p n lim h p h f S n,χ n (p h f) = lim h p h f S n,χ n (p h f) χ n (p)p n lim h p h f S n,χ n (p h f) p = lim h p h f S n,χ n (p h f) χ n (p)p n h f lim χ h p h n (a)a n f a= = lim h p h f = lim h p h f = lim h p h f = lim h q h f p h f a= p h f a= (a,p)= p h f a= (a,p)= q n f a= (a,p)= p χ n (a)a n h f lim χ h p h n (ap)(ap) n f a= χ n (a)a n a n χ(a) χ(a) a n. Recall that in Theorem 3.4, we have that n ( ) n c n = ( ) n i b i. i i=0 letting h h in second limit as χ n (a) = χ(a)ω(a) n

28 28 KARTIK PRASANNA Then for Then We claim that c n = n ( ) n ( ) n i i i=0 c n,h = q h f c n,h = χ(a) a= n i=0 lim h q h f a= (a,p)= q h f n ( n ( ) n i i i=0 c n,h q h f 0 q h f a= (a,p)= ( ) n i ( n i ) a i = qn mod q 2 f χ(a) a i c n,h = lim h q h f ) χ(a) a i. q h f a= (a,p)= χ(a)( a ) n. (for n large enough). In other words, induction on h. For h =, we have that c n,h q h f q n q 2 f is a p-adic integer. To show this, we proceed by c n, qf = qf qf a= (a,p)= χ(a)( a ) n which is congruent to 0 modulo qn, since a mod q. For h, recall that q 2 f c n,h+ q h+ f = q h+ f q h f a= (a,p)= χ(a)( a ) n. For a q n+ f, let a = u + q h fv for u q h f, 0 v < q. Then ω(a) = ω(u), so Then We claim that: a = ω(a) a = ω(u) (u + q h fv) = u + ω(u) q h fv. ( a ) n = n i=0 ( ) n ( u ) i (ω(u) q h fv) n i. i ( a ) n ( u ) n mod q n+h. This is because q i divides ( u ) i and q n i divides the second term, and i + h(n i) = i + (n i) + (h )(n i) = n + (h )(n i) n + h if n i. Thus, modulo

29 q n h, only the i = n term survives. We have that MATH 678: (MOTIVIC) L-FUNCTIONS 29 c n,h+ = q q h f a= (a,p)= q h f u= (u,p)= qc n,h χ(a)( a ) n χ(u)( u ) n mod q n+h so completing the proof. c n,h+ q h+ f c n,k q h f mod qn+h q n+ f, 3.3. p-adic logarithms, exponential functions, and power functions. We will next work towards Leopoldt s theorem about the value L p (, χ). Note that is outside of the normal interpolation range: n for n. We first recall some facts and p-adic logarithms, exponential functions, and power functions. Recall that the p-adic log is defined by the power series log p ( + x) = ( ) n n= n xn which converges for + ξ where ξ n n 0 as n, i.e. for ξ <. Then log p : {x C p x < } C }{{} p U is a continuous homomorphism (checking that log p (xy) = log p (x) + log p (y) is left as an exercise). Theorem 3.8. There is a unique extension of log p to a homomorphism log p : C p C p such that log p (p) = 0. Further, log p is continuous and for all σ Aut(C p /Q p ) = Gal(Q p /Q p ), log p (α) σ = log p (α σ ). The choice of log p (p) = 0 is equivalent to choosing a branch of the logarithm. Note that the last identity is not true for complex analytic logs. This will make working with the p-adic logarithm much easier. Proof. We have the following diagram

30 30 KARTIK PRASANNA C p Q p Q Q p Z where the map is given by α r if α = p r. Pick a section of this homomorphism, r P r Q p such that Pr P s = P r+s and if r Z, then P r = p r. We may let P r = p m/n = (p m ) /n for r = m/n, embedding Q Q p. Let P be the image of this section. We have that P Q p C p. We claim that C p = P W U where W is the set of roots of unity in Q p of order prime to p and U = {x C p x < }. For α C p, pick α Q p such that α is close enough to α. Then α = α = P r for some P r P. Therefore: α = and setting β = α P r, we can choose β Q p P r close enough to β such that β = β =. Consider Q p Q p (β ). Then β is a unit in O K and if we consider the reduction map β β p O K O K /p = F p f (p) Z p Z p /(p) = F p Since β n = for some (n, p), by Hensel s Lemma, we can lift β to some nth root of unity w O K such that ω β mod p. Then ω β <, and hence β /ω <, so β /ω U. We have hence written α α = P r = P r ω β P r ω. Since β is close to β, β U ω. We finally need to show that the product is direct. Suppose P r ω u =. Then P r =, so P r =, and hence ω u =. Since ω <, ω =, and hence also u =. We have hence shown that C p = P W U.

31 MATH 678: (MOTIVIC) L-FUNCTIONS 3 For r = m n, P n r = P m = p m, so n log p P r = m log p p. Setting log p = 0, we get that log p (P r ωu) = log p (u). This defines the logarithm on C p. We finally have to show that log p (α σ ) = log p (α) σ. Let α = P r ωu. Then α σ = Pr σ ω σ u σ. For r = m, P n n r = p m, so (Pr σ ) = p m. Hence Pr σ = P r (nth root of unity), showing that n log p Pr σ = m log p p = 0. Finally, this shows that log p (α σ ) = log(u σ ) = log(u) σ, completing the proof. We define exp(x) = so n=0 x n. To check convergence, recall that n! n! p n p, ξ n n! ξ n, p n p showing that the above power series converges for ξ < p p. Exercise. If x < p p, then logp (exp(x)) = x and exp(log p ( + x)) = + x. We want to define x s for s Z p as exp(s log p (x)). For x + pz p, log p (x) pz p, i.e. log p (x) p < p p if p 2. To deal with the case p = 2, we note that for x + qzp, we have that log p (x) qz p, and we may define x s = exp(s log p (x)). However, we would still like to extend this to + pz p for all p. Lemma 3.9. For x + qz p and s Z p, x s = n=0 ( ) s (x ) n. n Proof. We use the proof of Theorem 3.4. Here c n = (x ) n and we know that c n q n. Then the right hand side converges for s q p p.

32 32 KARTIK PRASANNA In particular, it converges for s Z p. To check that the left hand side is equal to the right hand side, it suffices to check at all the integers (which are dense in Z p ). We have that ( ) m m ( ) m (x ) n = (x ) n n n completing the proof. n=0 n=0 = (x + ) m = x m, We may hence define x s for x + 2Z p and s Z 2 by the right hand side in the above lemma. We now have defined x s for x + pz p and s Z p. We want to analyze it as a function of the two variables. We define a function by φ: Z p Z p Z p 0 if x Z p, φ(x, s) = x s if p 2, x Z p, x s if p = 2, x Z p. Consider a finite extension K of Q p. Then C K = {f : Z p K continuous}. Set Note that f = sup x Z p f(x) = max x Z p f(x). f + g max( f, g ) fg = f g, af = a f, and C K is complete for. Indeed, if f n is a Cauchy sequence in C k, for each s, f n (S) is a Cauchy sequence in K, so we may set f(s) = lim f n (s) and this function is continuous. n Note that K[x] C K, so, for example, the function ( s n) is in CK. We define γ n C K by Proposition 3.0. We have that We first state a lemma. γ n (s) = n ( ) n ( ) n i φ(i, s). i i=0 φ(x, s) = ( ) x γ n (s). n n=0

33 MATH 678: (MOTIVIC) L-FUNCTIONS 33 Lemma 3.. We have that is divisible by n!. n ( ) n ( ) n i i m i i=0 We will prove this later. Proof of Proposition 3.0. First, let us check this for x = m, Then the right hand side is m ( ) m m ( ) n ( ) m n γ n (s) = ( ) n i φ(i, s) n n i n=0 n=0 i=0 m m ( )( ) m n = φ(i, s) ( ) n i n i and we have that m ( )( ) m n ( ) n i = n i n=i Altogether, this shows that = m! i! n=i i=0 n=i m m! n! n!(m n)! (n i)!i! ( )n i n=i m ( n i (m n)!(n i)! = m! i! m i t=0 ( ) t (m i t)!t! = m! ( + ( ))m i i! = δ mi. m ( ) m γ n (s) = φ(m, s). n n=0 where n i = t We claim that γ n n!. Pick a large integer m such that p divides m and p m < n!. Then we have that n ( ) n γ n (m) = ( ) n i φ(i, m). i i=0 Now, for i pz p, φ(i, m) = 0 and otherwise φ(i, m) = i m = (iω(i) ) m = i m since p m. We then see that n ( ) n γ n (m) = ( ) n i φ(i, m) i i=0 n ( ) n ( ) n i i m mod p m. i i=0 By Lemma 3., we see that γ n (m) n!. Since such m are dense in Z p, we have that γ n n!.

34 34 KARTIK PRASANNA Now, we use Lemma 3.3 to conclude that γ n n! np p n p. For x Z p, we have that ( x n) is bounded and since γn (s) n! 0 as n, this shows convergence. Hence the two functions are continuous functions Z p Z p Z p which agree on a dense subset, and hence are equal. We now prove the lemma. Proof of Lemma 3.. We have that (e x ) n = = = n ( ) n ( ) n i (e x ) i i n ( ) n ( ) n i (xi) m i m! m=0 x m n ( ) n ( ) n i i m m! i i=0 }{{} d m,n i=0 i=0 m=0 We have that d m,n = dm dx m (ex ) n. x=0 We have that d m dx m (ex ) n = dm dx m (x(ex ) n e x ) = dm dx m (x(ex ) n + n(e x ) n ), showing that d m,n = nd m,n + nd m,n. By induction on m, we see that n! d m,n. Let C K = {f : Z p K continuous} and Q K = {formal power series A = x n a n K, a n n! } with the sup norm A = sup n a n n!. Then K[x] Q K is a dense subset. Theorem 3.2. There is a unique bounded linear map Γ: Q K C K such that Γ(x n ) = γ n (s). It satisfies Γ(( + x) n ) = φ(n, s). Proof. Define a map Γ: K[x] C K by Γ(x n ) = γ n (s). We have that x n Q = n!, γ n n!, so Γ(x n ) x n Q. Hence Γ(A) A Q for all A K[x]. n=0

35 MATH 678: (MOTIVIC) L-FUNCTIONS 35 Hence Γ is a continuous function K[x] C K and we may extend it uniquely to a bounded linear map Q K C K. Then ( n ( ) n Γ(( + n) n ) = Γ )x i i i=0 n ( ) n = γ(i, s) i i=0 ( ) n = γ(i, s) i by Proposition 3.0. i=0 = φ(n, s) Write Γ(A) = Γ A. We write down a formula for Γ A (s). Proposition 3.3. Define δ n (A) by A(e t ) = n=0 δ n (A) tn n!. Then Γ A (s) = lim ni δ ni (A), where (n i ) is a sequence of integers such that n i in Z, (p ) n i, and n i s in Z p. Proof. Recall that (e t ) n = t m m! d m,n m=0 and we showed in Lemma 3. that d m,n n!. If A = a n x n, then n=0 A(e t ) = a n (e t ) n = n=0 What is δ n (A)? We have that Since a i d n,i a i i!, we have that n=0 δ n (A) = a n m=0 a i d n,i. i=0 t m m! d m,n = δ n (A) sup a i i! A Q. 0 i n a i i=0 m=0 t m m! d m,i. To check the equality on polynomials K[x], it is enough to check it for A = ( + x) m. In that case, the left hand side is Γ A (s) = φ(m, s)

36 36 KARTIK PRASANNA and we need to check the right hand side is equal to this. We have that A(e t (mt) n ) =, n! so δ n (A) = m n. We need to evaluate lim i m n i. If p m, φ(m, s) = 0 and lim i m n i = 0 since n i as integers. In the case p not dividing m, we need to consider the two subcases p = 2 and p odd. We just deal with p odd, the other case is similar. Here, the left hand side is whereas the right hand side is m s n=0 lim i m n i = lim i m n i = m s since (p ) n i and n i s p-adically. Therefore, the formula holds for all polynomials. Finally, let A Q K. Fix ɛ > 0. Then, there exists B K[x] such that A B Q < ɛ. This implies that Γ A Γ B < ɛ. We then estimate Γ A (s) δ ni (A) = Γ A (s) Γ B (s) + Γ B (s) δ ni (B) + δ ni (B) δ ni (A) completing the proof. max{ Γ A (s) Γ B (s), Γ }{{} B (s) δ ni (B), δ }{{} ni (B) δ ni (A) } }{{} <ɛ <ɛ for i 0 A B Q <ɛ < ɛ, Proposition 3.4. Let A(x) = n=0 a nx n Q K and A (x) = na n x n Q K, Then DA Q K and Γ DA (s) = sγ A (s). Proof. We have that By Proposition 3.3, we have that n=0 (DA)(x) = ( + x) log( + x)a (x). (DA)(e t ) = te t A (e t ) ( = t d dt Γ DA (s) = lim i = = ) A(e t ) ( t d ) dt n=0 n=0 (nδ n (A)) tn n! δ n (A) tn n! (n i δ ni (A)) = s Γ A (s). This completes the proof.

37 Consider A(x) = n=0 MATH 678: (MOTIVIC) L-FUNCTIONS 37 a n x n Q K. Then a n n! 0 and for ξ p p, since n! p n p Lemma 3.3, so a n ξ n a n p n p an n! 0. This shows that A(ξ) = a n ξ n converges. n=0 For such ξ, A(ξ) is defined and A(ξ) A QK. In particular, this shows that ξ A(ξ) is continuous. Proposition 3.5. Let A Q K. Then Γ A (0) = A(0) A(ξ ). p Proof. Since both sides are continuous in A, it is enough to check equality for polynomials A K[x]. Let A(x) = ( + x) m. Then ξ p = Γ A (s) = φ(m, s). If p m, the left hand side is φ(m, s) = 0 and the right hand side is A(0) = ξ m = 0. p ξ p = If p does not divide m, the left hand side is and the right hand side is ξ m =. p This completes the proof. ξ p = 3.4. Leopoldt s formula for L p (, χ). The goal of this section is to prove a formula analogous to the one in section 2.7 about L p (, χ). Suppose now that χ is a primitive character of conductor f. Let ζ be a fixed primitive fth root of unity. We may take ζ = e 2πi/f. Let f h(z) = χ(a)z a, Writing we have that H(z) = z f = a= h(z) z f = H(z) = f a= f a= f a= χ(a)z a z f. ζ a (z ζ s ) f, h(ζ a ) z ζ a ζ a f. by

38 38 KARTIK PRASANNA Now, Finally, h(ζ a )ζ a = f χ(b)(ζ a ) b ζ a = b= H(z) = g χ f f a= f χ(b)(ζ a ) b = χ(a)g χ. b= χ(a) z ζ a. On the other hand, so we define H(e t ) = () F (t) = te t H(e t ) = f a= a= χ(a)e t(a ) e ft, f χ(a)te at e ft = t n B n,χ n!. Pick an auxiliary integer N such that (N, fp) =. Let {λ} be the set of all Nth roots of unity in Q p. Lemma 3.6. We have that g χ f a f z λζ = a Nχ(N)zN H(z N ). In particular, Proof. We have so g χ f g χ f f a z λζ = g χ a f a= λ a= f a= λ f a= which is what we claimed. H(z). λ n=0 χ(a) z λζ a = Nχ(N)zN H(z N ) H(z). z λα = χ(a)nz N z N ζ an = g χχ(n) f NzN Z N α N Theorem 3.7 (Leopoldt s theorem). We have that L p (, χ) = g χ f f a= χ(an)nz N z N ζ an = g χ Nχ(N)H(z N ), The second assertion follows from the above description of ( χ(p) ) f χ(a) log p p ( ζ a ). a=

39 MATH 678: (MOTIVIC) L-FUNCTIONS 39 Comparing this to the standard equation for for L(, χ), the only difference is the Euler factor at p which was taken out when we defined L p, so we need to put it back in, and the use of p-adic logarithm instead of regular logarithm. In the proof of the theorem, we will take K/Q p containing {λ}, ζ, and the values of χ. We will then show that for A(x) = g χ f f a= λ ( χ(a) log + ) x λζ a we have that Γ A (s) = ( χ(n) N s )L p ( s, χ) if p is odd (and similarly for p = 2), and use Proposition 3.5 to get the result. We first with proving the above formula for Γ A (s). Theorem 3.8. Let K/Q p be a finite extension containing {λ}, ζ, and the values of χ. Let A(x) = g χ f f a= λ ( χ(a) log + ) x. λζ a Then A(x) Q K and Γ A (s) = { ( χ(n) N s )L p ( s, χ) if p is odd, ( χ(n)n s )L p ( s, χ) if p = 2. ( ) Proof. Note that (! λζ a ) is a unit, so log + x Q λζ a K. This shows that A(x) Q K. We suppose p is odd. The case p = 2 is dealt with in the same way, omitting all the occurrences of. The idea is to use Proposition 3.3 to find Γ DA (s) and the apply Proposition 3.4. We have that DA(x) = g f χ f ( + x) log( + x) χ(a) + x a= λ a= λ λζ a λζ a = g f χ f ( + x) log( + x) χ(a) x + ( λζ a ).

40 40 KARTIK PRASANNA Then DA(e t ) = g χ f tet f χ(a) e t λζ a ) a= λ = te t ( Nχ(N)(e t ) N H(e tn ) H(e t ) ) by Lemma 3.6 = χ(n)nte tn H(e tn ) te t H(e t ) = χ(n)f (Nt) F (t) as F (t) = te t H(e t ) by definition (Nt) n t n = χ(n) B n,χ B n,χ n! n! by equation () = n=0 n=0 n=0 t n n! (χ(n)n n )B n,χ. Letting δ n (DA) = (χ(n)n n )B n,χ and applying Proposition 3.3, we get that Γ DA (s) = lim i δ ni (DA) where n i as integers, (p ) n i, and n i s p-adically. By definition of L p ( n, χ) (Theorem 3.5), we have that L p ( n, χ) = ( χω n (p)p n )L( n, χω n ). Since (p ) n i, ω n i (p) =, and we have that L p ( n i, χ) = ( χ(p)p n i )L( n i, χ) = ( χ(p)p n i ) B n i,χ n i. Taking the limit as i, this shows that Since (p ) n i, N n i = N n i, and hence Finally, this shows that lim B n i,χ = L p ( s, χ) s. i lim N n i = lim N n i = N s. i i Γ DA (s) = lim i δ ni (DA) = (χ(n) N s )L p ( s, χ). Applying Proposition 3.4, we see that completing the proof. Γ A (s) = ( χ(n), N s )L p ( s, χ), This allows to prove Leopoldt s formula 3.7 by applying Proposition 3.5. Proof of Theorem 3.7. By Proposition 3.5, we have that Γ A (0) = A(0) A(ξ ). p ξ p =

41 For A from Theorem 3.8, we have that MATH 678: (MOTIVIC) L-FUNCTIONS 4 A(0) = 0, A(ξ ) = g χ f f a= λ ( ) ξ λζ a χ(a) log p. λζ a Therefore: Γ A (0) = g χ pf = g χ pf = g χ pf a= f χ(a) log p ξ λζ a λζ a f λ ξ p = ( χ(a) log p a= λ f a= χ(a) log p ( λ λ p ζ ap ( λζ a ) p λζ ap ( λζ a ) p ) ). We now split the sum as follows: f a= χ(a) log p ( λ ) ( ) λζ ap f = χ(a) log λζ a p ( λζ ap ) a= λ }{{} S 2 p f ( λζ a ) ) ( χ(a) log p a= λ }{{} S. We claim that S 2 = χ(p)s. This is clear when (p, f) = because ζ a and ζ p are both primitive roots of unity. When p f, we may write f = pf. Since χ has conductor f, it does not factor as follows: (Z/fZ) C θ χ (Z/f Z)

42 42 KARTIK PRASANNA and hence ker χ ker θ, i.e. there is a b coprime to f such that χ(b) but b (f ). In that case, ab a (f ), so abp ap (f). Therefore: ( ) f S 2 = χ(a) log p ( λζ ap ) a= λ ( ) f = χ(ab) log p ( λζ abp ) a= λ ( ) f = χ(b) χ(a) log p ( λζ ap ) a= λ = χ(b)s 2. Since χ 0, this shows that S 2 = 0. In any case, we have shown that S 2 = χ(p)s. Hence: S 2 ps = (χ(p) p)s f ζ an = (χ(p) p) a log p ζ a a= ( f ) f = (χ(p) p) χ log p ( ζ an ) χ log p ( ζ a ) a= = (χ(p) p)(χ(n) ) Altogether, we have shown that: a= f χ(a) log p ( ζ a ). ( χ(n))l p (, χ) = Γ A (0) = g f χ fp (χ(p) p)(χ(n) ) χ(a) log p ( ζ a ). a= a= We are free to choose N. As long as we choose N so that χ(n), this shows that i.e. Leopoldt s formula 3.7. L p (, χ) = g χ f ( χ(p) ) f χ(a) log p p ( ζ a ), a= 3.5. p-adic class number formula. For the standard L-function, we proved that L(, χ) 0 for χ (Corollary 2.9) using the class number formula. We develop the analog of these notions in this section. Recall that L p ( n, χ) = ( ) L( n, χω n ). Suppose χ is odd. When n is even χω n is even and n is even, so L( n, χω n ) = 0. Similarly, when n is even, ωω n is odd and n is odd, so L( n, χω n ) = 0. Therefore L p (s, χ) 0 for χ odd.

43 MATH 678: (MOTIVIC) L-FUNCTIONS 43 This is consistent with Leopoldt s formula 3.7: for χ odd ( f ) L p (, χ) = f χ(a) log 2 p ( ζ a ) χ(a) log p ( ζ a ) = 0. a= Therefore, we only consider even characters. In this case, the analog of Dirichlet s Theorem 2.9 holds. Theorem 3.9. If χ is even, then L p (, χ) 0. Recall that in the proof of Dirichlet s Theorem 2.9, we considered the zeta function of K = Q(ζ f ). Because we only take even characters here, we need to consdier the totally real subfield K + = Q(ζ f ) + of K: a= K = Q(ζ f ) 2 K+ = Q(ζ f ) + Q φ(f)/2 We previously showed that ζ K +(s) = L(s, χ) χ character of G + and used the class number formula: 2 n h K +R K + 2 = L(, χ). d K + χ How is the regulator defined? For K, by Dirichlet s unit theorem, U K /torsion = Z r +r 2 where r is the number of real places and r 2 is the number of complex places. Then if σ,..., σ r, σ r +, σ r +,..., σ r +r 2, σ r +r 2 are all the embeddings of K into C, then (dropping one of the embeddings) where (u,..., u r ) is a basis for U K /torsion. R K = det(log(σ i (u j )) r r ) To get a p-adic analog for the class number formula, we want to define the p-adic regulator. Having fixed embeddings Q into C and C p, embeddings K C are in bijection with embeddings K C p, but it is difficult to distinguish between the real and complex embeddings in C p. However, if K is a totally real field, there is no problem. If σ,..., σ r+ are all the embeddings, then we define R K,p = det(log p (σ i (u j )) r r ) (dropping one of the embeddings.)