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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 2018, LATEX’edby Aleksander Horawa (who is the only person responsible for any mistakes that may befound in them).

This version is from January 15, 2019. Check for the latest version of these notes at

http://www-personal.umich.edu/~ahorawa/index.html

If you find any typos or mistakes, please let me know at [email protected]. Thanks toPeter Dillery for helping me catch up on anything I missed.

The class will involve 10 homeworks and a final project/talk. The homework, together withsome supporting papers, will be posted on the shared course file.

Contents

1. Overview 2

1.1. First example: the ζ-function 3

1.2. Brief overview of general conjectures 4

1.3. p-adic L-functions 6

2. L-functions of Dirichlet characters 7

2.1. Dirichlet series 7

2.2. Dirichlet characters 9

2.3. Analytic continuation 9

2.4. Functional equation 12

2.5. Special values of L(s, χ) 14

2.6. Zeta functions of number fields 16

2.7. Evaluation of L(1, χ) 18

3. Kubota–Leopoldt p-adic L-function 19

3.1. Power series over non-archimedean fields 19

3.2. Kubota–Leopoldt p-adic L-function 23

Date: January 15, 2019.1

http://www-personal.umich.edu/~ahorawa/index.html

mailto:[email protected]

2 KARTIK PRASANNA

3.3. p-adic logarithms, exponential functions, and power functions 29

3.4. Leopoldt’s formula for Lp(1, χ) 37

3.5. p-adic class number formula 42

4. p-adic measures and power series 47

4.1. Power series 48

4.2. Integration 52

4.3. Alternative construction of Lp(s, χ) 56

4.4. Applications to class numbers in cyclotomic towers 62

4.5. Main conjecture of Iwasawa theory 66

5. Serre’s construction of the p-adic L-function 67

5.1. Classical modular forms 67

5.2. Reduction modulo p 70

5.3. Serre’s p-adic modular forms 86

5.4. Hecke operators 89

6. Moduli-theoretic interpretation of modular forms 92

6.1. Modular forms over C 92

6.2. Modular forms over an arbitrary ring 94

7. p-adic L-functions associated to Hecke characters 101

7.1. Real analytic Eisenstein series 102

7.2. Algebraicity of L-values 106

7.3. p-adic interpolation 107

8. p-adic L-function for Rankin–Selberg L-function 114

8.1. Algebraicity of critical values 115

8.2. p-adic interpolation 120

Appendix A. Project topics 122

References 122

1. Overview

One can associate L-functions to varieties (or even pure motives) over number fields. Thequestion is: how do L-functions capture arithmetic of algebraic varieties (motives)?

MATH 678: (MOTIVIC) L-FUNCTIONS 3

In this class, we will use the language of motives very loosely. The reader interested in theprecise definitions and more details should consult the papers [Mil13, Kim10]

One can also associate p-adic L-functions to varieties (motives), which should conjecturallycapture similar arithmetic information.

L-functions L(M, s)

pure motives M

p-adic L-functions Lp(M, s)

This class will focus on the following cases:

(1) 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 wheresome of the conjectures are settled are those where one can associate an automorphic formto 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:

(1) GL(1)Q,(2) GL(1)k,(3) GL(2)Q.

Time permitting, we might consider other reductive groups at the end of the class.

1.1. First example: the ζ-function. We start with the simplest example of an L-function,the ζ-function. It is defined for Re(s) > 1 by the following Dirichlet series

ζ(s) =∞∑n=1

1

ns.

It admits an Euler product

ζ(s) =∏p

1

1− p−s.

In fact, it is the L-function of the trivial motive.

Analytic continuation. The function ζ(s) admits an analytic continuation to C with theexception of a pole of order one at s = 1.

4 KARTIK PRASANNA

Functional equation. Let ξ(s) = π−s/2Γ(s/2)ζ(s). Then

ξ(s) = ξ(1− s).

The surprising factor π−s/2Γ(s/2) should be thought of as the Euler factor at ∞. It isactually, in some sense, the most important Euler factor. Some basic properties of the Γfunction are recalled for the reader’s convenience: We recall some useful properties of the Γfunction:

(1) Γ(z + 1) = zΓ(z),(2) Γ(z)Γ(1− z) = π

sinπz,

(3) Γ(z)Γ(z + 1/2) = 21−2x√πΓ(2z),

(4) Γ(z) has poles at 0,−1,−2, . . ..

We are interested in the values of ζ(s) at integers. We know that ζ(2) = π2

6, ζ(4) = π4

90, and

in general

ζ(2k) ∈ π2k ·Q× for k ≥ 1.

Using the functional equation, we see that

ξ(1− 2k) = ξ(2k),

so

π−1−2k

2 Γ

(1− 2k

2

)︸︷︷︸∈√πQ×

ζ(1− 2k) = π−k Γ(k)︸︷︷︸∈Q×

ζ(2k).

This shows that

ζ(1− 2k) ∈ π−2kζ(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) forRe(s) > 1, because of the Euler product expansion. By the functional equation and knowingthe poles of Γ, we can hence conclude that ζ(s) does not have zeroes for Re(s) < 0 but itdoes have zeroes at −2,−4, . . ..

What about the values of ζ(s) at positive odd integers? Because of the poles of the Γ functionand zeros of the zeta function, we see here that ζ(2k+1) is related by the functional equationto 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) ≤ 1. The zeros there are conjectured tobe only at Re(s) = 1

2by the Riemann Hypothesis.

We finally summarize the situation in the following figure.

1.2. Brief overview of general conjectures. Recall that we can attach L-function topure motives. Conjecturally, the category of pure motives sits in a bigger category of mixedmotives. The zeros of the L-function of the pure motive are then related to Ext1 in thislarger category. In fact, the order r of vanishing of the L-function should be related to therank of Ext1. The arithmetic invariants of Ext are related to the values of L(r)(s).


−1−2−3−4 2 3 4

Figure 1. The green dots are the critical points, the red dots are values inthe critical strip and were omitted from the discussion (s = 1 is a pole of ζand ζ(0) ∈ Q×), the blue dots are zeros of the zeta function, the gray dots arethe mysterious points where not much is known.

Definition 1.1 (Deligne, [Del79]). A critical point of a (motivic) L-function is an integers = 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 negativenumbers. In these cases, the values of the ζ-function are known and can be written as aproduct of an explicit (transcendental) number and a rational number.

Sporadic results on special values.

Historically, people studied sums such as∑

m+ni∈Z+Zi

1(m+ni)4

. These are related to period of

CM elliptic curves and we will discuss them later in the course.

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 ofrationality. 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 KARTIK PRASANNA

The L-function of an elliptic curve is

L(E, s) =∏p

1

(1− αpp−s)(1− βpp−s)=∏p

1

(1− app−s + p1−2s,

where 1− ap = #E(Fp).

The factor at infinity is Γ(s) and the functional equation then also involves Γ(2− s). There-fore, the only critical point is 1. This is the point relevant to the Birch–Swinnerton-Dyerconjecture.

No other points are critical for this L-function, but one could still study its values at non-critical 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 1. Let n be a point which is not the center of the center plus 12. Then Beilinson defines

Reg(M,n) and conjectures that

L(M,n)

Ω(M,n) Reg(M,n)∈ Q×.

Case 2. The point is center plus 12. Beilinson states a similar conjecture in this case. Here,

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, forexample, BSD.)

The order in which one would study these conjectures is:

(1) critical points away from the center,(2) non-critical points when the order of vanishing is 1 and the center when the order of

vanishing is 1,(3) the case where the order of vanishing is ≥ 2 (very little is known here).

1.3. p-adic L-functions. Let us first go back to the ζ-function. Recall that ζ(1− n) for neven is rational. In fact, it is

ζ(1− n) = −Bn

nwhere Bn is the nth Bernoulli numbers, defined as

tet

et − 1=∞∑n=1

Bntn

n!.

Kummer congruences. When m ≡ n 6≡ 0 mod p− 1, then

Bm

m≡ Bn

nmod p

(both sides are p-adic integers). When m ≡ n 6≡ 0 mod pr(p− 1), then

(1− pm−1)Bm

m≡ (1− pn−1)

Bn

nmod pn.


Kubota and Leopoldt constructed a p-adic analytic function ζp(s) whose values interpolatethe values of ζ at negative integers.

In general, we already mentioned we can associate an L-function to a motive. We can alsoassociate to it a p-adic L-function. In general, there can be more than one p-adic L-functionand 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 follow-ing:

• analytic continuation,• functional equation,• formula for values at critical points,• L(1, χ) for χ 6= 1.

2.1. Dirichlet series. Since the L-function is defined as a Dirichlet series, let us start byreviewing the general theory. A Dirichlet series is a series of the form

∞∑n=1

anns.

We want to determine a convergence criterion for this.

Lemma 2.1 (Partial summation). Consider two sequence (bn), (cn) and let Bn = b1 + b2 +· · ·+ bn be the partial sum. Then∑

n=k

bncn = B`c` +∑n=k

Bn(cn − cn+1)−Bk−1ck.

Proof. We have that ∑n=k

bncn =`−1∑n=k

(Bn −Bn−1)cn

= B`c` +∑n=k

Bn(cn − cn+1)−Bk−1ck.

This completes the proof.

Proposition 2.2. If∞∑n=1

anns

converges at s0, then it converges for all s such that Re(s) >

Re(s0). Moreover, it converges uniformly on compact subsets, and hence the limit is ananalytic function in s for Re(s) > Re(s0).

Proof. Write∞∑n=1

anns

=∞∑n=1

anns0· 1

ns−s0

8 KARTIK PRASANNA

and let

P`(s0) =∑n=1

anns0

.

Using Partial summation 2.1, we have that∑n=k

anns

=∑n=k

anns0

1

ns−s0

= P`(s0)1

`s−s0+

`−1∑n=k

Pn(s0)

(1

ns−s0− 1

ns−s0−1

)− Pk−1(s0)

1

ks−s0.

Note that1

ns−s0− 1

(n+ 1)s−s0= (s− s0)

∫ n+1

n

1

xs−s0+1dx.

If Re(s) ≥ Re(s0) + δ, this shows that∣∣∣∣ 1

ns−s0− 1

(n+ 1)s−s0

∣∣∣∣ ≤ |s− s0|nδ+1

.

Applying this to the above sum, we have that∣∣∣∣∣∑n=k

anns

∣∣∣∣∣ ≤ |P`(s0)|`δ

+ c|s− s0| ·`−1∑n=k

1

nδ+1+|Pk−1(s0)|

`s.

Since the series∞∑n=1

1nδ+1 converges, this completes the proof.

Proposition 2.3. Consider∞∑n=1

ann−s and let

sn = a1 + · · ·+ an.

Let σ ≥ 0 and suppose |An| ≤ C · nσ for some C. Then∞∑n=1

ann−s converges for Re(s) > σ

and defines an analytic function.

Proof. The proof is similar to the proof above. Consider the partial sum

P`(s) =∑n=1

ann−s.

Then

P`(s)− Pk−1(s) =∑n=k

an1

ns

=A``s

+`−1∑n=k

an

(1

ns− 1

(n+ 1)s

)︸︷︷︸

≤n+1∫n

1xs+1 dx

−Ak−1

ks.


As in the previous proof, for Re(s) > σ, we take δ > 0 such that Re(s) ≥ σ+ δ and completethe proof similarly.

2.2. Dirichlet characters. A Dirichlet character is a homomorphism

χ :

(ZfZ

)×→ C×.

The smallest f such that χ factors through(

ZfZ

)×is the conductor of χ.

We may then define

χ(n) =

χ([n]) if (n, f) = 1

0 otherwise.

The L-function of χ is defined as

L(s, χ) =∞∑n=1

χ(n)

ns.

To consider its convergence, note that∣∣∣∣∣∑n=1

χ(n)

∣∣∣∣∣ ≤C · `0 if χ is non-trivial,

C · `1 if χ is trivial.

Then Proposition 2.3 yields the following result.

Proposition 2.4. If χ 6= 1, L(s, χ) converges to an analytic function for Re(s) > 0. Ifχ = 1, L(s, χ) converges to an analytic function for Re(s) > 1.

Exercise. Check that for Re(s) > 1, we have that

ζ(s) =∏p

1

1− p−s,

L(s, χ) =∏p

1

1− χ(p)p−s.

2.3. Analytic continuation. Fix a Dirichlet character χ of conductor f . Define

F (z) =

f∑a=1

χ(a) · zeaz

efz − 1,

G(z) =

f∑a=1

χ(a) · e−az

1− e−fz.

Note that F (−z) = z ·G(z). The poles of F are at 2πin/f .

10 KARTIK PRASANNA

Note also that for t ∈ R≥0:

G(t) =

f∑a=1

χ(a) · e−az

1− e−fz

=

f∑a=1

χ(a)e−at(1 + e−ft + e−2ft + · · · )

=∑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 Cε around 0:

Define

H(s) =

∫C

F (z)zs−1dz

z

where zs−1 = exp((s− 1) log(z)) and we take the principal branch of log.

The key observation is that for s ∈ Z we have that

H(s) =

∫Cε

F (z)zs−1dz

z.

Check that this integral converges absolutely and gives an analytic function of s.

Substituting s 7→ −s, we see that

H(s) =

∫−C

F (−z)(−z)s−1dz

z


where (−z)s = exp(s log(−z)) = zse−πis because log(−z) = log(z)− πi. Then

H(s) = −∫−C

F (−z)zse−πisz−1dz

z

= −e−πis∫−C

G(z)zs−1dz

= −e−πis(∫−Cε

G(z)zs−1dz + (e2πis − 1)

∫ ∞ε

G(t)ts−1dt

).

Now assume σ = Re(s) > 1. Note that on Cε, |G(z)| < C1

εon Cε. Also, |zs−1| < C2(s) · εσ−1.

Then we see that ∣∣∣∣∫−Cε

G(z)zs−1dz

∣∣∣∣ ≤ C1

εC2(s)εσ−1 · 2πε→ 0 as ε→ 0.

Therefore, letting ε→ 0, we see that

H(s) = −e−πis(e2πis − 1)

∫ ∞0

G(t)ts−1dt

= −(eπis − e−πis)∫ ∞

0

∞∑n=1

χ(n)e−ntts−1dt

= −(eπis − e−πis) ·∞∑n=1

χ(n)

∫ ∞0

e−ntts−1dt

= −(eπis − e−πis)∞∑n=1

χ(n)n−s∫ ∞

0

e−tts−1dt.

Altogether, we have shown that

H(s) = −(eπis − e−πis)Γ(s)L(s, χ)

and we know that H(s) is analytic. Finally,

−(eπis − e−πis)Γ(s) = −2i sin(πs)Γ(s) =−2πi

Γ(1− s).

This shows that

L(s, χ) = − 1

2πiΓ(1− s)H(s) for Re(s) > 1.

This will provide the analytic continuation. Indeed, the right hand side is defined for Re(s) ≤1, except for a possible pole at s = 1. To determine whether or not L(s, χ) has a pole ats = 1, we need to check if H(s) has a zero at s = 1.

We have that

H(1) =

∫Cε

F (z)dz

z= 2πiResz=0

F (z)

z.

Recall that

F (z) =

f∑a=1

χ(a) · zeaz

efz − 1.

12 KARTIK PRASANNA

Taking the power series expansions, we see that

Resz=0F (z)

z=

f∑a=1

χ(a)

f=

0 if χ 6= 1,

1 otherwise.

This completes the analytic continuation.

2.4. Functional equation. We will now establish the functional equation. Recall that

H(s) =

∫C

F (z)zs−1dz

z.

Let Re(s) < 0. Consider the keyhole contour C ′R of radius R:

Then

H(s) = − limR→∞

∫C′R

F (z)zs−1dz

z.

(Check that this is actually true.) Considering the residues, we have that

H(s) = −2πi∞∑

n=−∞,n6=0

Resz=2πin/f (F (z)zs−2).

We have that

Rn := Resz=2πin/f F (z)zs−2

=

f∑a=1

χ(a)e2πina/f

fe(s−1)(log(2πn/f)+π/2)

=1

f

f∑a=1

χ(a)e2πina/f

(2πn

f

)s−1

e(s−1)π2i.


Recall that the Gauss sum is defined as

gχ =

f∑a=1

χ(a)e2πia/f .

Then

Rn =χ(n)

fgχ

(2πn

f

)s−1

e(s−1)π2i

for n ≥ 1. Similarly,

R−n =1

f

f∑a=1

χ(a)e−2πian/fe(s−1)(log(2πn/f)−πi/2 =χ(−n)

fgχ

(2πn

f

)s−1

e−(s−1)π2i.

Then

H(s) = −2πi∞∑

n=−∞,n6=0

Resz=2πin/f (F (z)zs−2)

=∞∑n=1

−2πi

fgχ

(2πn

f

)s−1 (e(s−1)π

2i + χ(−1)e−(s−1)π

2i)χ(n)

= −igχ(

2π

f

)s (e(s−1)π

2i + χ(−1)e−(s−1)π

2i) ∞∑n=1

χ(n)ns−1

︸︷︷︸L(1−s,χ)

= −gχ(

2π

f

)s(es

π2i − χ(−1)e−s

π2i)L(1− s, χ)

since Re(s) < 0. Then

L(s, χ) =1

2πiΓ(1− s)

(2π

f

)s(es

π2i − χ(−1)es

π2i)L(1− s, χ).

Using Γ(1− s)Γ(s) = πsinπs

, we see that

L(s, χ) =1

2πi

π

sin(πs)Γ(s)

(2π

f

)s(es

π2i − χ(−1)es

π2i)L(1− s, χ).

Finally, this shows that

L(s, χ) =gχ

(2πf

)sL(1− s, χ)

(eπ2is + χ(−1)e−

π2is)Γ(s)

,

which is the functional equation.

Finally, we have that

eπ2is + χ(−1)e−

π2is =

2 cos sπ

2if χ is even

2i sin sπ2

if χ is odd.

14 KARTIK PRASANNA

Letting δ be 0 or 1 according to whether χ is even or odd, we can write

eπ2is + χ(−1)e−

π2is = 2iδ cos

π

2(s− δ).

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

Λ(s, χ) =

(π

f

)−( s+δ2 )Γ

(s+ δ

2

)L(s, χ).

Then

Λ(s, χ) =

(gχ√fiδ

)Λ(1− s, χ).

It is an exercise to show that |gχ| =√f , which will show that the term

(gχ√fiδ

)has absolute

value 1.

Proof. We have that

Λ(s, χ)

Λ(1− s, χ)=

(πf

)−( s+δ2 )Γ(s+δ

2

)L(s, χ)(

πf

)−( 1−s+δ2 )

Γ(

1−s+δ2

)L(1− s, χ)

=

(π

f

) 12−s Γ

(s+δ

2

)Γ(

1−s+δ2

) gχ

(2πf

)sΓ(s)2iδ cos π

2(s− δ)

=

(π

f

) 12 Γ

(s+δ

2

)Γ(

1−s+δ2

) gχ√π21−s2s

Γ(s2

)Γ(s+1

2

)2iδ cos π

2(s− δ)

by Γ(z)Γ

(z +

1

2

)= 21−2z

√πΓ(2z)

=gχ√fiδ

π

Γ(

1−s+δ2

)Γ(s+1−δ

2

) 1

cos π2(s− δ)

=gχ√fiδ

π sin π(s+1−δ

2

)π cos π

2(s− δ)

by Γ(1− s)Γ(s) =π

sin πs

=gχ√fiδ

,

completing the proof.

2.5. Special values of L(s, χ). Recall that we showed that

L(s, χ) = − 1

2πiΓ(1− s)H(s).


Therefore:

L(1− n, χ) = − 1

2πiΓ(n)H(1− n)

= − 1

2πiΓ(n)

∫Cε

F (z)z−n−1dz

= −Γ(n) Resz=0(F (z)z−n−1).

Let us take the equationf∑a=1

χ(a)teat

eft−1=∞∑n=0

Bn,χtn

n!

as the definition of Bn,χ, the generalized Bernoulli numbers. Note that Bn,χ ∈ Q(χ), thefield of definition of χ. We then see that

L(1− n, χ) = −Γ(n)Bn,χ

n!=−Bn,χ

n.

By the functional equation:

Bn,χ = 0 if n 6≡ δ (2)

except when χ = 1, n = 1.

Looking at the power series expansion of the terms in the definition of F , we see that

F (t) =

f∑a=1

χ(a)t(1 + at+ (at)2

2!+ · · · )

ft+ (ft)2

2!+ · · ·

=

f∑a=1

χ(a)

f(1 + at+ · · · )(1− ft

2+ · · · ).

Therefore,

B0,χ =

f∑a=1

χ(a)

f=

0 if χ 6= 1,

1 if χ = 1.

Similarly,

B1,χ =

f∑a=1

χ(a)a

f− 1

2

f∑a=1

χ(a) =

1f

f∑a=1

χ(a)a if χ 6= 1,

12

if χ = 1.

What about L(1, χ)? We will prove the following theorem in the next section (Corollary 2.9).

Theorem 2.6 (Dirichlet). For χ 6= 1, we have that

L(1, χ) 6= 0.

In particular, Dirichlet used this to show that there are infinitely many primes in arithmeticprogressions. See, for example, [Ser73a].

16 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

ζK(s) =∑a⊆OK

1

Nas=∏

p prime

1

1−Np−s.

To show convergence, we note that we may write

ζK(s) =∑n

n−s ·#a | Na = n,

which is a Dirichlet series. We have reduced the convergence problem to bounding

#a | Na = n.

We define

ρK =2r1(2π)r2hKRK

wK√|dK |

where r1, r2 are the numbers of real and complex embeddings of K, hK is the class number,RK is the regulator, wK is the number of roots of unity, and dK = discK/Q. Then one canshow that

#a | Na = n = ρKn+O(n1− 1[K:Q] ).

This implies that ζK(s) defines an analytic function for Re(s) > 1− 1[K:Q]

with a simple pole

at s = 1 of residue ρK . This is the analytic class number formula.

Similarly to the classical ζ-function, we have an analytic continuation and functional equa-tion. Define

ΓR(s) = π−s/2Γ(s/2),

ΓC(s) = (2π)−sΓ(s).

Then setting

Λ(s)√|dK |

sΓR(s)r1ΓC(s)r2ζK(s),

we have thatΓ(s) = Γ(1− s).

Example 2.7. Consider a primitive nth root of unity ζn and K = Q(ζn). Then

Gal(K/Q)∼=→(

ZnZ

)×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 7→ i(σ).

The injectivity of the map is clear. We prove surjectivity.

Suppose (p, n) = 1. We need to show that ζ 7→ ζp is an automorphism of K/Q. Let f(x) bethe irreducible polynomial of ζ. Then xn − 1 = f(x)h(x). If ζp is not a root of f , then it isa root of h(x), so ζ is a root of h(xp). Writing h(xp) = f(x)g(x) and reducing modulo p, wesee that h(x)p = f(x)g(x). This shows that f(x) has multiple roots. Since (n, p) = 1, xn− 1is separable modulo p, which is a contradiction. Thus ζp ∈ K.


Therefore, we have that

(ζ 7→ ζp) 7→ [p] ∈(

ZnZ

)×.

The element ζ 7→ ζp is the Frobenius at p, Frobp.

Therefore, any character of conductor f dividing n is a character of the Galois groupGal(K/Q).

Theorem 2.8. For K = Q(ζn), we have that

ζK(s) =∏

χ primitiveconductor dividing n

L(s, χ).

Proof. Write n = m · pr for (p,m) = 1. Then K = Q(ζn) = Q(ζm)Q(ζpr) and we have thefollowing diagram

K = Q(ζn)

K ′ = Q(ζm) Q(ζpr)

Q

Then K ′ is the maximal extension of Q unramified at p. Letting Q ⊆ F ⊆ K ′ be themaximal extension in which p splits completely, we know that K ′ is the fixed field of inertiaIp and F is the fixed field of the decomposition group Gp. In particular, if p factors as

(p) = (p1 . . . pg)e

in K, then the extensions have degrees

K

KIp = K ′

KGp = F

Q.

e

f

g

We consider the euler factor at p on both sides. On the left hand side, this isg∏i=1

(1− 1

Np−si

)=

g∏i=1

(1− 1

pfs) = (1− tf )g,

residue fields have size equal to f .

18 KARTIK PRASANNA

We now consider the right hand side. We first note that any Dirichlet character of conductordivisible 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 Frobp under this isomorphism, asmentioned above. In fact

Frobp ∈ Gal(K ′/F ) ∼= Gp/Ip

and # Gal(K ′/F ) = f . In particular, χ(Frobp) is an fth root of unity. Moreover, any choiceof fth root of unity determines the character χ of Gal(K ′/Q) and extends to exactly gcharacters of the whole Galois group Gal(K/Q).

This shows that∏χ

(1− χ(p)t) =∏χ

(1− χ(Frobp)t) =∏µf=1

(1− µt)g = (1− tf )g.

Since the Euler factors at p agree on both sides, this proves the theorem.

Corollary 2.9. For χ 6= 1, we have that

L(1, χ) 6= 0.

Finally, we have the following generalization. Consider Q ⊆ K ⊆ Q(ζn). Then

ζK(s) =∏χ

L(s, χ),

where the product is over characters of( ZnZ

)×= Gal(Q(ζn)/Q) which factor through Gal(K/Q).

2.7. Evaluation of L(1, χ). Suppose χ 6= 1. For χ odd, we have that

L(1, χ) =gχ

(2πf

)1

2iL(0, χ) =

gχπ

if(−B1,χ) =

iπgχf· 1

f

f∑a=1

χ(a)a.

Corollary 2.10. If χ is an odd character,

f∑a=1

χ(a)a 6= 0.

While this fact is very elementary, it would be really hard to prove it directly.

Note that when χ is odd, 1 is a critical point of L(s, χ), so we get this simple formula.

Application. Suppose K/Q is imaginary quadratic. Then K ⊆ Q(ζn) for some n by theKronecker–Weber theorem. Therefore,

ζK(s) = ζ(s)L(s, χ)

for some χ. What is χ? It is the unique non-trivial quadratic character of conductor equalto |dK |. The residue at s = 1 gives

2πhK

wK√|dK |

=πigχ|dK |

1

|dK |

|dK |∑a=1

χ(a)a

.


Therefore, taking absolute values

hK =wK2

1

|dK |

∣∣∣∣∣∣|dK |∑a=1

χ(a)a

∣∣∣∣∣∣ .This allows to compute the class number in some cases. For example, one can use this tocompute hQ(

√−23). This gives hK = 3. More examples are given in the homework.

For real quadratic thing, we would obtain a factor corresponding to the regulator whichcould be harder to compute.

Assume now χ is even and non-trivial. In this case, 1 is not a critical point of L(s, χ).Nonetheless, we have that:

L(1, χ) =∞∑n=1

χ(n)

n

=∞∑n=1

1

n

1

gχ

f∑a=1

χ(a)e2πian/f

=1

gχ

f∑a=1

χ(a)∞∑n=1

1

ne2πian/f

= − 1

gχ

f−1∑a=1

χ(a) log(1− e2πia/f ).

This is valid when χ is odd as well, but we will simply recover the previous formula Corol-lary 2.10.

3. Kubota–Leopoldt p-adic L-function

We will use the formulas for L(1−n, χ) to p-adically interpolate these values. This will givethe 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].

3.1. Power series over non-archimedean fields. We fix notation as follows:

Cp = completion of Qp

Qp = algebraic closure of Qp

K = finite extension of Qp

Qp

20 KARTIK PRASANNA

We will assume that the absolute value on Qp is normalized so that |p| = 1p.

Consider KJxK, the ring of power series with coefficients in K. If A(x) ∈ KJxK. Then forany ξ ∈ K, A(ξ) =

∑anξ

n converges if and only if |anξn| → 0.

Observe that if this power series converge for some ξ0, then it converges for all ξ with|ξ| ≤ |ξ0|.Lemma 3.1. Let A(x), B(x) ∈ KJxK, both convergent in some neighborhood of 0. Supposethere is a sequence (ξn) ∈ Cp of non-zero elements such that ξn → 0 as n → ∞, andA(ξn) = B(ξn) for all n. Then A = B.

Proof. By looking at A−B, we may assume that A(ξn) = 0 for all n and show that A = 0.Suppose

A(x) = amxm + am+1x

m+1 + · · · and am 6= 0.

Then we have that

0 = amξmk + am+1ξ

m+1k + · · · ,

−amξmk = am+1ξm+1k + · · · ,

−am = ξk(am+1 + am+2ξk + · · · ).

Since A(x) =∑anx

n converges in neighborhood of 0, there is an R such that |ai|Ri → 0, so|ai| ≤ C ·R−i. For k large enough, |ξk| < R` where we choose ` such that

|am+nξn−1k | ≤ C ·R−(m+n)(R`)n−1.

By choosing ` appropriately, the sum am+1 + am+2ξk + · · · is bounded as k → ∞. Henceletting k →∞ in the expression above, ξk → 0, so −am = 0, which is a contradiction.

For A =∑anx

n, we let ‖A‖ = supn|an|. Let

PK = A ∈ KJxK | ‖A‖ <∞.Note that:

• K[x] ⊂ PK ⊆ KJxK,• ‖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 PK is a Banach algebra, we just need to show it is complete.

Proposition 3.2. The algebra PK is complete for ‖−‖.

Proof. Suppose (Ak) is a Cauchy sequence in PK . Write

Ak(x) =∞∑n=1

an,kxn.

For ` ≥ k, we have that

A`(x)− Ak(x) =∑n

(an,` − an,k)xn.


Therefore, for each n, (an,k) is a Cauchy sequence. Let an = limk→∞

an,k and

A(x) =∞∑n=1

anxn.

Then |an,k| ≤ ‖Ak‖. To see that A(x) ∈ PK , note that |an| = limk |an,k| and so |an| ≤supk(‖Ak‖) <∞.

We now want to show that limk→∞

Ak = A. Given ε > 0, there is an N such that for any

` ≥ k ≥ N , ‖A` − Ak‖ < ε. Then

|an,` − an,k| < ε for all n and ` ≥ k ≥ N,

so|an − an,k| ≤ ε for all n and k ≥ N,

and hence‖A− Ak‖ ≤ ε for all k ≥ N.


We want an estimate for∥∥(X

n

)∥∥, where(X

n

)=X(X − 1) . . . (X − n+ 1)

n!.

For that, we need to estimate the p-adic absolute value of n!.

Lemma 3.3. We have that

|p|np−1 ≤ |p|

n−1p−1 ≤ |n!| ≤ np|p|

np−1 .

Proof. Write n = a0 + a1p + · · · + ampm where pm ≤ n < pm+1 and 0 ≤ ai ≤ p − 1. Let

r = a0 + a1 + · · ·+ am. The power of p dividing n! is[n

p

]+

[n

p2

]+ · · ·+

[n

pm

]= (a1 + · · ·+ amp

m−1) + (a2 + a3p+ · · · ) + · · ·+ am

= a1 + a2(p+ 1) + a3(p2 + p+ 1) + · · ·+ am(pm−1 + · · ·+ 1)

=1

p− 1

(a1(p− 1) + a1(p2 − 1) + · · ·+ am(pm − 1)

)=

1

p− 1(n− a0 − a1 − · · · − am)

=n− rp− 1

.

Finally, we obtain that

|n!| = |p|n−rp−1 ≥ |p|

n−1p−1 ≥ |p|

np−1 .

For the other bound on n!, we have that r ≤ (m+ 1)(p− 1), so

n− rp− 1

≥ n

p− 1− (m+ 1),

and hence|n!| = |p|

n−rp−1 ≤ |p|

np−1−(m+1) = pm+1|p|

np−1 ≤ np|p|

np−1 ,

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 (bn)n≥0 be a sequence in K and (cn)n≥0 be defined by

e−t∑n=0

bntn

n!=∞∑n=0

cntn

n!.

Suppose there is a real number R, 0 < R < |p|1p−1 < 1 such that |cn| ≤ C ·Rn. Then there is

a power series A(x) ∈ PK such that

(1) A(x) coverges for all ξ such that |ξ| < 1R|p|

1p−1 ,

(2) A(n) = bn.

Proof. Let Ak(x) =k∑i=0

ci(Xi

). Note that degAk ≤ k. We claim that Ak(x) is a Cauchy

sequence in PK . We have that

‖A`(x)− Ak(x)‖ =

∥∥∥∥∥ ∑i=k+1

ci

(X

i

)∥∥∥∥∥≤ max

k+1≤i≤`

∥∥∥∥ci(Xi)∥∥∥∥

≤ maxk+1≤i≤`

CRi · pi

p−1 .

Let R1 = Rp1p−1 . Note that R1 < 1. Then by the above estimates

‖A`(x)− Ak(x)‖ ≤ CRk+11 ,

showing that Ak(x) is Cauchy.

By Proposition 3.2, there Ak(x) converges, so let

A(x) = limk→∞

Ak(x) =∞∑n=0

anxn.

We may write

Ak(x) =∞∑n=0

an,kxn

where we note that an,k = 0 for n ≥ k. Then an = limk→∞

an,k. Since |an| = limk→∞|an,k|, the

estimate

|an,k| = |an,k − an,n−1| as an,n−1 = 0

≤ ‖Ak − An−1‖≤ C ·Rn

1

shows that |an| ≤ C ·Rn1 .


Note that A(x) converges for all ξ such that |ξ| < 1R1

which shows (1). We now need to show

that (2) A(n) = bn.

We claim that A(ξ) = limk→∞

Ak(ξ). We have that

A(ξ)− Ak(ξ) =∞∑n=0

(an − an,k)ξn.

For n > k, an,k = 0, so

|(an − an,k)ξn| = |anξn| ≤ C ·Rn1 |ξ|n ≤ C · (R1|ξ|)k.

For n ≤ k,

|(an − an,k)ξn| ≤ ‖A− Ak‖|ξ|n ≤ CRk+11 |ξ|n,

as ‖A− Ak‖ ≤ C ·Rk+11 . Finally, this shows that

|(an − an,k)ξn| ≤

C ·Rk+1

1 if |ξ| < 1,

C ·R1 · (R1|ξ|)k if |ξ| ≥ 1.

As k → ∞, these bounds show that limk→∞

(A(ξ) − Ak(ξ)) = 0. Finally, recall that Ak(x) =

k∑i=0

ci(Xi

), so

Ak(n) =k∑i=0

ci

(n

i

)= bn.

This shows existence and uniqueness follows from Lemma 3.1.

This is the main tool that we will use to construct p-adic L-functions.

3.2. Kubota–Leopoldt p-adic L-function. Let p be an odd prime. The short exactsequence

1 1 + pZp Z×p(

ZpZ

)×1

splits and the splitting is given by the map

ω :

(ZpZ

)×→ roots of unity in Z×p

defined by the relation ω(a) ≡ a (p). This is the Teichmuller character. We fix throughoutembeddings Q → C, Q → Qp. Then ω defines a Dirichlet character via the first embedding.

The split short exact sequence shows that there is an isomorphism

Z×p →(

ZpZ

)×× (1 + pZp)

a 7→ (ω(a), 〈a〉).

When p = 2, the situation is similar with p replaced by 4 where necessary:

24 KARTIK PRASANNA

1 1 + 4Z2 Z×2( Z

4Z

)×1.

We hence define

q =

p if p odd

4 if p = 2

to deal with both cases simultaneously.

Theorem 3.5 (Kubota–Leopoldt). Let χ be a Dirichlet character and K = Qp(χ). Then

there is a unique power series Aχ(x) ∈ PK such that Aχ converges for |ξ| < |q|−1|p|1p−1 and

Aχ(n) = (1− χω−n(p)pn−1)Bn,χω−n .

Note that for primitive characters χ1, χ2, by χ1χ2 we mean the primitive character associatedto the product.

We immediately note that the interpolation seems related to L(1−n, χω−n) and investigatethis further. First,

Aχ(0) = (1− χ(p)p−1)B0,χ =

0 if χ 6= 1,

(1− p−1) if χ = 1.,

so we let

Lp(s, χ) =1

s− 1Aχ(1− s).

Then

Lp(1− n, χ) =1

−n(1− χω−n(p)pn−1)Bn,χω−n = (1− χω−n(p)pn−1)L(1− n, χω−n).

Therefore, this p-adic function interpolates the L-values of L(s, χω−n).

The intuitive reason the factor (1 − χω−n(p)pn−1) comes up is that the sum∑

1nk

wouldbehave badly p-adically if we allow p in the denominator. The reason for the Teichmullercharacter is a little harder to describe, but it is related to congruence between special valuesof L-functions.

We will next work towards proving this theorem. Recall that we define Bernoulli numbersvia

tet

et − 1=∞∑n=0

Bntn

n!.

The Bernoulli polynomials are similarly defined by

etx · tet

et − 1=∞∑n=0

Bn(x)tn

n!.

Moreover, for a character χ, we defined the generalized Bernoulli numbers by

f∑a=1

χ(a)teat

eft − 1=∞∑n=0

Bn,χtn

n!.


Again, analogously, we define Bn,χ(x) the generalized Bernoulli polynomials by

etxf∑a=1

χ(a)teat

eft − 1=∞∑n=0

Bn,χ(x)tn

n!.

Proposition 3.6. We have that

(1) Bn,1(x) = Bn(x),(2) Bn,χ(0) = Bn,χ,

(3) Bn,χ(x) =n∑i=0

(ni

)Bi,χx

n−i,

(4) if χ 6= 1, Bn,χ(−x) = (−1)nχ(−1)Bn,χ(x) and if χ = 1, Bn,1(−x) = (−1)nBn,1(x−1),(5) Bn,χ = 0 if χ 6= 1 and n 6≡ δ (2),

(6) Bn,χ = fn−1f∑a=0

χ(a)Bn

(af− 1)

.

Proof. All of these properties are straightforward, so we just prove (6). We have that

∞∑n=0

Bn,χ(x)tn

n!= etx

f∑a=1

teatχ(a)

eft − 1

=

f∑a=1

χ(a)tet(a+x)

eft − 1

=1

f

f∑a=1

χ(a)(ft)eft

a+xf

eft − 1

=1

f

f∑a=1

χ(a)∞∑n=0

Bn

(a+X

f− 1

)(ft)n

n!.

Then Bn,χ = Bn,χ(0) = fn−1f∑a=1

χ(a)Bn

(af− 1)

.

We may use Taylor expansions in the equality

f∑a=1

χ(a)teat = (eft − 1)∑n=0

Bn,χtn

n!

to compute the Bernoulli numbers. In general, on the left side, we will get expressions ofthe form

Sn,χ(k) =k∑a=1

χ(a)an.


(1) Sn,χ(kf) = 1n+1

(Bn+1,χ(kf)−Bn+1,χ(0)),

(2) Bn,χ = limh→∞

1phf

Sn,χ(phf) for any prime p (the limit is a p-adic limit).

26 KARTIK PRASANNA

Proof. For (1), we let

Fχ(t, x) = etxf∑a=1

χ(a)teat

eft − 1.

Then

Fχ(t, x)− Fχ(t, x− f) = etxf∑a=1

χ(a)teat

eft − 1− et(x−f)

f∑a=1

χ(a)teat

eft − 1

=

f∑a=1

χ(a)teat

eft − 1etx(1− e−ft)

=

f∑a=1

χ(a)tet(x+a−f).

Taking coefficients of tn on both sides, we see that

Bn,χ(x)−Bn,χ(x− f)

n!=

f∑a=1

χ(a)(x+ a− f)n−1

(n− 1)!,

Bn,χ(x)−Bn,χ(x− f) = n

f∑a=1

χ(a)(x+ a− f)n−1.

Plugging in x = f, 2f, . . . , kf , we see that

Bn,χ(f)−Bn,χ(0) = n

f∑a=1

χ(a)(a)n−1

Bn,χ(2f)−Bn,χ(f) = n

f∑a=1

χ(a)(a+ f)n−1

...

Bn,χ(kf)−Bn,χ((k + 1)f) = n

f∑a=1

χ(a)(a+ (k − 1)f)n−1

and summing these together, we get

Bn,χ(kf)−Bn,χ(0) = n

kf∑a=1

χ(a)an−1.

This shows (1).

We will use (1) to show (2). We have that

Bn+1,χ(x) =n+1∑i=0

(n+ 1

i

)Bi,χx

n+1−i.

Thus

Bn+1,χ(kf)−Bn+1,χ(0) =n∑i=0

(n+ 1

i

)Bi,χ(kf)n+1−i.


Let k = ph. Then

Bn+1,χ(phf)−Bn+1,χ(0)

phf=

n∑i=0

(n+ 1

i

)Bi,χ(phf)n−i.

Now, take the p-adic limit as h→∞:

limh→∞

(Bn+1,χ(phf)−Bn+1,χ(0)

phf

)= (n+ 1)Bn,χ.

By (1), we have that

limh→∞

Sn,χ(phf)

phf= Bn,χ.

This shows (2).

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 χn = χω−n and

bn = (1− χn(p)pn−1)Bn,χn .

Observe that Bn,χn = limh→∞

1phf

Sh,χn(phf). This is true by Lemma 3.7 after noting that the

conductor of χn is f · pε where ε ∈ 0, 1,−1. Then

bn = limh→∞

1

phfSn,χn(phf)− χn(p)pn−1 lim

h→∞

1

phfSn,χn(phf)

= limh→∞

1


h→∞

1

ph−1fSn,χn(ph−1f) letting h 7→ h− 1 in second limit

= limh→∞

1


h→∞

1

ph−1f

ph−1f∑a=1

χn(a)an

= limh→∞

1

phf

phf∑a=1

χn(a)an − limh→∞

1

phf

ph−1f∑a=1

χn(ap)(ap)n

= limh→∞

1

phf

phf∑a=1

(a,p)=1

χn(a)an

= limh→∞

1

phf

phf∑a=1

(a,p)=1

〈a〉nχ(a) as χn(a) = χ(a)ω(a)−n

= limh→∞

1

qhf

qnf∑a=1

(a,p)=1

χ(a)〈a〉n.

Recall that in Theorem 3.4, we have that

cn =n∑i=0

(−1)n−i(n

i

)bi.

28 KARTIK PRASANNA

Then

cn =n∑i=0

(−1)n−i(n

i

)limh→∞

1

qhf

qhf∑a=1

(a,p)=1

χ(a)〈a〉i = limh→∞

cn,hqhf

for

cn,h =n∑i=0

qhf∑a=1

(a,p)=1

(−1)n−i(n

i

)χ(a)〈a〉i.

Then

cn,h =

qhf∑a=1

χ(a)n∑i=0

(−1)n−i(n

i

)〈a〉i =

qhf∑a=1

(a,p)=1

χ(a)(〈a〉 − 1)n.

We claim thatcn,hqhf≡ 0 mod

qn

q2f

(for n large enough). In other words,

cn,hqhfqn

q2f

is a p-adic integer. To show this, we proceed by

induction on h. For h = 1, we have that

cn,1qf

=1

qf

qf∑a=1

(a,p)=1

χ(a)(〈a〉 − 1)n

which is congruent to 0 modulo qn

q2f, since 〈a〉 ≡ 1 mod q. For h ≥ 1, recall that

cn,h+1

qh+1f=

1

qh+1f

qhf∑a=1

(a,p)=1

χ(a)(〈a〉 − 1)n.

For 1 ≤ a ≤ qn+1f , let a = u+ qhfv for 1 ≤ u ≤ qhf , 0 ≤ v < q. Then ω(a) = ω(u), so

〈a〉 = ω(a)−1a = ω(u)−1(u+ qhfv) = 〈u〉+ ω(u)−1qhfv.

Then

(〈a〉 − 1)n =n∑i=0

(n

i

)(〈u〉 − 1)i(ω(u)−1qhfv)n−i.

We claim that:

(〈a〉 − 1)n ≡ (〈u〉 − 1)n mod qn+h−1.

This is because qi divides (〈u〉 − 1)i and qn−i divides the second term, and i + h(n − i) =i + (n − i) + (h − 1)(n − i) = n + (h − 1)(n − i) ≥ n + h − 1 if n − i ≥ 1. Thus, modulo


qnh−1, only the i = n term survives. We have that

cn,h+1 =

qhf∑a=1

(a,p)=1

χ(a)(〈a〉 − 1)n

≡ q

qhf∑u=1

(u,p)=1

χ(u)(〈u〉 − 1)n

≡ qcn,h mod qn+h−1

so

cn,h+1

qh+1f≡ cn,kqhf

modqn+h−1

qn+1f,


3.3. p-adic logarithms, exponential functions, and power functions. We will nextwork towards Leopoldt’s theorem about the value Lp(1, χ). Note that 1 is outside of thenormal interpolation range: 1− n for n ≥ 1.

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

logp(1 + x) =∞∑n=1

(−1)n

nxn

which converges for 1 + ξ where |ξ|n

|n| → 0 as n→∞, i.e. for |ξ| < 1. Then

logp : x ∈ Cp | |x− 1| < 1︸︷︷︸U1

→ Cp

is a continuous homomorphism (checking that logp(xy) = logp(x) + logp(y) is left as anexercise).

Theorem 3.8. There is a unique extension of logp to a homomorphism

logp : C×p → Cp

such that logp(p) = 0. Further, logp is continuous and for all σ ∈ Aut(Cp/Qp) = Gal(Qp/Qp),logp(α)σ = logp(α

σ).

The choice of logp(p) = 0 is equivalent to choosing a branch of the logarithm. Note that thelast identity is not true for complex analytic logs. This will make working with the p-adiclogarithm much easier.

Proof. We have the following diagram

30 KARTIK PRASANNA

C×p

Qp× Q

Q×p Z

where the map is given by α 7→ r if |α| = p−r.

Pick a section of this homomorphism, r 7→ Pr ∈ Qp×

such that Pr × Ps = Pr+s and if r ∈ Z,

then Pr = pr. We may let Pr = pm/n = (pm)1/n for r = m/n, embedding Q → Qp.

Let P be the image of this section. We have that P ⊆ Qp× ⊆ C×p . We claim that

C×p = P ×W × U1

where W is the set of roots of unity in Qp of order prime to p and

U1 = x ∈ Cp | |x− 1| < 1.For α ∈ C×p , pick α1 ∈ Qp such that α1 is close enough to α. Then |α| = |α1| = |Pr| for somePr ∈ P . Therefore: ∣∣∣∣ αPr

∣∣∣∣ = 1

and setting β = αPr

, we can choose β ∈ Q×p close enough to β such that

|β1| = |β| = 1.

Consider Qp ⊆ Qp(β1). Then β1 is a unit in OK and if we consider the reduction map

β1 β1

p OK OK/p = Fpf

(p) Zp Zp/(p) = Fp

Since β1n

= 1 for some (n, p), by Hensel’s Lemma, we can lift β1 to some nth root of unityw ∈ O×K such that ω ≡ β1 mod p. Then |ω − β1| < 1, and hence |β1/ω − 1| < 1, soβ1/ω ∈ U1.

We have hence written

α = Prα

Pr= Prω

β

ω.

Since β1 is close to β, βω∈ U1. We finally need to show that the product is direct. Suppose

Pr · ω · u = 1. Then |Pr| = 1, so Pr = 1, and hence ω · u = 1. Since |ω − 1| < 1, ω = 1, andhence also u = 1. We have hence shown that

C×p = P ×W × U1.


For r = mn

, P nr = Pm = pm, so n logp Pr = m logp p. Setting log p = 0, we get that

logp(Prωu) = logp(u).

This defines the logarithm on C×p .

We finally have to show that logp(ασ) = logp(α)σ. Let α = Prωu. Then

ασ = P σr ω

σuσ.

For r = mn

, P nr = pm, so (P σ

r ) = pm. Hence P σr = Pr · (nth root of unity), showing that

n logp Pσr = m logp p = 0.


logp(ασ) = log(uσ) = log(u)σ,


We define exp(x) =∞∑n=0

xn

n!. To check convergence, recall that

|n!| ≥ |p|np−1 ,

so ∣∣∣∣ξnn!

∣∣∣∣ ≤ |ξ|n

|p|np−1

,

showing that the above power series converges for

|ξ| < |p|1p−1 .

Exercise. If |x| < |p|1p−1 , then logp(exp(x)) = x and exp(logp(1 + x)) = 1 + x.

We want to define xs for s ∈ Zp as exp(s logp(x)). For x ∈ 1 + pZp, logp(x) ∈ pZp, i.e.

| logp(x)| ≤ |p| < |p|1p−1 if p 6= 2. To deal with the case p = 2, we note that for x ∈ 1 + qZp,

we have that logp(x) ∈ qZp, and we may define

xs = exp(s logp(x)).

However, we would still like to extend this to 1 + pZp for all p.

Lemma 3.9. For x ∈ 1 + qZp and s ∈ Zp,

xs =∞∑n=0

(s

n

)(x− 1)n.

Proof. We use the proof of Theorem 3.4. Here cn = (x − 1)n and we know that |cn| ≤ |q|n.Then the right hand side converges for

|s| ≤ 1

|q||p|

1p−1 .

32 KARTIK PRASANNA

In particular, it converges for s ∈ Zp. To check that the left hand side is equal to the righthand side, it suffices to check at all the integers (which are dense in Zp). We have that

∞∑n=0

(m

n

)(x− 1)n =

m∑n=0

(m

n

)(x− 1)n

= (x− 1 + 1)m

= xm,


We may hence define xs for x ∈ 1 + 2Zp and s ∈ Z2 by the right hand side in the abovelemma.

We now have defined xs for x ∈ 1 + pZp and s ∈ Zp. We want to analyze it as a function ofthe two variables. We define a function

φ : Zp × Zp → Zpby

φ(x, s) =

0 if x 6∈ Z×p ,〈x〉s if p 6= 2, x ∈ Z×p ,xs if p = 2, x ∈ Z×p .

Consider a finite extension K of Qp. Then CK = f : Zp → K continuous. Set

‖f‖ = supx∈Zp|f(x)| = max

x∈Zp|f(x)|.

Note that

‖f + g‖ ≤ max(‖f‖, ‖g‖)

‖fg‖ = ‖f‖‖g‖,

‖af‖ = |a|‖f‖,and CK is complete for ‖−‖. Indeed, if fn is a Cauchy sequence in Ck, for each s, fn(S) is aCauchy sequence in K, so we may set f(s) = lim

nfn(s) and this function is continuous.

Note that K[x] → CK , so, for example, the function(sn

)is in CK .

We define γn ∈ CK by

γn(s) =n∑i=0

(−1)n−i(n

i

)φ(i, s).


φ(x, s) =∞∑n=0

γn(s)

(x

n

).

We first state a lemma.


Lemma 3.11. We have thatn∑i=0

(−1)n−i(n

i

)im

is divisible by n!.

We will prove this later.

Proof of Proposition 3.10. First, let us check this for x = m, Then the right hand side ism∑n=0

γn(s)

(m

n

)=

m∑n=0

(m

n

) n∑i=0

(−1)n−i(n

i

)φ(i, s)

=m∑i=0

φ(i, s)m∑n=i

(m

n

)(n

i

)(−1)n−i

and we have thatm∑n=i

(m

n

)(n

i

)(−1)n−i =

m∑n=i

m!

n!(m− n)!

n!

(n− i)!i!(−1)n−i

=m!

i!

m∑n=i

(−1n−i

(m− n)!(n− i)!

=m!

i!

m−i∑t=0

(−1)t

(m− i− t)!t!where n− i = t

=m!

i!(1 + (−1))m−i

= δmi.

Altogether, this shows thatm∑n=0

γn(s)

(m

n

)= φ(m, s).

We claim that ‖γn‖ ≤ |n!|. Pick a large integer m such that p− 1 divides m and |pm| < |n!|.Then we have that

γn(m) =n∑i=0

(−1)n−i(n

i

)φ(i,m).

Now, for i ∈ pZp, φ(i,m) = 0 and otherwise φ(i,m) = 〈i〉m = (iω(i)−1)m = im since p− 1|m.We then see that

γn(m) =n∑i=0

(−1)n−i(n

i

)φ(i,m)

≡n∑i=0

(−1)n−i(n

i

)im mod pm.

By Lemma 3.11, we see that |γn(m)| ≤ |n!|. Since such m are dense in Zp, we have that‖γn‖ ≤ |n!|.

34 KARTIK PRASANNA

Now, we use Lemma 3.3 to conclude that

‖γn‖ ≤ |n!| ≤ np|p|np−1 .

For x ∈ Zp, we have that∣∣(xn

)∣∣ is bounded and since |γn(s)| ≤ |n!| → 0 as n→∞, this showsconvergence.

Hence the two functions are continuous functions Zp × Zp → Zp which agree on a densesubset, and hence are equal.

We now prove the lemma.

Proof of Lemma 3.11. We have that

(ex − 1)n =n∑i=0

(−1)n−i(n

i

)(ex)i

=n∑i=0

(−1)n−i(n

i

) ∞∑m=0

(xi)m

m!

=∞∑m=0

xm

m!

n∑i=0

(−1)n−i(n

i

)im︸︷︷︸

dm,n

We have that

dm,n =dm

dxm(ex − 1)n

∣∣∣∣x=0

.

We have that

dm

dxm(ex − 1)n =

dm−1

dxm−1(x(ex − 1)n−1ex) =

dm−1

dxm−1(x(ex − 1)n + n(ex − 1)n−1),

showing thatdm,n = ndm−1,n + ndm−1,n−1.

By induction on m, we see that n!|dm,n.

Let CK = f : Zp → K‖ continuous and

QK = formal power series A =∞∑n=0

xn | an ∈ K, |ann!| → ∞

with the sup norm ‖A‖ = supn |ann!|. Then K[x] ⊆ QK is a dense subset.

Theorem 3.12. There is a unique bounded linear map

Γ: QK → CK

such that Γ(xn) = γn(s). It satisfies Γ((1 + x)n) = φ(n, s).

Proof. Define a map Γ: K[x]→ CK by Γ(xn) = γn(s).

We have that ‖xn‖Q = |n!|, ‖γn‖ ≤ |n!|, so ‖Γ(xn)‖ ≤ ‖xn‖Q. Hence

‖Γ(A)‖ ≤ ‖A‖Q for all A ∈ K[x].


Hence Γ is a continuous function K[x]→ CK and we may extend it uniquely to a boundedlinear map QK → CK . Then

Γ((1 + n)n) = Γ

(n∑i=0

(n

i

)xi

)

=n∑i=0

(n

i

)γ(i, s)

=∞∑i=0

(n

i

)γ(i, s)

= φ(n, s)

by Proposition 3.10.

Write Γ(A) = ΓA. We write down a formula for ΓA(s).

Proposition 3.13. Define δn(A) by

A(et − 1) =∞∑n=0

δn(A)tn

n!.

Then ΓA(s) = limniδni(A), where (ni) is a sequence of integers such that ni → ∞ in Z,

(p− 1)|ni, and ni → s in Zp.

Proof. Recall that

(et − 1)n =∞∑m=0

tm

m!dm,n

and we showed in Lemma 3.11 that |dm,n| ≤ |n!|.

If A =∞∑n=0

anxn, then

A(et − 1) =∞∑n=0

an(et − 1)n =∞∑n=0

an

∞∑m=0

tm

m!dm,n =

∞∑i=0

ai

∞∑m=0

tm

m!dm,i.

What is δn(A)? We have that

δn(A) =∞∑i=0

aidn,i.

Since |aidn,i| ≤ |aii!|, we have that

|δn(A)| ≤ sup0≤i≤n

|aii!| ≤ ‖A‖Q.

To check the equality on polynomials K[x], it is enough to check it for A = (1 + x)m. Inthat case, the left hand side is

ΓA(s) = φ(m, s)

36 KARTIK PRASANNA

and we need to check the right hand side is equal to this. We have that

A(et − 1) =∞∑n=0

(mt)n

n!,

so δn(A) = mn. We need to evaluate limimni . If p|m, φ(m, s) = 0 and lim

imni = 0 since

ni →∞ as integers.

In the case p not dividing m, we need to consider the two subcases p = 2 and p odd. Wejust deal with p odd, the other case is similar. Here, the left hand side is

〈m〉s

whereas the right hand side is

limimni = lim

i〈m〉ni = 〈m〉s

since (p− 1)|ni and ni → s p-adically. Therefore, the formula holds for all polynomials.

Finally, let A ∈ QK . Fix ε > 0. Then, there exists B ∈ K[x] such that ‖A− B‖Q < ε. Thisimplies that ‖ΓA − ΓB‖ < ε. We then estimate

|ΓA(s)− δni(A)| = |ΓA(s)− ΓB(s) + ΓB(s)− δni(B) + δni(B)− δni(A)|≤ max|ΓA(s)− ΓB(s)|︸︷︷︸

<ε

, |ΓB(s)− δni(B)|︸︷︷︸<ε for i0

, |δni(B)− δni(A)|︸︷︷︸≤‖A−B‖Q<ε

< ε,


Proposition 3.14. Let A(x) =∑∞

n=0 anxn ∈ QK and

A′(x) =∞∑n=0

nanxn−1 ∈ QK ,

(DA)(x) = (1 + x) log(1 + x)A′(x).

Then DA ∈ QK and ΓDA(s) = sΓA(s).

Proof. We have that

(DA)(et − 1) = tetA′(et − 1)

=

(td

dt

)A(et − 1)

=

(td

dt

) ∞∑n=0

δn(A)tn

n!

=∞∑n=0

(nδn(A))tn

n!

By Proposition 3.13, we have that

ΓDA(s) = limi

(niδni(A)) = s · ΓA(s).



Consider A(x) =∞∑n=0

anxn ∈ QK . Then |ann!| → 0 and for |ξ| ≤ |p|

1p−1 , since |n!| ≥ |p|

np−1 by

Lemma 3.3, so

|anξn| ≤ |an| · |p|np−1 ≤ |ann!| → 0.

This shows that A(ξ) =∞∑n=0

anξn converges.

For such ξ, A(ξ) is defined and |A(ξ)| ≤ ‖A‖QK . In particular, this shows that ξ 7→ A(ξ) iscontinuous.

Proposition 3.15. Let A ∈ QK. Then

ΓA(0) = A(0)− 1

p

∑ξp=1

A(ξ − 1).

Proof. Since both sides are continuous in A, it is enough to check equality for polynomialsA ∈ K[x]. Let A(x) = (1 + x)m. Then

ΓA(s) = φ(m, s).

If p|m, the left hand side is φ(m, s) = 0 and the right hand side is

A(0) = 1− 1

p

∑ξp=1

ξm = 0.

If p does not divide m, the left hand side is 1 and the right hand side is

1− 1

p

∑ξp=1

ξm = 1.


3.4. Leopoldt’s formula for Lp(1, χ). The goal of this section is to prove a formula anal-ogous to the one in section 2.7 about Lp(1, χ).

Suppose now that χ is a primitive character of conductor f . Let ζ be a fixed primitive fthroot of unity. We may take ζ = e2πi/f . Let

h(z) =

f∑a=1

χ(a)za−1,

H(z) =h(z)

zf − 1=

f∑a=1

χ(a)za−1

zf − 1.

Writing

1

zf − 1=

f∑a=1

1

(z − ζs)ζa

f,

we have that

H(z) =

f∑a=1

h(ζa)

z − ζaζa

f.

38 KARTIK PRASANNA

Now,

h(ζa)ζa =

f∑b=1

χ(b)(ζa)b−1ζa =

f∑b=1

χ(b)(ζa)b = χ(a)gχ.

Finally,

H(z) =gχf

f∑a=1

χ(a)

z − ζa.

On the other hand,

H(et) =

f∑a=1

χ(a)et(a−1)

eft − 1,

so we define

(1) F (t) = tetH(et) =

f∑a=1

χ(a)teat

eft − 1=∞∑n=0

Bn,χtn

n!.

Pick an auxiliary integer N such that (N, fp) = 1. Let λ be the set of all Nth roots ofunity in Qp.


gχf

f∑a=1

∑λ

a

z − λζa= Nχ(N)zN−1H(zN).

In particular,

gχf

f∑a=1

∑λ 6=1

χ(a)

z − λζa= Nχ(N)zN−1H(zN)−H(z).

Proof. We have ∑ 1

z − λα=

NzN−1

ZN − αNso

gχf

f∑a=1

∑λ

a

z − λζa=

gχf

f∑a=1

χ(a)NzN−1

zN − ζaN=

gχχ(N)

f

f∑a=1

χ(aN)NzN−1

zN − ζaN= gχNχ(N)H(zN),

which is what we claimed. The second assertion follows from the above description ofH(z).

Theorem 3.17 (Leopoldt’s theorem). We have that

Lp(1, χ) = −gχf

(1− χ(p)

p

) f∑a=1

χ(a) logp(1− ζa).


Comparing this to the standard equation for for L(1, χ), the only difference is the Eulerfactor at p which was taken out when we defined Lp, so we need to put it back in, and theuse of p-adic logarithm instead of regular logarithm.

In the proof of the theorem, we will take K/Qp containing λ, ζ, and the values of χ. Wewill then show that for

A(x) =gχf

f∑a=1

∑λ 6=1

χ(a) log

(1 +

x

1− λζa

)

we have that

ΓA(s) = (1− χ(N)〈N〉s)Lp(1− s, χ)

if p is odd (and similarly for p = 2), and use Proposition 3.15 to get the result.

We first with proving the above formula for ΓA(s).

Theorem 3.18. Let K/Qp be a finite extension containing λ, ζ, and the values of χ. Let

A(x) =gχf

f∑a=1

∑λ 6=1

χ(a) log

(1 +

x

1− λζa

).

Then A(x) ∈ QK and

ΓA(s) =

(1− χ(N)〈N〉s)Lp(1− s, χ) if p is odd,

(1− χ(N)N s)Lp(1− s, χ) if p = 2.

Proof. Note that (!− λζa) is a unit, so log(

1 + x1−λζa

)∈ QK . This shows that A(x) ∈ QK .

We suppose p is odd. The case p = 2 is dealt with in the same way, omitting all theoccurrences of 〈−〉. The idea is to use Proposition 3.13 to find ΓDA(s) and the applyProposition 3.14. We have that

DA(x) =gχf

(1 + x) log(1 + x)

f∑a=1

∑λ 6=1

χ(a)1

1 + x1−λζa

1

1− λζa

=gχf

(1 + x) log(1 + x)

f∑a=1

∑λ 6=1

χ(a)1

x+ (1− λζa).

40 KARTIK PRASANNA

Then

DA(et − 1) =gχftet

f∑a=1

∑λ 6=1

χ(a)1

et − λζa)

= tet(Nχ(N)(et)N−1H(etN)−H(et)

)by Lemma 3.16

= χ(N)NtetNH(etN)− tetH(et)

= χ(N)F (Nt)− F (t) as F (t) = tetH(et) by definition

= χ(N)∞∑n=0

Bn,χ(Nt)n

n!−∞∑n=0

Bn,χtn

n!by equation (1)

=∞∑n=0

tn

n!(χ(N)Nn − 1)Bn,χ.

Letting δn(DA) = (χ(N)Nn − 1)Bn,χ and applying Proposition 3.13, we get that

ΓDA(s) = limi→∞

δni(DA)

where ni → ∞ as integers, (p − 1)|ni, and ni → s p-adically. By definition of Lp(1 − n, χ)(Theorem 3.5), we have that

Lp(1− n, χ) = (1− χω−n(p)pn−1)L(1− n, χω−n).

Since (p− 1)|ni, ω−ni(p) = 1, and we have that

Lp(1− ni, χ) = (1− χ(p)pni−1)L(1− ni, χ) = −(1− χ(p)pni−1)Bni,χ

ni.

Taking the limit as i→∞, this shows that

limi→∞

Bni,χ = Lp(1− s, χ) · s.

Since (p− 1)|ni, Nni = 〈N〉ni , and hence

limi→∞

Nni = limi→∞〈N〉ni = 〈N〉s.


ΓDA(s) = limi→∞

δni(DA) = (χ(N)〈N〉s − 1)Lp(1− s, χ).

Applying Proposition 3.14, we see that

ΓA(s) = (1− χ(N), 〈N〉s)Lp(1− s, χ),


This allows to prove Leopoldt’s formula 3.17 by applying Proposition 3.15.

Proof of Theorem 3.17. By Proposition 3.15, we have that

ΓA(0) = A(0)− 1

p

∑ξp=1

A(ξ − 1).


For A from Theorem 3.18, we have that

A(0) = 0,

A(ξ − 1) =gχf

f∑a=1

∑λ 6=1

χ(a) logp

(ξ − λζa

1− λζa

).

Therefore:

ΓA(0) = −gχpf

f∑a=1

χ(a) logp

∏λ6=1ξp=1

ξ − λζa

1− λζa

= −gχ

pf

f∑a=1

χ(a) logp

(∏λ 6=1

1− λpζap

(1− λζa)p

)

= −gχpf

f∑a=1

χ(a) logp

(∏λ 6=1

1− λζap

(1− λζa)p

).

We now split the sum as follows:

f∑a=1

χ(a) logp

(∏λ 6=1

1− λζap

1− λζa

)=

f∑a=1

χ(a) logp

(∏λ 6=1

(1− λζap)

)︸︷︷︸

S2

−p·f∑a=1

χ(a) logp

(∏λ 6=1

(1− λζa)

)︸︷︷︸

S1

.

We claim that S2 = χ(p)S1. This is clear when (p, f) = 1 because ζa and ζp are bothprimitive roots of unity. When p|f , we may write f = pf ′. Since χ has conductor f , it doesnot factor as follows:

(Z/fZ)× C×

(Z/f ′Z)×

θ

χ

42 KARTIK PRASANNA

and hence kerχ 6⊇ ker θ, i.e. there is a b coprime to f such that χ(b) 6= 1 but b ≡ 1 (f ′). Inthat case, ab ≡ a (f ′), so abp ≡ ap (f). Therefore:

S2 =

f∑a=1

χ(a) logp

(∏λ 6=1

(1− λζap)

)

=

f∑a=1

χ(ab) logp

(∏λ 6=1

(1− λζabp)

)

= χ(b)

f∑a=1

χ(a) logp

(∏λ6=1

(1− λζap)

)= χ(b)S2.

Since χ 6= 0, this shows that S2 = 0.

In any case, we have shown that S2 = χ(p)S1. Hence:

S2 − pS1 = (χ(p)− p)S1

= (χ(p)− p)f∑a=1

a logp1− ζaN

1− ζa

= (χ(p)− p)

(f∑a=1

χ logp(1− ζaN)−f∑a=1

χ logp(1− ζa)

)

= (χ(p)− p)(χ(N)− 1)

f∑a=1


Altogether, we have shown that:

(1− χ(N))Lp(1, χ) = ΓA(0) = −gχfp

(χ(p)− p)(χ(N)− 1)

f∑a=1


We are free to choose N . As long as we choose N so that χ(N) 6= 1, this shows that

Lp(1, χ) = −gχf

(1− χ(p)

p

) f∑a=1

χ(a) logp(1− ζa),

i.e. Leopoldt’s formula 3.17.

3.5. p-adic class number formula. For the standard L-function, we proved that L(1, χ) 6=0 for χ 6= 1 (Corollary 2.9) using the class number formula. We develop the analog of thesenotions in this section.

Recall that Lp(1− n, χ) = (∗) · L(1− n, χω−n). Suppose χ is odd. When n is even χω−n iseven and 1− n is even, so L(1− n, χω−n) = 0. Similarly, when n is even, ωω−n is odd and1− n is odd, so L(1− n, χω−n) = 0. Therefore Lp(s, χ) ≡ 0 for χ odd.


This is consistent with Leopoldt’s formula 3.17: for χ odd

Lp(1, χ) =1

2

(f∑a=1

χ(a) logp(1− ζa)−f∑a=1

χ(a) logp(1− ζ−a)

)= 0.

Therefore, we only consider even characters. In this case, the analog of Dirichlet’s Theo-rem 2.9 holds.

Theorem 3.19. If χ is even, then Lp(1, χ) 6= 0.

Recall that in the proof of Dirichlet’s Theorem 2.9, we considered the zeta function ofK = Q(ζf ). Because we only take even characters here, we need to consdier the totally realsubfield K+ = Q(ζf )

+ of K:

K = Q(ζf )

K+ = Q(ζf )+

Q

2

φ(f)/2

We previously showed that

ζK+(s) =∏

χ characterof G+

L(s, χ)

and used the class number formula:2nhK+RK+

2√|dK+|

=∏χ 6=1

L(1, χ).

How is the regulator defined? For K, by Dirichlet’s unit theorem,

UK/torsion ∼= Zr1+r2−1

where r1 is the number of real places and r2 is the number of complex places. Then if

σ1, . . . , σr1 , σr1+1, σr1+1, . . . , σr1+r2 , σr1+r2

are all the embeddings of K into C, then (dropping one of the embeddings)

RK = det(log(σi(uj))r×r)

where (u1, . . . , ur) is a basis for UK/torsion.

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 Cp, embeddings K → C are in bijection with embed-dings K → Cp, but it is difficult to distinguish between the real and complex embeddingsin Cp. However, if K is a totally real field, there is no problem. If σ1, . . . , σr+1 are all theembeddings, then we define

RK,p = det(logp(σi(uj))r×r)

(dropping one of the embeddings.)

44 KARTIK PRASANNA

We know that for any K, RK 6= 0. The analogous statement for the p-adic regulator is onlyconjectural.

Conjecture. For any totally real field K, RK,p 6= 0.

It is known for K abelian.

Theorem 3.20. If K/Q is an abelian totally realy field, then RK,p 6= 0.

We omit the proof of this here. This could be one of the final projects for this class. Weinstead focus on proving the following theorem.

Theorem 3.21 (p-adic class number formula). We have that

2nhK+RK+,p

2√|dK |

=∏χ 6=1

(Lp(1, χ)

(1− χ(p)

p

)−1).

Combining these two theorem, we in particular get Theorem 3.19.

We only prove this theorem in the case K = Q(ζpm). The general case is more complicatedand we refer to [Was97] for that.

We need a way of producing units in K+ = Q(ζpm)+ in order to compute RK+,p. They willbe indexed by the integers in the set

1 < a < pm/2 | (a, p) = 1.

Note that there are φ(pm)2− 1 = n− 1 of them. For each such a, we consider

ζa = ζ−(a−1)/2 ζa − 1

ζ − 1,

where ζ = ζpm .

Remark 3.22. If n is odd, Q(ζn) = Q(ζ2n), so ζ12 ∈ K. Thus ζa ∈ K.

The map ζ 7→ ζ−1, generating Gal(K/K+), sends ζa to

ζ(a−1)/2 ζ−a − 1

ζ−1 − 1= ζ−(a−1)/2 1− ζa

1− ζ= ζa.

Hence ζa ∈ K+.

We finally check that ζa is a unit. It is clearly an integer, so we just need to check it has aninverse. Since (a, p) = 1, there exits a b such that ab ≡ 1 (pm). Then

ζa = (root of unity) · ζa − 1

ζab − 1,

which clearly shows it is a unit.

Define CK+ = 〈ζa,−1〉 ⊆ UK+ . Then CK+/(torsion) ⊆ UK+/(torsion).

Theorem 3.23. We have that CK+ is finite index in UK+ and |UK+/CK+ | = hK+.

Note that if G = Gal(K/Q), G+ = Gal(K+/Q), we have that


G σb | 1 ≤ b < pm, (b, p) = 1

G+ σb | 1 ≤ b < pm/2, (b, p) = 1.

∼=

∼=

We want to compute

det (log |σb(ζa)|)1<a,b<pm/2 .

We have that:

log |σb(ζa)| = log

∣∣∣∣σb(ζ(a−1)/2(ζa − 1)

ζ − 1

)∣∣∣∣= log

∣∣∣∣σb(ζa − 1)

σb(ζ − 1)

∣∣∣∣= log

∣∣∣∣σb(σa(ζ − 1))

σb(ζ − 1)

∣∣∣∣= log |σbσa(ζ − 1)| − log |σb(ζ − 1)|.

Lemma 3.24. Let G be a finite abelian group. Let f : G → k be any function, where k isan algebraically closed field of characteristic 0. Then

(1) det[f(στ−1)]σ,τ∈G =∏

χ : G→k×

(∑σ∈G

χ(σ)f(σ)

),

(2) det[f(στ−1)− f(σ)]σ,τ∈G\1 =∏

χ : G→k×χ 6=1

(∑σ∈G

χ(σ)f(σ)

).

Proof. For (1), let V = F : G tok. Then G acts on V by (g · F )(σ) = F (gσ). Consider

T =∑σ∈G

f(σ)σ.

We compute the determinant of T . Let eτ be the characteristic function of τ . Then

T (eτ )(α) =

(∑σ

f(σ)σ

)(eτ )(α)

=∑σ

f(σ)eτ (σα)

=∑σ

f(σ)eτσ−1(α)

=∑τ ′

f(τ(τ ′)−1)eτ ′(α),

and thus the matrix of T in this basis is (f(στ−1)).

46 KARTIK PRASANNA

Another basis for T is given by all the characters and

T (χ)(τ) =∑σ

f(σ)χ(στ) =

(∑σ

f(σ)χ(σ)

)χ(τ).

This completes the proof of (1).

Finally, for (2), consider the subspace of V given by

V 0 =

F : G→ k

∣∣∣∣∣ ∑σ

f(σ) = 0

.

Then (2) follows in the same as above.

This implies that

det (log |σb(ζa)|)1<a,b<pm/2 =∏

χ : G→k×χ 6=1

∑σ∈G+

χ(σ) log |σ(1− ζ)|.

To complete the proof of the theorem, we recall that ζ = ζpm , and we have that∑σ∈G+

χ(σ) log |σ(1− ζ)| =∑

1≤a<pm/2

χ(a) log |1− ζapm|

=1

2

∑1≤a<pm

χ(a) log |1− ζapm|

=1

2

pk∑b=1

(b,p)=1

χ(b)∑

1≤a<pma≡b (pk)

log |1− ζapm| where cond(χ) = pk|pm

=1

2

pk∑b=1

(p,b)=1

χ(b) log

∣∣∣∣∣∣∣∣∏

1≤a<pma≡b (pk)

(1− ζapm)

∣∣∣∣∣∣∣∣ .Lemma 3.25. We have that ∏

1≤a<pma≡b (pk)

(1− ζapm) = 1− ζbpk .

Proof. Note that

1− xpm−k =∏

1≤j≤pm−k(1− ζj

pm−kX).

Setting X = ζbpm , we get that

1− ζbpk =∏

1≤j≤pm−k(1− ζj

pm−kζbpm) =

∏1≤j≤pm−k

(1− ζjpk+b

pm ),

proving the lemma.


This gives

det (log |σb(ζa)|)1<a,b<pm/2 =∏χ 6=1

1

2

fχ∑b=1

(b,p)=1

χ(b) log |1− ζbfχ|

.

Recall that

L(1, χ) =−gχfχ

fχ∑b=1

χ(b) log |1− ζbfχ|.

Hence the determinant is equal to

± 1

2n−1

∏χ

fχgχL(1, χ) = ± 1

2n−1

∏χ

gχL(1, χ).

Finally, the class number formula says that

2nhK+RK+

2√|dK+|

=∏χ 6=1

L(1, χ) = 2n−1 1∏χ gχ

Reg(ζa).

This implies that

hK+RK+ = Reg(ζa)

and proves Theorem 3.23.

Let N = `m for some prime `. We have Q ⊆ K+ ⊆ K = Q(ζN). By Theorem 3.23, we havethat

RK+,p = Regp(ζa)/hK+ .

Then the same proof as above shows that

RK+,p =∏χ 6=1

1

2

N∑a=1


Leopoldt’s formula 3.17 then shows that

Lp(1, χ) = −gχfχ

(N∑a=1

χ(a) logp(1− ζa)

)·(

1− χ(p)

p

).

Tracing through the same argument as above completes the proof of Theorem 3.21.

4. p-adic measures and power series

So far, we constructed p-adic L-functions as p-adic analytic functions. In this section, weprovide a different construction using measures on a Galois group and power series. In theprevious method, all the L-functions Lp(1, χ) for different χ were seemingly unrelated, butnow we will see how we can work with all the characters simultaneously.

The Galois extensions in question are the following

48 KARTIK PRASANNA

Q(ζNp∞)

Q(ζNp)

Q

(1+pZp,·)∼=(Zp,+)

∆

Hence the group we want to construct measures on is actually Zp.

Definition 4.1. Consider a finite extension K of Qp with ring of integers O. An O-valuedmeasure on Zp is a collection of maps

µn : Z/pnZ→ Osuch that

µn(x) =∑y 7→x

y∈Z/pn+1Z

µn+1(y).

Theorem 4.2. There is a canonical bijection

O-valued measures on Zp ↔ OJxK.

Proof. Consider a measure given by µn ∈ O[Z/pnZ] and consider µn+1 ∈ O[Z/pn+1Z] suchthat µn 7→ µn+1. Then we have that

µ ∈ lim←−O[Z/pnZ] ∼= lim←−O[T ]

T pn − 1∼= lim←−

O[x]

(1 + x)pn − 1.

The assertion now follows from Theorem 4.3.

Theorem 4.3. There is a canonical isomorphism

OJxK = lim←−O[x]

(1 + x)pn − 1.

In order to prove this theorem, we need some preparation about the power series ring.

4.1. Power series. We start with a division algorithm in OJxK.

Theorem 4.4 (Division algorithm). Consider f(x) ∈ OJxK given by

f(x) = a0 + a1x+ · · ·with where the uniformizer π divides a0, . . . , ad−1, but π does not divide ad. Consider anyg(x) ∈ OJxK. Then there exist unique q(x) ∈ OJxK and r(x) ∈ O[x] such that deg r < d and

g(x) = q(x)f(X) + r(x).

Before we prove this, we will derive a corollary of this theorem.

Definition 4.5. A distinguished polynomial p(x) ∈ O[x] is a polynomial whose leadingcoefficient is a unit and all other coefficients are divisible by π.

Example 4.6. The polynomial (1 + x)pn − 1 above is a distinguished polynomial.


Corollary 4.7 (Weierstrass Preparation Theorem). Given any 0 6= f(x) ∈ OJxK, we canfactor it as

f(x) = πn · h(x) · u(x)

where h(x) is a distinguished polynomial and u(x) is a unit power series (i.e. the constantterm is a unit). Moreover, this is “unique” (up to obvious ambiguity by O×).

Proof. Write f(x) = πng(x) for g(x) ∈ OJxK and at least one coefficient is a unit. Writingg(x) = a0 + a1x+ · · · , we may assume that π|a0, . . . , ad−1 and π does not divide ad. By thedivision algorithm 4.4, we have that

xd = q(x)g(x) + r(x)

for r(x) < d. Write q(x) = b0 + b1x+ · · · to see that

1 = b0ad + b1ad−1 + · · ·

by looking at the coefficients of xd. This implies that b0 is a unit, so q(x) is invertible.Looking at the coefficient of xi for i < d, we see that the coefficients of r(x) are divisible byπ. Finally:

xd − r(x) = q(x)g(x)

g(x) =1

q(x)(xd − r(x)).

Taking u(x) = 1q(x)

and h(x) = xd − r(x), we get the desired factorization. Uniqueness is

easy to check.

Corollary 4.8. Any non-zero power series f(x) ∈ OJxK has only finitely many zeros in themaximal ideal of OCp.

Proof. Writing f(x) = πnh(x)u(x) as in the Weierstrass Preparation Theorem 4.7, we havethat

f(α) = πnh(α)u(α),

so the roots of f are the roots of h and there are clearly finitely many of them.

Lemma 4.9. Suppose h(x), g(x) ∈ O[x] with h(x) distinguished. If h(x) divides g(x) inOJxK, then h(x) divides g(x) in O[x].

Proof. We have that g(x) = h(x)f(x) in OJxK. Consider a finite extension K of K containing

all the roots of h(x). If O ⊆ K is the ring of integers, then the roots of h(x) lie in its maximalideal because h(x) is distinguished. Let α be such a root. Substituting α in g(x) = h(x)f(x),we get that g(α) = 0. Dividing both sides by x− α, we get an equation

g1(x) = h1(x)f(x) in OJxK.

Since h1(x) is still distinguished, we may continue this way to see that f(x) is a polynomial

in K[x], and hence also in K[x]. By Gauss’ Lemma, h(x)|g(x) in O[x].

50 KARTIK PRASANNA

Proof of the division algorithm 4.4. This proof is due to Manin. Define two linear mapsα, τ : OJxK→ OJxK:

α(b0 + b1x+ · · · ) = b0 + b1x+ · · ·+ bd−1xd−1

τ(b0 + b1x+ · · · ) = bd + bd+1x+ bd+2x2 + · · ·

Note that:

• α(F ) = F if and only if F is a polynomial of degree less than d if and only if τ(F ) = 0,• τ(xdF ) = F .

The idea of the proof is to “solve for q”. We want the equality:

g = qf + r.

This is equivalent to τ(g) = τ(gf). We may write

f = α(f) + xdτ(f)

and multiplying by q, we get that

qf = qα(f) + xdqτ(f).

Now,τ(g) = τ(qf) = τ(qα(f)) + qτ(f).

Letting z = qτ(f), we get

τ(g) = z + τ

(zα(f)

τ(f)

).

Let H be multiplication by α(f)τ(f)

. Note that α(f) is divisible by π and z(f) is a unit power

series. Then we may writeτ(g) = (I + τ H)(z).

The inverse of the operator I + z H is

(I + z H)−1 = I − (z H) + (z H)2 + · · · .Then

z = (I + z H)−1τ(g).

Finally, we get

q =1

z(f)(I + τ H)−1τ(g)

and this also gives the r. Uniqueness follows immediately from this proof.

We can finally go back to prove Theorem 4.3.

Proof of Theorem 4.3. Let hn(x) = (1 + x)pn − 1, so hn+1(x) = (1 + x)p

n+1 − 1. Then

hn+1(x)

hn(x)=

(1 + x)pn+1 − 1

(1 + x)pn − 1=tp − 1

t− 1= 1 + t+ t2 + · · ·+ tp−1,

where t = (1 + x)pn ∈ 1 + (p, x). Therefore:

hn+1(x)

hn(x)∈ (p, x) ⊆ (π, x) ⊆ O[x].


Starting with f(x) ∈ OJxK, we divide f(x) by hn(x) to get

f(x) = qn(x)hn(x) + γn(x).

We then define

OJxK φ→ O[x]

(1 + x)pn − 1

f(x) 7→ (γn(x))

To check that (γn(x)) is a compatible system, we note that

qn(x)h(x) + γn(x) = f(x) = qn+1hn+1 + γn+1(x)

so hn(x) divides γn+1 − γn(x) in OJxK, and hence hn(x) divides γn+1(x) − γn(x) in O[x] aswell by Lemma 4.9. Hence we have a well-defined homomorphism φ.

To show injectivity, suppose φ(x) = 0. Then f(x) is divisible by hn(x) for all n in O[x]. Buthn(x) ∈ (π,X)n, so f(x) ∈ (π, x)nOJxK for all n. By Krull’s Theorem,⋂

n

(π, x)n = 0,

so f = 0. Alternatively, one can note that hn(x)|f(x) for all n would imply that f(x) hasinfinitely many zeros, contradicting Corollary 4.8.

To show surjectivity, let

(fn(x)) ∈ lim←−O[x]

(hn(x)).

For m ≥ n, fm − fn ≡ 0 (hn), so fm − fn ∈ (π, x)n = mn, so lim fm exists and we can writef = lim fm. We use here that R = OJxK ∼= lim←−R/m

n, because R is complete with respect tothe m-adic topology.

We check that φ(f) = (fn). Write

fm − fn = qm,n · hn.

Then

qm,n =fm − fnhn

→ f − fnhn

.

Keeping n fixed and letting m → ∞, we define qn = limm→∞

qm,n ∈ OJxK. Then f = fn +

qnhn,showing that f 7→ fn under φ.

This finally completes the proof of Theorem 4.2, establishing the bijection between measuresand power series.

Example 4.10. Let s ∈ Zp The Dirac measure supported on S is given by

µn : Z/pnZ→ O

and

µn(x) =

1 if x ≡ s (pn),

0 if x 6≡ s (pn).

52 KARTIK PRASANNA

We will show that the power series associated to this measure is

(1 + x)s =∞∑i=0

(s

i

)xi.

Pick sn ∈ Z such that sn ≡ s (pnZ). We have the isomorphisms

O[Z/pnZ]∼=→ O[T ]

T pn − 1

∼=→ O[x]

(1 + x)pn − 1,

characteristic function of [sn] 7→ [T sn ] 7→ [(1 + x)sn ].

This suggests that the power series is

(1 + x)s =∞∑k=0

(s

k

)xk.

Indeed, the function

(1 + x)sn =sn∑k=0

(snk

)xk =

∞∑k=0

(snk

)xk

tends to (1 + x)s as sn 7→ s. This shows the assertion by the proof of Theorem 4.3.

4.2. Integration. We now discuss integration of functions with respect to p-adic measures.

Let C(Zp,Cp) = f : Zp → C | continuous (a Banach space for the sup-norm) and let µ bean O-valued p-adic measure on Zp. We want to define∫

fdµ.

As in real analysis, we approximate f by step functions, i.e. functions that factor throughZ/pnZ for large n. In other words, let (fn)n be a sequence of functions fn : Z/pnZ→ Cp besuch that fn → f . Then define ∫

fdµ = limn

∫fndµn,

where ∫fndµn =

∑x∈Z/pnZ

fn(x)µn(x).

It is easy to check this is well-defined. Also,∣∣∣∣∫ fdµ

∣∣∣∣ ≤ ‖f‖‖µ‖,where ‖f‖ = supx∈Zp |f(x)| and ‖µ‖ = sup

n∈Nx∈Z/pnZ

|µn(x)|.

Then the map

C(Zp,Cp)→ Cp

f 7→∫fdµ


is a bounded linear functional on C(Zp,Cp).

Consider now continuous functions from Zp to O, C(Zp,O). We have a functional

C(Zp,O)λ→ O

f 7→∫fdµ.

Conversely, any linear functional of this sort comes from an O-valued measure on Zp.Given λ, we can recover µ by setting µn(x) = λ(Ix+pnZp), where Ix+pnZp is the character-istic function of x+ pnZp.

Remark 4.11. Given µ and a continuous function ϕ : Zp → O, we can construct a newmeasure ϕdµ given by ∫

f(ϕdµ) =

∫fϕdµ.

Theorem 4.12. In the correspondence between power series and measures in Theorem 4.2,suppose f(x) = c0 + c1x+ c2x

2 + · · · corresponds to the measure µf . Then:

ck =

∫Zp

(x

k

)dµf (x).

Proof. Considering µn : Z/pnZ→ O, we have that

O[Z/pnZ]∼=→ O[T ]

T pn − 1

∼=→ O[x]

(1 + x)pn − 1,∑

r∈Z/pnZ

µn(r)r 7→pn−1∑r=0

µn(r)T r 7→p∑r=0

µn(r)(1 + x)r.

We then compute that

p∑r=0

µn(r)(1 + x)r =

pn−1∑r=0

r∑k=0

(r

k

)xk

=

pn−1∑r=0

µn(r)

pn−1∑k=0

(r

k

)xk

=

pn−1∑k=0

xk

(pn−1∑r=0

µn(r)

(r

k

)).

Therefore,

ck = limn→∞

pn−1∑r=0

µn(r)

(r

k

)=

∫Zp

(x

k

)dµ,

as required.

Corollary 4.13. Suppose ξ ∈ m, where m is the maximal ideal of O. Then∫Zp

(1 + ξ)xdµf (x) = f(ξ).

54 KARTIK PRASANNA

Proof. We have that:∫Zp

(∞∑k=0

(x

k

)ξk

)dµf =

∞∑k=0

ξk∫Zp

(x

k

)dµf (x) =

∞∑k=0

ξkck = f(ξ)

by Theorem 4.12.

Examples 4.14.

(1) We have that

∫Zpdµf (x) = f(0).

(2) Consider the measure (1 + ξ)xdµf (x). Suppose this measure corresponds to g. Wewill work out what g is by using Corollary 4.13. For ξ′ ∈ m,

g(ξ′) =

∫Zp

(1 + ξ′)xdµg(x)

=

∫Zp

(1 + ξ′)x(1 + ξ)xdµf (x)

= ((1 + ξ′)(1 + ξ))xdµf (x)

=

∫Zp

(1 + ξ + ξ′ + ξξ′)xdµf (x)

= f(ξ + ξ′ + ξξ′)

= f((1 + ξ)(1 + ξ′)− 1).

Therefore, g(x) = f((1 + ξ)(1 + x)− 1).(3) If ζ ∈ O is a p-power root of unity, then ξ = ζ − 1 ∈ m. Then the above example

shows that

ζxdµf (x)←→ f(ζ(1 + x)− 1).

(4) Let ϕ be a step function on Zp, i.e. ϕ factors through Z/pnZ for some n. We assumethat O contains all the pnth roots of unity. Then

ϕdµf ←→∑ζpn=1

ϕ(ζ)f(ζ(1 + x)− 1),

where for G = Z/pnZ, the character group is G ∼= µpn , and we define

ϕ(ζ) =1

pn

∑x∈Z/pnZ

ϕ(x)(ζ−1)x

to be the Fourier transform of ϕ. The Fourier inversion formula is

ϕ(x) =∑ζ∈µpn

ϕ(ζ)ζx.

Note that this holds for x ∈ Z/pnZ, but may also be viewed as a formula for x ∈ Zp.To show the above correspondence for ϕdµf , we now see that

ϕdµf =∑ζ∈µn

ϕ(ζ)ζxdµf ,


and by Example (3) above, we see this corresponds to the power series∑ζ∈µpn

ϕ(ζ)f(ζ(1 + x)− 1).

(5) Consider a power series f(x) with the corresponding measure µf . We then have ameasure µf |Z×p by restricting to Z×p . Note that

µf |Z×p = ϕµf

where ϕ is the characteristic function of Z×p , a step function factoring through Z/pZ.We need to apply (4). Note that

ϕ(ζ) =1

p

∑x∈Z/pZ

ϕ(X)(ζ−1)x.

For ζ = 1, we see that ϕ(1) = p−1p

. For ζ 6= 1, we have that

ϕ(ζ) =1

p

∑06=x∈Z/pZ

(ζ−1)x = −1

p.

Therefore, the power series corresponding to µf |Z×p is

p− 1

pf(x)− 1

p

∑ζ 6=1ζp=1

f(ζ(1 + x)− 1).

This is equal to

f(x)− 1

p

∑ζp=1

f(ζ(1 + x)− 1).

This is more or less the power series that came up in the proof of Leopoldt’s for-mula 3.17.

Before we move onto the construction of a measure that gives Lp(s, χ), we discuss Γ-transforms and Mellin transforms.

Definition 4.15. Let µ be an O-valued measure on Zp. The Γ-transform of µ is defined by

Γµ(s) =

∫Z×p〈x〉sdµ(x).

The Mellin transform is

Mµ(s) = Γx−1dµ(x) =

∫Z×p〈x〉sx−1dµ(x).

Theorem 4.16. The function Γµ(s) is a p-adic analytic function on Zp (in fact, on a largeropen disc).

56 KARTIK PRASANNA

Proof. By breaking up Z×p into costs, it suffices to prove this for a measure supported on1 + qZp. Then

Γµ(s) =

∫1+qZp

xsdµ(x)

=

∫1+qZp

∞∑k=0

(s

k

)(x− 1)kdµ(x)

=∞∑k=0

(s

k

)·∫

1+qZp

(x− 1)kdµ(x).

We have that |x− 1| ≤ |q| and∣∣ 1k!

∣∣ ≤ |p| −kp−1 , and hence∣∣∣∣(x− 1)k

k!

∣∣∣∣ ≤ |q|k|p|− kp−1 .

Thus the coefficients of sn is bounded by

|q|n|p|−np−1 ,

so the sum converges in the analytic disc |s| < |q|−1|p|1/p−1.

4.3. Alternative construction of Lp(s, χ). We’ll give an alternative construction of thep-adic L-function, as a measure.

Let χ be a character of conductor N = M · pr where (M, p) = 1. We consider the followingextensions:

Q(ζMp∞)

Q(ζMp)

Q

G

where we note that

G ∼= lim←−r

(Z/MprZ)×

∼= (Z/MZ)× × Z×p∼= (Z/MqZ)× × (1 + qZp).

We will construct a measure

dE1,c on lim←−Z/MprZ

where c ∈ Z, (c,Mp) = 1 and the 1 stands for the 1st Bernoulli polynomial.


Theorem 4.17. Let χ be any finite order character of conductor Mpr for r ≥ 0. Then∫G

χω−1(x)〈x〉sdE1,c(x) = −(1− χ(c)〈c〉s+1)Lp(−s, χ).

Note that this gives a construction of all the p-adic L-functions of Dirichlet characters χsimultaneously, through one measure.

Example 4.18 (M = 1). The above theorem gives the equality:∫Z×p

χω−1(x)〈x〉s−1dE1,c(x) = −(1− χ(c)〈c〉s)Lp(1− s, χ)

and the left hand side is simply∫Z×p

χ(x)〈x〉sx−1dE1,c(x) = MχdE1,c(s),

the Mellin transform of χdE1,c. One should note the analogy of this to classical L-functionswhich also have integral representations as a Mellin transform.

Recall that we defined Bernoulli polynomials as

tet

et − 1etx =

∞∑k=0

Bk(x)tk

k!.

We renormalize this slightly and define

t

et − 1etx =

∞∑k=0

Bk(x)tk

k!

so that Bk(x) = Bk(x + 1). In this notation, Proposition 3.6 becomes much nicer. If wedefine

t

et − 1=∞∑k=0

Bktk

k!

then B1 = −12

while previously we had B1 = +12.


(1) Bk(0) = Bk,

(2) Bk(x) =k∑i=0

(k

i

)Bix

k−i = xk − k

2xk−1 + (lower order terms),

(3) distribution relation: Bk(x) = Nk−1

N−1∑a=0

Bk

(x+ a

N

).

Proof. Since (1) and (2) are clear, we just prove (3). Note that

1

k!

N−1∑a=0

Bk

(X + a

N

)= coefficient of tk in

N−1∑a=0

t

et − 1etX+an .

58 KARTIK PRASANNA

Rearranging the power series, we get

t

et − 1etX/N

N−1∑a=0

eta/N =tetX/N

et − 1

(et/N)N − 1

et/N − 1= N

(t/N)e(t/N)x

et/N − 1=N ·N−k

k!Bk(x).

This shows (3).

Definition 4.20. We define a function

E(N)k : Z/NZ→ Q

by

E(N)k (x) =

Nk−1

kBk

( xN

),

where • denotes the fractional part.

This gives a distribution on Z = lim←−Z/nZ, i.e. if N |M , then

E(N)k (x) =

∑y∈Z/MZy 7→x

E(M)k (y).

To check this relation, write M = ND. We may assume that 0 ≤ x < N . The left hand sideis

E(N)k (x) =

Nk−1

kBk

( xN

)and the right hand side is

D∑i=0

(ND)k−1

kBk

(x+ iN

ND

)=Nk−1

kDk−1

D−1∑i=0

Bk

(x/N + i

D

)=Nk−1

kBk

( xN

)by Proposition 4.19 (3).

Finally, for N = Mpr and c ∈ Z such that (c,Mp) = 1, define

E(N)k,c (x) = E

(N)k (x)− ckE(N)

k (c−1x).

Theorem 4.21. We have that E(N)k,c (x) ∈ Zp.

Proof of Theorem 4.21 for k = 1. We first just show this for k = 1. We have that

E(N)1,c (x) = E

(N)1 (x)− cE(N)

1 (c−1x) = B1

( xN

)− cB1

(c−1x

N

).

Now, recall that B1(x) = x− 12. Let d ∈ Z be such that d = c−1 in (Z/NZ)×, i.e. cd ≡ 1 (N).

If dx = qN + r, x ≡ cdx ≡ cr (N). Then by the above:

E(N)1,c (x) = B1

( xN

)− cB1

( rn

)=

x

N− 1

2− c

(r

n− 1

2

)=x− crN

+c− 1

2∈ Zp.


This completes the proof for k = 1.

For k > 1, we need the following lemma.

Lemma 4.22. Let N = Mpr. Then

E(N)k,c (x) ≡ xk−1E

(N)1,c (x) mod

N

kD(k)Zp,

where D(k) is the least common multiple of the denominators of Bk(x).

Proof. Let 0 ≤ x < N . First, if dx = qN + r and cr = x+Nt, then it is easy to check that

E(N)1,c (x) =

c− 1

2− t.

Then

E(N)k,c (x) =

Nk−1

k

[Bk

( xN

)− ckBk

( rN

)]≡ Nk−1

k

[[( xN

)k− k

2

( xN

)k−1]− ck

[( rN

)k− k

2

( rN

)k−1]]

modN

kD(k)Zp

=1

k

[xk

N− k

2xk−1 − (x+Nt)k

N+k

2c(x+Nt)k−1

]≡ 1

k

[xk

N− k

2xk−1 − xk

N− kxk−1Nt

N+kc

2xk−1

]mod

N

kD(k)Zp

= xk−1

[−1

2− t+

c

2

]= xk−1E

(N)1,c (x).


Proof of Theorem 4.21, general case. If r 0, then the congruence in Lemma 4.22 shows

that E(N)k,c (x) ∈ Zp. From the distribution relation, we get integrality for all N .

We can finally prove Theorem 4.17. We only prove it in the case M = 1, where G = Z×p .The general case will be given as an exercise.

We want to show that∫Z×p

χω−1(x)〈x〉s−1dE1,c(x) = −(1− χ(c)〈c〉s)Lp(1− s, χ).

We first note that Lemma 4.22 gives the following corollary.

Corollary 4.23. We have that dEk,c = xk−1dE1,c.

60 KARTIK PRASANNA

Proof of Theorem 4.17 for M = 1. In this case N = pr is the conductor of χ. We have that∫Z×p

χω−1(x)〈x〉k−1dE1,c(x) =

∫Z×p

χω−k︸︷︷︸χ′

(x) · xk−1dE1,c(x)

=

∫Zp

χ′(x)xk−1dE1,c(x)−∫pZp

χ′(x)xk−1dE1,c(x)

=

∫Zp

χ′(x)xk−1dE1,c(x)−∫Zp

χ′(py)yk−1pk−1dE1,c(py) for x = py

= (1− χ′(p)pk−1)

∫Zp

χ′(x)xk−1dE1,c(x)

= (1− χ′(p)pk−1)

∫Zp

χ′(x)dEk,c(x) by Corollary 4.23.

We now compute the last integral:∫Zp

χ′(x)dEk,c(x) =Nk−1

k

(N−1∑a=0

Bk

( aN

)χ′(a)− ck

N−1∑a=0

Bk

(c−1a

N

)χ′(a)

)

=Nk−1

k

(N−1∑a=0

Bk

( aN

)χ′(a)

)(1− χ′(c)ck)

=Bk,χω−n

k(1− χω−k(c)ck) by Proposition 3.6.

Then ∫Z×pχω−1(x)〈x〉k−1dE1,c(x) = (1− χ(p)pk−1)

Bk,χω−n

k(1− χ(c)ω(c)−kck)

= −Lp(1− k, χ)(1− χ(c)〈c〉k),as claimed.

As mentioned above, the proof of the general case of Theorem 4.17 is left as an exercise.

Remark 4.24. The same construction of dE1,c works if we take c ∈ Z×p instead of c ∈ Zsuch that (c, p) = 1.

Recall that we showed∫Z×p

χ(x)〈x〉sx−1dE1,c(x) = −(1− χ(c)〈c〉s)Lp(1− s, χ).

Recall thatZ×p ∼= (Z/qZ)×︸︷︷︸

∆

×(1 + qZp).


An O-valued measure on Z×p is an element of

OJZ×p K ∼= O[∆]J1 + pZpK.

Let θ be a character of ∆. Then θ gives a map O[∆]→ O and for each θ we get a map

OJ1 + qZpK→ OJ1 + qZpK ∼= OJZpK ∼= OJxK,

where the isomorphism Zp ∼= 1 + qZp is given by sending 1 to a fixed topological generatorγ of 1 + qZp.

If µ = x−1dE1,c(x), let dµθ be the associated measure on 1 + qZp, dµθ be the associatemeasure on Zp, and finally gθ ∈ ZpJxK be the corresponding power series.

Write χ = θ · ψ, where

• θ is a character of the first kind, i.e. it factors through ∆,• ψ is a character of the second kind, i.e. it factors through 1 + qZp.

Therefore, we may rewrite the above integral as follows:∫Z×pχ(x)〈x〉sx−1dE1,c(x) =

∫1+qZp

ψ(x)xsdµθ

=

∫Zp

ψ(γt)(γt)sdµθ(t)

=

∫Zp

(ψ(γ)γs)tdµθ(t)

= gθ(ψ(γ)γs − 1).

We recall this integral was equal to −(1− χ(c)〈c〉s)Lp(1− s, χ), so we get the equation:

gθ(ψ(γ)γs − 1) = −(1− ψ(γ)γs)Lp(1− s, θψ) = (ψ(γ)γs − 1)Lp(1− s, θψ).

Finally, define

gθ(x) =gθ(x)

x.

Thengθ(ψ(γ)γs − 1) = Lp(1− s, θψ).

If θ 6= 1, then gθ(x) ∈ ZpJxK. Taking s = 0 and ψ = 1, the above equation gives:

gθ(0) = 0

since Lp(s, θ) is analytic at s = 1.

For the rest of this section, we assume that p is odd for simplicity.

If θ = 1, then gθ(x) is a unit power series in ZpJxK. Taking ψ = 1 and s = 1, we see that

g1(γ − 1) = −(1− γ)Lp(0,1).

By the interpolation formula

Lp(1− n, χ) = (1− χω−n(p))L(1− n, χω−n),

62 KARTIK PRASANNA

we see that

Lp(0,1) = L(0, ω−1) = −B0,ω−1

1= −1

p

p−1∑a=1

ω−1(a)a ≡ −p− 1

p≡ 1

pmod Zp.

This shows that

g1(γ − 1) = −(1− γ)Lp(0,1) ∈ Z×p ,

so g1 is a unit power series.

Now, define

f θ(χ) = gθ(

γ

1 + x− 1

).

Then

f θ(ψ−1(γ)γs − 1) = gθ(

γ

ψ−1(γ)γs− 1

)= gθ(ψ(γ)γ1−s − 1) = Lp(s, θψ).

Note that

f θ(x) ∈ ZpJxK if θ 6= 1

but for θ = 1, we have only have that

f1(x) ∈ 1γ

1+x− 1

ZpJxK.

4.4. Applications to class numbers in cyclotomic towers. We keep the simplifyingassumption that p 6= 2.

Let Kn = Q(ζpn+1) and K∞ = Q(ζp∞). Write K+ for the totally real subfield of K. Then wehave the following tower of extensions:


K∞

. ..

K+∞

Kn

. ..

K+n

K1

K0 K+1

Q

Iwasawa’s idea was to try to understand how the class number behaves in this tower.

Let hn be the class number of Kn and h+n be the class number of K+

n . We first note that h+n

divides hn. Indeed, if H is the maximal unramified abelian extension of K+n (at all places,

including infinite ones), then we have:

HKn

Kn H

K+n

h+n

2 h+n

Since HKn is unramified and abelian over Kn, it is contained in the maximal unramifiedabelian extension of Kn. This shows that h+

n |hn.

64 KARTIK PRASANNA

Let us write hn = h+nh−n . We will try to understand how the p-part of h−n grows with n. Let

K+ = K+n and K = Kn. Then we have that

(2π)dhkRK

wk√|dK |

=∏χ 6=1

L(1, χ) characters χ of Gal(K/Q),

2dhK+RK+

2√|dK+ |

=∏χ 6=1

L(1, χ) characters χ of Gal(K+/Q).

Dividing on equation by the other, we get that

πdh−KRK

RK+

=wk2

∏χ odd

L(1, χ)

√|dK |√|dK+|

.

Therefore:

h−KRK

RK+

=wK2id∏χ

gχfχB1,χ

√|dK |√|dK+|

using

L(1, χ) =πigχfχ

B1,χ.

One can also show that ∏χ

gχfχ

√|dK |√|dK+ |

= id

andRK

RK+

= 2d−1.

Altogether, we see that:

h−K = wk∏χ odd

(−1

2B1,χ

).

Recall that we had K = Kn, so we may write for any n:

h−n = wn∏χ odd

Gal(Kn/Q)

(−1

2B1,χ

)

h−0 = w0

∏χ odd

Gal(K0/Q)

(−1

2B1,χ

)

Note that wn = w0pn, so this shows that

h−n = h−0 · pn ·∏χ

(−1

2B1,χ

),

where the product is now over all odd characters χ of Gal(Kn/Q) that do not factor throughGal(K0/Q).


We may write χω = θψ for θ even and note that

Lp(0, χω) = (1− χ(p))L(0, χ) = L(0, χ) = −B1,χ.

Finally, we see that:

h−n ∼ h−0 · pn ·∏θ even,ψ 6=1

Lp(0, χω),

where ∼ means up to p-adic units (as elements of Zp). Recall that we’re assuming p 6= 2, so2 is a p-adic unit.

We will now use the power series that gives the p-adic L-function. Recall that χω = θψ.Then

h−0 · pn ·∏θ even,ψ 6=1

Lp(0, χω) = h−0 · pn ·∏θ even,ψ 6=1

f θ(ψ−1(γ)− 1)

= pn · h−0 ·∏θ even

∏ζpn

ζ 6=1

f θ(ζ − 1)

= pn · h−0 ·

∏ζpn

ζ 6=1

f1(ζ − 1)

·∏ζpn

ζ 6=1

A(ζ − 1)

where we defineA(x) =

∏θ 6=1

f θ(x).

Recall that

f1(x) =g1(

γ1+x− 1)

γ1+x− 1

.

Setting ζ − 1, we see that

f1(ζ − 1) =g1 (γζ−1 − 1)

γζ−1 − 1,

so

|f1(ζ − 1)| = 1

|γζ−1 − 1|=

1

|ζ−1 − 1|.

Finally, ∏ζpn

ζ 6=1

|f1(ζ − 1)| = 1

|pn|.

Therefore:

|h−n | = |h−0 |

∣∣∣∣∣∣∣∣∏ζpn

ζ 6=1

A(ζ − 1)

∣∣∣∣∣∣∣∣ .By Weierstrass preparation theorem 4.7, we may write

A(x) = pµP (x)U(x),

66 KARTIK PRASANNA

where P (x) is a distinguished polynomial and U(x) is a unit power series. Then:

|h−n | = |h−0 ||pµ|pn−1∏ζpn

ζ 6=1

|P (ζ − 1)|.

Recall thatP (X) = Xλ + aλ−1X

λ−1 + · · · .with aλ−1, . . . is divisible by p. For n 0, we hence have that

|P (ζ − 1)| = |(ζ − 1)λ|.Finally, this shows that

|h−n | = |h−0 ||pµ|pn−1|pn|λ

for n 0. In particular, for n 0,

vp(h−n ) = vp(h

−0 ) + µ(pn − 1) + nλ = µpn + λn+ C.

Therefore, the existence of the power series shows something very non-trivial about how h−nbehave in the tower of extensions.

Remark 4.25. Iwasawa proved that µ = 0 in this case. In particular, vp(h−n ) grows linearly

with n. The proof can be found in [Was97] and could be a potential project in this class.

4.5. Main conjecture of Iwasawa theory. We recall that we had an infinite tower ofextensions

Q ⊆ K0 ⊆ K1 ⊆ · · · ⊆ Kn ⊆ · · ·K∞.Let Ci be the p-part of the class group of Ki.

Let C = lim←−iCi, which is a ZpJGK-module, where G = ∆ × (1 + pZp). Let ε be a characterof ∆ and write

eε =1

|∆|∑a∈∆

ε−1(a)a.

Let Λ = ZpJxK ∼= ZpJΓK. Then Cε = eεC is a Λ-module. Here, Γ = 1 + pZp ∼= Zp.

Theorem 4.26. We have that Cε is a finitely-generated Λ-module.

We say that M and N are pseudoisomorphic— and write M ∼ N if there is a morphismϕM → N of Λ-modules with finite kernel and cokernel.

Theorem 4.27. Any finitely-generated Λ-module is pseudoisomorphic ton⊕i=1

Λ/(fi).

In that case, let char(M) be the ideal generated by the productn∏i=1

fi. We call this the

characteristic ideal of M .

Theorem 4.28 (Iwasawa’s Main Conjecture). Suppose ε is odd and ε 6= ω. Then

char(Cε) = (f ε−1ω(x)).


Remark 4.29. We give a brief overview of the history of this conjecture. First, Herbrandshowed that if p divides a particular Bernoulli number, then a particular part of the classgroup does not vanish. In 1976, Ribet [Rib76a] succeeded in proving the converse by astriking use of modular forms to construct unramified p-extensions of Q(ζp).

This inspired the work of Mazur and Wiles [MW84] who proved Iwasawa’s Main Conjecture(for all real abelian extensions of Q and p odd).

Later, a more elementary proof was found by Rubin using Kolyvagin’s Euler systems. Thisis the proof that can be found in [Was97] and Lang [Lan90, Appendix by Rubin].

5. Serre’s construction of the p-adic L-function

The goal of this section is to discuss an alternative construction of the p-adic L-functiondue to Serre [Ser73b] which uses congruences between modular forms and develops the firstapproach to p-adic modular forms.

5.1. Classical modular forms. We start with a review of some of the necessary notionsfrom the theory of modular forms. Some standard references for this topic are [Shi94] and[Miy06] and we refer to these two for a complete treatment of the topic. In particular, theproofs of the unproven results of this section can be found there.

Let Γ = SL2(Z). A modular form of weight k for Γ is a holomorphic function f : h→ C suchthat

(1) f(az+bcz+d

)= (cz + d)kf(z) where γ =

(a bc d

),

(2) holomorphic at cusps, i.e. since

(1 10 1

)∈ Γ, f(z + 1) = f(z), so we may write

f(z) =∑n∈Z

ane2πinz, and we require that requires an = 0 for n < 0.

We write q = e2πiz so that the Fourier expansion is

f(z) =∞∑n=0

anqn.

We let Mk(Γ) be the set of forms of weight k for Γ, Sk(Γ) ⊆ Mk(Γ) be the subset of cuspforms (i.e. forms with a0 = 0).

We know Mk(Γ) and Sk(Γ) are finite-dimensional. In fact, they can be expressed as globalsections of a line bundle, so we may compute their dimensions using the Riemann–RochTheorem.

Proposition 5.1.

(1) If k is odd, then Mk(Γ) = 0.

68 KARTIK PRASANNA

(2) For k ≥ 2 even,

dimSk(Γ) =

0 if k = 2,

bk2c if k ≡ 2 (12),

b k12c if k 6≡ 2 (12),

and

dimMk(Γ) =

b k

12c if k ≡ 2 (12),

b k12c+ 1 if k 6≡ 2 (12).

How does one construct examples of modular forms? We may define the Eisenstein seriesas

Ek(z) =∑

(m,n)∈Z2\(0,0)

1

(mz + n)k.

We write Ek, because we want to reserve Ek for a different normalization.

Proposition 5.2. For k ≥ 4 even, this series converges absolutely and uniformly on compact

subsets, so it defines a holomorphic function on h. Moreover, Ek ∈ Mk(Γ) and its Fourierexpansion is

2ζ(k)

(1 +−2k

Bk

∞∑n=1

σk−1(n)qn

).

Proof. For convergence, see [Shi94] or [Miy06]. For the last assertion, let γ =

(a bc d

). Then

Ek(γz) =∑(m,n)

1(maz+b

cz+d+ n)k

= (cz + d)k∑(m,n)

1

((ma+ nc)z + (mb+ nd))k

= (cz + d)kEk(z)

(m

n

)7→(a cb d

)−1(m

n

)To show that it is holomorphic at the cusps, we compute the Fourier expansion at∞. Recallthat

π cot(πz) =1

z+∞∑n=1

(1

z + n+

1

z − n

).


Also

π cot(πz) = (iπ)eπiz + e−πiz

eπiz − e−πiz

= iπe2πiz + 1

e2πiz − 1

= iπ

(1 +

2

e2πiz − 1

)= iπ

(1− 2

∞∑n=0

e2πinz

).

Differentiating this k − 1 times, we get that:

−2πi(2πi)k−1

∞∑n=0

nk−1e2πinz = (−1)k−1(k − 1)!

(1

zk+∞∑n=1

(1

(z + n)k+

1

(z − n)k

)).

Finally,

Ek(z) =∑(m,n)

1

(mz + n)k

=∑n6=0

1

nk+∑m 6=0

1

(mz + n)k

= 2ζ(k) + 2∞∑m=0

(−1)k(2πi)k

(k − 1)!

∞∑n=1

nk−1e2πinmz

= 2ζ(k) + 2(2πi)k

(k − 1)!

∞∑n=1

∑d|n

dk−1

︸︷︷︸

σk−1(n)

e2πinz

= 2ζ(k)

(1 +−2k

Bk

∞∑n=1

σk−1(n)qn

).

In particular, this shows that Ek ∈Mk(Γ).

We define

Ek(z) = 1 +−2k

Bk

∞∑n=1

σk−1(n)qn.

This is the normalized Eisenstein series. Note that it has rational Fourier coefficients.

Note that B4 = − 130

and B6 = 142

. Therefore,

E4 = 1 + 240∞∑n=1

σ3(n)qn,

E6 = 1− 504∞∑n=1

σ5(n)qn.

70 KARTIK PRASANNA

Ramanujan studied these functions, denoting E4 by Q and E6 by R.

Theorem 5.3. There is an isomorphism of graded algebras⊕k≥0

Mk(Γ) = C[E4, E6],

where E4 has degree 4 and E6 has degree 6. In particular, for any f ∈Mk(Γ), we can writef = F (E4, E6) where F is isobaric of weight k.

The proof of this is standard and can be find in textbooks such as Shimura, Miyake.

Example 5.4. The smallest k where there exists a cusp form is k = 12. We let

∆ = 12−3[E3

4 − E26

].

Then

• ∆(q) = q∞∏n=1

(1− qn)24,

• ∆ has no zeros on h,

• if J(z) =E3

4

∆, then J(τ) = j(C/(Zτ + Z)) is the j-invariant of the elliptic curve.

Corollary 5.5. Let Mk(Γ,Q) be the subspace of Mk(Γ), consisting of forms with rationalFourier coefficients in Q. Then

M(Γ,Q) =⊕k

Mk(Γ,Q) = Q[E4, E6].

5.2. Reduction modulo p. Fix a prime p and write

Mk =f =

∑anq

n ∈Mk(Γ) | an ∈ Q, vp(an) ≥ 0 for all n.

(We omit the p from the notation.) We then have a map

Mk → Mk ⊆ FpJqK

f 7→ f =∑

anqn ∈ FpJqK.

Similarly, for M =⊕k≥0

Mk, we consider the quotient map

M M.

Assume p ≥ 5.

Exercise. Show that M = Z(p)[E4, E6]. (Hint: use induction on the weight using ∆.)

Consider

Z(p)[x, y]∼=→M,

x 7→ E4,

y 7→ E6.

We want to study the map ϕ given by


Z(p)[x, y] M

Fp[x, y] M

∼=

ϕ

The main goal of this section is to prove the following theorem.

Theorem 5.6. Write Ep−1 = A(Q,R), A ∈ Z(p)[x, y]. Then

ker(ϕ) = (A(x, y)− 1).

Recall that Q = E4, R = E6.

Before we prove this, we need to discuss some preliminaries. We introduce

P = E2 = 1− 24∞∑n=1

σ1(n)qn

by analogy with the definition of E2. The − 124

comes from a Bernoulli number. This wouldlike to be a modular forms of weight 2, but it cannot be, because there is no weight 2 modularforms.

We know that SL2(Z) is generated by

(1 10 1

)and

(0 1−1 0

). Since E2 is given by a q-series,

it is immediately invariant under the first matrix, but it will fail to be invariant under thesecond. The precise formula is given in Lemma 5.9.

Define the differential operator

θ = qd

dqon q-expansions. In terms of z, we have that

θ =1

2πi

d

dz.

Theorem 5.7. Let f ∈Mk(Γ). Then θf − k12Pf ∈Mk+2(Γ).

Definition 5.8. We define ∂ = 12θ − kP : Mk(Γ)→Mk+2(Γ).


E2

(−1

z

)= z2E2(z) +

12z

2πi.

We will prove this lemma later. For now, we show how it implies Theorem 5.7.

Proof of Theorem 5.7. We have that f(−1z

)= zkf(z) because f ∈ Mk(Γ). Differentiating

this, we get:

f ′(−1

z

)1

z2= zkf ′(z) + kzk−1f(z).

Let

g(z) =1

2πi

d

dzf − k

12Pf.

72 KARTIK PRASANNA

Then

g

(−1

z

)=

1

2πiz2[zkf ′(z) + kzk−1f(z)]− k

12P

(−1

z

)f

(−1

z

)=

1

2πi[z2+kf ′(z) + kz1+kf(z)]− k

12zkf(z)[zkP (z) +

12z

2πi]

=1

2πiz2+kf ′(z)− k

12z2+kf(z)P (z)

= zk+2g(z).

This shows that θf − k12Pf ∈Mk+2(Γ).

Remark 5.10. The map ∂ is a derivation on⊕

Mk(Γ), i.e.

∂(αβ) = α∂(β) + β∂(α).

Indeed, for modular forms f and g of weights k and `, we see that

∂(fg) = 12θ(fg)− (k + `)Pfg = 12(fθ(g) + gθ(f))− (k + `)Pfg = f∂g + g∂f.

Example 5.11. We compute ∂ applied in a few examples. We know that ∂Q is a multipleof R. We figure out what multiple, we just consider the constant term in the q-expansions.We had

Q = 1 + 240∑

σ3(n)qn, R = 1− 504∑

σ5(n)qn.

Then

120Q− 4PQ = 12[240∑

nσ3(n)qn]− 4[1− 24∑

σ1(n)qn][1 + 240∑

σ3(n)qn]

= −4R.

Therefore:∂(Q) = −4R.

One could similarly compute that ∂(R) = −6Q2, but we omit this here.

We still have to prove Lemma 5.9. We will do this using Dirichlet series, so we first prove ageneral result.

Theorem 5.12. Suppose f(z) =∞∑n=0

ane2πinz, g(z) =

∞∑n=0

bne2πinz satisfy an, bn = O(nc).

Consider

φ(s) =∞∑n=1

ann−s, Φ(s) = (2π)−sΓ(s)φ(s),

ψ(s) =∞∑n=1

bnn−s, Ψ(s) = (2π)−sΓ(s)ψ(s).

Let k > 0 be a positive integer. Then the following are equivalent:

(1) Φ(s) + a0s

+ b0k−s admits an analytic continuation to all s ∈ C, which is bounded in

vertical strips, andΦ(k − s) = Ψ(s),

(2) f(−1z

)=(zi

)kg(z).


Proof. We have that

Φ(s) = (2π)−sΓ(s)∞∑n=1

ann−s

=∞∑n=1

an(2πn)−s∞∫

0

tse−tdt

t

=

∞∫0

(∞∑n=1

ane−2πnt

)tsdt

t

=

∞∫0

(f(it)− a0)tsdt

t

=

1∫0

(f(it)− a0)tsdt

t+

∞∫1

(f(it)− a0)tsdt

t.

The second integral

∞∫1

(f(it) − a0)tsdt

tconverges uniformly on vertical strips, and hence

defines an analytic function bounded in vertical strips. The key will be to deal with the firstintegral.

Let us prove that (2) implies (1). In the first variable, we make the change of variables t 7→ 1t

to get

1∫0

(f(it)− a0)tsdt

t=

1∫∞

(f(i/t)− a0)t−s−1/t2

1/tdt

=

∞∫1

(f(i/t)− a0)t−sdt

t

= −a0

∞∫1

t−1−sdt+

∫ ∞1

tkg(it)t−st−sdt

tby (2)

=−a0

s+

∞∫1

tk−s(g(it)− b0)dt

t+

∫ ∞1

tk−sb0dt

t

= −a0

s− b0

k − s+

∞∫1


t.

Therefore,

Φ(s) +a0

s+

b0

k − s=

∞∫1


t+

∞∫1

(f(it)− a0)tsdt

t,

74 KARTIK PRASANNA

which gives an analytic continuation to the complex plane, bounded in vertical strips (EBV).

Replacing g by f in the above argument and noting that g(−1z

)=(zi

)kf(z), we get that

Ψ(s) +b0

s+

a0

k − s=

∞∫1

tk−s(f(it)− a0)dt

t+

∞∫1

(g(it)− b0)tsdt

t.

Therefore, Φ(k − s) = Ψ(s).

We now show that (1) implies (2). We have that

Φ(s) =

∫ ∞0

(f(it)− a0)tsdt

t,

is the Mellin transform of f(it)− a0. The inverse Mellin transform shows that

f(it)− a0 =1

2πi

∫Re(s)=σ0

t−sΦ(s)ds.

Considering a large rectangular contour, we obtain

f(it)− a0 =1

2πi

∫Re(s)=σ0

t−sΦ(s)ds

= −a0 + b0t−k +

1

2πi

∫Re(s)=σ0

t−sΦ(s)ds

= −a0 + b0t−k +

1

2πi

∫Re(s)=σ0

t−(k−s)Φ(k − s)ds letting s 7→ k − s

= −a0 + b0t−k +

1

2πi

∫Re(s)=σ0

t−(k−s)Ψ(s)ds using Ψ(k − s) = Φ(s)

= −a0 + b0t−k + t−k

(g

(i

t

)− b0

)same argument for g

This shows that

f(it) = t−kg(i/t),

and hence

f(z) =(zi

)−kg

(−1

z

).


Recall that we defined

Ek = 1− 2k

Bk

∞∑n=1

σk−1(n)qn.

for k ≥ 2 even. We renormalize it as follows:

Gk = −Bk

2k+∞∑n=1

σk−1(n)qn.


If φk(s) is the associated Dirichlet series, then

φk(s) =∞∑n=1

σk−1(n)n−s

=∞∑n=1

∑d|n

dk−1

n−s

=∞∑n=1

∞∑m=1

mk−1(mn)−s

=

(∑n

n−s

)(∑m

mk−1−s

)= ζ(s)ζ(s+ 1− k).

LetΦk(s) = (2π)−sΓ(s)ζ(s)ζ(s+ 1− k).

Note that

(2π)−sΓ(s)ζ(s) =2−sπ−s/2π−s/2Γ(s/2)Γ((s+ 1)/2)

21−s√π

=1√ππ−s/2Γ((s+ 1)/2)π−(1−s)/2Γ((1− s)/2)ζ(1− s) functional equation for ζ

=1

2πΓ((s+ 1)/2)Γ((1− s)/2)ζ(1− s)

=1

2

ζ(1− s)sin(π(s+1

2

))=ζ(1− s)2 cos πs

2

.

This gives

Φk(s) = (2π)−sΓ(s)ζ(s)ζ(s+ 1− k)

=ζ(1− s)ζ(s+ 1− k)

2 cos πs2

= (−1)k/2Φ(k − s),which is the functional equation.

In the notation of Theorem 5.12, we have that

a0 = −Bk

2k, b0 = (−1)k/2+1Bk

2k.

If we can show that

(2π)−sΓ(s)ζ(s)ζ(s+ 1− k) +a0

s+

b0

k − sis analytic, then we can prove the modularity of Gk(z).

We make a table of potential poles and zeros of each of the factors

76 KARTIK PRASANNA

Γ(s) ζ(s) ζ(s+ 1)− knegative even integers pole, order 1 zero holomorphicnegative odd integers pole, order 1 holomorphic zero

0 pole, order 1 holomorphic holomorphic1 holmorphic pole, order 1 ζ(2− k)k Γ(k) ζ(k) pole, order 1

We check the residue at s = 0 to make sure that the pole is cancelled:

1−1

2

(−Bk

k

)+ a0 = 0.

We do the same at s = k:

(2π)−kΓ(k)ζ(k)− (−1)k/2+1Bk

k= 0.

When k 6= 2, we have that ζ(2 − k) = 0, so the pole of order 1 of ζ(s) cancels, and hencethis function is holomorphic. Overall, this shows that

Gk

(−1

z

)=(zi

)k(−1)k/2Gk(z) = zkGk(z),

as we already knew by Proposition 5.2

When k = 2, ζ(2 − k) = ζ(0) does not have a zero, so this method does not apply. Weconsider this case separately. We have to trace through the proof of Theorem 5.12 and seewhat happens in this case. We have that

Φ2(s) = −Φ2(2− s)where

Φ2(s) = (2π)−sΓ(s)ζ(s)ζ(s− 1).

Then

G2(it)− a0 =1

2πi

∫Re(s)=σ0

t−sΦ2(s)ds

=2∑i=0

Ress=i +1

2πi

∫Re(s)=σ0

t−sΦ2(s)ds

=2∑i=0

Ress=i−1

2πi

∫Re(s)=σ0

t−(2−s)Φ2(s)ds 2 7→ 2− s

=2∑i=0

Ress=i−t−2 1

2πi

∫Re(s)=σ0

tsΦ2(s)ds

=2∑i=0

Ress=i−(G2(i/2)− a0),

where the residues are of the function

Φ2(s)t−s = (2π)−sΓ(s)ζ(s)ζ(s− 1)t−s.


Finally, we do the residue calculation

Ress=0 =

(−1

2

)(− 1

12

)=

1

24,

Ress=1 = (2π)−1

(−1

2

)t−1 = − 1

4πt,

Ress=2 = (2π)−2π2

6t−2 =

1

24t−2.

Therefore:

G2(it) +1

24=

1

24− 1

4πt+

1

24t−2 − t−2

(G2(i/t) +

1

24

).

This gives

G2

(−1

z

)= − z

4πi+ z2G2(z).

Coming back to the other normalization of the Eisenstein series, we get that

E2

(−1

z

)=

12z

2πi+ z2E2(z),

completing the proof of Lemma 5.9.

We now prove an analog of Theorem 5.7 for P = E2.

Proposition 5.13. We have that (12θ − P )P is a modular forms of weight 4. Specifically:

(12θ − P )P = −Q.

Proof. Differentiating the equation from Lemma 5.9, we see that

P ′(−1

z

)1

z2= z2P ′(z) + 2zP (z) +

12

2πi,

so

P ′(−1

z

)= z4P ′(z) + 2z3P (z) +

12

2πiz2.

Therefore:

(12θ − P )P

(−1

z

)=

12

2πi

[z4P ′(z) + 2z3P (z) +

12

2πiz2

]−[z2P (z) +

12

2πiz

]2

=12

2πiz4P ′(z)− z4P (z)2

= z4(12θP − P 2)(z).

This shows that 12θ − P is a modular form of weight 4. To check it is −Q, compare q-expansions.

Recall that M = Z(p)[E4, E6]. Consider

Z(p)[x, y]∼=→M,

x 7→ E4,

y 7→ E6.

78 KARTIK PRASANNA

We want to study the kernel of the map ϕ given by

Z(p)[x, y] M

Fp[x, y] M

∼=

ϕ

Write Ep−1 = A(Q,R), A ∈ Z(p)[x, y]. Then Theorem 5.6 states that

ker(ϕ) = (A(x, y)− 1).

We will finally want to prove this.

Recall that ∂y = −6x2 and ∂x = −4y. The derivation ∂ makes sense as a derivation on Mbut also on Fp[x, y].

Theorem 5.14 (Kummer congruences). If m ≡ n (p− 1) but m 6≡ 0 (p− 1). Then

Bm

m≡ Bn

nmod p.

Instead of presenting the classical proof, we will use the slightly roundabout way that usesp-adic L-functions. We do this in the interest of time.

Proof. Let c be a primitive root modulo p, i.e. a generator of (Z/pZ)×. Then, by a homeworkproblem,

(1− cm)Bm

m=

∫Zpxm−1dE1,c

(1− cn)Bn

n=

∫Zpxn−1dE1,c

Since cm − 1 and cn − 1 are p-units and cm ≡ cn (p), we have that

(1− cm)

(Bm

m− Bn

n

)− (cm − cn)

Bn

n︸︷︷︸∈pZp

=

∫Zp

(xm−1 − xn−1)dE1,c︸︷︷︸∈pZp

.


Theorem 5.15 (Van Staudt). If k ≡ 0 (p− 1), then

Bk ≡ −1

pmod Zp.

Proof. Assume p 6= 2. Take c = 1 + p. We have that

(1− ck)Bk

k=

∫Zpxk−1dE1,c.


We claim that ck ≡ 1 + pk (pkkZp). We prove this by induction on vp(k). For vp(k) = 0, wehave that (1 + p)k = 1 + pk (p2Zp), so the assertion follows. Now, assuming the claim for k,we prove it for pk. We have that

cpk ≡ (1 + pk)p = 1 + p2k +

(p

2

)(pk)2 + · · · ≡ 1 + p(pk) mod (p3kZp).

This proves that ck ≡ 1 + pk (pkkZp). Then

1

1− ck=

1

1− (1 + pk + pkt)

=1

−pk(1 + pt)

= − 1

pk(1 + ps) for s ∈ Zp.

Therefore,

Bk = −1

p(1 + ps)

∫Zpxk−1dE1,c

≡ −1

p(1 + ps)

p−1∑i=0

ik−1

∫i+pZp

dE1,c(x) mod (Zp)

= −1

p(1 + ps)

p−1∑i=0

ik−1

(i

p

− c

c−1i

p

+c− 1

2

)

= −1

p(1 + ps)

p−1∑i=0

ik−1

(i

p− (1 + p)

i

p+p

2

)

= −1

p(1 + ps)

p−1∑i=0

(−ik +p

2ik−1)

≡ −1

p(1 + ps)

p−1∑i=0

(−ik) mod (Zp)

≡ (p− 1)

p(1 + ps) mod (Zp)

≡ −1

pmod (Zp)

This completes the proof when p 6= 2. We omit the case p = 2, since we assume p ≥ 5 lateron anyway.

Now recall that

Ep+1 = 1− 2(p+ 1)

Bp+1

∞∑n=1

σp(n)qn

Ep−1 = 1− 2(p− 1)

Bp−1

∞∑n=1

σp−2(n)qn.

80 KARTIK PRASANNA

Since Bp−1 ≡ −1p

(Zp), we have that

2(p− 1)

−1p

+ t=

2p(p− 1)

(−1 + pt)≡ 0 mod p.

Therefore, Ep−1 ≡ 1 (p). This shows that Ep−1 = 1.

Moreover,Bp+1

p+ 1≡ B2

2≡ 1

12mod p,

soBp+1

p+ 1∈ Z×p ,

as p ≥ 5. Moreover,

1− 2(p+ 1)

Bp+1

∞∑n=1

σp(n)qn ≡ 1− 24∞∑n=1

σ1(n)qn = E2 mod p,

becauseσp(n) =

∑d|n

dp ≡ σ1(n) mod p.

Note that Ep−1 = A(Q,R), Ep+1 = B(Q,R), where A,B are isobaric polynomials ofweight p − 1 and p + 1, respectively. We may think of A(x, y), B(x, y) ∈ Z(p)[x, y] and

A(x, y), B(x, y) ∈ Fp[x, y].


(1) A(Q, R) = 1, B(Q, R) = P ,

(2) ∂A(x, y) = B(x, y), ∂B(x, y) = −xA(x, y).

Proof. The above discussion already shows (1). For (2), we compute

˜(∂A−B)(Q,R) = ˜(12θ − (p− 1)P )A(Q,R)− B(Q,R)

= P − B(Q, R)

= 0.

Thus (∂A − B)(x, y) is p times a polynomial in (x, y), so ∂A = B. The second equality isproved similarly:

˜(∂B + xA)(Q,R) = ˜(12θ − (p+ 1)P )B(Q,R) + ˜QA(Q,R) = (12θ − P )P + Q = 0,

as required.

Lemma 5.17. We have that A(x, y) has no repeated factors and is prime to B(x, y).

Proof. Write A(x, y) = xayb · p(x, y), where p(x, y) is an isobaric polynomial with not allterms involving both x and y. Over Fp, it factors as

A(x, y) = xayb∏i

(x3 − ciy2)ni .


Suppose A(x, y) is divisible exactly by (x3 − cy2)n with n ≥ 2. Note that x3 − y2 does not

divide A(X, Y ) (which we see by plugging in x = Q, y = R). Now,

∂(x3 − cy2) = 3x2(−4y)− c(2y)(−6x2) = 12(c− 1)x2y,

which is coprime to x3 − cy2. Also note that n < p. Then

A(x, y) = (x3 − cy2)n · (coprime to it)

and

B(x, y) = ∂A(x, y) = n(x3 − cy2)n−1 · (coprime to (x3 − cy2))

is divisible exactly by (x3 − cy2)n−1. Since n ≥ 2, we can repeat the same argument to see

that −xA = ∂B is divisible exactly by (x3 − cy2)n−2. This is a contradiction. One can usethe same trick to show that a, b ≤ 1.

The fact that A is prime to B follows by a similar argument.

Proof of Theorem 5.6. We want to compute the kernel of the map

Fp[x, y] M.

Let I be the kernel. Since M is a domain which is not a field, I is prime, but it is not

maximal. Moreover, (A(x, y)− 1) ⊆ I by Lemma 5.16 (1). Then

(A(x, y)− 1) ⊆ I ( m

and Fp[x, y] has dimension 2, so it suffices to show that

A(x, y)− 1

is irreducible.

Suppose to the contrary that and irreducible factor φ(x, y) divides it and

A(x, y)− 1 = (φn(x, y) + φn−1(x, y) + · · ·+ 1)︸︷︷︸φ(x,y)

·(∗)

with φn(x, y) isobaric of degree n. Let c ∈ F×p whcih is a primitive (p − 1)st root of unity.Then

A(c4x, c6y) = cp−1A(x, y) = A(x, y).

Therefore, φ(c4x, c6y) also divides A and is not equal to φ(x, y) since n < p − 1, so it iscoprime to it. Therefore, we must have

A(x, y)− 1 = (φn(x, y) + φn−1(x, y) + · · ·+ 1) ·(φn(c4x, c6y) + · · ·+ 1

)· (∗).

Looking at the leading degree terms, we see that φn(x, y) · φn(c4x, c6y) divides A(x, y),

and hence φn(x, y)2 divides A(x, y) (because φn(x, y) was isobaric of degree n). This is acontradiction with Lemma 5.17.

Examples 5.18. For p = 11, E10 = QR and

M ∼=F11[Q,R]

(QR− 1).

82 KARTIK PRASANNA

For p = 13, E12 =441Q3 + 250R2

691, so

M ∼=F13[Q,R]

(Q3 + 3R2 + 11).

Recall that we had the following picture

Z(p)[x, y] M = Z(p)[Q,R] Z(p)JqK

Fp[x,y]

(A(x,y)−1)M FpJqK

∼=

∼=

where Ep−1 = A(Q,R).

As a corollary to Theorem 5.6, we get the diagram

Mk Mk+(p−1)

Mk Mk+(p−1)

·Ep−1

For α ∈ Z/(p− 1)Z, we define

M (α) = lim−→k≡α (p−1)

Mk =⋃

k≡α (p−1)

Mk.

Remark 5.19. If α is not even, then M (α) = 0. Also,

M =∑

M (α).

Corollary 5.20. We have that

M =⊕

α∈Z/(p−1)Z

M (α)

and

M (α) · M (β) ⊆ M (α+β).

In other words, M is graded with respect to Z/(p− 1)Z.

Proof. One can check this using the fact that A(Q, R) = 1 is the only relation.

Corollary 5.21. If f ∈Mk and g ∈Mk′ satisfy f ≡ g 6≡ 0 (p), then k ≡ k′ (p− 1).

Proof. Since f = g 6= 0, they belong to the same graded piece, so k ≡ k′ (p − 1) usingCorollary 5.20.

The next goal will be to prove a generalization of this, replacing p with a power of p. Wewill work towards this by studying the above diagram in more detail.


Proposition 5.22. The operator θ preserves M . In fact, if f ∈ Mk, then θf ∈ Mk+p+1.

Proof. Lift f to some f ∈Mk and write f = φ(Q,R). Then

∂f = (12θ − kP )f,

so12θf = ∂f + kPf = ∂φ(Q,R) + kPf.

Reducing this modulo p, we get that

12θf = ∂φ(Q, R) + kP f

= A(Q, R)∂φ(Q, R) + kB(Q, R)f

= ˜(Ep−1∂φ+ kEp+1f)

Therefore, ∂f is a reduction of a modular form of weight k + p+ 1.

Note, however, that θf could live in M` for ` < k + p + 1. This motivates the followingdefinition.

Definition 5.23 (Weight filtration). Let f be a graded element in M , i.e. it lives in M (α)

for some α. Then w(f) is the smallest integer k ≡ α (p− 1) such that f ∈ Mk.

Remark 5.24. For f ∈ Mk, f = Φ(Q,R) where Φ is isobaric of degree k. Then f =

Φ(Q, R) ∈ Mk. Then w(f) ≤ k and w(f) < k if and only if A divides Φ.

Proposition 5.25. Let f ∈ M be a graded element. Then:

(1) w(θf) ≤ w(f) + p+ 1,

(2) w(θf) = w(f) + p+ 1 if and only if w(f) 6≡ 0 (p).

Proof. Part (1) is clear. For (2), let k = w(f). Then f lifts to f ∈Mk. We use the equality

12θf = A(Q, R)∂φ(Q, R) + kB(Q, R)f

from the proof of Proposition 5.22.

If p|k, then θf = ∂φ, so w(θf) ≤ k + 2 < k + (p+ 1).

If k 6≡ 0 (p), then A does not divide A∂φ+ kBφ, so w(θf) = k + (p+ 1).

Theorem 5.26. Suppose f ∈Mk, f′ ∈Mk′ such that f ≡ f (pm) but f, f ′ 6≡ 0 (p). Then

k ≡ k′ (pm−1(p− 1)).

Proof. For m = 1, this is simply Corollary 5.21. We may hence assume m ≥ 2.

Let h = k′ − k. We have that

Epn(p−1) = 1− 2pn(p− 1)

Bpn(p−1)

·∑k

σpn(p−1)−1(k)qk ≡ 1 (pn+1)

by the van Staudt congruence 5.15.

84 KARTIK PRASANNA

Multiplying f ′ by Epn(p−1) for n 0, we may assume that h ≥ 4. Since (p− 1)|h, we havethat

Eh ≡ 1 mod p.

We want to show that vp(h) ≥ m− 1, i.e. vp(h) + 1 ≥ m. Write r = vp(h) + 1 and supposer < m. Then look at

fEh − f ′ = (f − f ′)︸︷︷︸≡0 (pm)

+ f(Eh − 1)︸︷︷︸≡0 (pr)

.

Then

fEh − f ′ ≡ 0 (pr).

Now, we have that

p−r (fEh − f ′)︸︷︷︸∈Mk′

≡ p−rf(Eh − 1) (p).

Now,

p−r(Eh − 1) = λ

∞∑n=1

σh−1(n)qn

where λ is a unit. Let

φ =∞∑n=1

σh−1(n)qn.

Then

fφ ≡ g (p)

where

g = λ−1p−r(fEh − f ′) ∈Mk′ .

Reducing this equations modulo p, we get that f φ = g and f 6= 0, so

φ =g

f∈ fraction field of M.

Since k′ ≡ k (p− 1),

φ ∈ fraction field of M (0).

Now, we recall that

φ =∞∑n=1

σh−1(n)qn.

Since h ≡ 0 (p− 1), h− 1 ≡ p− 2 (p− 2), and hence

φ ≡∞∑n=1

σp−2(n)qn (p).

Therefore,

φp ≡∞∑n=1

σh−1(pn)qnp

and σh−1(np) = σh−1(n), so

φ− φp = ψ


where

ψ =∞∑n=1

(p,n)=1

σp−2(n)qn.

We claim that

ψ ≡ θp−2

(∞∑n=1

σ1(n)qn

).

This is clear since

σp−2(n) =∑d|n

dp−2 ≡∑d|n

d−1

np−2σ1(n) = np−2

∑d|n

d

≡ n−1∑d|n

d

when p does not divide n.

Now,

ψ ≡ θp−2

(∞∑n=1

σ1(n)qn

)= − 1

24θp−2(P ) = − 1

24θp−2(Ep+1).

Hence

ψ ∈ M (0)

since ψ ∈ M(p+1)+(p−2)(p+1) = M(p+1)(p−1).

Fact. The space M (0) is integrally closed. This will be Theorem 6.7.

We assume this fact for now and prove it later. Then the above argument shows that

φ ∈ M (0).

Let ` = w(φ). We may then write

φ = Φ(Q, R)

where Φ is isobaric of degree ` and A does not divide Φ. Then

φp = Φp(Q, R)

and A does not divide Φp. Therefore, w(φp) = `p, which shows that

w(ψ) = w(φ− φp) = `p.

Now, note that w(Ep+1) = p + 1 (because P is not modular). By Proposition 5.25, we getthat

w(θp−2(Ep+1)) = p+ 1 + (p− 2)(p+ 1) = (p+ 1)(p− 1).

This cannot be equal to `p, which is a contradiction.

We still need to prove the fact that M (0) is integrally closed. This is quite difficult, so wedelay the proof of this further (Theorem 6.7).

86 KARTIK PRASANNA

5.3. Serre’s p-adic modular forms. The goal of this section is to prove the existence ofthe p-adic L-functions using congruences between modular forms. The references are [SD73]and [Ser73b]. This is the most classical approach to p-adic modular forms.

We will later present a different, more geometric approach to p-adic modular forms due toKatz [Kat73].

Recall that M is the set of forms with Z(p)-coefficients. Then M ⊗ Q are the forms withQ-coefficients. Let

A = f ∈ QpJqK | vp(f) = infnvp(an(f)) > −∞ = ZpJqK⊗Zp Qp ⊆ QpJqK.

Then we have maps

M →M ⊗Q q-expansion−→ A ⊆ QpJqK.

Definition 5.27. If fn and f are elements of QpJqK, we say that fn → f if the coefficientsof fn tend to the coefficients of f uniformly.

Definition 5.28. A p-adic modular form (in the sense of Serre) is a formal power series

f =∞∑n=0

anqn ∈ QpJqK

such that there is a sequence of fi ∈M ⊗Q such that fi → f .

Remark 5.29. Note that a p-adic modular form f is an element of A.

What is the weight of a p-adic modular form? Let

Xm = Z/pm−1(p− 1)Z ∼= Z/pm−1Z× Z/(p− 1)Zand

X = lim←−Xm = Zp × Z/(p− 1)Z.Note that there is a natural diagonal embedding ∆: Z → X.

Theorem 5.30. Let f be a p-adic modular form. Then f has a well-defined weight κ ∈ Xdefined by κ = lim

iki where fi → f and fi has weight ki.

Proof. By multiplying f by a power of p, we may assume that vp(f) = 0. Taking fi → f ,eventually vp(fi) = vp(f) = 0, so fi ∈M and fi 6≡ 0 (p).

Given m, for any i, j 0, we have that vp(fi − fj) ≥ m. By Theorem 5.26,

ki ≡ kj (pm−1(p− 1)).

Therefore, limi ki is well-defined in X. It is also easy to see this is independent of the choiceof fi.

Lemma 5.31. If f, f ′ are two p-adic modular forms of the same weight, then f + f ′ is alsoa p-adic modular form of that weight.

Proof. Choose convergent sequences fi → f of weights ki and f ′i → f of weights k′i. Byassumption limi ki = limi k

′i. Then multiply the terms of the sequences by appropriate

Eisenstein series E(p−1)pm to get new sequences fi → f , f ′i → f ′ of the same weights k′′i → κ.

Then fi + f ′′i is a sequence of modular forms of weights k′′i converging to f + f ′.


We remark that Theorem 5.26 has the following generalization.

Corollary 5.32. If vp(f − f ′) ≥ vp(f) +m, then k ≡ k′ (pm−1(p− 1)).

Proof. If vp(f) = r, then vp(p−rf) = 0, so vp(p

−rf ′− p−rf) ≥ m. Then k′ ≡ k (pm−1(p− 1))by Theorem 5.26.

This has the following generalization for p-adic modular forms.

Theorem 5.33. Suppose f ′ and f are p-adic modular forms of weight κ, κ′ ∈ X, and supposethat vp(f

′ − f) ≥ vp(f) +m. Then κ′ = κ in Xm.

Proof. Write f = limi fi, f′ = limi f

′i . Then vp(f) = vp(fi), vp(f

′) = vp(f′i) for i 0.

Similarly,

vp(fi) +m ≤ vp(f − f ′) = vp(fi − f ′i) for i 0.

Hence k′i ≡ ki (pm−1(p− 1)) for i 0 by Corollary 5.32. Then κ′ = κ in Xm.

Corollary 5.34. Let f = a0 + a1q + · · · be a p-adic modular form of weight κ. Supposeκ 6= 0 in Xm+1. Then vp(a0) +m ≥ infi vp(ai).

Proof. This is equivalent to saying that if ai ∈ Zp for all i ≥ 1 and at least one is a unit,then vp(a0) ≥ −m. We may assume that a0 6= 0. Suppose vp(a0) = −r. We want to showthat r ≤ m. Suppose contrary that r > m. We have that

pm+1f = pm+1a0 +∞∑i=1

(pm+1ai)qi.

Then

v(pm+1f) = v(pm+1a0) ≤ 0.

Let f ′ = pm+1a0. Note, moreover, that

vp(pm+1f − f ′) ≥ m+ 1 ≥ m+ 1 + vp(f

′).

By Theorem 5.33, this shows that κ = 0 in Xm+1, which is a contradiction. Therefore,r ≤ m.

Theorem 5.35. Let f (i) =∞∑n=0

a(i)n q

n be a sequence of p-adic modular forms of weights κ(i)

such that

(a) a(i)n → an ∈ Qp uniformly for n ≥ 1,

(b) κ(i) → κ 6= 0 in X.

Then a(i)0 also converges to some a0 ∈ Qp and

f = a0 + a1q + a2q2 + · · ·

is a p-adic modular form of weight κ.

88 KARTIK PRASANNA

Proof. By (b), there exists m such that κ(i) 6= 0 in Xm for i 0. Also, there exists t ∈ Zsuch that vp(a

(i)n ) ≥ t for all n ≥ 1 and all i 0. By Corollary 5.34,

vp(a(i)0 ) +m ≥ t for i 0.

Therefore, a(i)0 for i 0 belong to a closed ball, and hence there exists a convergent subse-

quence a(ij)0 → a0. Then

f (ij) → f = a0 + a1q + · · · .Since each f (ij) are all p-adic modular forms, f is also a p-adic modular form of weight κ.

Suppose a(i′j)

0 → a′0 was a different convergent subsequence, we would get that

f (ij) → f ′ = a′0 + a1q + · · ·

which is also a p-adic modular form of weight κ. Then f − f ′ = a0 − a′0 is a p-adic modularform of weight κ by Lemma 5.31. But it is a constant a0− a′0 so it is a p-adic modular formof weight 0. Since κ 6= 0, this is a contradiction.

Take κ ∈ X = Zp × Z/(p− 1)Z and let κ = (s, u). Let

σ∗κ−1(n) =∑d|n

(p,d)=1

dκ−1

where

dκ = 〈d〉sω(d)u.

Observe that if ki → κ in X and ki →∞ as integers, then

σ∗κ−1(n) = limiσki−1(n),

because

limiσki−1(n) = lim

i

∑d|n

(p,d)=1

dki−1 = σ∗κ−1(n)

uniformly.

Recall that

Gki = −Bki

2ki+∞∑n=1

σki−1(n)qn.

Since∞∑n=1

σki−1(n)qn →∞∑n=1

σκ−1(n)qn

Theorem 5.35 shows that

limi

−Bki

kiexists and we define it to be

ζ∗(1− κ).


If 1− κ = (s, u), we defineζ∗(s, u) := lim

ki→κki→∞

ζ(1− ki),

which is defined except when s = 1 and u = 1. For fixed u, this is a continuous functionin s.

Definition 5.36 (p-adic Eisenstein series). For a weight κ ∈ X, κ 6= 0, we set

G∗κ =1

2ζ∗(1− κ) +

∞∑n=1

σ∗κ−1(n)qn.

Theorem 5.37. We have that ζ∗(s, u) = Lp(s, ω1−u).

Since this is a continuous function on Zp, we just need to evaluate ζ∗(s, u) for s = 1 − k.This is not easy to do. On the other hand, knowing that the p-adic L-function exists, it isnot so hard to check that the two definitions agree. We will prove that this is true withoutassuming the existence of the p-adic L-function. In other words, we will show directly that

ζ∗(1− k) = −Bk

k(1− pk−1)

for k ∈ Z, k ≥ 2.

5.4. Hecke operators. Let f =∑n≥0

anqn. We define operators U , V , and T` for ` 6= p:

f |U =∞∑n=0

anpqn

f |V =∞∑n=0

anqpn

f |κT` =∞∑n=0

a`nqn + `κ−1

∞∑n=0

anq`n for κ ∈ X

Theorem 5.38. If f is a p-adic modular form of weight κ, then so are f |U , f |V , and f |κT`.

Proof. Consider modulars forms fi of weight ki such that fi → f . Note that ki → κ in X.Replacing fi by fi · Epi(p−1), we may assume that ki →∞.

Fact. The Hecke operator |kT` preserves the space of usual modular forms of weight k.

Now,

fi|kiT` =∞∑n=0

a`n,iqn + `ki−1

∞∑n=0

an,iq`n.

Suppose ` 6= p. Then the above expression tends to∞∑n=0

a`nqn + `κ−1

∞∑n=0

anq`n = f |κT`.

Hence, f |κT` is a p-adic modular form, since it is the limit of regular modular forms fi|kiT`.

90 KARTIK PRASANNA

Now, suppose ` = p. Then the above expression tends to∞∑n=0

apnqn = f |U.

Hence f |U is a p-adic modular form of weight κ.

Applying this to the modular form fi, we see that fi|U is a p-adic modular form of weight ki.Moreover,

fi|kiTp =∞∑n=0

anp,iqn

︸︷︷︸fi|U

+ `ki−1

∞∑n=0

an,iqnp

︸︷︷︸fi|V

.

Hence fi|V is a p-adic modular form of weight ki by Lemma 5.31.

Finally, f |V = limifi|V is also a p-adic modular form.

Suppose k ≥ 4 in an integer. Then

Gk =1

2ζ(1− k) +

∞∑n=1

σk−1(n)qn.

We claim that Gk is an eigenfunction of all the T` with

Gk|kT` = (1 + `k−1)Gk.

To see this, we just compute:

Gk|kT` =1

2ζ(1− k) +

∞∑n=1

σk−1(n`)qn + `k−1

[∞∑n=1

σk−1(n)q`n +1

2ζ(1− k)

]

=1

2ζ(1− k)(1 + `k−1) +

∞∑n=1

qn(σk−1(n`) + `k−1σk−1(n/`)).

We need to check that

σk−1(n`) + `k−1σk−1(n/`) = (1 + `k−1)σk−1(n),

i.e.

σk−1(n`)− σk−1(n) = `k−1(σk−1(n)− σk−1(n/`)).

When ` does not divide n, this is clear. Otherwise, write n = `rm for (`,m) = 1. Then theexpression above is:

σk−1(`r+1)σk−1(m)− σk−1(`r)σk−1(m) = `k−1(σk−1(`r)− σk−1(`r−1))σk−1(m).

Finally, this expression simplifies to

(`r+1)k−1 = `k−1(`r)k−1,

which is clearly true.

Recall that

G∗κ = limki→κki→∞

Gki .


Proposition 5.39. If k ≥ 4, then

G∗k = Gk − pk−1Gk|V.

Proof. We have that

Gk − pk−1Gk|V = −Bk

2k+∑n≥1

σk−1(n)qn − pk−1

[−Bk

2k+∑n≥1

σk−1(n)qnp

]

= −(1− pk−1)Bk

2k+∞∑n=1

(σk−1(n)− pk−1σk−1(n/p))qn

We claim that

σk−1(n)− pk−1σ(n/p) = σ∗k−1(n).

If p does not divide n, this is clear. If p|n,

σ∗k−1(n) =∑d|n

dk−1 −∑d|np|d

dk−1 = σk−1(n)− pk−1σ(n/p).

Therefore,

Gk − pk−1Gk|V = −(1− pk−1)Bk

2k+∞∑n=1

σ∗k−1(n)qn.

Since Gk − pk−1Gk|V and G∗k are both p-adic modular forms of the same weight and alltheir terms agree except possibly the constant term. By Theorem 5.35, this implies thatGk − pk−1Gk|V = G∗k.

Remark 5.40. This finishes the proof of Theorem 5.37.

Assume for now that this proposition holds also for k = 2.

Corollary 5.41. The Eisenstein series G2 is a p-adic modular form.

Proof. We have that G∗k = Gk|(1− pk−1V ). Therefore,

Gk = G∗k|(1 + pk−1V + ()pk−1V )2 + · · · )is a limit of p-adic modular forms, and hence a p-adic modular form.

However, to prove the above proposition for k = 2, we need to check that

ζ(−1)(1− p) = limki→2

ζ(1− ki).

Recall that the Kummer congruences 5.14 say that

m ≡ n 6≡ 0 (p− 1) implies thatBm

m≡ Bn

nmod (p).

Exercise. If m ≡ n (pr−1(p− 1)), m 6≡ 0 (p− 1), then

(1− pm−1)Bm

m≡ (1− pn−1)

Bn

nmod (pr).

92 KARTIK PRASANNA

Therefore, modulo large powers of p,

(1− pki−1)Bki

ki≡ (1− p)B2

2.

Since ki →∞, this shows that

ζ(−1)(1− p) = limki→2

ζ(1− ki).

We have hence just proven the following proposition.

Proposition 5.42. The Eisenstein series P is a p-adic modular form of weight 2.

Theorem 5.43. Let f =∑n≥0

anqn be a p-adic modular form of weight κ. Then

θf =∑n≥0

nanqn

is a p-adic modular form of weight κ+ 2.

Proof. Take classical modular forms fi of weight ki such that fi → f . Recall that weintroduce the operator ∂ such that

∂fi = (12θ − kiP )fi

is a modular form of weight ki + 2. Hence θfi is a p-adic modular form of weight ki + 2 byProposition 5.42. Hence θf = lim

iθfi is a p-adic modular form of weight κ+ 2.

6. Moduli-theoretic interpretation of modular forms

Note that the results of Section 5 assume the fact that M (0) in integrally closed. The proofof this relies on a geometric reinterpretation of modular forms due to Katz [Kat73].

6.1. Modular forms over C. Recall that, over C, we have the following diagram:

SL2(Z)\h elliptic curves over C/ ∼=

lattices in C/homothety

For an element τ ∈ h, the corresponding lattice is Zτ + Z. Changing the lattice Zτ + Zby a homothety corresponds to changing τ by an element of SL2(Z). Give a lattice L, thecorresponding elliptic curve is C/L. Indeed, the elliptic curve is given by the equation

E : y2 = 4x3 − g2(L)x− g3(L)

where

g2(L) = 60∑

`∈L\0

1

`4,

g3(L) = 140∑

`∈L\0

1

`6.


Then the isomorphism is defined using the Weierstrass ℘ function

C/L∼=→ E

z 7→ (℘(z;L), ℘′(z;L)).

Note also that the canonical differential dxy

on E pulls back to dz on C/L.

A modular form of weight k satisfies

f

(az + b

cz + d

)= (cz + d)kf(z).

Therefore, we can think of it as a function on lattices

F (L) = f(ω1/ω2)ω−k2 where L = Zω1 + Zω2, ω1/ω2 ∈ h.

Then, note that F (λL) = λ−kF (L).

Finally, we may consider it as a function on pairs (E,ω) where E is an elliptic curve and ωis a differential on E. Then

f(E, λω) = λ−kF (E,ω).

Now, consider the change of variables q = e2πiτ . First note that

F (C/(Zτ + Z), dz) = F (C/(2πi(Zτ + Z)), 2πidz).

Moreover, applying the exponential map

exp: C/(2πi(Zτ + Z))→ C∗/qZ,

we get that

F (C/(Zτ + Z), dz) = F

(C∗/qZ,

dt

t

),

where the variable on C∗/qZ is denoted by t.

Finally, we may write

y2 = 4x3 − E4

12x+

E6

216

with the differential dxy

. Then

F

(y2 = 4x3 − E4

12x+

E6

216,dx

y

)∈ C((q)),

considering the q-expansions of the Eisenstein series. The modular form is holomorphic atinfinity if this belongs to CJqK.

Consider the change of variables X = x+ 112

and Y = x+ 2y. Then the elliptic curve is

Y 2 +XY = X3 +B(q)X + C(Q)

94 KARTIK PRASANNA

where

B(q) = −5

(E4 − 1

240

),

C(q) =−5(E4−1240

)− 7

(E6−1−504

)12

.

One can check that this is an elliptic curve over Z((q)). It is called the Tate curve. Thedifferential on it is given by

dX

2Y +X.

We denote it by Tate(q).

6.2. Modular forms over an arbitrary ring.

Definition 6.1. A modular form of weight k defined over a ring R is a function on pairs(E/R′, ω) where p : E → Spec(R′) is an elliptic curve where R′ is an R-algebra and ω isa nowhere vanishing differential form, i.e. a generator of ωE/R = p∗(Ω

1E/R), satisfying the

following properties:

(1) f(E/R′, ω) ∈ R′ only depends on the isomorphism class of (E/R′, ω),(2) f(E/R′, λω) = λ−kf(E/r′, ω) for λ ∈ (R′)×,(3) given a pullback of E ′ → SpecR′ along ϕ : R′ → R′′:

E ′′ E ′

Spec(R′′) Spec(R′)

and if ω′ pulls back to ω′′, then

f(E ′′/R′′, ω′′) = ϕ(f(E ′/R′, ω′)).

Recall that we defined an elliptic curve Tate(q) over Z((q)). Base changing it to Z((q)) ⊗ Rgives an elliptic curve over Z((q))⊗R with a canonical differential. We may then evaluate

f(TateR, ωcan) ∈ Z((q))⊗R.This is the q-expansion of f . We will denote it by qexp(f). We say that f is holomorphic if

f(TateR, ωcan) ∈ ZJqK⊗R.

Let Modk(R) be the R-modular of holomorphic modular forms of weight k, defined over R.

For R→ R′, we have the diagram

Modk(R) Modk(R′)

ZJqK⊗R ZJqK⊗R′qexp qexp

Theorem 6.2 (Deligne–Rapoport/Katz, q-expansion principle).


(1) If f ∈ Modk(R) and qexp(f) = 0, then f = 0.(2) If R → R′ and if f ∈ Modk(R

′) is such that qexp(f) ∈ ZJqK⊗R, then f ∈ Modk(R).

Exercise. If f ∈Mk(SL2(Z),C), then f is a modular form defined over C, i.e.

Mk(SL2(Z),C) = Modk(C).

Fix a prime p ≥ 5. Recall that

Mk = fourier expansions of holomorphic modular forms with coefficients in Z(p) ⊆ Z(p)JqK.

Via the q-expansion, using the q-expansion principle,

Modk(Z(p)) ∼= Mk.

We then have

Modk(Z(p)) Mk Z(p)JqK

Modk(Fp) Mk FpJqK

∼=

∼=

Consider Ep−1 ∈ Modk(Z(p)) and recall that its q-expansion is congruent to 1 modulo p.Therefore, we get an element

Ep−1 ∈ Modp−1(Fp)whose q-expansion is 1.

We will construct explicitly an element of Modp−1(Fp), called the Hasse invariant A andcheck that Ep−1 = A.

Definition 6.3. Let R be an Fp-algebra and E/R be an elliptic curve. The absolute Frobe-nius

Fabs : E → E

defines a pullbackF ∗abs : H1(E,OE)→ H1(E,OE)

which is not R-linear, but rather p-linear:

F ∗abs(αx) = αpF ∗abs(x)

for α ∈ R. Picking a differential ω on E/R, we get η which is a dual basis of H1(E,OE).Then we define the Hasse invariant A to be:

F ∗abs(η) = A(E,ω) · η.

Note that taking ω 7→ λω gives η 7→ λ−1η, then

F ∗abs(λ−1η) = λ−pF ∗abs(η) = λ−pA(E,ω)η = λ−(p−1)A(E,ω)(λ−1η).

This shows thatA(E, λω) = λ−(p−1)A(E,ω),

so A is a modular form of weight p− 1.

Theorem 6.4. We have that A = Ep−1.

96 KARTIK PRASANNA

We defer the proof of this theorem until later. For now, we discuss its applications.

Definition 6.5. Let K be a field of characteristic p. We say that an elliptic curve E/K issupersingular if

Fabs : H1(E,OE)→ H1(E,OE)

is the zero map.

Therefore, an elliptic curve is supersingular if and only if its Hasse invariant vanishes.

Theorem 6.6. There are only finitely many supersingular j-invariants; in fact all of themlie in Fp2.

Exercise. Pick your favorite modular form f and an elliptic curve with a differential (E,ω).Compute f(E,ω). You can use Magma or Sage for this. If you pick a modular form withrational coefficient and a rational elliptic curve, your answer will be a rational number.

Theorem 6.7. Recall thatM (0) = lim−→

k≡0 (p−1)

Mk.

ThenM (0) ∼= O(P1

Fp \ supersingular points),so it is integrally closed.

Sketch of proof. Take f ∈ M (0). Then consider

f ∈ Mh(p−1) = Modk(p−1)(Fp).Note that also

Ep−1h ∈ Modh(p−1)(Fp).

Then

f

/Ep−1

h

is a function on non-supersingular elliptic curves over Fp. Check that this is a rationalfunction on P1

Fp with poles only at supersingular points.

Exercise. The map

M (0) → O(P1Fp \ supersingular points)

is injective and surjective.

We still need to prove Theorem 6.4:A = Ep−1.

The original proof due to Deligne is to compute the q-expansion of A. We will instead presenta proof due to Kaneko–Zagier [KZ98]. We will only show the equality up to a non-zero scalar,but one can trace through the details of the proof to get the actual equality.

We need to checkA(E,ω) = Ep−1(E,ω)

for every (E,ω) over Fp.


Let us think how to evaluate the modular form Ep−1 at an elliptic curve E/Fp. First, we lift

the elliptic curve to E/Z(p), then evaluate Ep−1 at this lift of the elliptic curve, and reducethe answer modulo p.

Remark 6.8. An elliptic curve E over Fp is supersingular if and only if E[p](Fp) = 0.

Suppose E is given by the equation y2 = f(x) where f(x) is a cubic polynomial.

Proposition 6.9. Let E/Fq be defined by the equation y2 = f(x). Let ap be the coefficient

of xp−1 in f(x)p−12 . Then

|E(Fq)| = 1−NFq/Fp(ap) in Fp.

Proof. We count the number of solutions to y2 = f(x) for fixed x:

• if f(x) 6∈ F2p, then there are 0 solutions,

• if f(x) = 0, then there is 1 solution,• if f(x) ∈ (F×p )2, then there are 2 solutions.

This may be written concisely as

(1 + f(x))q−12 in Fq.

Therefore,

|E(Fq)| = 1 +∑x∈Fq

(1 + f(x)q−12 ) in Fq.

Hence

|E(Fq)| = 1 +∑x∈Fq

f(x)q−12 in Fq.

The degrees of the monomials appearing in this expression are between 0 and 3(q−1)2

. Notethat ∑

x∈Fq

=

0 if j 6= q − 1,

−1 if j = 1.

Therefore:

|E(Fq)| = 1−(

coefficient of xq−1 in f(x)q−12

)in Fq.

Finally,

f(x)q−12 = f(x)

pr−12

= f(x)p−12

(1+p+···+pr−1)

= f(x)p−12 f (p)(xp)

p−12 · · · f (pr−1)(xp

r−1

)p−12 .

Write

xq−1 = xa0xpa1 · · ·xpr−1ar−1

according to the above decomposition. Then

q − 1 = a0 + pa1 + · · ·+ pr−1ar−1 = (p− 1) + p(p− 1) + · · ·+ pr−1(p− 1)

98 KARTIK PRASANNA

we see that

0 ≤ ai ≤ 3(p− 1)

2.

In particular,

|ai − (p− 1)| ≤ p− 1.

Then induction on i gives ai = p− 1. Hence the coefficient of xq−1 in f(x)q−12 is

apapp · · · ap

r−1p = NFq/Fp(ap).


Corollary 6.10. An elliptic curve is supersingular if and only if ap = 0.

Proof. Suppose ap 6= 0. Then there is an n such that NFq/Fp(ap)n = 1. Then

#E(Fqn) = 1−NFqn/Fp(ap) = 1−NFq/Fp(ap)n = 0 in Fp

by Proposition 6.9. This shows that E(Fq)[p] 6= ∅, so the curve is not supersingular.

Let Eλ : y2 = x(x−1)(x−λ) be an elliptic curve in Legandre form. When is Eλ supersingular?

The coefficient of xp−1 in (x(x− 1)(x−λ))p−12 is the coefficient of x

p−12 in ((x− 1)(x−λ))

p−12

which is

r=(p−1)/2∑i=0

(−1)r−i(

r

r − i

)(−λ)i

(r

i

)= (−1)r

r∑i=0

(r

i

)2

λi.

We call this polynomial H(λ). Then the corollary says that Eλ is supersingular if and onlyif H(λ) = 0.

Observation. Since λ 7→ j(Eλ) is a rational function of degree 6 is roughly 6-to-1. There-fore, there are roughly p−1

12supersingular j-invariants (up to a constant).

In fact, one can prove a precise result of this form by analyzing when two j-invariants canbe equal. The details of this are in Silverman’s book [Sil09].

Proposition 6.11 (Ihara). The polynomial H(λ) has no multiple roots.

Proof. Consider the differential operator

D = 4λ(1− λ)d2

dλ2+ 4(1− 2λ)

d

dλ− 1.


We claim that DH(λ) = 0.

(−1)rDH(λ) = 4λ(1− λ)r∑i=0

(r

i

)2

i(i− 1)λi−2 + 4(1− 2λ)r∑i=0

(r

i

)2

iλi−1 −r∑i=0

(r

i

)2

λi

=r∑i=0

(r

i

)2 [4(i)(i− 1)λ(1− λ)λi−2 + 4(1− 2λ)iλi−1 − λi

]=

r∑i=0

(r

i

)2 [λi(−4i(i− 1)− 8i− 1) + λi−1(4i(i− 1) + 4i)

]=

r∑i=0

(r

i

)2 [λi(−4i2 − 4i− 1) + λi−1(4i2)

]=

r∑i=0

λi

[(r

i

)2

(−4i2 − 4i− 1) +

(r

i+ 1

)2

4(i+ 1)2

]

=r∑i=0

λi(r

i

)2 [(−4i2 − 4i− 1) + (r − i)2 · 4

]=

r∑i=0

λi(r

i

)2

[−4i2 − 4i− 1 + 4r2 − 8ri+ 4i2 − 4i− 1 + (p− 1)2 − 4(p− 1)i]

= 0

in Fp. For λ 6= 0, 1, this shows that the polynomial H(λ) has no multiple roots. For λ = 0, 1,one can check this by hand.

Since the roots of H(λ) are at λ corresponding to supersingular elliptic curves Eλ, we im-mediately get the following corollary.

Corollary 6.12. The numerator1 of the rational function A(Eλ,

dxy

)in λ is divisible by

H(λ).

We now reach the idea of Kaneko and Zagier. They consider

Ek = usual Eisenstein series

Gk = coefficient of xk in (1− 3E4(τ)x4 + 2E6(τ)x6)−12

Hk = coefficient of xk in (1− 3E4(τ)x4 + 2E6(τ)x6)k2 ,

which are all modular forms of weight k.


Gp−1 ≡ Hp−1 ≡ ±Ep−1 mod p.

1Note that A(Eλ,

dxy

)is a rational function in λ with only possibly λ or λ− 1 in the denominator.

100 KARTIK PRASANNA

Proof. It is easy to see that Gp−1 ≡ Hp−1 mod p. Note that

Gk = Resx=0dx

xk+1√

1− 3E4x4 + 2E6x6.

The elliptic curve given by the equation

y2 = x3 − 3E4(τ)x+ 2E6(τ)

is isomorphic to Eτ = C/(Zτ + Z) defined before. In fact, it is parameterized by

x = ℘(z), y = −1

2℘′(z)

where we recall that

℘(z) =1

z2+

∑`∈(Zτ+Z)′

(1

(z − `)2− 1

`2

).

Writing1

(z − `)2− 1

`2=

1

`2

[2z

`+

3z2

`2+ · · ·

],

we get that

℘(z) =1

z2−∑n≥4even

12n/2Bn

(n− 2)!nEn(τ)zn−2.

Consider the substitution X = ℘(z)−12 . Then

1− 3E4X4 + 2E6X

6 = 1− 3E4℘(z)−2 + 2E6℘(z)−3.

The residue expression for Gk above becomes

Gk = Resz=0

−12℘(z)−

32℘′(z)dz

℘(z)−( k+12 )√1− 3E4℘(z)−2 + 2E6℘(z)−3

= −1

2Resz=0

℘′(z)dz√℘(z)3 − 3E4℘(z) + 2E6

℘(z)k+12

= Resz=0 ℘(z)k+12 dz

= coefficient of zk in

1−∑n≥4even

12n/2Bn

(n− 2)!nEn(τ)zn

k+12

.

For k = p− 1, we get that

Gp−1 = coefficient of zk in

1−∑n≥4even

12n/2Bn

(n− 2)!nEn(τ)zn

p2

.

Note that for 4 ≤ n < p− 1,Bn

(n− 2)!nEn(τ)


is p-integral. Note that(1− 12

p−12Bp−1

(p− 3)!(p− 1)Ep−1(z)zp−1

) p2

≈ −p2

12p−12 Bp−1

(p− 3)!(p− 1)Ep−1(τ)

≡ 12p−12 Ep−1(τ) as

pBp−1

(p− 1)!≡ 1 (p)

≡ ±Ep−1(τ) mod p.


We therefore see that

±Ep−1

(Eλ,

dx

y

)= Hp−1

(Eλ,

dx

y

)= (constant) ·H(λ)

where the last inequality is given by noticing that Hp−1 is the coefficient of xp−1 in

(x3 − 3Qx+ 2R)p−12 ,

evaluating it at the Legendre family. and checking it agree with the definition of H(λ) afterconverting from Legandre form to Weiestrass form. This can be found in Hartshorne [Har77].

Combining this with Corollary 6.12, we have shown that

±Ep−1

A

(Eλ,

dx

y

)= λα(λ− 1)β

for some α, β. We just need to show that α = β = 0. Since the ratio

Ep−1

A

is a function of elliptic curves (invariant scaling the differential forms), it must be a rationalfunction in j. Moreover,

j(Eλ) = 256(λ2 − λ+ 1)3

λ2(λ− 1)2

and one can verify by an elementary calculation that the maps

λ 7→ 1− λ(2)

λ 7→ 1

λ(3)

preserve this expression. Then (2) shows that α = β and (3) shows that α = β = 0. Wehave hence finally proved Theorem 6.4.

7. p-adic L-functions associated to Hecke characters

We will be interested in p-adically interpolating special values of L-functions of Hecke char-acters of imaginary quadratic fields, We will first have to establish algebraicity results.

102 KARTIK PRASANNA

Let K be an imaginary quadratic field. Assume that it has class number 1 for simplicity.Unramified (algebraic) Hecke character are multiplicative functions

χ : I → C×

where I is the set of ideal of K such that

χ((α)) = αkαj = αk−j(αα)j

where k and j are integers and the number of roots of unity divides k − j (so that this iswell-defined). We then say that χ has infinity type (k, j).

For definition of L-functions, let us assume χ has infinity type (k,−j). The L-function sucha character χ is

L(χ, s) =∑a⊆OK

χ(a)Na−s =∏p

(1− χ(p)Np−s)−1.

Note that replacing s by s+1 scales the Dirichlet L-funtion by Na. For a = (α), Na = (αα).Therefore, it is enough to compute the L-function at 0 and the other values at integers willbe obtained by changing the character by the norm. We have

L(χ−1, 0) =1

wK

∑α∈OK

αj

αk

=1

wK

∑α∈OK

(αα)j

αj+k

=1

wK

∑α∈OK

1

αj+k|α|−2j

=1

wK

∑(m,n)∈Z2

6=(0,0)

1

(mτ0 + n)j+k|mτ0 + n|−2jwhere OK = Z + Zτ0 for τ0 ∈ h.

7.1. Real analytic Eisenstein series. The above calculation motivates the definition ofreal analytic Eisenstein series.

Definition 7.1. The real analytic Eisenstein series is defined as

Ek(z, s) =∑(m,n)

1

(mz + n)k|mz + n|2s.

This is absolutely convergent when k + 2s > 2, but we will ignore this issue for now.

Note that Ek(z, 0) = Ek(z). Also,

Ek(z,−r) · Im(z)−r

is a modular form of weight k, but it is not holomorphic (merely analytic).

Strategy. To relate Ek(z,−r) to the usual holomorphic Eisenstein series via differentialoperators.


Definition 7.2. The Shimura–Maass operator is defined as

δk =1

2πi

(∂

∂z+

k

2iy

)=

(θ − k

4πy

).

Note that

δk =1

2πiy−k

∂

∂z(yk · f).

Remark 7.3. This is a level raising operator on the Lie algebra of GL2 written out explicitly.

Proposition 7.4. If f is modular of weight k, then δkf is modular of weight k + 2.

Proof. Recall that ∂ = θ− k12P takes modular forms of weight k to modular forms of weight

k + 2 by Theorem 5.7. It is hence enough to check that

P

12− 1

4πy=E2(z)

12− 2i

4π(z − z)

is a real analytic modular form of weight 2. We just need to check invariance under z 7→ −1z

Recall that by Lemma 5.9, we have that

E2

(−1

z

)= z2E2(z) +

12z

2πi.

Then the expression above after letting z 7→ −1z

is

z2E2(z)

12+

z

2πi+

zz

2πi(z − z)= z2

(E2(z)

12+

1

(2πi)(z − z)

).


Definition 7.5 (Shimura). A nearly holomorphic modular form of weight k is

f(z) =

p∑j=0

fj(z)y−j

where fj are holomorphic modular forms, f transforms like a modular form of weight k, andthe finiteness condition at the cusp is satisfies. The space of such forms is denoted by Np

k (Γ),where p is the maximal power of y−1 allowed.

In this notation, Proposition 7.4 implies that the map δk on Npk (Γ) is

δk : Npk (Γ)→ Np+1

k+2 (Γ).

Definition 7.6. We let δrk = δk+2r−2 · · · δk+2 δk. If k is known, we simply write δr.


δrkEk(z, s) =1

(2πi)r(z − z)−rEk+2r(z, s− r) ·

Γ(k + r)

Γ(k)+ s · (holomorphic function in s).

Letting s = 0, we obtain the following corollary.

Corollary 7.8. Setting s = 0,

δrkEk(z) =

(− 1

4πy

)rΓ(k + r)

Γ(k)Ek+2r(z,−r).

104 KARTIK PRASANNA

Proof of Lemma 7.7. We proceed by induction on r. We have that

(2πiδk)r+1Ek(z, s) =

Γ(k + r)

Γ(k)(z − z)−(k+2r) ∂

∂z

(z − z)k+r ·∑(m,n)

1

(mz + n)k+2r|mz + n|2(s−r)

=

Γ(k + r)

Γ(k)· (z − z)−(k+2r)

[∑(k + r)(z − z)k+r−1(mz + n)−(k+2r)|mz + n|−2s+2r

− (z − z)k+r(k + 2r)(mz + n)−(k+2r+1)|mz + n|−2(s−r)m

+ (z − z)k+r(mz + n)−(k+2r)(|mz + n|2)r−s−1(mz + n)m(r − s)]

= (z − z)−(r+1)

[∑(k + r)(mz + n)−(k+zr)|mz + n|−2s+2r

− (k + 2r)(mz + n)−(k+2r+1)|mz + n|−2s+2r((mz + n)− (mz + n))

+ (mz + n)−(k+2r)|mz + n|2r−2s−2(mz + n)[(mz + n)− (mz + n)](r − s)]

where we note that (z − z)m = (mz + n)− (mz + n). Multiplying this out, we get a sum of5 terms:

(1): + (k + r)(mz + n)−(k+2r)|mz + n|−2s+2r

(2):− (k + 2r)(mz + n)−(k+2r)|mz + n|−2s+2r

(3): + (k + 2r)(mz + n)−(k+2r+2)|mz + n|−2s+2r+2

(4): + (r − s)(mz + n)−(k+2r)|mz + n|−2s+2r

(5):− (r + s)(mz + n)−(k+2r+2)|mz + n|2r−2s+2

Combining (1), (2), and (4), we get something which only involves something which is stimes a holomorphic function in s, so we may ignore that. Combining (3) and (5), we finallyget that the part not equal to s times a holomorphic function in s is equal to:

(z − z)−(r+1) Γ(k + r)

Γ(k)(k + r)Ek+2r+2(z, s− r − 1).

This completes the proof by induction on r.

Take Γ ⊆ Γ(N) ⊆ SL2(Z).

Definition 7.9 (Shimura). Let k ≥ 0. Define

Ak(Q) = meromorphic functions F/G | F ∈Mk+r(Γ,Q), G ∈Mr(Γ,Q)

Theorem 7.10 (Shimura). Let f ∈ Ak(Q), k ≥ 0. Then δrf behaves like an element ofAk+2r(Q) at all CM points. More precisely, if τ ∈ h belongs to an imaginary quadratic field,then

δrf(τ)

g(τ)∈ Q

for all g ∈ Ak+2r(Q) such that g(τ) 6= 0.


Proof. For τ ∈ K and any α ∈ K×, we note that

α

[τ1

]=

[a bc d

] [τ1

]for some

γα =

[a bc d

]∈ GL2(Q)+.

This defines a mapK →M2(Q)

such that γα(τ) = τ . As always, we define

(f |kγα)(z) = f(γα(z))j(γα, z)−k.

Exercise. If f ∈Mk(Γ,Q), then

(f |kγα) ∈Mk(Γ′,Q)

for some Γ′ ⊆ Γ.

This shows thatf |kγαf

= h ∈ A0(Q).

Explicitly:h = j(γα, τ)−k.

Thenδk(f |kγα) = δk(fh) = δkfh+ fδ0h.

Exercise. Check thatδk(f |kγ) = (δkf)|k+2γ

(in general).

Therefore:(δkf)(γα(τ))j(γα, τ)−(k+2) = (δkf)(τ)h(τ) + (fδ0(h))(τ).

By definition of γα:

(δkf)(τ) · α−(k+2) = (δkf)(τ)α−k + (f · δ0(h))(τ).

Dividing by g(z), we get that

δk(f)(τ) · α−k(α−2 − 1)

g(z)=

(f · δ0(h))(τ)

g(τ)∈ Q,

sincef · δ0(h) ∈ Ak+2(Q).

This settles the case r = 1. The general case follows by induction:

δr(f |kγα) = δr(fh) = δr(f)h+∑

δa(f)δb(h) · (coefficient) + δr(h)

andδr(h) = δr−1( δ0h︸︷︷︸

∈A2(Q)

).

Filling in the details of this step is left as an exercise.

106 KARTIK PRASANNA

7.2. Algebraicity of L-values. We renormalize the infinity type of the character χ asfollows:

χ((α)) = αk+jα−j = αk+2j(αα)j.

Then

L(χ−1, 0) =∑(α)

(αα)

αk+2j

= w−1K ·

∑α∈OK\0

(αα)j

αk+2j

= w−1K ·

∑ 1

(m+ nτ0)k+2j|m+ nτ0|−2j

∼Q× Ek+2j(τ0,−j)

∼Q× δjEk(τ0)πj by Corollary 7.8

∼Q× πj+kEk as Ek ∼ πkEk

∼Q× πj+kg(τ0),

where g is any holomorphic modular form of weight k + 2j with algebraic coefficients thatis non-zero at τ0; it exists by Shimura’s Theorem 7.10.

The elliptic curve C/Zτ0 +Z has CM by K and is defined over Q. Pick any elliptic curve Eover Q with CM by K. Then the period lattice of E is

Ω · (Zτ ′ + Z)

for τ ′ ∈ K, where for a non-vanishing differential ω ∈ H0(E,Ω1) and a K-basis γ ofH1(E,Q) ∼= K,

Ω =

∫γ

ω ∈ C×/Q×.

One can then easily show that

g(τ0) ∼Q×

(Ω

π

)k+2j

.

Finally, we have shown that

L(χ−1, 0) ∼ πj+k(

Ω

π

)k+2j

∼ π−jΩk+2j.

Example 7.11. Let K = Q(i). For j = 0, k = 4`, one can pick the elliptic curve y2 = 1−x4

to obtain that ∑(m,n)∈Z2\(0,0)

1

(mi+ n)4`∼Q× Ω4`

for

Ω =

∫ 1

0

dx√1− x4

.

One can get a more precise rationality result, but one has to pick the elliptic curve morecarefully.


Remark 7.12. So far, we need to assume that k ≥ 4 to use the Eisenstein series Ek. But,in fact, the result also holds for k = 2.

(1) We may defined E2(z) by analytically continuing∑ 1

(mz + n)2|mz + n|2s

in s. In fact, if we let

E[2(z) =

E2(z)

12− 1

4πy,

which is a modular form of weight 2, which appeared in the proof of Proposition 7.4,then

E2(z) ∼ π2E[2.

(2) At this point we cannot use δjE[2(z), because E[

2(z) is not holomorphic. We may,however, rewrite it as

E[2(z) =

δ(∆)

∆.

The details of this can be found in [Kat76].

7.3. p-adic interpolation. Recall that if χ((α)) = αk+jα−j, then we have shown that

L(χ−1, 0) = (−4πy0)jΓ(k)

Γ(k + j)δjEk(τ0)

where OK = Z + Zτ for τ0 ∈ h and y0 = Im(τ0). Recall that

Ek = 2ζ(k)Ek =2(2πi)k

(k − 1)!Gk.

Supposing that K has odd discriminant −D, we have that

τ0 =1 +√−D

2.

Then

L(χ−1, 0) = 2(2πi)j√Dj Γ(k)

Γ(k + j)

(2πi)k

Γ(k)δjGk(τ0).

Recall that we fixed an isomorphism (E,ω) ∼= (C/OK ,Ωdz). Then(2πi

Ω

)k+2j

δjGk(τ0) = δjGk(E,ω) ∈ Q.

The value we will interpolate is hence

1

2

(2πi)jΓ(k + j)√D−jL(χ−1, 0)

Ωk+2j= δjGk(E,ω).

We will eventually get a 2-variable p-adic L-functions (after taking out an Euler factor atp), where we allow both j and k to vary. However, we will first let j = 0 and just let k vary.

When j = 0, the character is χk((α)) = αk. We assume that k is divisible by the number ofroots of unity, as before. The weight space is W = Zp × Z/(p− 1)Z.

108 KARTIK PRASANNA

Suppose p = pp is split in K. Recall that we have fixed embeddings

Q → C, Q → Qp.

We normalize so that p is the prime corresponding to this fixed embedding.

Theorem 7.13. There is a continuous function

Lp : W \ 0 → Cp

such that

Lp(k) = Gk(E,ω) ·(

1− χk(p)

p

)for k ≥ 2.

We only prove this for k ≥ 4.

Remark 7.14. There is a more general statement for ramified characters, but we restrictto this simpler case.

The idea is to construct the function as follows

W \ 0 p-adic modular forms

Cp

Lpevaluate at (E,ω)

We first discuss evaluating p-adic modular forms. Fix some model for (E,ω) defined overthe ring of integers of a number field with good reduction at p. View (E,ω) as defined overR = OQp .

The key point is that E has ordinary reduction at primes above p. This is because (by thetheory of complex multiplication)

ap = ξ + ξ

where ξ is a generator of p. Therefore, p does not divide ap.

If (E,ω) is defined over R and has ordinary reduction, we may define f(E,ω) for any p-adicmodular form f . We write f = limi fi for fi of weight ki such that ki →∞. Then

f(E,ω) = limifi(E,ω).

To check convergence, taking fi and fj, we know that fi ≡ fj (pm), so ki = kj +rpm−1(p−1),and hence

fiEpm−1rp−1 ≡ fj

as modular forms. This shows convergence.

If f is a usually modular form, want f(E,ω) to be the same as before. If we take

f · Epm

p−1 → f

and then we easily check that the limit of the evaluations is the evaluation.


Finally, the other map is simple to define

W \ 0 → p-adic modular forms

κ 7→ G∗κ =1

2ζ∗(1− κ) +

∑n≥1

σ∗κ−1(n)qn.

In particular, we defineLp(κ) = G∗κ(E,ω).

Proof of Theorem 7.13. We just need to show that for k ≥ 4,

G∗k(E,ω) = Gk(E,ω) ·(

1− χk(p)

p

).

Recall thatG∗k = Gk − pk−1Gk|V,

and Tp = U + pk−1V . We hence need a modular interpretation of Tp, U, V . Recall thatE[p] ∼= (Z/pZ)2, and hence it has (p + 1) subgroup schemes of order p, C0, . . . , Cp. If wedefine

φi : E → Ei = E/Ci

then for a modular form f of weight k:

(f |Tp)(E,ω) = pk−1∑i

f(Ei, φ∗i (ω)).

When (E,ω) is define over R and has ordinary reduction, there is a canonical subgroup, C0.It is the kernel of the reduction modulo p map from E[p].

Then the modular interpretation of V is

(f |V )(E,ω) = f(E0, φ∗0ω).

One can check these identities using q-expansions.

Altogether, we see that

G∗k(E,ω) = Gk(E,ω)− pk−1(GK |V )(E,ω).

The mapE → E/C0

corresponds toC/OK → C/p−1OK .

The dual isogeny is multiplication by p, so the differential dz pulls back to pdz. Altogether,if

(E,ω) ∼= (C/OK ,Ωdz),

then(E0, φ

∗0ω) ∼= (C/p−1OK ,Ωpdz).

Multiplication by ξ, where p = (ξ), gives a further isomorphism

(C/p−1OK ,Ωpdz) ∼=(C/OK ,

Ωp

ξdz

).

110 KARTIK PRASANNA

Altogether, this shows that

G∗k(E,ω) = Gk(E,ω)(1− pk−1ξ−k

).

Finally, (1− χk(p)

p

)= 1− ξkp−1 = 1− pk−1(ξ)−k.


We would like to define the two variable p-adic L-function as

θjG∗k(E,ω).

However, when we differentiate the q-expansion of G∗k, we get

∞∑n=1

nσ∗k−1(n)qn−1

so the n do not interpolate p-adically well when p|n. Therefore, we will get rid of qn for p|n.

Recall that ∑anq

n U7→∑

anpqn,∑

anqn V7→

∑anq

np.

Then

f |V U = f,

f |UV =∑

anpqnp.

Then

f |k(V U − UV ) = f |k(I − (Tp − pk−1V )V ) = f |k(I − TpV + pk−1V 2)

is the p-deprived version of f . Define

G∗∗k = Gk|k(I − TpV + pk−1V 2) = Gk|(I − (1 + pk−1)V + pk−1V 2) =∑

(p,n)=1

σk−1(n)qn.

Then we define a function as above

W ×W p-adic modular forms

Cp

with the vertical map sending (k, j) to

θjG∗∗k =∑n≥1

(p,n)=1

njσ∗k−1(n)qn.


We have that:

θjG∗∗k (E,ω) = θj(Gk|(I − (1 + pk−1)V + pk−1V 2))(E,ω)

= (θjGk)|(I − (1 + pk−1)pjV + pk−1+2jV 2)(E,ω)

= θjGk(E,ω)

(1− (1 + pk−1)pj

ξk+2j

+ pk−1+2j 1

ξ2k+4j

)by the proof of Theorem 7.13

= θjGk(E,ω)

(1− pk−1pj

ξk+2j

)(1− pj

ξk+2j

)= θjGk(E,ω)(1− ξk+j−1ξ

−j−1)(1− ξjξ−k−j) as p = ξξ

= θjGk(E,ω)

(1− χ(p)

p

)(1− χ−1(p)

)χk(p) = ξk+jξ

j

Remark 7.15. Note that the factor (1− χ(p)

p

)is not an Euler factor. However,

(1− χ−1(p))

is the Euler factor of L(χ−1, 0) at p.

Theorem 7.16 (Katz). We have that

δjGk(E,ω) = θjGk(E,ω).

More generally, if f ∈Mk and (E,ω) is an ordinary CM point, then

δjf(E,ω) = θjf(E,ω)

Sketch of proof. Suppose j = 1. Recall that

δ = θ − k

4πy,

∂ = 12θ − kP,

θ =1

12∂ +

kP

12,

δ =1

12∂ +

kP

12− k

4πy.

We therefore just need to show that the p-adic modular form P of weight 2 is equal to theC∞ modular form S = P − 3

πyof weight 2 at ordinary CM points.

Then for j = 1, proving the theorem reduces to checking that P (E,ω) = S(E,ω) for ordinaryCM points.

Suppose E/L has good reduction at primes above p, where L/K is a finite extension. Wehave the short exact sequence

112 KARTIK PRASANNA

0 H0(E,Ω1E) H1

dR(E) H1(E,OE) 0

ThenH1

dR(E/L) = L · ω + L · ηwhere ω is a basis of H0(E,Ω1

E) and η is the dual basis of H1(E,OE). Note that (ω, η) is asymplectic basis for H1

dR, i.e. 〈ω, η〉dR = 1.

If E is given by the equation y2 = x3 + ax+ b, then we can write

ω =dx

y, η =

xdx

y.

Note that we have a Hodge decomposition:

H1dR(E/C) = H1,0 ⊕H0,1.

Here ω generates H1,0 and ω generates H0,1. We may hence write

ω = aω + bη, with b 6= 0.

We then define the slope to be ab.

Exercise. S(E,ω) = ab.

For the p-adic side, note that we have a Frobenius action on

H1dR(E/L)⊗ LP,

where the prime P is induced by the fixed embedding Q → Cp.

Since E is ordinary, one eigenvalue of the Froebnius is a p-adic unit and the other is not.Let u be the eigenvector with unit eigenvalue. Then

u = a′ω + b′η

and the slope isa′

b′.

Fact. P (E,ω) = a′

b′. This requires some computations with the Tate curves.

Now, for an ordinary elliptic curve with complex multiplciation i : K → EndL(E), we havethat

H1dR = Lω + Lω′

with i(α)ω = αω and i(α)ω′ = αω′.

Fact. Then

ω ∈Cω′,u ∈LPω

′.

The details of the proof can be found in [Kat76].

We have hence constructed the 2-variable p-adic L-function for χ with infinity type (k+j,−j)and k ≥ 2 and j ≥ 0. Working with higher level modular forms, one can get k ≥ 1.Altogether, this gives the following theorem.


Theorem 7.17. Fix some conductor f (ideal in OK). Consider the extension K(fp∞) of Kand write G for the Galois group. Then there exists a measure dµ on G such that for k ≥ 1and j ≥ 0:

1

Ωk+2jp

∫χdµ =

(1− χ(p)

p

)Lfp(χ

−1, 0)

Ωk+2j

(2π√d

)jgχ,

where Ωp is some p-adic period.

Let us write (`1, `2) = (k+ j,−j). We do not want to restrict to the case k ≥ 2, because thepoint (`1, `2) = (1, 0) is relevant to the Birch–Swinnerton-Dyer conjecture.

Since k ≥ 1 and j ≥ 0, `1 ≥ 1, `2 ≤ 0, and `1 + `2 ≥ 1. Then the functional equation givesa similar interpolation property for points with `1 + `2 ≤ 1.

`1

`2

BSD point

Rubin’s point

interpolation

functional equation

The blue region is the interpolation region. The green region is the region where thefunctional equation gives the interpolation. Rubin’s point refers to Theorem 7.19 below.

Example 7.18 (A grossencharacter of type (1, 0)). Consider K = Q(√−7) and note that

j(OK) ∈ Q. We have that

(OK/(√−7))× ∼= (Z/7Z)×

ε→ ±1with ε(−1) = −1. One can the define a Hecke character ψ by

ψ((α)) = α · ε(α), for α prime to 7.

The corresponding elliptic curve E has conductor 49. Moreover,

L(E, s) = L(ψ, s) = L(ψ, s)

by the theory of complex multiplication. Then

Lp(ψ) = L(E, 1)(∗).

114 KARTIK PRASANNA

Therefore, to study L(E, 1) (which is the goal of the Birch–Swinnerton-Dyer conjecture),the p-adic L-function is useful.

Theorem 7.19 (Rubin [Rub92]). Let E/Q be an elliptic curve with CM by K and letψ : A×K → C× be the associated Hecke character. Suppose the sign of L(E, s) is equal to −1and suppose L(E, s) vanishes to order one at s = 1. (In that case, the theorem of Gross–Zagier and Kolyvagin implies that E(Q) has rank one.) Suppose E(Q)⊗Q = Q ·P (so Q isa Heegner point). Then

Lp(ψ∗).= logp(P )2,

where ψ∗ is ψ composed with complex conjugation,

logp : E(Qp)→ Qp

is the formal group logarithm, and.= means ‘up to explicit constants’.

Remark 7.20. Note that Lp(ψ).= L(E, 1) = 0. This is why we look at ψ∗ instead of ψ.

However, ψ∗ has infinity type (0, 1) which is outside of the range of interpolation.

This should, therefore, be though of as an analog of Leopoldt’s formula 3.17 which evaluatesthe Kubota–Leopoldt p-adic L-function outside of the range of interpolation.

An alternative proof of this theorem without the assumption that K is CM was given in[BDP12], [BDP13]. We will discuss this next.

8. p-adic L-function for Rankin–Selberg L-function

Let f , g be two modular forms on SL2(Z) with weights k, `, respectively.

Assume f, g are cusp forms and eigenfunctions for all the Hecke operators Tp. Write

f =∑

anqn, g =

∑bnq

n

and normalize them so that a1 = 1, b1 = 1. Then

L(s, f) =∑ an

ns=∏p

1

(1− app−s + pk−1−2s)=∏p

1

(1− αpp−s)(1− α′pp−s).

Here, αp and α′p are two roots of Frobp acting on the Galois representation ρf,` associatedto f . For g, we have a similar equation with βp and β′p.

It is well known that L(s, f), L(s, g) admit analytic continuation and functional equation.

We will be interested in the Rankin–Selberg L-functions associated to f and g:

L(s, f × g) =∏p

1

(1− αpβpp−s)(1− αpβ′pp−s)(1− α′pβpp−s)(1− α′pβ′p).

Note that

L(s, f × g) = L(s, ρf ⊗ ρg).

We assume that k > ` and k ≡ 2 (2). The center of the functional equation is s = k+`−12

.The critical points are:


` k + `− 1

2

k − 1

k + `

2

We will be interested in the critical points to the right of the center.

Consider the Dirichlet series:

D(s, f, g) =∑ anbn

ns.


D(s, f, g) =L(s, f × g)

ζ(2s+ 2− k − `).

Proof. We have that

D(s, f, g) =∑

anbnn−s

=∏p

∑n

apnbpn(pn)−s

Note that

apn =αn+1p − (α′p)

n+1

αp − α′p,

bpn =βn+1p − (β′p)

n+1

βp − β′p.

Therefore, the Euler factor at p is:∑n

αn+1p − (α′p)

n+1

αp − α′pβn+1p − (β′p)

n+1

βp − β′p(p−s)n

=

∑n

(αpβp)(αpβpp−s)n − (α′βp)(α

′pβpp

−s)n − (αpβ′p)(αpβ

′pp−s)n + (α′pβ

′pp−s)n(α′pβ

′p)

(αp − α′p)(βp − β′p)

=1

(αp − α′p)(βp − β′p)

[αpβp

1− αpβp−

α′pβp

1− α′pβpp−s−

αpβ′p

1− αpβ′pp−s+

α′pβ′p

1− α′pβ′pp−s

]=

1− αpα′pβpβ′pp−2s

(1− αpβp) . . . (1− α′pβ′p)

Therefore, the Euler factors at p agree for all p.

8.1. Algebraicity of critical values. We follow [Shi76b] to prove algebraicity of criticalvalues of the Rankin–Selberg convolution of two cusp forms f and g.

Definition 8.2. Let fρ(z) = f(−z) =∞∑n=0

ane2πinz.

116 KARTIK PRASANNA

Note that1∫

0

fρ(z)g(z)dx =

∫ 1

0

(∞∑n=1

ane−2πnye−2πinx

)(∞∑n=1

bne2πnye2πinx

)dx

=∞∑n=1

anbne−4πny.

Therefore,

∞∫0

ys−1

1∫0

fρ(z)g(z)dxdy =

∫ ∞0

ys−1

∞∑n=1

anbne−4πnydy = (4π)−sΓ(s)D(s, f, g)

for Re(s) 0. The left hand side is equal to∫S

fρ(z)g(z)ys+1dxdy

y2

where S = z = x+ iy | 0 ≤ 0 ≤ 1, y > 0. For each

γ =

(a bc d

)∈ SL(2,Z).

Note that

fρ(z)g(z)ys+1 γ = (cz + d)k(cz + d)`|cz + d|−2s−2

= (cz + d)`−k|cz + d|2k−2s−2.

To make the integrand invariant under Γ, we need to multiply by something. Note thatthe integrand is already invariant under x 7→ x + 1. Let R be a set of representatives forΓ∞\ SL2(Z), where

Γ∞ =

±(

1 m0 1

) ∣∣∣∣ m ∈ Z.

Then let

Eλ(z, s) =∑r∈R

1

(cz + d)λ|cz + d|2s.

Note that

Γ∞\ SL2(Z) ∼= (c, d) ∈ Z2 | (c, d) = 1/± 1.

Exercise. Check that

Eλ(z, s) = 2ζ(2s+ λ)Eλ(z, s),

where the left hand side is the real analytic Eisenstein series from Section 7.1.

Altogether, we have that

(4π)−sΓ(s)D(s, f, g) =

∫SL2(Z)\h

fρ(z)g(z)ys+1Ek−`(z, s+ 1− k)dxdy

y2.

Recall that the critical points were given by the following picture


` k + `− 1

2

k − 1

k + `

2

and we were interested in studying the points to the right of the center, i.e.

D(k − 1− r, f, g) for 0 ≤ r ≤ k − `2− 1.

According to the above formula:

(4π)−(k−1−r)Γ(k − 1− r)D(k − 1− r, f, g) =

∫SL2(Z)\h

fρgEk−`(z,−r)ys−1dxdy.

Now, similarly as for E, we have that

δrEλ(z) =

(− 1

4πy

)rΓ(λ+ r)

Γ(λ)Eλ+2r(z,−r).

Here, we want to take λ = k − `− 2r, to get

D(k − 1− r, f, g) = (4π)k−1 Γ(k − `− 2r) · (−1)r

Γ(k − `− r)Γ(k − 1− r)

∫SL2(Z)\h

fρgδrEk−`−2r(z)yk

dxdy

y2

Definition 8.3. If Γ ⊆ SL2(Z) and φ, ψ are modular forms of weight k for Γ, then thePetersson inner product is defined as

〈φ, ψ〉 =1

vol(Γ\h)

∫Γ\h

φ · ψ · yk dxdyy2

.

Write

C =4k−1

3

Γ(k − `− 2r)

Γ(k − `− r)Γ(k − 1− r)(−1)r.

Altogether, we conclude that:

D(k − 1− r, f, g) = C · πk · 〈fρ, g · δrEk−`−2r〉.

Remark 8.4. Note that the s = k i.e. the rightmost point (r = k), is the easiest point tostudy. Indeed, in the case we just have a holomorphic Eisenstein series.

We first deal with the case r = 0. We have that

D(k − 1, f, g) = C · πk〈fρ, gEk−`〉.

Here, G = g · Ek−` is a holomotphic modular form of weight k, defined over Q.

118 KARTIK PRASANNA

Theorem 8.5 (Shimura). For 0 ≤ r ≤ k−`2− 1,

π−kD(k − 1− r, f, g)

〈f, f〉is algebraic.

Proof of Theorem 8.5, r = 0. Take an orthonormal basis for Sk(Γ) where the first vector is fρand all other elements in this basis are defined over Q. Then

G = α · fρ +H

where H is orthogonal to fρ. Then

〈fρ, G〉 = α〈fρ, fρ〉 = α〈f, f〉.

However, α ∈ Q, because both f and G are defined over Q.

To prove the general case, we need the following theorem. Recall that N rk (Γ) is the space

of nearly holomorphic modular forms of weight k such that the polynomial in 1y

is degree atmost r.

Theorem 8.6. Let h ∈ N tk(Γ) and suppose k > 2r. Then

h =r∑

ν=0

δνhν , hν ∈Mk−2ν .

Moreover, this decomposition is unique.

Note that the condition k > 2r is necessary: indeed, the form E2 has k = 2, r = 1, but thereare no modular forms of lower level.

Proof. Write

h = g0 + y−1g1 + · · ·+ y−rgr.

Note that

y−1(γ(z)) =|cz + d|2

y=

(cz + d)(cz + d)

y=

(cz + d)(cz + d+ c(z − z))

y=

(cz + d)2

y−2ic(cz+d).

Therefore:

h(z) = h(γ(z))(cz + d)−k

= (cz + d)−k[g0(γ(z)) + y−1

1 g1(γ(z)) + · · ·+ y−1(γ(z))rgr(γ(z))]

so the coefficient of y−r in this expression is

gr(γ(z))(cz + d)2r−k.

Since the expression above is unique, this shows that

gr(γ(z))(cz + d)2r−k = gr(z),

so gr is a modular form of weight k− 2r. As long as k > 2r, there is a constant C such that

h− C · δrgr ∈ N r−1k (Γ),


because

δ =1

2πi

[d

dz+

weight

2iy

].

This completes the proof by induction. Exercise. Check uniqueness.

Recall that we showed that

(4π)−sΓ(s)D(s, f, g) =

∫Γ\h

fρ(z)g(z)ys+1Ek−`(z, s+ 1− k)dxdy

y2

for k > `. In fact, this still holds for k = `, where E0(z, s) is defined by analytic continuation.

Fact 8.7. The Eisenstein series E0(z, s) is holomorphic in s for Re(s) > 1 and has a simplepole at s = 1 with residue 3

πy.

Therefore:

(4π)−kΓ(k) Ress=kD(s, f, g) =

∫Γ\h

fρ(z)g(z)3

πyyk−1dxdy = 〈fρ, g〉.

Note that this is just 0 = 0 when fρ is in a different isotypic component than g. Theinteresting case is hence g = fρ. Rewriting this in that case, we get

〈f, f〉 = (4π)−kΓ(k) Ress=kD(s, f, fρ)

=(4π)−kΓ(k)

ζ(2)Ress=k L(s, f × fρ).

If f had Hecke eigenvalues α and β, then fρ has Hecke eigenvalues α, β. If f has Nebentypus ε,then

α · α = pk−1,

α · β = ε(p)pk−1,

α = β · ε(p),

β = α · ε(p).Computing the pairwise products, we get:

(α2ε(p), αβε(p), αβε(p), β2ε(p)).

By the above, αβε(p) = pk−1. Now, note that∏(1− pk−1

ps

)−1

= ζ(s− k + 1).

Using the formulaL(s, f × fρ) = D(s, f, g)ζ(2s− 2k + 2).

Now, we also get a factorization

L(s, f × fρ) = L(s, adj0f)ζ(s− k + 1)ζ(2s− 2k + 2).

This finally shows that

(4) (4π)kΓ(k)−1〈f, f〉 = Ress=k L(s, f × fρ) = L(twist on sym2f, k) = L(adj0f, 1),

120 KARTIK PRASANNA

because the ζ factor has no pole at s = k.

This is the representation comes up in Wiles’ paper proving modularity.

Proposition 8.8. If f ∈ Sk, g ∈ S`, k = `+ 2r, then

〈f, δrg〉 = 0.

Proof. If g(z) =∞∑n=0

bne2πinz, then

δrg = (2πi)−rr∑

ν=0

cν(2iy)ν−r(d

dz

)νg

=r∑

ν=0

(−4πy)ν−rcν

∞∑n=0

bnnνe2πinz.

As in section 8.1, we can easily show that:∫ ∞0

ys−1

∫ 1

0

fρδrgdxdy = (4π)−sD(s− r, f, g)

r∑ν=0

(−1)ν−rcνΓ(s+ ν − r).

Moreover, the left hand side is∫Γ\h

fρδrgE0(z, s+ 1− k)ys−1dxdy.

The residue of this as s = k is

〈fρ, δrg〉.The residue of the right hand side above is 0, which completes the proof.

Exercise. Use Theorem 8.6 together with Proposition 8.8 to complete the proof of Theo-rem 8.5.

8.2. p-adic interpolation. Let f be a modular form of weight k, K be an imaginary qua-dratic field, and χ be a Hecke character of K of infinity type (`, 0).

Consider the L-function L(s, fK × χ). For example, if f corresponds to an elliptic curveE, then base change E to K and twist the Galois representation by χ. We may it as theRankin–Selberg L-function as

L(s, f × θχ)

where

θχ =∑a

χ(a)e2πi(Na)z ∈ S`+1.

Recall that the period that makes this algebraic depends on which weight is dominant.Therefore, we have the following dichotomy.

• When k > `, the period is π(∗)〈f, f〉.


• When k ≤ `, the period is π〈θχ, θχ〉. Taking the residue at s = `+ 1, we get that

〈θχ, θχ〉 ∼ π(∗) Ress=`+1L(χ, χρ, s)ζK(s− `)

ζQ(2s− 2`)by equation (4)

= π(∗) 1

ζ(2)L(εK , 1)L(χχρ, `+ 1) as ζK(s) = ζQ(s)L(s, εK)

∼ π(∗′)Ω(∗),

where Ω is the CM period.

Suppose f corresponds to an elliptic curve E. We have the following diagram for the p-adicL-function interpolating L(f, χ−1, 0), according to the weight for χ:

`1

`2

BSD point

interpolation

functional equation

anticyclotomic direction

The dashed line `2 = 2− `1 is the line of symmetry, which is also the cyclotomic direction.The other dashed line `2 = `1 is the anticyclotomic direction.

The blue region is the interpolation region. The green region is the region where thefunctional equation gives the interpolation. The BSD point is the point relevant to the

three formulas below. The red region is where evaluating the p-adic L-function should giveinteresting arithmetic information, but we currently do not know anything about.

We have three formulas:

(1) Gross–Zagier [GZ86]: L′(E, 1).= 〈PK , PK〉,

(2) Perrin-Riou [PR87]: L1p(N)

.= L(E, 1) = 0; then (L1

p)′ = 〈PK , PK〉p-adic height, where

the p-adic L-function is in the cyclotomic direction,(3) Bertolini–Darmon–Prasanna [BDP13]: (L2

p)′(N)

.= log2

p(PK), where p-adic L-functionis in the anticyclotomic direction.

122 KARTIK PRASANNA

Appendix A. Project topics

(1) Herbrand–Ribet Theorem. The goal of this project is to work through the proofs ofHerbrand’s theorem (see, for example, [Was97]) and Ribet’s converse to Herbrandtheorem [Rib76b].

(2) Gross’s paper [Gro80] on factorization of p-adic L-functions of imaginary quadraticfields.

(3) Deligne–Ribet’s paper [DR80] on p-adic L-functions of totally real number fields.(4) Gekeler’s paper [Gek88] on Drinfeld modular forms (i.e. the function field case).

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contentsahorawa/math_678.pdf2.1. dirichlet series. since the l-function is de ned as a dirichlet...

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