A trailer for p-adic analysis, second half: Mahler coefficients

In the previous post we defined {p}-adic numbers. This post will state (mostly without proof) some more surprising results about continuous functions {f \colon \mathbb Z_p \rightarrow \mathbb Q_p}. Then we give the famous proof of the Skolem-Mahler-Lech theorem using {p}-adic analysis.

1. Digression on {\mathbb C_p}

Before I go on, I want to mention that {\mathbb Q_p} is not algebraically closed. So, we can take its algebraic closure {\overline{\mathbb Q_p}} — but this field is now no longer complete (in the topological sense). However, we can then take the completion of this space to obtain {\mathbb C_p}. In general, completing an algebraically closed field remains algebraically closed, and so there is a larger space {\mathbb C_p} which is algebraically closed and complete. This space is called the {p}-adic complex numbers.

We won’t need {\mathbb C_p} at all in what follows, so you can forget everything you just read.

2. Mahler coefficients: a description of continuous functions on {\mathbb Z_p}

One of the big surprises of {p}-adic analysis is that we can concretely describe all continuous functions {\mathbb Z_p \rightarrow \mathbb Q_p}. They are given by a basis of functions

\displaystyle  \binom xn \overset{\mathrm{def}}{=} \frac{x(x-1) \dots (x-(n-1))}{n!}

in the following way.

Theorem 1 (Mahler; see Schikhof Theorem 51.1 and Exercise 51.B)

Let {f \colon \mathbb Z_p \rightarrow \mathbb Q_p} be continuous, and define

\displaystyle  a_n = \sum_{k=0}^n \binom nk (-1)^{n-k} f(n).  \ \ \ \ \ (1)

Then {\lim_n a_n = 0} and

\displaystyle  f(x) = \sum_{n \ge 0} a_n \binom xn.

Conversely, if {a_n} is any sequence converging to zero, then {f(x) = \sum_{n \ge 0} a_n \binom xn} defines a continuous function satisfying (1).

The {a_i} are called the Mahler coefficients of {f}.

Exercise 2

Last post we proved that if {f \colon \mathbb Z_p \rightarrow \mathbb Q_p} is continuous and {f(n) = (-1)^n} for every {n \in \mathbb Z_{\ge 0}} then {p = 2}. Re-prove this using Mahler’s theorem, and this time show conversely that a unique such {f} exists when {p=2}.

You’ll note that these are the same finite differences that one uses on polynomials in high school math contests, which is why they are also called “Mahler differences”.

\displaystyle  \begin{aligned} a_0 &= f(0) \\ a_1 &= f(1) - f(0) \\ a_2 &= f(2) - 2f(1) - f(0) \\ a_3 &= f(3) - 3f(2) + 3f(1) - f(0). \end{aligned}

Thus one can think of {a_n \rightarrow 0} as saying that the values of {f(0)}, {f(1)}, \dots behave like a polynomial modulo {p^e} for every {e \ge 0}. Amusingly, this fact was used on a USA TST in 2011:

Exercise 3 (USA TST 2011/3)

Let {p} be a prime. We say that a sequence of integers {\{z_n\}_{n=0}^\infty} is a {p}-pod if for each {e \geq 0}, there is an {N \geq 0} such that whenever {m \geq N}, {p^e} divides the sum

\displaystyle  \sum_{k=0}^m (-1)^k \binom mk z_k.

Prove that if both sequences {\{x_n\}_{n=0}^\infty} and {\{y_n\}_{n=0}^\infty} are {p}-pods, then the sequence {\{x_n y_n\}_{n=0}^\infty} is a {p}-pod.

3. Analytic functions

We say that a function {f \colon \mathbb Z_p \rightarrow \mathbb Q_p} is analytic if it has a power series expansion

\displaystyle  \sum_{n \ge 0} c_n x^n \quad c_n \in \mathbb Q_p \qquad\text{ converging for } x \in \mathbb Z_p.

As before there is a characterization in terms of the Mahler coefficients:

Theorem 4 (Schikhof Theorem 54.4)

The function {f(x) = \sum_{n \ge 0} a_n \binom xn} is analytic if and only if

\displaystyle  \lim_{n \rightarrow \infty} \frac{a_n}{n!} = 0.

Just as holomorphic functions have finitely many zeros, we have the following result on analytic functions on {\mathbb Z_p}.

Theorem 5 (Strassmann’s theorem)

Let {f \colon \mathbb Z_p \rightarrow \mathbb Q_p} be analytic. Then {f} has finitely many zeros.

4. Skolem-Mahler-Lech

We close off with an application of the analyticity results above.

Theorem 6 (Skolem-Mahler-Lech)

Let {(x_i)_{i \ge 0}} be an integral linear recurrence. Then the zero set of {x_i} is eventually periodic.

Proof: According to the theory of linear recurrences, there exists a matrix {A} such that we can write {x_i} as a dot product

\displaystyle  x_i = \left< A^i u, v \right>.

Let {p} be a prime not dividing {\det A}. Let {T} be an integer such that {A^T \equiv \mathbf{1} \pmod p}.

Fix any {0 \le r < N}. We will prove that either all the terms

\displaystyle  f(n) = x_{nT+r} \qquad n = 0, 1, \dots

are zero, or at most finitely many of them are. This will conclude the proof.

Let {A^T = \mathbf{1} + pB} for some integer matrix {B}. We have

\displaystyle  \begin{aligned} f(n) &= \left< A^{nT+r} u, v \right> = \left< (\mathbf1 + pB)^n A^r u, v \right> \\ &= \sum_{k \ge 0} \binom nk \cdot p^n \left< B^n A^r u, v \right> \\ &= \sum_{k \ge 0} a_n \binom nk \qquad \text{ where } a_n = p^n \left< B^n A^r u, v \right> \in p^n \mathbb Z. \end{aligned}

Thus we have written {f} in Mahler form. Initially, we define {f \colon \mathbb Z_{\ge 0} \rightarrow \mathbb Z}, but by Mahler’s theorem (since {\lim_n a_n = 0}) it follows that {f} extends to a function {f \colon \mathbb Z_p \rightarrow \mathbb Q_p}. Also, we can check that {\lim_n \frac{a_n}{n!} = 0} hence {f} is even analytic.

Thus by Strassman’s theorem, {f} is either identically zero, or else it has finitely many zeros, as desired. \Box

A trailer for p-adic analysis, first half: USA TST 2003

I think this post is more than two years late in coming, but anywhow…

This post introduces the {p}-adic integers {\mathbb Z_p}, and the {p}-adic numbers {\mathbb Q_p}. The one-sentence description is that these are “integers/rationals carrying full mod {p^e} information” (and only that information).

The first four sections will cover the founding definitions culminating in a short solution to a USA TST problem.

In this whole post, {p} is always a prime. Much of this is based off of Chapter 3A from Straight from the Book.

1. Motivation

Before really telling you what {\mathbb Z_p} and {\mathbb Q_p} are, let me tell you what you might expect them to do.

In elementary/olympiad number theory, we’re already well-familiar with the following two ideas:

  • Taking modulo a prime {p} or prime {p^e}, and
  • Looking at the exponent {\nu_p}.

Let me expand on the first point. Suppose we have some Diophantine equation. In olympiad contexts, one can take an equation modulo {p} to gain something else to work with. Unfortunately, taking modulo {p} loses some information: (the reduction {\mathbb Z \twoheadrightarrow \mathbb Z/p} is far from injective).

If we want finer control, we could consider instead taking modulo {p^2}, rather than taking modulo {p}. This can also give some new information (cubes modulo {9}, anyone?), but it has the disadvantage that {\mathbb Z/p^2} isn’t a field, so we lose a lot of the nice algebraic properties that we got if we take modulo {p}.

One of the goals of {p}-adic numbers is that we can get around these two issues I described. The {p}-adic numbers we introduce is going to have the following properties:

  1. You can “take modulo {p^e} for all {e} at once”. In olympiad contexts, we are used to picking a particular modulus and then seeing what happens if we take that modulus. But with {p}-adic numbers, we won’t have to make that choice. An equation of {p}-adic numbers carries enough information to take modulo {p^e}.
  2. The numbers {\mathbb Q_p} form a field, the nicest possible algebraic structure: {1/p} makes sense. Contrast this with {\mathbb Z/p^2}, which is not even an integral domain.
  3. It doesn’t lose as much information as taking modulo {p} does: rather than the surjective {\mathbb Z \twoheadrightarrow \mathbb Z/p} we have an injective map {\mathbb Z \hookrightarrow \mathbb Z_p}.
  4. Despite this, you “ignore” some “irrelevant” data. Just like taking modulo {p}, you want to zoom-in on a particular type of algebraic information, and this means necessarily losing sight of other things. (To draw an analogy: the equation { a^2 + b^2 + c^2 + d^2 = -1} has no integer solutions, because, well, squares are nonnegative. But you will find that this equation has solutions modulo any prime {p}, because once you take modulo {p} you stop being able to talk about numbers being nonnegative. The same thing will happen if we work in {p}-adics: the above equation has a solution in {\mathbb Z_p} for every prime {p}.)

So, you can think of {p}-adic numbers as the right tool to use if you only really care about modulo {p^e} information, but normal {\mathbb Z/p^e} isn’t quite powerful enough.

To be more concrete, I’ll give a poster example now:

Example 1 (USA TST 2002/2)

For a prime {p}, show the value of

\displaystyle f_p(x) = \sum_{k=1}^{p-1} \frac{1}{(px+k)^2} \pmod{p^3}

does not depend on {x}.

Here is a problem where we clearly only care about {p^e}-type information. Yet it’s a nontrivial challenge to do the necessary manipulations mod {p^3} (try it!). The basic issue is that there is no good way to deal with the denominators modulo {p^3} (in part {\mathbb Z/p^3} is not even an integral domain).

However, with {p}-adic analysis we’re going to be able to overcome these limitations and give a “straightforward” proof by using the identity

\displaystyle \left( 1 + \frac{px}{k} \right)^{-2} = \sum_{n \ge 0} \binom{-2}{n} \left( \frac{px}{k} \right)^n.

Such an identity makes no sense over {\mathbb Q} or {\mathbb R} for converge reasons, but it will work fine over the {\mathbb Q_p}, which is all we need.

2. Algebraic perspective

We now construct {\mathbb Z_p} and {\mathbb Q_p}. I promised earlier that a {p}-adic integer will let you look at “all residues modulo {p^e}” at once. This definition will formalize this.

2.1. Definition of {\mathbb Z_p}

Definition 2 (Introducing {\mathbb Z_p})

A {p}-adic integer is a sequence

\displaystyle x = (x_1 \bmod p, \; x_2 \bmod{p^2}, \; x_3 \bmod{p^3}, \; \dots)

of residues {x_e} modulo {p^e} for each integer {e}, satisfying the compatibility relations {x_i \equiv x_j \pmod{p^i}} for {i < j}.

The set {\mathbb Z_p} of {p}-adic integers forms a ring under component-wise addition and multiplication.

Example 3 (Some {3}-adic integers)

Let {p=3}. Every usual integer {n} generates a (compatible) sequence of residues modulo {p^e} for each {e}, so we can view each ordinary integer as {p}-adic one:

\displaystyle 50 = \left( 2 \bmod 3, \; 5 \bmod 9, \; 23 \bmod{27}, \; 50 \bmod{81}, \; 50 \bmod{243}, \; \dots \right).

On the other hand, there are sequences of residues which do not correspond to any usual integer despite satisfying compatibility relations, such as

\displaystyle \left( 1 \bmod 3, \; 4 \bmod 9, \; 13 \bmod{27}, \; 40 \bmod{81}, \; \dots \right)

which can be thought of as {x = 1 + p + p^2 + \dots}.

In this way we get an injective map

\displaystyle \mathbb Z \hookrightarrow \mathbb Z_p \qquad n \mapsto \left( n \bmod p, n \bmod{p^2}, n \bmod{p^3}, \dots \right)

which is not surjective. So there are more {p}-adic integers than usual integers.

(Remark for experts: those of you familiar with category theory might recognize that this definition can be written concisely as

\displaystyle \mathbb Z_p \overset{\mathrm{def}}{=} \varprojlim \mathbb Z/p^e \mathbb Z

where the inverse limit is taken across {e \ge 1}.)

Exercise 4

Check that {\mathbb Z_p} is an integral domain.

2.2. Base {p} expansion

Here is another way to think about {p}-adic integers using “base {p}”. As in the example earlier, every usual integer can be written in base {p}, for example

\displaystyle 50 = \overline{1212}_3 = 2 \cdot 3^0 + 1 \cdot 3^1 + 2 \cdot 3^2 + 1 \cdot 3^3.

More generally, given any {x = (x_1, \dots) \in \mathbb Z_p}, we can write down a “base {p}” expansion in the sense that there are exactly {p} choices of {x_k} given {x_{k-1}}. Continuing the example earlier, we would write

\displaystyle \begin{aligned} \left( 1 \bmod 3, \; 4 \bmod 9, \; 13 \bmod{27}, \; 40 \bmod{81}, \; \dots \right) &= 1 + 3 + 3^2 + \dots \\ &= \overline{\dots1111}_3 \end{aligned}

and in general we can write

\displaystyle x = \sum_{k \ge 0} a_k p^k = \overline{\dots a_2 a_1 a_0}_p

where {a_k \in \{0, \dots, p-1\}}, such that the equation holds modulo {p^e} for each {e}. Note the expansion is infinite to the left, which is different from what you’re used to.

(Amusingly, negative integers also have infinite base {p} expansions: {-4 = \overline{\dots222212}_3}, corresponding to {(2 \bmod 3, \; 5 \bmod 9, \; 23 \bmod{27}, \; 77 \bmod{81} \dots)}.)

Thus you may often hear the advertisement that a {p}-adic integer is an “possibly infinite base {p} expansion”. This is correct, but later on we’ll be thinking of {\mathbb Z_p} in a more and more “analytic” way, and so I prefer to think of this as a “Taylor series with base {p}. Indeed, much of your intuition from generating functions {K[[X]]} (where {K} is a field) will carry over to {\mathbb Z_p}.

2.3. Constructing {\mathbb Q_p}

Here is one way in which your intuition from generating functions carries over:

Proposition 5 (Non-multiples of {p} are all invertible)

The number {x \in \mathbb Z_p} is invertible if and only if {x_1 \ne 0}. In symbols,

\displaystyle x \in \mathbb Z_p^\times \iff x \not\equiv 0 \pmod p.

Contrast this with the corresponding statement for {K[ [ X ] ]}: a generating function {F \in K[ [ X ] ]} is invertible iff {F(0) \neq 0}.

Proof: If {x \equiv 0 \pmod p} then {x_1 = 0}, so clearly not invertible. Otherwise, {x_e \not\equiv 0 \pmod p} for all {e}, so we can take an inverse {y_e} modulo {p^e}, with {x_e y_e \equiv 1 \pmod{p^e}}. As the {y_e} are themselves compatible, the element {(y_1, y_2, \dots)} is an inverse. \Box

Example 6 (We have {-\frac{1}{2} = \overline{\dots1111}_3 \in \mathbb Z_3})

We claim the earlier example is actually

\displaystyle \begin{aligned} -\frac{1}{2} = \left( 1 \bmod 3, \; 4 \bmod 9, \; 13 \bmod{27}, \; 40 \bmod{81}, \; \dots \right) &= 1 + 3 + 3^2 + \dots \\ &= \overline{\dots1111}_3. \end{aligned}

Indeed, multiplying it by {-2} gives

\displaystyle \left( -2 \bmod 3, \; -8 \bmod 9, \; -26 \bmod{27}, \; -80 \bmod{81}, \; \dots \right) = 1.

(Compare this with the “geometric series” {1 + 3 + 3^2 + \dots = \frac{1}{1-3}}. We’ll actually be able to formalize this later, but not yet.)

Remark 7 ({\frac{1}{2}} is an integer for {p > 2})

The earlier proposition implies that {\frac{1}{2} \in \mathbb Z_3} (among other things); your intuition about what is an “integer” is different here! In olympiad terms, we already knew {\frac{1}{2} \pmod 3} made sense, which is why calling {\frac{1}{2}} an “integer” in the {3}-adics is correct, even though it doesn’t correspond to any element of {\mathbb Z}.

Fun (but trickier) exercise: rational numbers correspond exactly to eventually periodic base {p} expansions.

With this observation, here is now the definition of {\mathbb Q_p}.

Definition 8 (Introducing {\mathbb Q_p})

Since {\mathbb Z_p} is an integral domain, we let {\mathbb Q_p} denote its field of fractions. These are the {p}-adic numbers.

Continuing our generating functions analogy:

\displaystyle \mathbb Z_p \text{ is to } \mathbb Q_p \quad\text{as}\quad K[[X]] \text{ is to } K((X)).

This means {\mathbb Q_p} is “Laurent series with base {p}”, and in particular according to the earlier proposition we deduce:

Proposition 9 ({\mathbb Q_p} looks like formal Laurent series)

Every nonzero element of {\mathbb Q_p} is uniquely of the form

\displaystyle p^k u \qquad \text{ where } k \in \mathbb Z, \; u \in \mathbb Z_p^\times.

Thus, continuing our base {p} analogy, elements of {\mathbb Q_p} are in bijection with “Laurent series”

\displaystyle \sum_{k \ge -n} a_k p^k = \overline{\dots a_2 a_1 a_0 . a_{-1} a_{-2} \dots a_{-n}}_p

for {a_k \in \left\{ 0, \dots, p-1 \right\}}. So the base {p} representations of elements of {\mathbb Q_p} can be thought of as the same as usual, but extending infinitely far to the left (rather than to the right).

(Fair warning: the field {\mathbb Q_p} has characteristic zero, not {p}.)

Remark 10 (Warning on fraction field)

This result implies that you shouldn’t think about elements of {\mathbb Q_p} as {x/y} (for {x,y \in \mathbb Z_p}) in practice, even though this is the official definition (and what you’d expect from the name {\mathbb Q_p}). The only denominators you need are powers of {p}.

To keep pushing the formal Laurent series analogy, {K((X))} is usually not thought of as quotient of generating functions but rather as “formal series with some negative exponents”. You should apply the same intuition on {\mathbb Q_p}.

(At this point I want to make a remark about the fact {1/p \in \mathbb Q_p}, connecting it to the wish-list of properties I had before. In elementary number theory you can take equations modulo {p}, but if you do the quantity {n/p \bmod{p}} doesn’t make sense unless you know {n \bmod{p^2}}. You can’t fix this by just taking modulo {p^2} since then you need {n \bmod{p^3}} to get {n/p \bmod{p^2}}, ad infinitum. You can work around issues like this, but the nice feature of {\mathbb Z_p} and {\mathbb Q_p} is that you have modulo {p^e} information for “all {e} at once”: the information of {x \in \mathbb Q_p} packages all the modulo {p^e} information simultaneously. So you can divide by {p} with no repercussions.)

3. Analytic perspective

3.1. Definition

Up until now we’ve been thinking about things mostly algebraically, but moving forward it will be helpful to start using the language of analysis. Usually, two real numbers are considered “close” if they are close on the number of line, but for {p}-adic purposes we only care about modulo {p^e} information. So, we’ll instead think of two elements of {\mathbb Z_p} or {\mathbb Q_p} as “close” if they differ by a large multiple of {p^e}.

For this we’ll borrow the familiar {\nu_p} from elementary number theory.

Definition 11 ({p}-adic valuation and absolute value)

We define the {p}-adic valuation {\nu_p : \mathbb Q_p^\times \rightarrow \mathbb Z} in the following two equivalent ways:

  • For {x = (x_1, x_2, \dots) \in \mathbb Z_p} we let {\nu_p(x)} be the largest {e} such that {x_e \equiv 0 \pmod{p^e}} (or {e=0} if {x \in \mathbb Z_p^\times}). Then extend to all of {\mathbb Q_p^\times} by {\nu_p(xy) = \nu_p(x) + \nu_p(y)}.
  • Each {x \in \mathbb Q_p^\times} can be written uniquely as {p^k u} for {u \in \mathbb Z_p^\times}, {k \in \mathbb Z}. We let {\nu_p(x) = k}.

By convention we set {\nu_p(0) = +\infty}. Finally, define the {p}-adic absolute value {\left\lvert \bullet \right\rvert_p} by

\displaystyle \left\lvert x \right\rvert_p = p^{-\nu_p(x)}.

In particular {\left\lvert 0 \right\rvert_p = 0}.

This fulfills the promise that {x} and {y} are close if they look the same modulo {p^e} for large {e}; in that case {\nu_p(x-y)} is large and accordingly {\left\lvert x-y \right\rvert_p} is small.

3.2. Ultrametric space

In this way, {\mathbb Q_p} and {\mathbb Z_p} becomes a metric space with metric given by {\left\lvert x-y \right\rvert_p}.

Exercise 12

Suppose {f \colon \mathbb Z_p \rightarrow \mathbb Q_p} is continuous and {f(n) = (-1)^n} for every {n \in \mathbb Z_{\ge 0}}. Prove that {p = 2}.

In fact, these spaces satisfy a stronger form of the triangle inequality than you are used to from {\mathbb R}.

Proposition 13 ({\left\lvert \bullet \right\rvert_p} is an ultrametric)

For any {x,y \in \mathbb Z_p}, we have the strong triangle inequality

\displaystyle \left\lvert x+y \right\rvert_p \le \max \left\{ \left\lvert x \right\rvert_p, \left\lvert y \right\rvert_p \right\}.

Equality holds if (but not only if) {\left\lvert x \right\rvert_p \neq \left\lvert y \right\rvert_p}.

However, {\mathbb Q_p} is more than just a metric space: it is a field, with its own addition and multiplication. This means we can do analysis just like in {\mathbb R} or {\mathbb C}: basically, any notion such as “continuous function”, “convergent series”, et cetera has a {p}-adic analog. In particular, we can define what it means for an infinite sum to converge:

Definition 14 (Convergence notions)

Here are some examples of {p}-adic analogs of “real-world” notions.

  • A sequence {s_1}, \dots converges to a limit {L} if {\lim_{n \rightarrow \infty} \left\lvert s_n - L \right\rvert_p = 0}.
  • The infinite series {\sum_k x_k} converges if the sequence of partial sums {s_1 = x_1}, {s_2 = x_1 + x_2}, \dots, converges to some limit.
  • \dots et cetera \dots

With this definition in place, the “base {p}” discussion we had earlier is now true in the analytic sense: if {x = \overline{\dots a_2 a_1 a_0}_p \in \mathbb Z_p} then

\displaystyle \sum_{k=0}^\infty a_k p^k \quad\text{converges to } x.

Indeed, the {n}th partial sum is divisible by {p^n}, hence the partial sums approach {x} as {n \rightarrow \infty}.

While the definitions are all the same, there are some changes in properties that should be true. For example, in {\mathbb Q_p} convergence of partial sums is simpler:

Proposition 15 ({|x_k|_p \rightarrow 0} iff convergence of series)

A series {\sum_{k=1}^\infty x_k} in {\mathbb Q_p} converges to some limit if and only if {\lim_{k \rightarrow \infty} |x_k|_p = 0}.

Contrast this with {\sum \frac1n = \infty} in {\mathbb R}. You can think of this as a consequence of strong triangle inequality. Proof: By multiplying by a large enough power of {p}, we may assume {x_k \in \mathbb Z_p}. (This isn’t actually necessary, but makes the notation nicer.)

Observe that {x_k \pmod p} must eventually stabilize, since for large enough {n} we have {\left\lvert x_n \right\rvert_p < 1 \iff \nu_p(x_n) \ge 1}. So let {a_1} be the eventual residue modulo {p} of {\sum_{k=0}^N x_k \pmod p} for large {N}. In the same way let {a_2} be the eventual residue modulo {p^2}, and so on. Then one can check we approach the limit {a = (a_1, a_2, \dots)}. \Box

Here’s a couple exercises to get you used to thinking of {\mathbb Z_p} and {\mathbb Q_p} as metric spaces.

Exercise 16 ({\mathbb Z_p} is compact)

Show that {\mathbb Q_p} is not compact, but {\mathbb Z_p} is. (For the latter, I recommend using sequential continuity.)

Exercise 17 (Totally disconnected)

Show that both {\mathbb Z_p} and {\mathbb Q_p} are totally disconnected: there are no connected sets other than the empty set and singleton sets.

3.3. More fun with geometric series

While we’re at it, let’s finally state the {p}-adic analog of the geometric series formula.

Proposition 18 (Geometric series)

Let {x \in \mathbb Z_p} with {\left\lvert x \right\rvert_p < 1}. Then

\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots.

Proof: Note that the partial sums satisfy {1 + x + x^2 + \dots + x^n = \frac{1-x^n}{1-x}}, and {x^n \rightarrow 0} as {n \rightarrow \infty} since {\left\lvert x \right\rvert_p < 1}. \Box

So, {1 + 3 + 3^2 + \dots = -\frac{1}{2}} is really a correct convergence in {\mathbb Z_3}. And so on.

If you buy the analogy that {\mathbb Z_p} is generating functions with base {p}, then all the olympiad generating functions you might be used to have {p}-adic analogs. For example, you can prove more generally that:

Theorem 19 (Generalized binomial theorem)

If {x \in \mathbb Z_p} and {\left\lvert x \right\rvert_p < 1}, then for any {r \in \mathbb Q} we have the series convergence

\displaystyle \sum_{n \ge 0} \binom rn x^n = (1+x)^r.

(I haven’t defined {(1+x)^r}, but it has the properties you expect.) The proof is as in the real case; even the theorem statement is the same except for the change for the extra subscript of {p}. I won’t elaborate too much on this now, since {p}-adic exponentiation will be described in much more detail in the next post.

3.4. Completeness

Note that the definition of {\left\lvert \bullet \right\rvert_p} could have been given for {\mathbb Q} as well; we didn’t need {\mathbb Q_p} to introduce it (after all, we have {\nu_p} in olympiads already). The big important theorem I must state now is:

Theorem 20 ({\mathbb Q_p} is complete)

The space {\mathbb Q_p} is the completion of {\mathbb Q} with respect to {\left\lvert \bullet \right\rvert_p}.

This is the definition of {\mathbb Q_p} you’ll see more frequently; one then defines {\mathbb Z_p} in terms of {\mathbb Q_p} (rather than vice-versa) according to

\displaystyle \mathbb Z_p = \left\{ x \in \mathbb Q_p : \left\lvert x \right\rvert_p \le 1 \right\}.

(Remark for experts: {\mathbb Q_p} is a field with {\nu_p} a non-Arcihmedian valuation; then {\mathbb Z_p} is its valuation ring.)

Let me justify why this definition is philosophically nice.

Suppose you are a numerical analyst and you want to estimate the value of the sum

\displaystyle S = \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{10000^2}

to within {0.001}. The sum {S} consists entirely of rational numbers, so the problem statement would be fair game for ancient Greece. But it turns out that in order to get a good estimate, it really helps if you know about the real numbers: because then you can construct the infinite series {\sum_{n \ge 1} n^{-2} = \frac16 \pi^2}, and deduce that {S \approx \frac{\pi^2}{6}}, up to some small error term from the terms past {\frac{1}{10001^2}}, which can be bounded.

Of course, in order to have access to enough theory to prove that {S = \pi^2/6}, you need to have the real numbers; it’s impossible to do serious analysis in the non-complete space {\mathbb Q}, where e.g. the sequence {1}, {1.4}, {1.41}, {1.414}, \dots is considered “not convergent” because {\sqrt2 \notin \mathbb Q}. Instead, all analysis is done in the completion of {\mathbb Q}, namely {\mathbb R}.

Now suppose you are an olympiad contestant and want to estimate the sum

\displaystyle f_p(x) = \sum_{k=1}^{p-1} \frac{1}{(px+k)^2}

to within mod {p^3} (i.e. to within {p^{-3}} in {\left\lvert \bullet \right\rvert_p}). Even though {f_p(x)} is a rational number, it still helps to be able to do analysis with infinite sums, and then bound the error term (i.e. take mod {p^3}). But the space {\mathbb Q} is not complete with respect to {\left\lvert \bullet \right\rvert_p} either, and thus it makes sense to work in the completion of {\mathbb Q} with respect to {\left\lvert \bullet \right\rvert_p}. This is exactly {\mathbb Q_p}.

4. Solving USA TST 2002/2

Let’s finally solve Example~1, which asks to compute

\displaystyle f_p(x) = \sum_{k=1}^{p-1} \frac{1}{(px+k)^2} \pmod{p^3}.

Armed with the generalized binomial theorem, this becomes straightforward.

\displaystyle \begin{aligned} f_p(x) &= \sum_{k=1}^{p-1} \frac{1}{(px+k)^2} = \sum_{k=1}^{p-1} \frac{1}{k^2} \left( 1 + \frac{px}{k} \right)^{-2} \\ &= \sum_{k=1}^{p-1} \frac{1}{k^2} \sum_{n \ge 0} \binom{-2}{n} \left( \frac{px}{k} \right)^{n} \\ &= \sum_{n \ge 0} \binom{-2}{n} \sum_{k=1}^{p-1} \frac{1}{k^2} \left( \frac{x}{k} \right)^{n} p^n \\ &\equiv \sum_{k=1}^{p-1} \frac{1}{k^2} - 2x \left( \sum_{k=1}^{p-1} \frac{1}{k^3} \right) p + 3x^2 \left( \sum_{k=1}^{p-1} \frac{1}{k^4} \right) p^2 \pmod{p^3}. \end{aligned}

Using the elementary facts that {p^2 \mid \sum_k k^{-3}} and {p \mid \sum k^{-4}}, this solves the problem.


Vinogradov’s Three-Prime Theorem (with Sammy Luo and Ryan Alweiss)

This was my final paper for 18.099, seminar in discrete analysis, jointly with Sammy Luo and Ryan Alweiss.

We prove that every sufficiently large odd integer can be written as the sum of three primes, conditioned on a strong form of the prime number theorem.

1. Introduction

In this paper, we prove the following result:

Theorem 1 (Vinogradov)

Every sufficiently large odd integer {N} is the sum of three prime numbers.

In fact, the following result is also true, called the “weak Goldbach conjecture”.

Theorem 2 (Weak Goldbach conjecture)

Every odd integer {N \ge 7} is the sum of three prime numbers.

The proof of Vinogradov’s theorem becomes significantly simpler if one assumes the generalized Riemann hypothesis; this allows one to use a strong form of the prime number theorem (Theorem 9). This conditional proof was given by Hardy and Littlewood in the 1923’s. In 1997, Deshouillers, Effinger, te Riele and Zinoviev showed that the generalized Riemann hypothesis in fact also implies the weak Goldbach conjecture by improving the bound to {10^{20}} and then exhausting the remaining cases via a computer search.

As for unconditional proofs, Vinogradov was able to eliminate the dependency on the generalized Riemann hypothesis in 1937, which is why the Theorem 1 bears his name. However, Vinogradov’s bound used the ineffective Siegel-Walfisz theorem; his student K. Borozdin showed that {3^{3^{15}}} is large enough. Over the years the bound was improved, until recently in 2013 when Harald Helfgott claimed the first unconditional proof of Theorem 2, see here.

In this exposition we follow Hardy and Littlewood’s approach, i.e. we prove Theorem 1 assuming the generalized Riemann hypothesis, following the exposition of Rhee. An exposition of the unconditional proof by Vinogradov is given by Rouse.

2. Synopsis

We are going to prove that

\displaystyle  	\sum_{a+b+c = N} \Lambda(a) \Lambda(b) \Lambda(c) \asymp \frac12 N^2 \mathfrak G(N) 	 \ \ \ \ \ (1)


\displaystyle  \mathfrak G(N) 	\overset{\text{def}}{=} \prod_{p \mid N} \left( 1 - \frac{1}{(p-1)^2} \right) 	\prod_{p \nmid N} \left( 1 + \frac{1}{(p-1)^3} \right)

and {\Lambda} is the von Mangoldt function defined as usual. Then so long as {2 \nmid N}, the quantity {\mathfrak G(N)} will be bounded away from zero; thus (1) will imply that in fact there are many ways to write {N} as the sum of three distinct prime numbers.

The sum (1) is estimated using Fourier analysis. Let us define the following.

Definition 3

Let {\mathbb T = \mathbb R/\mathbb Z} denote the circle group, and let {e : \mathbb T \rightarrow \mathbb C} be the exponential function {\theta \mapsto \exp(2\pi i \theta)}. For {\alpha\in\mathbb T}, {\|\alpha\|} denotes the minimal distance from {\alpha} to an integer.

Note that {|e(\theta)-1|=\Theta(\|\theta\|)}.

Definition 4

For {\alpha \in \mathbb T} and {x > 0} we define

\displaystyle  S(x, \alpha) = \sum_{n \le x} \Lambda(n) e(n\alpha).

Then we can rewrite (1) using {S} as a “Fourier coefficient”:

Proposition 5

We have

\displaystyle  		\sum_{a+b+c = N} \Lambda(a) \Lambda(b) \Lambda(c) 		= \int_{\alpha \in \mathbb T} S(N, \alpha)^3 e(-N\alpha) \; d\alpha. 		 	\ \ \ \ \ (2)

Proof: We have

\displaystyle S(N,\alpha)^3=\sum_{a,b,c\leq N}\Lambda(a)\Lambda(b)\Lambda(c)e((a+b+c)\alpha),


\displaystyle  \begin{aligned} \int_{\alpha \in \mathbb T} S(N, \alpha)^3 e(-N\alpha) \; d\alpha &= \int_{\alpha \in \mathbb T} \sum_{a,b,c\leq N}\Lambda(a)\Lambda(b)\Lambda(c)e((a+b+c)\alpha) e(-N\alpha) \; d\alpha \\ &= \sum_{a,b,c\leq N}\Lambda(a)\Lambda(b)\Lambda(c)\int_{\alpha \in \mathbb T}e((a+b+c-N)\alpha) \; d\alpha \\ &= \sum_{a,b,c\leq N}\Lambda(a)\Lambda(b)\Lambda(c)I(a+b+c=N) \\ &= \sum_{a+b+c=N}\Lambda(a)\Lambda(b)\Lambda(c), \end{aligned}

as claimed. \Box

In order to estimate the integral in Proposition 5, we divide {\mathbb T} into the so-called “major” and “minor” arcs. Roughly,

  • The “major arcs” are subintervals of {\mathbb T} centered at a rational number with small denominator.
  • The “minor arcs” are the remaining intervals.

These will be made more precise later. This general method is called the Hardy-Littlewood circle method, because of the integral over the circle group {\mathbb T}.

The rest of the paper is structured as follows. In Section 3, we define the Dirichlet character and other number-theoretic objects, and state some estimates for the partial sums of these objects conditioned on the Riemann hypothesis. These bounds are then used in Section 4 to provide corresponding estimates on {S(x, \alpha)}. In Section 5 we then define the major and minor arcs rigorously and use the previous estimates to given an upper bound for the integral over both areas. Finally, we complete the proof in Section 6.

3. Prime number theorem type bounds

In this section, we collect the necessary number-theoretic results that we will need. It is in this section only that we will require the generalized Riemann hypothesis.

As a reminder, the notation {f(x)\ll g(x)}, where {f} is a complex function and {g} a nonnegative real one, means {f(x)=O(g(x))}, a statement about the magnitude of {f}. Likewise, {f(x)=g(x)+O(h(x))} simply means that for some {C}, {|f(x)-g(x)|\leq C|h(x)|} for all sufficiently large {x}.

3.1. Dirichlet characters

In what follows, {q} denotes a positive integer.

Definition 6

A Dirichlet character modulo {q} {\chi} is a homomorphism {\chi : (\mathbb Z/q)^\times \rightarrow \mathbb C^\times}. It is said to be trivial if {\chi = 1}; we denote this character by {\chi_0}.

By slight abuse of notation, we will also consider {\chi} as a function {\mathbb Z \rightarrow \mathbb C^\ast} by setting {\chi(n) = \chi(n \pmod q)} for {\gcd(n,q) = 1} and {\chi(n) = 0} for {\gcd(n,q) > 1}.

Remark 7

The Dirichlet characters form a multiplicative group of order {\phi(q)} under multiplication, with inverse given by complex conjugation. Note that {\chi(m)} is a primitive {\phi(q)}th root of unity for any {m \in (\mathbb Z/q)^\times}, thus {\chi} takes values in the unit circle.

Moreover, the Dirichlet characters satisfy an orthogonality relation

Experts may recognize that the Dirichlet characters are just the elements of the Pontryagin dual of {(\mathbb Z/q)^\times}. In particular, they satisfy an orthogonality relationship

\displaystyle  	\frac{1}{\phi(q)} 	\sum_{\chi \text{ mod } q} \chi(n) \overline{\chi(a)} 	= \begin{cases} 		1 & n = a \pmod q \\ 		0 & \text{otherwise} 	\end{cases} 	 \ \ \ \ \ (3)

and thus form an orthonormal basis for functions {(\mathbb Z/q)^\times \rightarrow \mathbb C}.

3.2. Prime number theorem for arithmetic progressions

Definition 8

The generalized Chebyshev function is defined by

\displaystyle  \psi(x, \chi) = \sum_{n \le x} \Lambda(n) \chi(n).

The Chebyshev function is studied extensively in analytic number theory, as it is the most convenient way to phrase the major results of analytic number theory. For example, the prime number theorem is equivalent to the assertion that

\displaystyle  \psi(x, \chi_0) = \sum_{n \le x} \Lambda(n) \asymp x

where {q = 1} (thus {\chi_0} is the constant function {1}). Similarly, Dirichlet’s theorem actually asserts that any {q \ge 1},

\displaystyle  	\psi(x, \chi) 	= \begin{cases} 		x + o_q(x) & \chi = \chi_0 \text{ trivial} \\ 		o_q(x) & \chi \neq \chi_0 \text{ nontrivial}. 	\end{cases}

However, the error term in these estimates is quite poor (more than {x^{1-\varepsilon}} for every {\varepsilon}). However, by assuming the Riemann Hypothesis for a certain “{L}-function” attached to {\chi}, we can improve the error terms substantially.

Theorem 9 (Prime number theorem for arithmetic progressions)

Let {\chi} be a Dirichlet character modulo {q}, and assume the Riemann hypothesis for the {L}-function attached to {\chi}.

  1. If {\chi} is nontrivial, then

    \displaystyle  \psi(x, \chi) \ll \sqrt{x} (\log qx)^2.

  2. If {\chi = \chi_0} is trivial, then

    \displaystyle  \psi(x, \chi_0) = x + O\left( \sqrt x (\log x)^2 + \log q \log x \right).

Theorem 9 is the strong estimate that we will require when putting good estimates on {S(x, \alpha)}, and is the only place in which the generalized Riemann Hypothesis is actually required.

3.3. Gauss sums

Definition 10

For {\chi} a Dirichlet character modulo {q}, the Gauss sum {\tau(\chi)} is defined by

\displaystyle \tau(\chi)=\sum_{a=0}^{q-1}\chi(a)e(a/q).

We will need the following fact about Gauss sums.

Lemma 11

Consider Dirichlet characters modulo {q}. Then:

  1. We have {\tau(\chi_0) = \mu(q)}.
  2. For any {\chi} modulo {q}, {\left\lvert \tau(\chi) \right\rvert \le \sqrt q}.

3.4. Dirichlet approximation

We finally require Dirichlet approximation theorem in the following form.

Theorem 12 (Dirichlet approximation)

Let {\alpha \in \mathbb R} be arbitrary, and {M} a fixed integer. Then there exists integers {a} and {q = q(\alpha)}, with {1 \le q \le M} and {\gcd(a,q) = 1}, satisfying

\displaystyle  \left\lvert \alpha - \frac aq \right\rvert \le \frac{1}{qM}.

4. Bounds on {S(x, \alpha)}

In this section, we use our number-theoretic results to bound {S(x,\alpha)}.

First, we provide a bound for {S(x,\alpha)} if {\alpha} is a rational number with “small” denominator {q}.

Lemma 13

Let {\gcd(a,q) = 1}. Assuming Theorem 9, we have

\displaystyle  S(x, a/q) 		= \frac{\mu(q)}{\phi(q)} x + O\left( \sqrt{qx} (\log qx)^2 \right)

where {\mu} denotes the Möbius function.

Proof: Write the sum as

\displaystyle  S(x, a/q) = \sum_{n \le x} \Lambda(n) e(na/q).

First we claim that the terms {\gcd(n,q) > 1} (and {\Lambda(n) \neq 0}) contribute a negligibly small {\ll \log q \log x}. To see this, note that

  • The number {q} has {\ll \log q} distinct prime factors, and
  • If {p^e \mid q}, then {\Lambda(p) + \dots + \Lambda(p^e) 			= e\log p = \log(p^e) < \log x}.

So consider only terms with {\gcd(n,q) = 1}. To bound the sum, notice that

\displaystyle  \begin{aligned} 		e(n \cdot a/q) &= \sum_{b \text{ mod } q} e(b/q) \cdot \mathbf 1(b \equiv an) \\ 		&= \sum_{b \text{ mod } q} e(b/q) \left( \frac{1}{\phi(q)} 			\sum_{\chi \text{ mod } q} \chi(b) \overline{\chi(an)} \right) 	\end{aligned}

by the orthogonality relations. Now we swap the order of summation to obtain a Gauss sum:

\displaystyle  \begin{aligned} 		e(n \cdot a/q) &= \frac{1}{\phi(q)} \sum_{\chi \text{ mod } q} \overline{\chi(an)} 			\left( \sum_{b \text{ mod } q} \chi(b) e(b/q) \right) \\ 		&= \frac{1}{\phi(q)} \sum_{\chi \text{ mod } q} \overline{\chi(an)} \tau(\chi). 	\end{aligned}

Thus, we swap the order of summation to obtain that

\displaystyle  \begin{aligned} 		S(x, \alpha) &= \sum_{\substack{n \le x \\ \gcd(n,q) = 1}} 			\Lambda(n) e(n \cdot a/q) \\ 		&= \frac{1}{\phi(q)} \sum_{\substack{n \le x \\ \gcd(n,q) = 1}} 			\sum_{\chi \text{ mod } q} \Lambda(n) \overline{\chi(an)} \tau(\chi) \\ 		&= \frac{1}{\phi(q)} \sum_{\chi \text{ mod } q} \tau(\chi) 			\sum_{\substack{n \le x \\ \gcd(n,q) = 1}} \Lambda(n) \overline{\chi(an)} \\ 		&= \frac{1}{\phi(q)} \sum_{\chi \text{ mod } q} \overline{\chi(a)} \tau(\chi) 			\sum_{\substack{n \le x \\ \gcd(n,q) = 1}} \Lambda(n)\overline{\chi(n)} \\ 		&= \frac{1}{\phi(q)} \sum_{\chi \text{ mod } q} \overline{\chi(a)} 			\tau(\chi) \psi(x, \overline\chi) \\ 		&= \frac{1}{\phi(q)} \left( \tau(\chi_0) \psi(x, \chi_0) 			+ \sum_{1 \neq \chi \text{ mod } q} \overline{\chi(a)} \tau(\chi) 			\psi(x, \overline\chi) \right). 	\end{aligned}

Now applying both parts of Lemma 11 in conjunction with Theorem 9 gives

\displaystyle  \begin{aligned} 		S(x,\alpha) 		&= \frac{\mu(q)}{\phi(q)} 			\left( x + O\left( \sqrt x (\log qx)^2 \right) \right) 			+ O\left( \sqrt x (\log x)^2 \right) \\ 		&= \frac{\mu(q)}{\phi(q)} x + O\left( \sqrt{qx} (\log qx)^2 \right) 	\end{aligned}

as desired. \Box

We then provide a bound when {\alpha} is “close to” such an {a/q}.

Lemma 14

Let {\gcd(a,q) = 1} and {\beta \in \mathbb T}. Assuming Theorem 9, we have

\displaystyle  		S(x, a/q + \beta) = 		\frac{\mu(q)}{\phi(q)} \left( \sum_{n \le x} e(\beta n) \right) 		+ O\left( (1+\|\beta\|x) \sqrt{qx} (\log qx)^2 \right).

Proof: For convenience let us assume {x \in \mathbb Z}. Let {\alpha = a/q + \beta}. Let us denote {\text{Err}(x, \alpha) 		= S(x,\alpha) - \frac{\mu(q)}{\phi(q)} x}, so by Lemma 13 we have {\text{Err}(x,\alpha) \ll \sqrt{qx}(\log x)^2}. We have

\displaystyle  \begin{aligned} 		S(x, \alpha) &= \sum_{n \le x} \Lambda(n) e(na/q) e(n\beta) \\ 		&= \sum_{n \le x} e(n\beta) \left( S(n, a/q) - S(n-1, a/q) \right) \\ 		&= \sum_{n \le x} e(n\beta) \left( 			\frac{\mu(q)}{\phi(q)} 			+ \text{Err}(n, \alpha) - \text{Err}(n-1, \alpha) \right) \\ 		&= \frac{\mu(q)}{\phi(q)} \left( \sum_{n \le x} e(n\beta) \right) 			+ \sum_{1 \le m \le x-1} \left( e( (m+1)\beta) - e( m\beta ) \right) 			\text{Err}(m, \alpha) \\ 		&\qquad + e(x\beta) \text{Err}(x, \alpha) - e(0) \text{Err}(0, \alpha) \\ 		&\le \frac{\mu(q)}{\phi(q)} \left( \sum_{n \le x} e(n\beta) \right) 			+ \left( \sum_{1 \le m \le x-1} \|\beta\| \text{Err}(m, \alpha) \right) 			+ \text{Err}(0, \alpha) + \text{Err}(x, \alpha) \\ 		&\ll \frac{\mu(q)}{\phi(q)} \left( \sum_{n \le x} e(n\beta) \right) 			+ \left( 1+x\left\| \beta \right\| \right) 			O\left( \sqrt{qx} (\log qx)^2 \right) 	\end{aligned}

as desired. \Box

Thus if {\alpha} is close to a fraction with small denominator, the value of {S(x, \alpha)} is bounded above. We can now combine this with the Dirichlet approximation theorem to obtain the following general result.

Corollary 15

Suppose {M = N^{2/3}} and suppose {\left\lvert \alpha - a/q \right\rvert \le \frac{1}{qM}} for some {\gcd(a,q) = 1} with {q \le M}. Assuming Theorem 9, we have

\displaystyle  S(x, \alpha) \ll \frac{x}{\varphi(q)} + x^{\frac56+\varepsilon}

for any {\varepsilon > 0}.

Proof: Apply Lemma 14 directly. \Box

5. Estimation of the arcs

We’ll write

\displaystyle  f(\alpha) \overset{\text{def}}{=} S(N,\alpha)=\sum_{n \le N} \Lambda(n)e(n\alpha)

for brevity in this section.

Recall that we wish to bound the right-hand side of (2) in Proposition 5. We split {[0,1]} into two sets, which we call the “major arcs” and the “minor arcs.” To do so, we use Dirichlet approximation, as hinted at earlier.

In what follows, fix

\displaystyle  \begin{aligned} 	M &= N^{2/3} \\ 	K &= (\log N)^{10}. \end{aligned}

5.1. Setting up the arcs

Definition 16

For {q \le K} and {\gcd(a,q) = 1}, {1 \le a \le q}, we define

\displaystyle  		\mathfrak M(a,q) = \left\{ \alpha \in \mathbb T 		\mid \left\lvert \alpha - \frac aq \right\rvert \le \frac 1M \right\}.

These will be the major arcs. The union of all major arcs is denoted by {\mathfrak M}. The complement is denoted by {\mathfrak m}.

Equivalently, for any {\alpha}, consider {q = q(\alpha) \le M} as in Theorem 12. Then {\alpha \in \mathfrak M} if {q \le K} and {\alpha \in \mathfrak m} otherwise.

Proposition 17

{\mathfrak M} is composed of finitely many disjoint intervals {\mathfrak M(a,q)} with {q \le K}. The complement {\mathfrak m} is nonempty.

Proof: Note that if {q_1, q_2 \le K} and {a/q_1 \neq b/q_2} then {\left\lvert \frac{a}{q_1} - \frac{b}{q_2} \right\rvert 	\ge \frac{1}{q_1q_2} \gg \frac{3}{qM}}. \Box

In particular both {\mathfrak M} and {\mathfrak m} are measurable. Thus we may split the integral in (2) over {\mathfrak M} and {\mathfrak m}. This integral will have large magnitude on the major arcs, and small magnitude on the minor arcs, so overall the whole interval {[0,1]} it will have large magnitude.

5.2. Estimate of the minor arcs

First, we note the well known fact {\phi(q) \gg q/\log q}. Note also that if {q=q(\alpha)} as in the last section and {\alpha} is on a minor arc, we have {q > (\log N)^{10}}, and thus {\phi(q) \gg (\log N)^{9}}.

As such Corollary 3.3 yields that {f(\alpha) \ll \frac{N}{\phi(q)}+N^{.834} \ll \frac{N}{(\log N)^9}}.


\displaystyle  \begin{aligned} 	\left\lvert \int_{\mathfrak m}f(\alpha)^3e(-N\alpha) \; d\alpha \right\rvert 	&\le \int_{\mathfrak m}\left\lvert f(\alpha)\right\rvert ^3 \; d\alpha \\ 	&\ll \frac{N}{(\log N)^9} \int_{0}^{1}\left\lvert f(\alpha)\right\rvert ^2 \;d\alpha \\ 	&=\frac{N}{(\log N)^9}\int_{0}^{1}f(\alpha)f(-\alpha) \; d\alpha \\ 	&=\frac{N}{(\log N)^9}\sum_{n \le N} \Lambda(n)^2 \\ 	&\ll \frac{N^2}{(\log N)^8}, \end{aligned}

using the well known bound {\sum_{n \le N} \Lambda(n)^2 \ll \frac{N}{\log N}}. This bound of {\frac{N^2}{(\log N)^8}} will be negligible compared to lower bounds for the major arcs in the next section.

5.3. Estimate on the major arcs

We show that

\displaystyle  \int_{\mathfrak M}f(\alpha)^3e(-N\alpha) d\alpha \asymp \frac{N^2}{2} \mathfrak G(N).

By Proposition 17 we can split the integral over each interval and write

\displaystyle  \int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha 	= \sum_{q \le (\log N)^{10}}\sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} 	\int_{-1/qM}^{1/qM}f(a/q+\beta)^3e(-N(a/q+\beta)) \; d\beta.

Then we apply Lemma 14, which gives

\displaystyle  \begin{aligned} 	f(a/q+\beta)^3 	&= \left(\frac{\mu(q)}{\phi(q)}\sum_{n \le N}e(\beta n) \right)^3 \\ 	&+\left(\frac{\mu(q)}{\phi(q)}\sum_{n \le N}e(\beta n)\right)^2 		O\left((1+\|\beta\|N)\sqrt{qN} \log^2 qN\right) \\ 	&+\left(\frac{\mu(q)}{\phi(q)}\sum_{n \le N}e(\beta n)\right) 		O\left((1+\|\beta\|N)\sqrt{qN} \log^2 qN\right)^2 \\ 	&+O\left((1+\|\beta\|N)\sqrt{qN} \log^2 qN\right)^3. \end{aligned}

Now, we can do casework on the side of {N^{-.9}} that {\|\beta\|} lies on.

  • If {\|\beta\| \gg N^{-.9}}, we have {\sum_{n \le N}e(\beta n) \ll \frac{2}{|e(\beta)-1|} 	\ll \frac{1}{\|\beta\|} \ll N^{.9}}, and {(1+\|\beta\|N)\sqrt{qN} \log^2 qN \ll N^{5/6+\varepsilon}}, because certainly we have {\|\beta\|<1/M=N^{-2/3}}.
  • If on the other hand {\|\beta\|\ll N^{-.9}}, we have {\sum_{n \le N}e(\beta n) \ll N} obviously, and {O(1+\|\beta\|N)\sqrt{qN} \log^2 qN) \ll N^{3/5+\varepsilon}}.

As such, we obtain

\displaystyle  f(a/q+\beta)^3 \ll \left( \frac{\mu(q)}{\phi(q)}\sum_{n \le N}e(\beta n) \right)^3 	+ O\left(N^{79/30+\varepsilon}\right)

in either case. Thus, we can write

\displaystyle  \begin{aligned} 	&\qquad \int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha \\ 	&= \sum_{q \le (\log N)^{10}} \sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} 	\int_{-1/qM}^{1/qM} f(a/q+\beta)^3e(-N(a/q+\beta)) \; d\beta \\ 	&= \sum_{q \le (\log N)^{10}} \sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} 		\int_{-1/qM}^{1/qM}\left[\left(\frac{\mu(q)}{\phi(q)}\sum_{n \le N}e(\beta n)\right)^3 		+ O\left(N^{79/30+\varepsilon}\right)\right]e(-N(a/q+\beta)) \; d\beta \\ 	&=\sum_{q \le (\log N)^{10}} \frac{\mu(q)}{\phi(q)^3} S_q 		\left(\sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} e(-N(a/q))\right) 		\left( \int_{-1/qM}^{1/qM}\left(\sum_{n \le N}e(\beta n)\right)^3e(-N\beta) 		\; d\beta \right ) \\ 	&\qquad +O\left(N^{59/30+\varepsilon}\right). \end{aligned}

just using {M \le N^{2/3}}. Now, we use

\displaystyle  \sum_{n \le N}e(\beta n) = \frac{1-e(\beta N)}{1-e(\beta)} 	\ll \frac{1}{\|\beta\|}.

This enables us to bound the expression

\displaystyle  \int_{1/qM}^{1-1/qM}\left (\sum_{n \le N}e(\beta n)\right) ^ 3 e(-N\beta)d\beta 	\ll \int_{1/qM}^{1-1/qM}\|\beta\|^{-3} d\beta = 2\int_{1/qM}^{1/2}\beta^{-3} d\beta 	\ll q^2M^2.

But the integral over the entire interval is

\displaystyle  \begin{aligned} 	\int_{0}^{1}\left(\sum_{n \le N}e(\beta n) \right)^3 e(-N\beta)d\beta 	&= \int_{0}^{1} \sum_{a,b,c \le N} e((a+b+c-N)\beta) \\ 	&\ll \sum_{a,b,c \le N} \mathbf 1(a+b+c=N) \\ 	&= \binom{N-1}{2}. \end{aligned}

Considering the difference of the two integrals gives

\displaystyle  \int_{-1/qM}^{1/qM}\left(\sum_{n \le N}e(\beta n) \right)^3 	e(-N\beta) \; d\beta - \frac{N^2}{2} \ll q^2 M^2 + N 	\ll (\log N)^c N^{4/3},

for some absolute constant {c}.

For brevity, let

\displaystyle  S_q = \sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} e(-N(a/q)).


\displaystyle  \begin{aligned} 	\int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha &= \sum_{q \le (\log N)^{10}} \frac{\mu(q)}{\phi(q)^3}S_q 		\left( \int_{-1/qM}^{1/qM}\left(\sum_{n \le N}e(\beta n)\right)^3e(-N\beta) 		\; d\beta \right ) \\ 	&\qquad +O\left(N^{59/30+\varepsilon}\right) \\ &= \frac{N^2}{2}\sum_{q \le (\log N)^{10}} 	\frac{\mu(q)}{\phi(q)^3}S_q + O((\log N)^{10+c} N^{4/3}) 		+ O(N^{59/30+\varepsilon}) \\ &= \frac{N^2}{2}\sum_{q \le (\log N)^{10}} \frac{\mu(q)}{\phi(q)^3} 		+ O(N^{59/30+\varepsilon}). \end{aligned}


The inner sum is bounded by {\phi(q)}. So,

\displaystyle \left\lvert \sum_{q>(\log N)^{10}} 	\frac{\mu(q)}{\phi(q)^3} S_q \right\rvert 	 \le \sum_{q>(\log N)^{10}} \frac{1}{\phi(q)^2},

which converges since {\phi(q)^2 \gg q^c} for some {c > 1}. So

\displaystyle  \int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha 	= \frac{N^2}{2}\sum_{q = 1}^\infty \frac{\mu(q)}{\phi(q)^3}S_q 	+ O(N^{59/30+\varepsilon}).

Now, since {\mu(q)}, {\phi(q)}, and {\sum_{\substack{1 \le a \le q \\ \gcd(a,q)=1}} e(-N(a/q))} are multiplicative functions of {q}, and {\mu(q)=0} unless {q} is squarefree,

\displaystyle  \begin{aligned} \sum_{q = 1}^\infty \frac{\mu(q)}{\phi(q)^3} S_q 	&= \prod_p \left(1+\frac{\mu(p)}{\phi(p)^3}S_p \right) \\ 	&= \prod_p \left(1-\frac{1}{(p-1)^3} 		\sum_{a=1}^{p-1} e(-N(a/p))\right) \\ 	&= \prod_p \left(1-\frac{1}{(p-1)^3}\sum_{a=1}^{p-1} 		(p\cdot \mathbf 1(p|N) - 1)\right) \\ 	&= \prod_{p|N}\left(1-\frac{1}{(p-1)^2}\right) 		\prod_{p \nmid N}\left(1+\frac{1}{(p-1)^3}\right) \\ 	&= \mathfrak G(N). \end{aligned}


\displaystyle \int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha = \frac{N^2}{2}\mathfrak{G}(N) + O(N^{59/30+\varepsilon}).

When {N} is odd,

\displaystyle  \mathfrak{G}(N) = \prod_{p|N}\left(1-\frac{1}{(p-1)^2}\right)\prod_{p \nmid N}\left(1+\frac{1}{(p-1)^3}\right)\geq \prod_{m\geq 3}\left(\frac{m-2}{m-1}\frac{m}{m-1}\right)=\frac{1}{2},

so that we have

\displaystyle \int_{\mathfrak M} f(\alpha)^3e(-N\alpha) \; d\alpha \asymp \frac{N^2}{2}\mathfrak{G}(N),

as desired.

6. Completing the proof

Because the integral over the minor arc is {o(N^2)}, it follows that

\displaystyle \sum_{a+b+c=N} \Lambda(a)\Lambda(b)\Lambda(c) = \int_{0}^{1} f(\alpha)^3 e(-N\alpha) d \alpha \asymp \frac{N^2}{2}\mathfrak{G}(N) \gg N^2.

Consider the set {S_N} of integers {p^k\leq N} with {k>1}. We must have {p \le N^{\frac{1}{2}}}, and for each such {p} there are at most {O(\log N)} possible values of {k}. As such, {|S_N| \ll\pi(N^{1/2}) \log N\ll N^{1/2}}.


\displaystyle \sum_{\substack{a+b+c=N \\ a\in S_N}} \Lambda(a)\Lambda(b)\Lambda(c) \ll (\log N)^3 |S|N \ll\log(N)^3 N^{3/2},

and similarly for {b\in S_N} and {c\in S_N}. Notice that summing over {a\in S_N} is equivalent to summing over composite {a}, so

\displaystyle  \sum_{p_1+p_2+p_3=N} \Lambda(p_1)\Lambda(p_2)\Lambda(p_3) 	=\sum_{a+b+c=N} \Lambda(a)\Lambda(b)\Lambda(c) + O(\log(N)^3 N^{3/2}) 	\gg N^2,

where the sum is over primes {p_i}. This finishes the proof.

Linnik’s Theorem for Sato-Tate Laws on CM Elliptic Curves

\title{A Variant of Linnik for Elliptic Curves} \maketitle

Here I talk about my first project at the Emory REU. Prerequisites for this post: some familiarity with number fields.

1. Motivation: Arithemtic Progressions

Given a property {P} about primes, there’s two questions we can ask:

  1. How many primes {\le x} are there with this property?
  2. What’s the least prime with this property?

As an example, consider an arithmetic progression {a}, {a+d}, \dots, with {a < d} and {\gcd(a,d) = 1}. The strong form of Dirichlet’s Theorem tells us that basically, the number of primes {\equiv a \pmod d} is {\frac 1d} the total number of primes. Moreover, the celebrated Linnik’s Theorem tells us that the first prime is {O(d^L)} for a fixed {L}, with record {L = 5}.

As I talked about last time on my blog, the key ingredients were:

  • Introducing Dirichlet characters {\chi}, which are periodic functions modulo {q}. One uses this to get the mod {q} into the problem.
  • Introducing an {L}-function {L(s, \chi)} attached to {\chi}.
  • Using complex analysis (Cauchy’s Residue Theorem) to boil the proof down to properties of the zeros of {L(s, \chi)}.

With that said, we now move to the object of interest: elliptic curves.

2. Counting Primes

Let {E} be an elliptic curve over {\mathbb Q}, which for our purposes we can think of concretely as a curve in Weirestrass form

\displaystyle y^2 = x^3 + Ax + B

where the right-hand side has three distinct complex roots (viewed as a polynomial in {x}). If we are unlucky enough that the right-hand side has a double root, then the curve ceases to bear the name “elliptic curve” and instead becomes singular.


Here’s a natural number theoretic question: for any rational prime {p}, how many solutions does {E} have modulo {p}?

To answer this it’s helpful to be able to think over an arbitrary field {F}. While we’ve written our elliptic curve {E} as a curve over {\mathbb Q}, we could just as well regard it as a curve over {\mathbb C}, or as a curve over {\mathbb Q(\sqrt 2)}. Even better, since we’re interested in counting solutions modulo {p}, we can regard this as a curve over {\mathbb F_p}. To make this clear, we will use the notation {E/F} to signal that we are thinking of our elliptic curve over the field {F}. Also, we write {\#E(F)} to denote the number of points of the elliptic curve over {F} (usually when {F} is a finite field). Thus, the question boils down to computing {\#E(\mathbb F_p)}.

Anyways, the question above is given by the famous Hasse bound, and in fact it works over any number field!

Theorem 1 (Hasse Bound)

Let {K} be a number field, and let {E/K} be an elliptic curve. Consider any prime ideal {\mathfrak p \subseteq \mathcal O_K} which is not ramified. Then we have

\displaystyle \#E(\mathbb F_\mathfrak p) = \mathrm{N}\mathfrak p + 1 - a_\mathfrak p

where {\left\lvert a_\mathfrak p \right\rvert \le 2\sqrt{\mathrm{N}\mathfrak p}}.

Here {\mathbb F_\mathfrak p = \mathcal O_K / \mathfrak p} is the field of {\mathrm{N}\mathfrak p} elements. The extra “{+1}” comes from a point at infinity when you complete the elliptic curve in the projective plane.

Here, the ramification means what you might guess. Associated to every elliptic curve over {\mathbb Q} is a conductor {N}, and a prime {p} is ramified if it divides {N}. The finitely many ramified primes are the “bad” primes for which something breaks down when we take modulo {p} (for example, perhaps the curve becomes singular).

In other words, for the {\mathbb Q} case, except for finitely many bad primes {p}, the number of solutions is {p + 1 + O(\sqrt p)}, and we even know the implied {O}-constant to be {2}.

Now, how do we predict the error term?

3. The Sato-Tate Conjecture

For elliptic curves over {\mathbb Q}, we the Sato-Tate conjecture (which recently got upgraded to a theorem) more or less answers the question. But to state it, I have to introduce a new term: an elliptic curve {E/\mathbb Q}, when regarded over {\mathbb C}, can have complex multiplication (abbreviated CM). I’ll define this in just a moment, but for now, the two things to know are

  • CM curves are “special cases”, in the sense that a randomly selected elliptic curve won’t have CM.
  • It’s not easy in general to tell whether a given elliptic curve has CM.

Now I can state the Sato-Tate result. It is most elegantly stated in terms of the following notation: if we define {a_p = p + 1 - \#E(\mathbb F_p)} as above, then there is a unique {\theta_p \in [0,\pi]} which obeys

\displaystyle a_p = 2 \sqrt p \cos \theta_p.

Theorem 2 (Sato-Tate)

Fix an elliptic curve {E/\mathbb Q} which does not have CM (when regarded over {\mathbb C}). Then as {p} varies across unramified primes, the asymptotic probability that {\theta_p \in [\alpha, \beta]} is

\displaystyle \frac{2}{\pi} \int_{[\alpha, \beta]} \sin^2\theta_p.

In other words, {\theta_p} is distributed according to the measure {\sin^2\theta}.

Now, what about the CM case?

4. CM Elliptic Curves

Consider an elliptic curve {E/\mathbb Q} but regard it as a curve over {\mathbb C}. It’s well known that elliptic curves happen to have a group law: given two points on an elliptic curve, you can add them to get a third point. (If you’re not familiar with this, Wikipedia has a nice explanation). So elliptic curves have more structure than just their set of points: they form an abelian group; when written in Weirerstrass form, the identity is the point at infinity.


Letting {A = (A, +)} be the associated abelian group, we can look at the endomorphisms of {E} (that is, homomorphisms {A \rightarrow A}). These form a ring, which we denote {\text{End }(E)}. An example of such an endomorphism is {a \mapsto n \cdot a} for an integer {n} (meaning {a+\dots+a}, {n} times). In this way, we see that {\mathbb Z \subseteq \text{End }(E)}.

Most of the time we in fact have {\text{End }(E) \cong \mathbb Z}. But on occasion, we will find that {\text{End }(E)} is congruent to {\mathcal O_K}, the ring of integers of a number field {K}. This is called complex multiplication by {K}.

Intuitively, this CM is special (despite being rare), because it means that the group structure associated to {E} has a richer set of symmetry. For CM curves over any number field, for example, the Sato-Tate result becomes very clean, and is considerably more straightforward to prove.

Here’s an example. The elliptic curve

\displaystyle E : y^2 = x^3 - 17 x

of conductor {N = 2^6 \cdot 17^2} turns out to have

\displaystyle \text{End }(E) \cong \mathbb Z[i]

i.e. it has complex multiplication has {\mathbb Z[i]}. Throwing out the bad primes {2} and {17}, we compute the first several values of {a_p}, and something bizarre happens. For the {3} mod {4} primes we get

\displaystyle \begin{aligned} a_{3} &= 0 \\ a_{7} &= 0 \\ a_{11} &= 0 \\ a_{19} &= 0 \\ a_{23} &= 0 \\ a_{31} &= 0 \end{aligned}

and for the {1} mod {4} primes we have

\displaystyle \begin{aligned} a_5 &= 4 \\ a_{13} &= 6 \\ a_{29} &= 4 \\ a_{37} &= 12 \\ a_{41} &= -8 \end{aligned}

Astonishingly, the vanishing of {a_p} is controlled by the splitting of {p} in {\mathbb Z[i]}! In fact, this holds more generally. It’s a theorem that for elliptic curves {E/\mathbb Q} with CM, we have {\text{End }(E) \cong \mathcal O_K} where {K} is some quadratic imaginary number field which is also a PID, like {\mathbb Z[i]}. Then {\mathcal O_K} governs how the {a_p} behave:

Theorem 3 (Sato-Tate Over CM)

Let {E/\mathbb Q} be a fixed elliptic curve with CM by {\mathcal O_K}. Let {\mathfrak p} be a unramified prime of {\mathcal O_K}.

  1. If {\mathfrak p} is inert, then {a_\mathfrak p = 0} (i.e. {\theta_\mathfrak p = \frac{1}{2}\pi}).
  2. If {\mathfrak p} is split, then {\theta_\mathfrak p} is uniform across {[0, \pi]}.

I’m told this is much easier to prove than the usual Sato-Tate.

But there’s even more going on in the background. If I look again at {a_p} where {p \equiv 1 \pmod 4}, I might recall that {p} can be written as the sum of squares, and construct the following table:

\displaystyle \begin{array}{rrl} p & a_p & x^2+y^2 \\ 5 & 4 & 2^2 + 1^2 \\ 13 & 6 & 3^2 + 2^2 \\ 29 & 4 & 2^2 + 5^2 \\ 37 & 12 & 6^2 + 1^2 \\ 41 & -8 & 4^2 + 5^2 \\ 53 & 14 & 7^2 + 2^2 \\ 61 & 12 & 6^2 + 5^2 \\ 73 & -16 & 8^2 + 3^2 \\ 89 & -10 & 5^2 + 8^2 \\ \end{array}

Each {a_p} is double one of the terms! There is no mistake: the {a_p} are also tied to the decomposition of {p = x^2+y^2}. And this works for any number field.

What’s happening? The main idea is that looking at a prime ideal {\mathfrak p = (x+yi)}, {a_\mathfrak p} is related to the argument of the complex number {x+yi} in some way. Of course, there are lots of questions unanswered (how to pick the {\pm} sign, and which of {x} and {y} to choose) but there’s a nice way to package all this information, as I’ll explain momentarily.

(Aside: I think the choice of having {x} be the odd or even number depends precisely on whether {p} is a quadratic residue modulo {17}, but I’ll have to check on that.)

5. {L}-Functions

I’ll just briefly explain where all this is coming from, and omit lots of details (in part because I don’t know all of them). Let {E/\mathbb Q} be an elliptic curve with CM by {\mathcal O_K}. We can define an associated {L}-function

\displaystyle L(s, E/K) = \prod_\mathfrak p \left( 1 - \frac{a_\mathfrak p}{(\mathrm{N}\mathfrak p)^{s+\frac{1}{2}}} + \frac{1}{(\mathrm{N}\mathfrak p)^{2s}} \right)

(actually this isn’t quite true actually, some terms change for ramified primes {\mathfrak p}).

At the same time there’s a notion of a Hecke Grössencharakter {\xi} on a number field {K} — a higher dimensional analog of the Dirichlet charaters we used on {\mathbb Z} to filter modulo {q}. For our purposes, think of it as a multiplicative function which takes in ideals of {\mathcal O_K} and returns complex numbers of norm {1}. Like Dirichlet characters, each {\xi} gets a Hecke {L}-function

\displaystyle L(s, \xi) = \prod_\mathfrak p \left( 1 - \frac{\xi(\mathfrak p)}{(\mathrm{N}\mathfrak p)^s} \right)

which again extends to a meromorphic function on the entire complex plane.

Now the great theorem is:

Theorem 4 (Deuring)

Let {E/\mathbb Q} have CM by {\mathcal O_K}. Then

\displaystyle L(s,E/K) = L(s, \xi)L(s, \overline{\xi})

for some Hecke Grössencharakter {\xi}.

Using the definitions given above and equating the Euler products at an unramified {\mathfrak p} gives

\displaystyle 1 - \frac{a_\mathfrak p}{(\mathrm{N}\mathfrak p)^{s+\frac{1}{2}}} + \frac{1}{(\mathrm{N}\mathfrak p)^{2s}} = \left( 1 - \frac{\xi(\mathfrak p)}{(\mathrm{N}\mathfrak p)^s} \right) \left( 1 - \frac{\overline{\xi(\mathfrak p)}}{(\mathrm{N}\mathfrak p)^s} \right)

Upon recalling that {a_\mathfrak p = 2 \sqrt{\mathrm{N}\mathfrak p} \cos \theta_\mathfrak p}, we derive

\displaystyle \xi(\mathfrak p) = \exp(\pm i \theta_\mathfrak p).

This is enough to determine the entire {\xi} since {\xi} is multiplicative.

So this is the result: let {E/\mathbb Q} be an elliptic curve of conductor {N}. Given our quadratic number field {K}, we define a map {\xi} from prime ideals of {\mathcal O_K} to the unit circle in {\mathbb C} by

\displaystyle \mathfrak p \mapsto \begin{cases} \exp(\pm i \theta_\mathfrak p) & \gcd(\mathrm{N}\mathfrak p, N) = 1 \\ 0 & \gcd(\mathrm{N}\mathfrak p, N) > 1. \end{cases}

Thus {\xi} is a Hecke Grössencharakter for some choice of {\pm} at each {\mathfrak p}.

It turns out furthermore that {\xi} has frequency {1}, which roughly means that the argument of {\xi\left( (\pi) \right)} is related to {1} times the argument of {\pi} itself. This fact is what explains the mysterious connection between the {a_p} and the solutions above.

6. Linnik-Type Result

With this in mind, I can now frame the main question: suppose we have an interval {[\alpha, \beta] \subset [0,\pi]}. What’s the first prime {p} such that {\theta_p \in [\alpha, \beta]}? We’d love to have some analog of Linnik’s Theorem here.

This was our project and the REU, and Ashvin, Peter and I proved that

Theorem 5

If a rational {E} has CM then the least prime {p} with {\theta_p \in [\alpha,\beta]} is

\displaystyle \ll \left( \frac{N}{\beta-\alpha} \right)^A.

I might blog later about what else goes into the proof of this. . . but Deuring’s result is one key ingredient, and a proof of an analogous theorem for non-CM curves would have to be very different.

Proof of Dirichlet’s Theorem on Arithmetic Progressions

In this post I will sketch a proof Dirichlet Theorem’s in the following form:

Theorem 1 (Dirichlet’s Theorem on Arithmetic Progression)


\displaystyle \psi(x;q,a) = \sum_{\substack{n \le x \\ n \equiv a \mod q}} \Lambda(n).

Let {N} be a positive constant. Then for some constant {C(N) > 0} depending on {N}, we have for any {q} such that {q \le (\log x)^N} we have

\displaystyle \psi(x;q,a) = \frac{1}{\phi(q)} x + O\left( x\exp\left(-C(N) \sqrt{\log x}\right) \right)

uniformly in {q}.

Prerequisites: complex analysis, previous two posts, possibly also Dirichlet characters. It is probably also advisable to read the last chapter of Hildebrand first, since this contains a much more thorough version of an easier version in which the zeros of {L}-functions are less involved.

Warning: I really don’t understand what I am saying. It is at least 50% likely that this post contains a major error, and 90% likely that there will be multiple minor errors. Please kindly point out any screw-ups of mine; thanks!

Throughout this post: {s = \sigma + it} and {\rho = \beta + i \gamma}, as always. All {O}-estimates have absolute constants unless noted otherwise, and {A \ll B} means {A = O(B)}, {A \asymp B} means {A \ll B \ll A}. By abuse of notation, {\mathcal L} will be short for either {\log q \left( \left\lvert t \right\rvert + 2 \right)} or {\log q \left( \left\lvert T \right\rvert + 2 \right)}, depending on context.

1. Outline

Here are the main steps:

  1. We introduce Dirichlet character {\chi : \mathbb N \rightarrow \mathbb C} which will serves as a roots of unity filter, extracting terms {\equiv a \pmod q}. We will see that this reduces the problem to estimating the function {\psi(x,\chi) = \sum_{n \le x} \chi(n) \Lambda(n)}.
  • Introduce the {L}-function {L(s, \chi)}, the generalization of {\zeta} for arithmetic progressions. Establish a functional equation in terms of {\xi(\chi,s)}, much like with {\zeta}, and use it to extend {L(s,\chi)} to a meromorphic function in the entire complex plane.
  • We will use a variation on the Perron transformation in order to transform this sum into an integral involving an {L}-function {L(\chi,s)}. We truncate this integral to {[c-iT, c+iT]}; this introduces an error {E_{\text{truncate}}} that can be computed immediately, though in this presentation we delay its computation until later.
  • We do a contour as in the proof of the Prime Number Theorem in order to estimate the above integral in terms of the zeros of {L(\chi, s)}. The main term emerges as a residue, so we want to show that the integral {E_{\text{contour}}} along this integral goes to zero. Moreover, we get some residues {\sum_\rho \frac{x^\rho}{\rho}} related to the zeros of the {L}-function.
  • By using Hadamard’s Theorem on {\xi(\chi,s)} which is entire, we can write {\frac{L'}{L}(s,\chi)} in terms of its zeros. This has three consequences:
    1. We can use the previous to get bounds on {\frac{L'}{L}(s, \chi)}.
    2. Using a 3-4-1 trick, this gives us information on the horizontal distribution of {\rho}; the dreaded Siegel zeros appear here.
    3. We can get an expression which lets us estimate the vertical distribution of the zeros in the critical strip (specifically the number of zeros with {\gamma \in [T-1, T+1]}).

    The first and third points let us compute {E_{\text{contour}}}.

  • The horizontal zero-free region gives us an estimate of {\sum_\rho \frac{x^\rho}{\rho}}, which along with {E_{\text{contour}}} and {E_{\text{truncate}}} gives us the value of {\psi(x,\chi)}.
  • We use Siegel’s Theorem to handle the potential Siegel zero that might arise.

Possibly helpful diagram:


The pink dots denote zeros; we think the nontrivial ones all lie on the half-line by the Generalized Riemann Hypothesis but they could actually be anywhere in the green strip.

2. Dirichlet Characters

2.1. Definitions

Recall that a Dirichlet character {\chi} modulo {q} is a completely multiplicative function {\chi : \mathbb N \rightarrow \mathbb C} which is also periodic modulo {q}, and vanishes for all {n} with {\gcd(n,q) > 1}. The trivial character (denoted {\chi_0}) is defined by {\chi_0(n) = 1} when {\gcd(n,q)=1} and {\chi_0(n) = 0} otherwise.

In particular, {\chi(1)=1} and thus each nonzero {\chi} value is a {\phi(q)}-th primitive root of unity; there are also exactly {\phi(q)} Dirichlet characters modulo {q}. Observe that {\chi(-1)^2 = \chi(1) = 1}, so {\chi(-1) = \pm 1}. We shall call {\chi} even if {\chi(1) = +1} and odd otherwise.

If {\tilde q \mid q}, then a character {\tilde\chi} modulo {\tilde q} induces a character {\chi} modulo {q} in a natural way: let {\chi = \tilde\chi} except at the points where {\gcd(n,q)>1} but {\gcd(n,\tilde q)=1}, letting {\chi} be zero at these points instead. (In effect, we are throwing away information about {\tilde\chi}.) A character {\chi} not induced by any smaller character is called primitive.

2.2. Orthogonality

The key fact about Dirichlet characters which will enable us to prove the theorem is the following trick:

Theorem 2 (Orthogonality of Dirichlet Characters)

We have

\displaystyle \sum_{\chi \mod q} \chi(a) \overline{\chi}(b) = \begin{cases} \phi(q) & \text{ if } a \equiv b \pmod q, \gcd(a,q) = 1 \\ 0 & \text{otherwise}. \end{cases}

(Here {\overline{\chi}} is the conjugate of {\chi}, which is essentially a multiplicative inverse.)

This is in some senses a slightly fancier form of the old roots of unity filter. Specifically, it is not too hard to show that {\sum_{\chi} \chi(n)} vanishes for {n \not\equiv 1 \pmod q} while it is equal to {\phi(q)} for {n \equiv 1 \pmod q}.

2.3. Dirichlet {L}-Functions

Now we can define the associated {L}-function by

\displaystyle L(\chi, s) = \sum_{n \ge 1} \chi(n) n^{-s} = \prod_p \left( 1-\chi(p) p^{-s} \right)^{-1}.

The properties of these {L}-functions are that

Theorem 3

Let {\chi} be a Dirichlet character modulo {q}. Then

  1. If {\chi \ne \chi_0}, {L(\chi, s)} can be extended to a holomorphic function on {\sigma > 0}.
  2. If {\chi = \chi_0}, {L(\chi, s)} can be extended to a meromorphic function on {\sigma > 0}, with a single simple pole at {s=1} of residue {\phi(q) / q}.

The proof is pretty much the same as for zeta.

Observe that if {q=1}, then {L(\chi, s) = \zeta(s)}.

2.4. The Functional Equation for Dirichlet {L}-Functions

While I won’t prove it here, one can show the following analog of the functional equation for Dirichlet {L}-functions.

Theorem 4 (The Functional Equation of Dirichlet {L}-Functions)

Assume that {\chi} is a character modulo {q}, possibly trivial or imprimitive. Let {a=0} if {\chi} is even and {a=1} if {\chi} is odd. Let

\displaystyle \xi(s,\chi) = q^{\frac{1}{2}(s+a)} \gamma(s,\chi) L(s,\chi) \left[ s(1-s) \right]^{\delta(x)}


\displaystyle \gamma(s,\chi) = \pi^{-\frac{1}{2}(s+a)} \Gamma\left( \frac{s+a}{2} \right)

and {\delta(\chi) = 1} if {\chi = \chi_0} and zero otherwise. Then

  1. {\xi} is entire.
  2. If {\chi} is primitive, then {\xi(s,\chi) = W(\chi)\xi(1-s, \overline{\chi})} for some complex number {\left\lvert W(\chi) \right\rvert = 1}.

Unlike the {\zeta} case, the {W(\chi)} is nastier to describe; computing it involves some Gauss sums that would be too involved for this post. However, I should point out that it is the Gauss sum here that requires {\chi} to be primitive. As before, {\xi} gives us an meromorphic continuation of {L(\chi, s)} in the entire complex plane. We obtain trivial zeros of {L(\chi, s)} as follows:

  • For {\chi} even, we get zeros at {-2}, {-4}, {-6} and so on.
  • For {\chi \neq \chi_0} even, we get zeros at {0}, {-2}, {-4}, {-6} and so on (since the pole of {\Gamma(\frac{1}{2} s)} at {s=0} is no longer canceled).
  • For {\chi} odd, we get zeros at {-1}, {-3}, {-5} and so on.

3. Obtaining the Contour Integral

3.1. Orthogonality

Using the trick of orthogonality, we may write

\displaystyle \begin{aligned} \psi(x;q,a) &= \sum_{n \le x} \frac{1}{\phi(q)} \sum_{\chi \mod q} \chi(n)\overline{\chi}(a) \Lambda(n) \\ &= \frac{1}{\phi(q)} \sum_{\chi \mod q} \overline{\chi}(a) \left( \sum_{n \le x} \chi(n) \Lambda(n) \right). \end{aligned}

To do this we have to estimate the sum {\sum_{n \le x} \chi(n) \Lambda(n)}.

3.2. Introducing the Logarithmic Derivative of the {L}-Function

First, we realize {\chi(n) \Lambda(n)} as the coefficients of a Dirichlet series. Recall last time we saw that {-\frac{\zeta'}{\zeta}} gave {\Lambda} as coefficients. We can do the same thing with {L}-functions: put

\displaystyle \log L(s, \chi) = -\sum_p \log \left( 1 - \chi(p) p^{-s} \right).

Taking the derivative, we obtain

Theorem 5

For any {\chi} (possibly trivial or imprimitive) we have

\displaystyle -\frac{L'}{L}(s, \chi) = \sum_{n \ge 1} \Lambda(n) \chi(n) n^{-s}.


\displaystyle \begin{aligned} -\frac{L'}{L}(s, \chi) &= \sum_p \frac{\log p}{1-\chi(p) p^{-s}} \\ &= \sum_p \log p \cdot \sum_{m \ge 1} \chi(p^m) (p^m)^{-s} \\ &= \sum_{n \ge 1} \Lambda(n) \chi(n) n^{-s} \end{aligned}

as desired. \Box

3.3. The Truncation Trick

Now, we unveil the trick at the heart of the proof of Perron’s Formula in the last post. I will give a more precise statement this time, by stating where this integral comes from:

Lemma 6 (Truncated Version of Perron Lemma)

For any {c,y,T > 0} define

\displaystyle I(y,T) = \frac{1}{2\pi i} \int_{c-iT}^{c+iT} \frac{y^s}{s} \; ds

Then {I(y,T) = \delta(y) + E(y,T)} where {\delta(y)} is the indicator function defined by

\displaystyle \delta(y) = \begin{cases} 0 & 0 < y < 1 \\ \frac{1}{2} & y=1 \\ 1 & y > 1 \end{cases}

and the error term {E(y,T)} is given by

\displaystyle \left\lvert E(y,T) \right\rvert < \begin{cases} y^c \min \left\{ 1, \frac{1}{T \left\lvert \log y \right\rvert} \right\} & y \neq 1 \\ cT^{-1} & y=1. \end{cases}

In particular, {I(y,\infty) = \delta(y)}.

In effect, the integral from {c-iT} to {c+iT} is intended to mimic an indicator function. We can use it to extract the terms of the Dirichlet series of {-\frac{L'}{L}(s, \chi)} which happen to have {n \le x}, by simply appealing to {\delta(x/n)}. Unfortunately, we cannot take {T = \infty} because later on this would introduce a sum which is not absolutely convergent, meaning we will have to live with the error term introduced by picking a particular finite value of {T}.

3.4. Applying the Truncation

Let’s do so: define

\displaystyle \psi(x;\chi) = \sum_{n \ge 1} \delta\left( x/n \right) \Lambda(n) \chi(n)

which is almost the same as {\sum_{n \le x} \Lambda(n) \chi(n)}, except that if {x} is actually an integer then {\Lambda(x)\chi(x)} should be halved (since {\delta(\frac{1}{2}) = \frac{1}{2}}). Now, we can substitute in our integral representation, and obtain

\displaystyle \begin{aligned} \psi(x;\chi) &= \sum_{n \ge 1} \Lambda(n) \chi(n) \cdot \left( E(x/n,T) + \int_{c-iT}^{c+iT} \frac{(x/n)^s}{s} \; ds \right) \\ &= \sum_{n \ge 1} \Lambda(n) \chi(n) \cdot E(x/n, T) + \int_{c-iT}^{c+iT} \sum_{n \ge 1} \left( \Lambda(n)\chi(n) n^{-s} \right) \frac{x^s}{s} \; ds \\ &= E_{\text{truncate}} + \int_{c-iT}^{c+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds \end{aligned}


\displaystyle E_{\text{truncate}} = \sum_{n \ge 1} \Lambda(n) \chi(n) \cdot E(x/n, T).

Estimating this is quite ugly, so we defer it to later.

4. Applying the Residue Theorem

4.1. Primitive Characters

Exactly like before, we are going to use a contour to estimate the value of

\displaystyle \int_{c-iT}^{c+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds.

Let {U} be a large half-integer (so no zeros of {L(\chi,s)} with {\text{Re } s = U}). We then re-route the integration path along the contour integral

\displaystyle c-iT \rightarrow -U-iT \rightarrow -U+iT \rightarrow c+iT.

During this process we pick up residues, which are the interesting terms.

First, assume that {\chi} is primitive, so the functional equation applies and we get the information we want about zeros.

  • If {\chi = \chi_0}, then so we pick up a residue of {+x} corresponding to

    \displaystyle (-1) \cdot -x^1/1 = +x.

    This is the “main term”. Per laziness, {\delta(\chi) x} it is.


  • Depending on whether {\chi} is odd or even, we detect the trivial zeros, which we can express succinctly by

    \displaystyle \sum_{m \ge 1} \frac{x^{a-2m}}{2m-a}

    Actually, I really ought to truncate this at {U}, but since I’m going to let {U \rightarrow \infty} in a moment I really don’t want to take the time to do so; the difference is negligible.

  • We obtain a residue of {-\frac{L'}{L}(s, \chi)} at {s = 0}, which we denote {b(\chi)}, for {s=0}. Observe that if {\chi} is even, this is the constant term of {-\frac{L'}{L}(s, \chi)} near {s=0} (but there is a pole of the whole function at {s=0}); otherwise it equals the value of {-\frac{L'}{L}(0, \chi)} straight-out.
  • If {\chi \ne \chi_0} is even then {L(s, \chi)} itself has a zero, so we are in worse shape. We recall that

    \displaystyle \frac{L'}{L}(s, \chi) = \frac 1 s + b(\chi) + \dots

    and notice that

    \displaystyle \frac{x^s}{s} = \frac 1s + \log x + \dots

    so we pick up an extra residue of {-\log x}. So, call this a bonus of {-(1-a) \log x}


  • Finally, the hard-to-understand zeros in the strip {0 < \sigma < 1}. If {\rho = \beta+i\gamma} is a zero, then it contributes a residue of {-\frac{x^\rho}{\rho}}. We only pick up the zeros with {\left\lvert \gamma \right\rvert < T} in our rectangle, so we get a term

    \displaystyle -\sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho}.

Letting {U \rightarrow \infty} we derive that

\displaystyle \begin{aligned} &\phantom= \int_{c-iT}^{c+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds \\ &= \delta(\chi) x + E_{\text{contour}} + \sum_{m \ge 1} \frac{x^{a-2m}}{2m-a} - b(\chi) - (1-a) \log x - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho} \end{aligned}

at least for primitive characters. Note that the sum over the zeros is not absolutely convergent without the restriction to {\left\lvert \gamma \right\rvert < T} (with it, the sum becomes a finite one).

4.2. Transition to nonprimitive characters

The next step is to notice that if {\chi} modulo {q} happens to be not primitive, and is induced by {\tilde\chi} with modulus {\tilde q}, then actually {\psi(x,\chi)} and {\psi(x,\tilde\chi)} are not so different. Specifically, they differ by at most

\displaystyle \begin{aligned} \left\lvert \psi(x,\chi)-\psi(x,\tilde\chi) \right\rvert &\le \sum_{\substack{\gcd(n,\tilde q)=1 \\ \gcd(n,q) > 1 \\ n \le x}} \Lambda(n) \\ &\le \sum_{\substack{\gcd(n,q) > 1 \\ n \le x}} \Lambda(n) \\ &\le \sum_{p \mid q} \sum_{\substack{p^k \le x}} \log p \\ &\le \sum_{p \mid q} \log x \\ &\le (\log q)(\log x) \end{aligned}

and so our above formula in fact holds for any character {\chi}, if we are willing to add an error term of {(\log q)(\log x)}. This works even if {\chi} is trivial, and also {\tilde q \le q}, so we will just simplify notation by omitting the tilde’s.

Anyways {(\log q)(\log x)} is piddling compared to all the other error terms in the problem, and we can swallow a lot of the boring residues into a new term, say

\displaystyle E_{\text{tiny}} \le (\log q + 1)(\log x) + 2.

Thus we have

\displaystyle \psi(x, \chi) = \delta(\chi) x + E_{\text{contour}} + E_{\text{truncate}} + E_{\text{tiny}} - b(\chi) - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho}.

Unfortunately, the constant {b(\chi)} depends on {\chi} and cannot be absorbed. We will also estimate {E_{\text{contour}}} in the error term party.

5. Distribution of Zeros

In order to estimate

\displaystyle \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho}

we will need information on both the vertical and horizontal distribution of the zeros. Also, it turns out this will help us compute {E_{\text{contour}}}.

5.1. Applying Hadamard’s Theorem

Let {\chi} be primitive modulo {q}. As we saw,

\displaystyle \xi(s,\chi) = (q/\pi)^{\frac{1}{2} s + \frac{1}{2} a} \Gamma\left( \frac{s+a}{2} \right) L(s, \chi) \left( s(1-s) \right)^{\delta(\chi)}

is entire. It also is easily seen to have order {1}, since no term grows much more than exponentially in {s} (using Stirling to handle the {\Gamma} factor). Thus by Hadamard, we may put

\displaystyle \xi(s, \chi) = e^{A(\chi)+B(\chi)z} \prod_\rho \left( 1-\frac{z}{\rho} \right) e^{\frac{z}{\rho}}.

Taking a logarithmic derivative and cleaning up, we derive the following lemma.

Lemma 7 (Hadamard Expansion of Logarithmic Derivative)

For any primitive character {\chi} (possibly trivial) we have

\displaystyle \begin{aligned} -\frac{L'}{L}(s, \chi) &= \frac{1}{2} \log\frac{q}{\pi} + \frac{1}{2}\frac{\Gamma'(\frac{1}{2} s + \frac{1}{2} a)}{\Gamma(\frac{1}{2} s + \frac{1}{2} a)} \\ &- B(\chi) - \sum_{\rho} \left( \frac{1}{s-\rho} + \frac{1}{\rho} \right) + \delta(\chi) \cdot \left( \frac{1}{s-1} + \frac 1s \right). \end{aligned}

Proof: One one hand, we have

\displaystyle \log \xi(s, \chi) = A(\chi) + B(\chi) s + \sum_\rho \left( \log \left( 1-\frac{s}{\rho} \right) + \frac{s}{\rho} \right).

On the other hand

\displaystyle \log \xi(s, \chi) = \frac{s+a}{2} \cdot \log \frac{q}{\pi} + \log \Gamma\left( \frac{s+a}{2} \right) + \log L(s, \chi) + \delta\chi(\log s + \log (1-s)).

Taking the derivative of both sides and setting them equal: we have on the left side

\displaystyle B(\chi) + \sum_{\rho} \left( \frac{1}{1-\frac{s}{\rho}} \cdot \frac{1}{-\rho} + \frac{1}{\rho} \right) = B(\chi) + \sum_\rho \left( \frac{1}{s-\rho} + \frac{1}{\rho} \right)

and on the right-hand side

\displaystyle \frac{1}{2} \log\frac{q}{\pi} + \frac{1}{2}\frac{\Gamma'}{\Gamma}\left( \frac{s+a}{2} \right) + \frac{L'}{L} (s, \chi) + \delta_\chi \left( \frac 1s + \frac{1}{s-1} \right).

Equating these gives the desired result. \Box

This will be useful in controlling things later. The {B(\chi)} is a constant that turns out to be surprisingly annoying; it is tied to {b(\chi)} from the contour, so we will need to deal with it.

5.2. A Bound on the Logarithmic Derivative

Frequently we will take the real part of this. Using Stirling, the short version of this is:

Lemma 8 (Logarithmic Derivative Bound)

Let {\sigma \ge 1} and {\chi} be primitive (possibly trivial). Then

\displaystyle \text{Re } \left[ -\frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} \right] = \begin{cases} O(\mathcal L) - \text{Re } \sum_\rho \frac{1}{s-\rho} + \text{Re } \frac{\delta(\chi)}{s-1} & 1 \le \sigma \le 2 \\ O(\mathcal L) - \text{Re } \sum_\rho \frac{1}{s-\rho} & 1 \le \sigma \le 2, \left\lvert t \right\rvert \ge 2 \\ O(1) & \sigma \ge 2. \end{cases}

Proof: The claim is obvious for {\sigma \ge 2}, since we can then bound the quantity by {\frac{\zeta'(\sigma)}{\zeta(\sigma)} \le \frac{\zeta'(2)}{\zeta(2)}} due to the fact that the series representation is valid in that range. The second part with {\left\lvert t \right\rvert \ge 2} follows from the first line, by noting that {\text{Re } \frac{1}{s-1} < 1}. So it just suffices to show that

\displaystyle O(\mathcal L) - \text{Re } \sum_\rho \frac{1}{s-\rho} + \text{Re } \frac{\delta(\chi)}{s-1}

where {1 \le \sigma \le 2} and {\chi} is primitive.

First, we claim that {\text{Re } B(\chi) = - \text{Re } \sum \frac{1}{\rho}}. We use the following trick:

\displaystyle B(\chi) = \frac{\xi'(0,\chi)}{\xi(0,\chi)} = -\frac{\xi'(1,\overline{\chi})}{\xi(1,\overline{\chi})} = \overline{B(\chi)} - \sum_{\overline{\rho}} \left( \frac{1}{1-\overline{\rho}} + \frac{1}{\overline{\rho}} \right) + \frac{\delta(\chi)}{s-1}

where the ends come from taking the logarithmic derivative directly. By switching {1-\overline{\rho}} with {\rho}, the claim follows.

Then, the lemma follows rather directly; the {\text{Re } \sum_\rho \frac{1}{\rho}} has miraculously canceled with {\text{Re } B(\chi)}. To be explicit, we now have

\displaystyle - \text{Re } \frac{L'(s, \chi)}{L(s, \chi)} = \frac{1}{2} \log\frac{q}{\pi} + \frac{1}{2} \text{Re } \frac{\Gamma'(\frac{1}{2} s + \frac{1}{2} a)}{\Gamma(\frac{1}{2} s + \frac{1}{2} a)} - \sum_{\rho} \text{Re } \frac{1}{s-\rho} + \frac{\delta(\chi)}{s} + \frac{\delta(\chi)}{s-1}

and the first two terms contribute {\log q} and {\log t}, respectively; meanwhile the term {\frac{\delta(\chi)}{s}} is at most {1}, so it is absorbed. \Box

Short version: our functional equation lets us relate {L(s, \chi)} to {L(1-s, \chi)} for {\sigma \le 0} (in fact it’s all we have!) so this gives the following corresponding estimate:

Lemma 9 (Far-Left Estimate of Log Derivative)

If {\sigma \le -1} and {t \ge 2} we have

\displaystyle \frac{L'(s, \chi)}{L(s, \chi)} = O\left[ \log q\left\lvert s \right\rvert \right].


We have

\displaystyle L(1-s, \chi) = W(\chi) 2^{1-s} \pi^{-s} q^{s-\frac{1}{2}} \cos \frac{1}{2} \pi (s-a) \Gamma(s) L(s, \overline{\chi})

(the unsymmetric functional equation, which can be obtained from Legendre’s duplication formula). Taking a logarithmic derivative yields

\displaystyle \frac{L'}{L}(s, \chi) = \log \frac{q}{2\pi} - \frac{1}{2} \pi \tan \frac{1}{2} \pi(1-s-a) + \frac{\Gamma'}{\Gamma}(1-s) + \frac{L'}{L}(1-s, \overline{\chi}).

Because we assumed {\left\lvert t \right\rvert \ge 2}, the tangent function is bounded as {s} is sufficiently far from any of its poles along the real axis. Also since {\text{Re }(1-s) \ge 2} implies the {\frac{L'}{L}} term is bounded. Finally, the logarithmic derivative of {\Gamma} contributes {\log \left\lvert s \right\rvert} according to Stirling. So, total error is {O(\log q) + O(1) + O(\log \left\lvert s \right\rvert) + O(1)} and this gives the conclusion. \Box

5.3. Horizontal Distribution

I claim that:

Theorem 10 (Horizontal Distribution Bound)

Let {\chi} be a character, possibly trivial or imprimitive. There exists an absolute constant {c_1} with the following properties:

  1. If {\chi} is complex, then no zeros are in the region {\sigma \ge 1 - \frac{c_1}{\mathcal L}}.
  2. If {\chi} is real, there are no zeros in the region {\sigma \ge 1 - \frac{c_1}{\mathcal L}}, with at most one exception; this zero must be real and simple.

Such bad zeros are called Siegel zeros, and I will denote them {\beta_S}. The important part about this estimate is that it does not depend on {\chi} but rather on {q}. We need the relaxation to non-primitive characters, since we will use them in the proof of Landau’s Theorem.

Proof: First, assume {\chi} is both primitive and nontrivial.

By the 3-4-1 lemma on {\log L(\chi, s)} we derive that

\displaystyle 3 \text{Re } \left[ -\frac{L'(\sigma, \chi_0)}{L(\sigma, \chi_0)} \right] + 4 \text{Re } \left[ -\frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} \right] + \text{Re } \left[ -\frac{L'(\sigma+2it, \chi^2)}{L(\sigma+2it, \chi^2)} \right] \ge 0.

This is cool because we already know that

\displaystyle \text{Re } \left[ -\frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} \right] < O(\mathcal L) - \text{Re } \sum_\rho \frac{1}{s-\rho}

We now assume {\sigma > 1}.

In particular, we now have (since {\text{Re } \rho < 1} for any zero {\rho})

\displaystyle \text{Re } \frac{1}{s-\rho} > 0.

So we are free to throw out as many terms as we want.

If {\chi^2} is primitive, then everything is clear. Let {\rho = \beta + i \gamma} be a zero. Then

\displaystyle \begin{aligned} \text{Re } \left[ -\frac{L'(\sigma, \chi_0)}{L(\sigma, \chi_0)} \right] &\le \frac{1}{\sigma-1} + O(1) \\ \text{Re } \left[ -\frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} \right] &\le O(\mathcal L) - \frac{1}{s-\rho} \\ \text{Re } \left[ -\frac{L'(\sigma+2it, \chi^2)}{L(\sigma+2it, \chi^2)} \right] &\le O(\mathcal L) \end{aligned}

where we have dropped all but one term for the second line, and all terms for the third line. If {\chi^2} is not primitive but at least is not {\chi_0}, then we can replace {\chi^2} with the inducing {\tilde\chi_2} for a penalty of at most

\displaystyle \begin{aligned} \text{Re } \frac{L'}{L}(s, \tilde\chi) - \text{Re } \frac{L'}{L}(s, \chi^2) &< \text{Re } \sum_{p^k \mid q} \tilde\chi(p^k) \log p \cdot \text{Re } (p^k)^{-s} \\ &< \sum_{p \mid q} \log p \cdot (p^{-\sigma} + p^{-2\sigma} + \dots) \\ &< \sum_{p \mid q} \log p \cdot 1 \\ &\le \log q \end{aligned}

just like earlier: {\Lambda} is usually zero, so we just look at the differing terms! The Dirichlet series really are practically the same. (Here we have also used the fact that {\sigma > 1}, and {p \ge 2}.)

Consequently, we derive using {3-4-1} that

\displaystyle \frac{3}{\sigma-1} - \frac{4}{s-\rho} + O(\mathcal L) \ge 0.

Selecting {s = \sigma + i \gamma} so that {s - \rho = \sigma-\beta}, we thus obtain

\displaystyle \frac{4}{\sigma-\beta} \le \frac{3}{\sigma-1} + O(\mathcal L).

If we select {\sigma = 1 + \frac{\varepsilon}{\mathcal L}}, we get

\displaystyle \frac{4}{1 + \frac{\varepsilon}{\mathcal L} - \beta} \le O(\mathcal L)


\displaystyle \beta < 1 - \frac{c_2}{\mathcal L}

for some constant {c_2}, initially only for primitive {\chi}.

But because the Euler product of the {L}-function of an imprimitive character versus its primitive inducing character differ by a finite number of zeros on the line {\sigma=0} it follows that this holds for all nontrivial complex characters.

Unfortunately, if we are unlucky enough that {\tilde\chi_2} is trivial, then replacing {\chi^2} causes all hell to break loose. (In particular, {\chi} is real in this case!) The problem comes in that our new penalty has an extra {\frac{1}{s-1}}, so

\displaystyle \left\lvert \text{Re } \frac{L'}{L}(s, \chi^2) - \text{Re } \frac{\zeta'}{\zeta}(s) \right\rvert < \frac{1}{s-1} + \log q

Applied with {s = \sigma + 2it}, we get the weaker

\displaystyle \frac{3}{\sigma-1} - \frac{4}{s-\rho} + O(\mathcal L) + \frac{1}{\sigma - 1 + 2it} \ge 0.

If {\left\lvert t \right\rvert > \frac{\delta}{\log q}} for some {\delta} then the {\frac{1}{\sigma-1+2it}} term will be at most {\frac{\log q}{\delta} = O(\mathcal L)} and we live to see another day. In other words, we have unconditionally established a zero-free region of the form

\displaystyle \sigma > 1 - \frac{c(\delta)}{\mathcal L} \quad\text{and}\quad \left\lvert t \right\rvert > \frac{\delta}{\log q}

for any {\delta > 0}.

Now let’s examine {\left\lvert t \right\rvert < \frac{\delta}{\log q}}. We don’t have the facilities to prove that there are no bad zeros, but let’s at least prove that the zero must be simple and real. By Hadamard at {t=0}, we have

\displaystyle -\frac{L'(\sigma, \chi)}{L(\sigma, \chi)} < O(\mathcal L) - \sum_\rho \frac{1}{\sigma-\rho}

where we no longer need the real parts since {\chi} is real, and in particular the roots of {L(s,\chi)} come in conjugate pairs. The left-hand side can be stupidly bounded below by

\displaystyle -\frac{L'(\sigma, \chi)}{L(\sigma, \chi)} \ge - \sum_{n \ge 1} (-1) \cdot \log n \cdot n^{-\sigma} = \frac{\zeta'(\sigma)}{\zeta(\sigma)} > -\frac{1}{\sigma-1} - O(1).


\displaystyle -\frac{1}{\sigma-1} < O(\mathcal L) - \sum_\rho \frac{1}{\sigma-\rho}.

In other words,

\displaystyle \sum_\rho \text{Re } \frac{\sigma-\rho}{\left\lvert \sigma-\rho \right\rvert^2} < \frac{1}{\sigma-1} + O(\mathcal L).

Then, let {\sigma = 1 + \frac{2\delta}{\log q}}, so

\displaystyle \sum_\rho \text{Re } \frac{\sigma-\rho}{\left\lvert \sigma-\rho \right\rvert^2} < \frac{\log q}{2\delta} + O(\mathcal L).

The rest is arithmetic; basically one finds that there can be at most one Siegel zero. In particular, since complex zeros come in conjugate pairs, that zero must be real.

It remains to handle the case that {\chi = \chi_0} is the constant function giving {1}. For this, we observe that the {L}-function in question is just {\zeta}. Thus, we can decrease the constant {c_2} to some {c_1} in such a way that the result holds true for {\zeta}, which completes the proof. \Box

5.4. Vertical Distribution

We have the following lemma:

Lemma 11 (Sum of Zeros Lemma)

For all real {t} and primitive characters {\chi} (possibly trivial), we have

\displaystyle \sum_\rho \frac{1}{4+(t-\gamma)^2} = O(\mathcal L).

Proof: We already have that

\displaystyle \text{Re } -\frac{L'}{L}(s, \chi) = O(\mathcal L) - \sum_\rho \text{Re } \frac{1}{s-\rho}

and we take {s = 2 + it}, noting that the left-hand side is bounded by a constant {\frac{\zeta'}{\zeta}(2) = -0.569961}. On the other hand, { \text{Re } \frac{1}{2+it-\rho} = \frac{\text{Re }(2+it-\rho)}{\left\lvert (2-\beta) + (t-\gamma)i \right\rvert^2} = \frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2}} and

\displaystyle \frac{1}{4+(t-\gamma)^2} \le \frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2} \le \frac{2}{1+(t-\gamma)^2}

as needed. \Box
From this we may deduce that

Lemma 12 (Number of Zeros Nearby {T})

For all real {t} and primitive characters {\chi} (possibly trivial), the number of zeros {\rho} with {\gamma \in [t-1, t+1]} is {O(\mathcal L)}.

In particular, we may perturb any given {T} by {\le 2} so that the distance between it and the nearest zero is at least {c_0 \mathcal L^{-1}}, for some absolute constant {c_0}.

From this, using an argument principle we can actually also obtain the following: For a real number {T > 0}, we have { N(T, \chi) = \frac{T}{\pi} \log \frac{qT}{2\pi e} + O(\mathcal L)} is the number of zeros of {L(s, \chi)} with imaginary part {\gamma \in [-T, T]}. However, we will not need this fact.

6. Error Term Party

Up to now, {c} has been arbitrary. Assume now {x \ge 6}; thus we can now follow the tradition

\displaystyle c = 1 + \frac{1}{\log x} < 2

so {c} is just to the right of the critical line. This causes {x^c = ex}. We assume also for convenience that {T \ge 2}.

6.1. Estimating the Truncation Error

Recall that

\displaystyle \left\lvert E(y,T) \right\rvert < \begin{cases} y^c \min \left\{ 1, \frac{1}{T \left\lvert \log y \right\rvert} \right\} & y \neq 1 \\ cT^{-1} & y=1. \end{cases}

We need to bound the right-hand side of

\displaystyle \left\lvert E_{\text{truncate}} \right\rvert \le \sum_{n \ge 1} \left\lvert \Lambda(n) \chi(n) \cdot E(x/n, T) \right\rvert = \sum_{n \ge 1} \Lambda(n) \left\lvert E(x/n, T) \right\rvert.

If {\frac34 x \le n \le \frac 54x}, the log part is small, and this is bad. We have to split into three cases: {\frac34 x \le n < x}, {n = x}, and {x < n \le \frac 54x}. This is necessary because in the event that {\Lambda(x) \neq 0} ({x} is a prime power), then {E(x/n,T) = E(1,T)} needs to be handled differently.

We let {x_{\text{left}}} and {x_{\text{right}}} be the nearest prime powers to {x} other than {x} itself. Thus this breaks our region to conquer into

\displaystyle \frac 34 x \le x_{\text{left}} < x < x_{\text{right}} \le \frac 54 x.

So we have possibly a center term (if {x} is a prime power, we have a term {n=x}), plus the far left interval and the far right interval. Let {d = \min\left\{ x-x_{\text{left}}, x_{\text{right}}-x \right\}} for convenience.

  • In the easy case, if {n = x} we have a contribution of {E(1,T) \log x < \frac{c}{T}\log x}, which is piddling (less than {\log x}).
  • Suppose {\frac 34x \le n \le x_{\text{left}} - 1}. If {n = x_{\text{left}} - a} for some integer {1 \le a \le \frac 14x}, then

    \displaystyle \log \frac xn \ge \log \frac{x_{\text{left}}}{x_{\text{left}}-a} = -\log\left( 1 - \frac{a}{x_{\text{left}}} \right) \ge \frac{a}{x_{\text{left}}}

    by using the silly inequality {-\log(1-t) \ge t} for {t < 1}. So the contribution in total is at most

    \displaystyle \begin{aligned} \sum_{1 \le a \le \frac 14 x} \Lambda(n) \cdot (x/n)^c \cdot \frac{1}{T \cdot \frac{a}{x_{\text{left}}}} &\le \frac{x_{\text{left}}}{T} \sum_{1 \le a \le \frac 14 x} \Lambda(n) \cdot \left( \frac 43 \right)^2 \frac 1a \\ &\le \frac{16}{9} \frac{x_{\text{left}}}{T} \log x \sum_{1 \le a \le \frac 14 x} \frac 1a \\ &\le \frac{16}{9} \frac{(x-1) (\log x)(\log \frac 14 x + 2)}{T} \\ &\le \frac{1.9x (\log x)^2}{T} \end{aligned}

    provided {x \ge 7391}.


  • If {n = x_{\text{left}}}, we have

    \displaystyle \log \frac xn = -\log\left( 1 - \frac{x-x_{\text{left}}}{x} \right) > \frac{d}{x}

    Hence in this case, we get an error at most

    \displaystyle \begin{aligned} \Lambda(x_{\text{left}}) \left( \frac{x}{x_{\text{left}}} \right)^c \min \left\{ 1, \frac{x}{Td} \right\} &< \Lambda(x_{\text{left}}) \left( \frac 43 \right)^2 \min \left\{ 1, \frac{x}{Td} \right\} \\ &\le \frac{16}{9} \log x \min \left\{ 1, \frac{x}{Td} \right\}. \end{aligned}


  • The cases {n = x_{\text{right}}} and {x_{\text{right}} + 1 \le n < \frac 54x} give the same bounds as above, in the same way.

Finally, if for {x} outside the interval mentioned above, we in fact have {\left\lvert \log x/n \right\rvert > \frac{1}{5}}, say, and so all terms contribute at most

\displaystyle \begin{aligned} \sum_n \Lambda(n) \cdot (x/n)^c \cdot \frac{1}{T \log \left\lvert x/n \right\rvert} &\le \frac{5x^c}{T} \sum_n \Lambda(n) \cdot n^{-c} \\ &= \frac{5ex}{T} \cdot \left\lvert -\frac{\zeta'}{\zeta} (c) \right\rvert \\ &< \frac{5ex}{T} \cdot \left( \frac{1}{c-1} + 0.5773 \right) \\ &\le \frac{14x \log x}{T}. \end{aligned}

(Recall {\zeta'/\zeta} had a simple pole at {s=1}, so near {s=1} it behaves like {\frac{1}{s-1}}.)

The sum of everything is {\le \frac{3.8x(\log x)^2+14x\log x}{T} + \frac{32}{9} \log x \min \left\{ 1, \frac{x}{Td} \right\}}. Hence, the grand total across all these terms is the horrible

\displaystyle \boxed{ E_{\text{truncate}} \le \frac{5x(\log x)^2}{T} + 3.6\log x \min \left\{ 1, \frac{x}{Td} \right\}}

provided {x \ge 1.2 \cdot 10^5}.

6.2. Estimating the Contour Error

We now need to measure the error along the contour, taken from {U \rightarrow \infty}. Throughout assume {U \ge 3}. Naturally, to estimate the integral, we seek good estimates on

\displaystyle \left\lvert \frac{L'}{L}(\sigma) \right\rvert.

For this we appeal to the Hadamard expansion. We break into a couple cases.

  • First, let’s look at the integral when {-1 \le \sigma \le 2}, so {s = \sigma \pm iT} with {T} large. We bound the horizontal integral along these regions; by symmetry let’s consider just the top

    \displaystyle \int_{-1+iT}^{c+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds.

    Thus we want an estimate of {-\frac{L'}{L}}.

    Lemma 13

    Let {s} be such that {-1 \le \sigma \le 2}, {\left\lvert t \right\rvert \ge 2}. Assume {\chi} is primitive (possibly trivial), and that {t} is not within {c_0\mathcal L^{-1}} of any zeros of {L(s, \chi)}. Then

    \displaystyle \frac{L'(s, \chi)}{L(s, \chi)} = O(\mathcal L^2)

    Proof: Since we assumed that {T \ge 2} we need not worry about {\frac{\delta(\chi)}{s-1}} and so we obtain

    \displaystyle \frac{L'(s, \chi)}{L(s, \chi)} = -\frac{1}{2} \log\frac{q}{\pi} - \frac{1}{2}\frac{\Gamma'(\frac{1}{2} s + \frac{1}{2} a)}{\Gamma(\frac{1}{2} s + \frac{1}{2} a)} + B(\chi) + \sum_{\rho} \left( \frac{1}{s-\rho} + \frac{1}{\rho} \right).

    and we eliminate {B(\chi)} by computing

    \displaystyle \frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} - \frac{L'(2+it, \chi)}{L(2+it, \chi)} = E_{\text{gamma}} + \sum_{\rho} \left( \frac{1}{\sigma+it-\rho} - \frac{1}{2+it-\rho} \right).


    \displaystyle E_{\text{gamma}} = \frac{1}{2}\frac{\Gamma'(\frac{1}{2} (2+it) + \frac{1}{2} a)}{\Gamma(\frac{1}{2} (2+it) + \frac{1}{2} a)} - \frac{1}{2}\frac{\Gamma'(\frac{1}{2} (\sigma+it) + \frac{1}{2} a)}{\Gamma(\frac{1}{2} (\sigma+it) + \frac{1}{2} a)} \ll \log T

    by Stirling (here we use the fact that {-1 \le \sigma \le 2}). For the terms where {\gamma \notin [t-1, t+1]} we see that

    \displaystyle \begin{aligned} \left\lvert \frac{1}{\sigma+it-\rho} - \frac{1}{2+it-\rho} \right\rvert &= \frac{2-\sigma}{\left\lvert \sigma+it-\rho \right\rvert \left\lvert 2+it-\rho \right\rvert} \\ &\le \frac{2-\sigma}{\left\lvert \gamma-t \right\rvert^2} \le \frac{3}{\left\lvert \gamma-t \right\rvert^2} \\ &\le \frac{6}{\left\lvert \gamma-t \right\rvert^2+1}. \end{aligned}

    So the contribution of the sum for {\left\lvert \gamma-t \right\rvert \ge 1} can be bounded by {O(\mathcal L)}, via the vertical sum lemma.

    As for the zeros with smaller imaginary part, we at least have {\left\lvert 2+it-\rho \right\rvert = \left\lvert 2-\beta \right\rvert > 1} and thus we can reduce the sum to just

    \displaystyle \frac{L'(\sigma+it, \chi)}{L(\sigma+it, \chi)} - \frac{L'(2+it, \chi)}{L(2+it, \chi)} = \sum_{\gamma\in[t-1,t+1]} \frac{1}{\sigma+it-\rho} + O(\mathcal L).

    Now by the assumption that {\left\lvert \gamma-t \right\rvert \ge c\mathcal L^{-1}}; so the terms of the sum are all at most {O(\mathcal L)}. Also, there are {O(\mathcal L)} zeros with imaginary part in that range. Finally, we recall that {\frac{L'(2+it, \chi)}{L(2+it, \chi)}} is bounded; we can write it using its (convergent) Dirichlet series and then note it is at most {\frac{\zeta'(2+it)}{\zeta(2+it)} \le \frac{\zeta'(2)}{\zeta(2)}}. \Box
    At this point, we perturb {T} as described in vertical distribution so that the lemma applies, and use can then compute

    \displaystyle \begin{aligned} \left\lvert \int_{-1+iT}^{c+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds \right\rvert &< O(\mathcal L^2) \cdot \int_{-1+iT}^{c+iT} \left\lvert \frac{x^s}{s} \right\rvert \; ds \\ &< O(\mathcal L^2) \int_{-1}^c \frac{x^\sigma}{2T} \; d\sigma \\ &< O(\mathcal L^2) \cdot \frac{x^{c+1}-1}{T \log x} \\ &< O\left(\frac{\mathcal L^2 x}{T \log x}\right). \end{aligned}


  • Next, for the integral {-U \le \sigma \le 1}, we use the “far-left” estimate to obtain

    \displaystyle \begin{aligned} \left\lvert \int_{-U+iT}^{-1+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds \right\rvert &\ll \int_{-\infty+iT}^{-1+iT} \left\lvert \frac{x^s}{s} \right\rvert \cdot \log q \left\lvert s \right\rvert \; ds \\ &\ll \int_{-\infty+iT}^{-1+iT} \left\lvert \frac{x^s}{s} \right\rvert \cdot \log q \left\lvert s \right\rvert \; ds \\ &\ll \log q \int_{-\infty+iT}^{-1+iT} \left\lvert \frac{x^s}{s} \right\rvert \; ds + \int_{-\infty+iT}^{-1+iT} \left\lvert \frac{x^s \log \left\lvert s \right\rvert}{s} \right\rvert \; ds \\ &< \log q \int_{-\infty+iT}^{-1+iT} \left\lvert \frac{x^s}{T} \right\rvert \; ds + \int_{-\infty}^{-1} \left\lvert \frac{x^s \log T}{T} \right\rvert \; ds \\ &\ll \frac{\log q}{T} \int_{-\infty}^{-1} x^\sigma \; d\sigma + \frac{\log T}{T} \int_{-\infty}^{-1} x^\sigma \; d\sigma \\ &< \frac{\mathcal L}{T} \left( \frac{x^{-1}}{\log x} \right) = \frac{\mathcal L}{T x \log x}. \end{aligned}

    So the contribution in this case is {O\left( \frac{\mathcal L}{T x \log x} \right)}.


  • Along the horizontal integral, we can use the same bound

    \displaystyle \begin{aligned} \left\lvert \int_{-U-iT}^{-U+iT} -\frac{L'}{L}(s, \chi) \frac{x^s}{s} \; ds \right\rvert &\ll \int_{-U-iT}^{-U+iT} \left\lvert \frac{x^s}{s} \right\rvert \cdot \log q \left\lvert s \right\rvert \; ds \\ &= x^{-U} \cdot \int_{-U-iT}^{-U+iT} \frac{\log q \left\lvert s \right\rvert}{\left\lvert s \right\rvert} \; ds \\ &= x^{-U} \cdot \int_{-U-iT}^{-U+iT} \frac{\log q + \log U}{U} \; ds \\ &= \frac{2T(\log q + \log U)}{Ux^U} \end{aligned}

    which vanishes as {U \rightarrow \infty}.

So we only have two error terms, {O\left( \frac{\mathcal L^2 x}{T \log x} \right)} and {O\left( \frac{\mathcal L}{Tx\log x} \right)}. The first is clearly larger, so we end with

\displaystyle \boxed{E_{\text{contour}} \ll \frac{\mathcal L^2x}{T \log x}}.

6.3. The term {b(\chi)}

We can estimate {b(\chi)} as follows:

Lemma 14

For primitive {\chi}. we have

\displaystyle b(\chi) = O(\log q) - \sum_{\left\lvert \gamma \right\rvert < 1} \frac{1}{\rho}

Proof: The idea is to look at {\frac{L'}{L}(s,\chi)-\frac{L'}{L}(2,\chi)}. By subtraction, we obtain

\displaystyle \begin{aligned} \frac{L'}{L}(s, \chi) -\frac{L'}{L}(2, \chi) &= - \frac{\Gamma'}{\Gamma} \left( \frac{s+a}{2} \right) + \frac{\Gamma'}{\Gamma} \left( \frac{2+a}{2} \right) \\ &- \frac rs - \frac r{s-1} + \frac r2 + \frac r1 \\ &+ \sum_\rho \left( \frac{1}{s-\rho} - \frac{1}{2-\rho} \right) \end{aligned}

Then at {s=0} (eliminating the poles), we have

\displaystyle \frac{L'}{L}(s, \chi) = O(1) - \sum_{\rho} \left( \frac{1}{\rho}+\frac{1}{2-\rho} \right)

where the {O(1)} is {\frac{L'}{L}(2,\chi) + \frac r2 + \gamma + \frac{\Gamma'}{\Gamma}(1)} if {a=0} and {\frac{L'}{L}(2,\chi) + \frac r2 - \frac{\Gamma'}{\Gamma}(\frac{1}{2}) + \frac{\Gamma'}{\Gamma}(\frac32)} for {a=1}. Furthermore,

\displaystyle \sum_{\rho, \left\lvert \gamma \right\rvert > 1} \left( \frac{1}{\rho}+\frac{1}{2-\rho} \right) \le \sum_{\rho, \left\lvert \gamma \right\rvert > 1} \frac{2}{\left\lvert \rho(2-\rho) \right\rvert} < 2 \sum_{\rho, \left\lvert \gamma \right\rvert > 1} \frac{1}{\left\lvert 2-\rho \right\rvert^2}

which is {O(\log q)} by our vertical distribution results, and similarly

\displaystyle \sum_{\rho, \left\lvert \gamma \right\rvert < 1} \frac{1}{2-\rho} = O(\log q).

This completes the proof. \Box

Let {\beta_1} be a Siegel zero, if any; for all the other zeros, we have that {\left\lvert \frac{1}{\rho} \right\rvert = \frac{1}{\beta^2+\gamma^2}}. We now have two cases.

  • {\overline{\chi} \neq \chi}. Then {\overline{\chi}} is complex and thus has no exceptional zeros; hence each of its zeros has {\beta < 1 - \frac{c}{\log q}}; since {\overline{\rho}} is a zero of {\overline{\chi}} if and only if {1-\rho} is a zero of {\chi}, it follows that all zeros of {\chi} are have {\left\lvert \frac{1}{\rho} \right\rvert < O(\log q)}. Moreover, in the range {\gamma \in [-1,1]} there are {O(\log q)} zeros (putting {T=0} in our earlier lemma on vertical distribution).

Thus, total contribution of the sum is {O\left( (\log q)^2 \right)}.


  • If {\overline{\chi} = \chi}, then {\chi} is real. The above argument goes through, except that we may have an extra Siegel zero at {\beta_S}; hence there will also be a special zero at {1 - \beta_S}. We pull these terms out separately.Consequently,

    \displaystyle \boxed{b(\chi) = O\left( (\log q)^2 \right) - \frac{1}{\beta_S} - \frac{1}{1-\beta_S}}.

    By adjusting the constant, we may assume {\beta_S > \frac{2014}{2015}} if it exists.

    7. Computing {\psi(x,\chi)} and {\psi(x;q,a)}

    7.1. Summing the Error Terms

    We now have, for any {T \ge 2}, {x \ge 6}, and {\chi} modulo {q} possibly primitive or trivial, the equality

    \displaystyle \psi(x, \chi) = \delta(\chi) x + E_{\text{contour}} + E_{\text{truncate}} + E_{\text{tiny}} - b(\chi) - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho}.


    \displaystyle \begin{aligned} E_{\text{contour}} &\ll \frac{x(\log x)^2}{T} + \log x \min \left\{ 1, \frac{x}{Td} \right\} \\ E_{\text{truncate}} &\ll \frac{\mathcal L^2 x}{T \log x} \\ E_{\text{tiny}} &\ll \log x \log q \\ b(\chi) &= O\left( (\log q)^2 \right) - \frac{1}{\beta_S} - \frac{1}{1-\beta_S}. \end{aligned}

    Assume now that {T \le x}, and {x} is an integer (hence {d \ge 1}). Then aggregating all the errors gives

    \displaystyle \psi(x, \chi) = \delta(\chi) x - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho} - \frac{x^{\beta_S}-1}{\beta_S} - \frac{x^{1-\beta_S}-1}{1-\beta_S} + O\left( \frac{x (\log qx)^2}{T} \right).

    where the sum over {\rho} now excludes the Siegel zero. We can omit the terms {\beta_S^{-1} < \frac{2015}{2014} = O(1)}, and also

    \displaystyle \frac{x^{1-\beta_S}-1}{1-\beta_S} < x^{\frac{1}{2015}} \log x.

    Absorbing things into the error term,

    \displaystyle \psi(x, \chi) = \delta(\chi) x - \frac{x^{\beta_S}}{\beta_S} - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho} + O\left( \frac{x (\log qx)^2}{T} + x^{\frac{1}{2015}} \log x \right).

    7.2. Estimating the Sum Over Zeros

    Now we want to estimate

    \displaystyle \sum_{\rho, \left\lvert \gamma \right\rvert < T} \left\lvert \frac{x^\rho}{\rho} \right\rvert.

    We do this is the dumbest way possible: putting a bound on {x^\rho} and pulling it out.

    For any non-Siegel zero, we have a zero-free region {\beta < 1 - \frac{c_1}{\mathcal L}}, whence

    \displaystyle \left\lvert x^\rho \right\rvert < x^{\beta} = x \cdot x^{\beta-1} = x \exp\left( \frac{-c_1 \log x}{\mathcal L} \right).

    Pulling this out, we can then estimate the reciprocals by using our differential:

    \displaystyle \begin{aligned} \sum_{\rho, \left\lvert \gamma \right\rvert < T} \left\lvert \frac{1}{\rho} \right\rvert < \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{1}{\gamma} < \sum_{t=1}^T \frac{\log q(t+2)}{t} \ll (\log qT)^2 \le (\log qx)^2. \end{aligned}


    \displaystyle \psi(x, \chi) = \delta(\chi) x - \frac{x^{\beta_S}}{\beta_S} + O\left( \frac{x (\log qx)^2}{T} + x^{\frac{1}{2015}} \log x + (\log qx)^2 \cdot x \exp\left( \frac{-c_1 \log x}{\mathcal L} \right) \right).

    We select

    \displaystyle T = \exp\left(c_3 \sqrt{\log x}\right)

    for some constant {c_3}, and moreover assume {q \le T}, then we obtain

    \displaystyle \psi(x, \chi) = \delta(\chi) x - \frac{x^{\beta_S}}{\beta_S} + O\left( x \exp\left( -c_4 \sqrt{\log x} \right) \right).

    7.3. Summing Up

    We would like to sum over all characters {\chi}. However, we’re worried that there might be lots of Siegel zeros across characters. A result of Landau tells us this is not the case:

    Theorem 15 (Landau)

    If {\chi_1} and {\chi_2} are real nontrivial primitive characters modulo {q_1} and {q_2}, then for any zeros {\beta_1} and {\beta_2} we have

    \displaystyle \min \left\{ \beta_1, \beta_2 \right\} < 1 - \frac{c_5}{\log q_1q_2}

    for some fixed absolute {c_5}. In particular, for any fixed {q}, there is at most one {\chi \mod q} with a Siegel zero.

    Proof: The character {\chi_1\chi_2} is not trivial, so we can put

    \displaystyle \begin{aligned} -\frac{\zeta'}{\zeta}(\sigma) &= \frac{1}{\sigma-1} + O(1) \\ -\frac{L'}{L}(\sigma, \chi_1\chi_2) &= O(\log q_1q_2) \\ -\frac{L'}{L}(\sigma, \chi_1) &= O(\log q_1) - \frac{1}{\sigma-\beta_i} \\ -\frac{L'}{L}(\sigma, \chi_2) &= O(\log q_2) - \frac{2}{\sigma-\beta_i}. \end{aligned}

    Now we use a silly trick:

    \displaystyle 0 \le -\frac{\zeta'}{\zeta}(\sigma) -\frac{L'}{L}(\sigma, \chi_1) -\frac{L'}{L}(\sigma, \chi_2) -\frac{L'}{L}(\sigma, \chi_1\chi_2)

    by “Simon’s Favorite Factoring Trick” (we use the deep fact that {(1+\chi_1)(1+\chi_2) \ge 0}, the analog of {3-4-1}). The upper bounds give now

    \displaystyle \frac{1}{\sigma-\beta_1} + \frac{1}{\sigma-\beta_2} < \frac{1}{\sigma-1} + O(\log q_1 \log q_2).

    and one may deduce the conclusion from here. \Box

    We now sum over all characters {\chi} as before to obtain

    \displaystyle \psi(x; q, a) = \frac{1}{\phi(q)} x - \frac{\chi_S(a)}{\phi(q)} \frac{x^{\beta_S}}{\beta_S} + O \left( x \exp\left(-c_6 \sqrt{\log x}\right) \right)

    where {\chi_S = \overline{\chi}_S} is the character with a Siegel zero, if it exists.

    8. Siegel’s Theorem, and Finishing Off

    The term with {x^{\beta_S} / \beta_S} is bad, and we need some way to get rid of it. We now appeal to Siegel’s Theorem:

    Theorem 16 (Siegel’s Theorem)

    For any {\varepsilon > 0} there is a {C_1(\varepsilon) > 0} such that any Siegel zero {\beta_S} satisfies

    \displaystyle \beta_S < 1-C_1(\varepsilon) q^{-\varepsilon}.

    Thus for a positive constant {N}, assuming {q \le (\log x)^N}, letting {\varepsilon = (2N)^{-1}} means {q^{-\varepsilon} > \frac{1}{\sqrt{\log x}}}, so we obtain

    \displaystyle x^{\beta_S} < x \exp\left( -C_1(\varepsilon) \log x q^{-\varepsilon} \right) < x \exp\left( -C_1(\varepsilon) \sqrt{\log x} \right).


    \displaystyle \begin{aligned} \psi(x; q, a) &= \frac{1}{\phi(q)} x - \frac{\chi_S(a)}{\phi(q)} \frac{x^{\beta_S}}{\beta_S} + O \left( x \exp\left(-c_6 \sqrt{\log x}\right) \right) \\ &\le \frac{1}{\phi(q)} x + O\left( x \exp\left( -C_1(\varepsilon) \sqrt{\log x} \right) \right) + O \left( x \exp\left(-c_6 \sqrt{\log x}\right) \right) \\ &\le \frac{1}{\phi(q)} x + O\left( x \exp\left( -C(N) \sqrt{\log x} \right) \right) \end{aligned}

    where {C(N) = \min \left\{ C_1(\varepsilon), c_6 \right\}}. This completes the proof of Dirichlet’s Theorem.