From 8131fb40625f5c20f06a60e15789af02f9883217 Mon Sep 17 00:00:00 2001 From: user202729 <25191436+user202729@users.noreply.github.com> Date: Fri, 16 Aug 2024 22:25:22 +0700 Subject: [PATCH] Write Dirichlet's theorem on arithmetic progression --- Napkin.tex | 1 + references.bib | 11 ++ tex/alg-NT/arithmetic.tex | 357 ++++++++++++++++++++++++++++++++++++++ 3 files changed, 369 insertions(+) create mode 100644 tex/alg-NT/arithmetic.tex diff --git a/Napkin.tex b/Napkin.tex index af92041a..59791dee 100644 --- a/Napkin.tex +++ b/Napkin.tex @@ -150,6 +150,7 @@ \part{Algebraic NT I: Rings of Integers} \include{tex/alg-NT/classgrp} \include{tex/alg-NT/discriminant} \include{tex/alg-NT/pell} +\include{tex/alg-NT/arithmetic} \part{Algebraic NT II: Galois and Ramification Theory} \label{part:algnt2} diff --git a/references.bib b/references.bib index fe822b13..1b05c65a 100644 --- a/references.bib +++ b/references.bib @@ -72,6 +72,17 @@ @book{ref:hatcher isbn = {0-521-79160-X}, year = {2002} } +@book{ref:neukirch, + title = {Algebraic Number Theory}, + ISBN = {9783662039830}, + ISSN = {0072-7830}, + url = {http://dx.doi.org/10.1007/978-3-662-03983-0}, + DOI = {10.1007/978-3-662-03983-0}, + journal = {Grundlehren der mathematischen Wissenschaften}, + publisher = {Springer Berlin Heidelberg}, + author = {Neukirch, J\"{u}rgen}, + year = {1999} +} @book{ref:miranda, author={Miranda, R.}, title={Algebraic Curves and Riemann Surfaces}, diff --git a/tex/alg-NT/arithmetic.tex b/tex/alg-NT/arithmetic.tex new file mode 100644 index 00000000..eb655b2e --- /dev/null +++ b/tex/alg-NT/arithmetic.tex @@ -0,0 +1,357 @@ +\chapter{Dirichlet's theorem on arithmetic progressions} + +\section{Introduction} + +In this chapter, we will sketch a proof of the following theorem: +\begin{theorem}[Dirichlet's theorem on arithmetic progressions] + For every positive integers $a, n > 0$ such that $\gcd(a, n) = 1$, then there are infinitely + many primes coprime to $a$ modulo $n$. +\end{theorem} + +For an even more concrete example: +\begin{corollary} +There are infinitely many primes $p$ that is congruent to $3$ modulo $4$. +\end{corollary} + +You can see why this is called a theorem on arithmetic progressions: +the example above is equivalent to saying that there are infinitely many primes in the arithmetic +progression +\[ 3, 7, 11, 15, 19, 23, \dots \] + +If you haven't noticed, the condition $\gcd(a, n) = 1$ is also very natural: +look at the arithmetic progression $2, 6, 10, \dots$ where $a = 2$ and $n = 4$ are not coprime, +and you will see that it obviously has only finitely many primes. + +(By the way, using Chebotarev density theorem which will be mentioned in the next section, +you can even deduce something stronger: for example, a ``randomly chosen'' prime has probability +$\frac{1}{2}$ to be congruent to $3$ modulo $4$. But here we will give a simple sketch of proof.) + +\section{The $L$-series} + +Actually, this chapter is just an excuse to introduce you to something called the $L$-series. +Here is a quote from \cite{ref:neukirch} (emphasis mine): + +\begin{quote} +One of the most astounding phenomena in number theory consists in the +fact that a great number of deep arithmetic properties of a number field are +hidden within a single analytic function, its \vocab{zeta function}. This function has +a simple shape, but it is unwilling to yield its mysteries. Each time, however, +that we succeed in stealing one of these well-guarded truths, we may expect to +be rewarded by the revelation of some \emph{surprising} and \emph{significant} relationship. +This is why zeta functions, as well as their generalizations, the $L$-series, +have increasingly moved to the foreground of the arithmetic scene, and today +are more than ever the focus of number-theoretic research. +\end{quote} + +Surprisingly, you will see $L$-series in many other places as well --- an elliptic curve over $\QQ$ +has a $L$-series, and a modular form also have a $L$-series!\todo{really? +And is there more examples?} + +So what is a $L$-series? In one sentence:\footnote{In my (limited) understanding, at the time of +writing this.} +\begin{moral} + A $L$-series bridges between an infinite sum (complex analysis) + and an infinite product over primes (number theory) + using unique factorization, + thus using tools from complex analysis to deduce hidden facts about the primes. +\end{moral} + +Okay, that is a long sentence, sorry. + +\subsection{Riemann zeta function} + +Let's look at the Riemann zeta function, which is the prototype for a $L$-series: + +\begin{definition}[Riemann zeta function] + When $\Re(s) > 1$, we define the \vocab{Riemann zeta function} by the following infinite sum + \[ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots. \] + We can prove that the series converges whenever $\Re(s) > 1$, and that there exists an unique + meromorphic function on $\CC$ that is equal to the series above. + For $\Re(s) \leq 1$ define $\zeta(s)$ using that meromorphic function. +\end{definition} + +This is called an \vocab{analytic continuation} of the series to the whole complex plane. +(If you recalled: +the identity theorem in complex analysis guarantees that if such a meromorphic function exists, +it must be unique. The difficult part is to prove it exists.) + +So far, so good --- we have seen the infinite sum (complex analysis) part. How about the primes +(number theory) part? + +For the Riemann zeta function, the idea has been discovered centuries ago by Euler +(though it wasn't formalized until much later). +Thus such product over primes are aptly called \vocab{Euler products}. +The \href{https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes}{% +Wikipedia page} has a good explanation. + +Here are some manipulations that is not fully justified. +We can factorize each integer in the denominator as follows: +\[ \zeta(s) = 1 + \frac{1}{2^s} + + \frac{1}{3^s} + + \frac{1}{2^{2s}} + + \frac{1}{5^s} + + \frac{1}{2^s} \cdot \frac{1}{3^s} + + \cdots +\] +Then, group like terms: +\[ \zeta(s) = + \left(1 + \frac{1}{2^s} + \frac{1}{2^{2s}} + \cdots \right) + \left(1 + \frac{1}{3^s} + \frac{1}{3^{2s}} + \cdots \right) + \left(1 + \frac{1}{5^s} + \frac{1}{5^{2s}} + \cdots \right) \cdots +\] +By unique factorization, this grouping is indeed correct. +So, our infinite product is: +\[ \zeta(s) = \prod_{p\text{ prime}} \left(1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots\right). \] + +\begin{theorem} + \label{thm:riemann_zeta_product_rep} + For $\Re(s) > 1$, this infinite product converges to the same value as $\zeta(s)$. + (That is, our questionable manipulations can be proven correct.) +\end{theorem} +The proof is omitted. + +Because the ``$\cdots$'' are verbose, +we note that the series $1 + x + x^2 + \cdots$ is equal to $\frac{1}{1-x}$. +So, for brevity we write +\[ \zeta(s) = \prod_{p\text{ prime}} \frac{1}{1-p^{-s}}. \] + +\subsection{Example: Euclid's theorem} + +We will use \Cref{thm:riemann_zeta_product_rep} to give a roundabout proof of the following: +\begin{theorem}[Euclid's theorem] + There are infinitely many primes. +\end{theorem} + +\begin{proof}[Sketch of Proof] + Note that, by the infinite sum representation, + $\zeta(1) = 1+\frac{1}{2}+\frac{1}{3}+\cdots$ is the harmonic series and diverges. + + Now look at the Euler product representation, + $\zeta(1) = \prod_{p\text{ prime}} \frac{1}{1-p^{-1}}$. + For this to diverge, there must be infinitely many primes $p$. +\end{proof} + +\section{The actual proof} + +Here comes the actual proof, or at least a sketch of it. +Just as above, the plan is to do the following: +\begin{itemize} + \ii Write down some infinite sum. + \ii Prove that it is equal to some Euler product. + \ii Use the infinite sum to determine the behavior of the function at $s = 1$. + \ii Use the Euler product to say something about the primes. +\end{itemize} + +Easier said than done. +You might guess we could try something like +\[ + \frac{1}{3^s} + \frac{1}{7^s} + \frac{1}{11^s} + \cdots = + \prod_{p\text{ prime},\ p \equiv 3 \ (\opname{mod} 4)} \frac{1}{1-p^{-s}} +\] +Unfortunately, this fails miserably. (Try to expand out a few terms of the product.) + +How can we fix it? Look at this: +\[ + \frac{1}{1^s} - \frac{1}{3^s} + \frac{1}{5^s} - \frac{1}{7^s} + \frac{1}{9^s} - \cdots +\] +\begin{exercise} + Do the ``non-rigorous'' factorization and grouping of like terms as above. It should work out! +\end{exercise} + +If you did the exercise correctly, you should get the following: +\[ + \left( \frac{1}{1+3^{-s}}\right) + \left( \frac{1}{1-5^{-s}}\right) + \left( \frac{1}{1+7^{-s}}\right) + \left( \frac{1}{1+11^{-s}}\right) + \left( \frac{1}{1-13^{-s}}\right) \cdots +\] +In order to save us from writing down a lot of ``$\cdots$'', for now, define the following function +on positive integers integers $n$: +\[ \chi(n) = \begin{cases} + 1 &\text{if }n \equiv 1 \pmod{4} \\ + -1&\text{if }n \equiv 3 \pmod{4} \\ + 0 &\text{otherwise.} +\end{cases} \] +Then we have +\[ \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \prod_{p\text{ prime}} \frac{1}{1-\chi(p) \cdot p^{-s}}. \] + +Of course, what we have seen deserves a name: +\begin{definition}[Dirichlet character, informal] + A \vocab{Dirichlet character} is a function $\chi$ as above that we are interested in and makes + the equality works. +\end{definition} + +\begin{exercise} + Notice that the sum and product in our broken attempt also comes from a function as follows: + \[ \chi(n) = \begin{cases} + \textcolor{red}{0} &\text{if }n \equiv 1 \pmod{4} \\ + \textcolor{red}{1} &\text{if }n \equiv 3 \pmod{4} \\ + 0 &\text{otherwise.} + \end{cases} \] + What makes this function $\chi$ not work (so, it's not a Dirichlet character)? +\end{exercise} + +So what's the point? Turns out there's \emph{a lot} of Dirichlet characters --- +enough for us to write down (almost) the product we want! + +Look: if we consider +\[ \chi_2(n) = \begin{cases} + 1 &\text{if }n \equiv 1 \pmod{4} \\ + 1 &\text{if }n \equiv 3 \pmod{4} \\ + 0 &\text{otherwise.} +\end{cases} \] +then we also have +\[ \sum_{n \geq 1} \frac{\chi_2(n)}{n^s} = +\prod_{p\text{ prime}} \frac{1}{1-\chi_2(p) \cdot p^{-s}}. \] + +Wait, writing down all these $\sum$ signs are boring. Let us define: +\begin{definition}[$L$-series] + For $\chi$ a Dirichlet character, define the \vocab{$L$-series} + \[ L(\chi, s) \sum_{n \geq 1} \frac{\chi(n)}{n^s} + = \prod_{p\text{ prime}} \frac{1}{1-\chi(p) \cdot p^{-s}}. \] +\end{definition} + +Okay, back to the proof. Writing things down concretely, +\begin{align*} + L(\chi, s) &= + \left( \frac{1}{1+3^{-s}} \right) + \left( \frac{1}{1-5^{-s}} \right) + \left( \frac{1}{1+7^{-s}} \right) + \left( \frac{1}{1+11^{-s}} \right) + \left( \frac{1}{1-13^{-s}} \right) \cdots + \\ + L(\chi_2, s) &= + \left( \frac{1}{1-3^{-s}} \right) + \left( \frac{1}{1-5^{-s}} \right) + \left( \frac{1}{1-7^{-s}} \right) + \left( \frac{1}{1-11^{-s}} \right) + \left( \frac{1}{1-13^{-s}} \right) \cdots +\end{align*} +We want to ``forget'' about the primes in the arithmetic progression $1, 5, 9, 13, \dots$ +so obviously we're going to divide the two series: +\[ + \frac{L(\chi, s)}{L(\chi_2, s)} = + \left( \frac{1-3^{-s}}{1+3^{-s}} \right) + \left( \frac{1-7^{-s}}{1+7^{-s}} \right) + \left( \frac{1-11^{-s}}{1+11^{-s}} \right) \cdots +\] +You can easily guess what are the remaining steps: +\begin{itemize} + \ii The numerator $L(\chi, s)$ is finite at $s = 1$. + \ii The denominator $L(\chi_2, s)$ diverges at $s = 1$. + \ii So the left hand side is $0$. + \ii This cannot happen if there are only finitely many terms on the right hand side. +\end{itemize} + +\section{Generalization} + +It is not a coincidence that the $\chi$ above are called \emph{characters} --- functions sure, but +characters? + +Turns out, if you have read the chapter on representation theory, +it has a lot in common with characters of representations: +\begin{itemize} + \ii they're independent, + \ii they're orthogonal, + \ii they span the whole space.\todo{double check. Besides, what space?} +\end{itemize} + +Abstract algebra aside, how are we going to construct the product in such a way that only the primes +congruent to $a$ modulo $m$ remains? +Look: + +\begin{theorem} + Fix a modulus $m$. + Let $\chi_1, \chi_2, \dots$ be all the Abelian group homomorphism + $\Zm{m} \to \CC^{\times}$. + For each of them, define a $\NN \to \CC$ function the obvious way: + \[ \chi_i(n) = \begin{cases} + \chi_i(n \bmod m)&\text{if }\gcd(n, m) = 1 \\ + 0&\text{otherwise.} + \end{cases} \] + Then there are exactly $|\Zm{m}|$ characters, they're Dirichlet characters, + and most importantly, you can extract out \emph{only} the primes congruent to $a$ modulo $m$ by: + \[ + \prod_{i=1}^{|\Zm{m}|} L(\chi_i, s)^{1/\chi_i(a)}. + \] +\end{theorem} +\todo{this is a lie, explain what the product expand to. See \cite[page 469]{ref:neukirch}} + +And: +\begin{proposition} % Neukirch chapter 7, proposition 5.13 + \label{prop:l1_finite_nonzero} + With notation as above, unless $\chi_i$ is the trivial character which assigns $1$ to every + element of $\Zm{m}$, then + \[ L(\chi_i, 1) \in \CC \setminus \{0\}. \] +\end{proposition} + +\begin{remark}[Anecdote] + $|\Zm{m}|$ is equal to $\phi(m)$, where $\phi$ is the Euler's totient function. +\end{remark} + +\section{More proof} + +In this section, we will prove \Cref{prop:l1_finite_nonzero}. + +\begin{exercise} + Recall the Dirichlet character we used in the example above. + \[ \chi(n) = \begin{cases} + 1 &\text{if }n \equiv 1 \pmod{4} \\ + -1&\text{if }n \equiv 3 \pmod{4} \\ + 0 &\text{otherwise.} + \end{cases} \] + Try to prove $L(\chi, 1)$ is finite. +\end{exercise} + +That one was easy --- the sum the series is decreasing and alternating. +But what if the modulus is $5$? + +Let us start with writing down all Dirichlet characters modulo $5$: +\[ + \begin{array}{|r|rrrr|} + \hline + & 1 & 2 & 3 & 4 \\ \hline + \chi_1 & 1 & 1 & 1 & 1 \\ + \chi_2 & 1 & i & -i & -1 \\ + \chi_3 & 1 & -i & i & -1 \\ \hline + \end{array} +\] +Great, we got some complex numbers. +No worries --- complex numbers need complex tools to deal with it. + +(Recall that a Dirichlet character can be defined from a group homomorphism $\Zm{5} \to +\CC^{\times}$. Of course the value $\chi(n)$ would be $0$ at values $5 \mid n$.) + +Consider the cyclotomic field extension $K = \QQ(\zeta_5)$. +We are going to \emph{generalize the Riemann zeta function $\zeta$ to $\zeta_K$}! + +How? To start with, there is no unique factorization of elements in arbitrary number fields, +and there is no notation of \emph{positive} or \emph{negative} values either. + +We do have unique factorization of ideals though, so maybe we can make something out of ideals? +Let's see: +\begin{definition} % Neukirch chapter 7, definition 5.1 + Define the Dedekind zeta function by + \[ \zeta_K(s) = \sum_{\ka\text{ ideal of }\OO_K} \frac{1}{\Norm(\ka)^s}. \] + When $\Re(s) \leq 1$, do analytic continuation as in the case of Riemann zeta function. +\end{definition} +\begin{exercise} + Verify that when $K = \QQ$ then $\zeta_\QQ = \zeta$. +\end{exercise} + +Of course we have the following ``Euler product'': +\begin{proposition} + \[ \zeta_K(s) = \prod_{\kp\text{ prime ideal of }\OO_K} \frac{1}{1-\Norm(\kp)^{-s}}. \] +\end{proposition} + +So?\todo{double check (how do you numerically evaluate $\zeta_K$ for $K \neq \QQ$ anyway?)} +\begin{proposition} % Neukirch chapter 7, proposition 5.12, specialized to m = 5 + For $K = \QQ(\zeta_5)$ and $\chi_i$ the Dirichlet characters modulo $5$ defined above, then + \[ \zeta_K(s) = \frac{1}{1-5^{-s}} \cdot \prod_{i=1}^{3} L(\chi_i, s). \] +\end{proposition} +We can prove that at $s = 1$, $\zeta_K(s)$ has a pole of order exactly $1$, +which means only $L(\chi_1, s)$ is allowed to have a pole there. +The other $L(\chi_i, s)$ are thus not allowed to have one.\todo{why can't it be that one has a pole +and the other have a root to cancel it out?} +