\classheader{2013-01-25}

\subsection*{Introduction to class field theory and Langlands correspondence}

Fermat $\to$ Gauss $\to$ $\cdots$

Gauss proved the quadratic reciprocity law in 1796.

For $p$ an odd prime number and $a\in\Z$, $p\nmid a$, the Legendre symbol $\legendre{a}{p}$ is defined by
\[\legendrebig{a}{p}=\begin{cases}
1 & \text{ if }a\bmod p \text{ has a square root in }\F_p,\\
-1 & \text{ otherwise.}
\end{cases}\]
We have
\[\legendrebig{ab}{p}=\legendrebig{a}{p}\legendrebig{b}{p}\]
because $\F_p^\times$ is a cyclic group of order $p-1$.

When we fix $p$ and let $a$ vary, it is easy to understand:
\[\legendrebig{a}{5}=\begin{cases}
1 & \text{ if }a\equiv 1,4\bmod 5,\\
-1 & \text{ if }a\equiv 2,3\bmod 5.
\end{cases}\]
This is because $1=1^2$, $4=2^2$ in $\F_5$, but 2 and 3 are not squares.

But if we fix $a$ and let $p$ vary, it seems like this would be very hard to understand:
\[\legendrebig{5}{p}=?\]
How can we figure out when $\F_p$ has a solution to $x^2=5$?

(The case of $x^3=5$ is harder; this is a case where we need Langlands correspondence.)

\subsection*{Quadratic Reciprocity Law}

Complementary Laws:
\begin{center}
\begin{minipage}{0.55\textwidth}
\begin{itemize}
\item[I:] $\displaystyle\legendrebig{-1}{p}=(-1)^{\frac{p-1}{2}}=\begin{cases}
1 & \text{ if }p\equiv 1\bmod 4,\\
-1 & \text{ if }p\equiv 3\bmod 4.
\end{cases}$\vspace{0.1in}
\item[II:] $\displaystyle\legendrebig{2}{p}=(-1)^{\frac{p^2-1}{8}}=\begin{cases}
1 & \text{ if }p\equiv 1,7\bmod 8,\\
-1 & \text{ if }p\equiv 3,5\bmod 8.
\end{cases}$
\end{itemize}
\end{minipage}
\end{center}

Love song in the land of prime numbers:
\[\legendrebig{q}{p}=\legendrebig{p}{q}\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\]
for distinct odd prime numbers $p$ and $q$.

$\legendre{q}{p}$ is how the girl $q$ is reflected in the heart of the boy $p$, and $\legendre{p}{q}$ is how the boy is reflected in the heart of the girl $q$. As you may some experience with, these are sometimes not related in our world. But in the world of prime numbers, they are related - this is very mysterious.

We can relate the quadratic reciprocity law as
\[\legendrebig{m}{p}=\chi(p)\]
where $m\in\Z$, $m$ is squarefreee. Let
\[N=\begin{cases}
|m| & \text{ if }m\equiv 1\bmod 4,\\
4|m| & \text{ otherwise.}
\end{cases}\]
There is a unique homomorphism $\chi:(\Z/N\Z)^\times\to\{\pm 1\}$ such that
\begin{itemize}
\item[(i)] $\chi(-1)=\begin{cases}
1 & \text{ if }m>0,\\
-1 & \text{ if }m<0
\end{cases}$, and
\item[(ii)] $\chi$ does not factor as
\begin{center}
\begin{tikzcd}[column sep=0.1in]
(\Z/N\Z)^\times \ar{rr} \ar{rd} & & \{\pm 1\}\\
 &(\Z/N'\Z)^\times \ar{ru}
\end{tikzcd}
\end{center}
for any proper divisor $N'$ of $N$.
\end{itemize}
When $m=2$, we have $N=8$, and $\chi:(\Z/8\Z)^\times\to\{\pm 1\}$ sends 1 and 7 to 1, and 3 and 5 to $-1$.
\begin{landscape}
$\text{}$\vfill
\begin{center}
\begin{tabular}{p{4cm}|p{5.5cm}p{5.5cm}p{6.5cm}}
& \parbox{5.5cm}{\centering What happens in finite fields} & \parbox{5.5cm}{\centering How primes decompose} & \parbox{6.5cm}{\centering Class field theory}\\\hline
\parbox{4cm}{\centering \rule[-.3\baselineskip]{0pt}{1.2\baselineskip}Quadratic reciprocity\\ (Gauss, 1796) \rule[-.3\baselineskip]{0pt}{0.7\baselineskip}} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip}  $x^{2}=m$ has a solution in $\F_p$\\ $p\nmid m$, $p\neq 2$} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $p\ringofintegersof{K}=\frak{p}_1\frak{p}_2$ for distinct\\$\frak{p}_i\subset\ringofintegersof{K}$, where $K=\Q(\sqrt{m})$} & 
\parbox{6.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $\chi(p)=1$, where $\chi$ is as above\rule[-.3\baselineskip]{0pt}{0.7\baselineskip}}
\\\hline
\parbox{4cm}{\centering \rule[-.3\baselineskip]{0pt}{1.2\baselineskip}Generalized reciprocity\\\rule[-.3\baselineskip]{0pt}{0.7\baselineskip}(Kummer, $19^{\text{th}}$ century)} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $x^n=1$ has $n$ solutions in $\F_p$\\$p\nmid n$} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $p\ringofintegersof{K}=\frak{p}_1\cdots\frak{p}_r$ in $\ringofintegersof{K}=\Z[\zeta_n]$,\\ where $r=\varphi(n)=[\Q(\zeta_n):\Q]$} &
\parbox{6.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $\chi(p)=1$ for all $\chi:(\Z/n\Z)^\times\to\C^\times$\\ ($\iff$ $p\equiv 1\bmod n$)\rule[-.3\baselineskip]{0pt}{0.7\baselineskip}}
\\\hline
\parbox{4cm}{\centering \rule[-.3\baselineskip]{0pt}{1.2\baselineskip}Class field theory\\ (Tagaki, Artin) \rule[-.3\baselineskip]{0pt}{0.7\baselineskip}} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $\frak{p}$ a maximal ideal of $\Z[\zeta_3]$;\\ $x^3=2$ has a solution in $\Z[\zeta_3]/\frak{p}$} & \parbox{5.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} $\frak{p}\ringofintegersof{L}=\frak{P}_1\frak{P}_2\frak{P}_3$ for distinct\\ $\frak{P}_i\subset\ringofintegersof{L}$, where $L=\Q(\sqrt[3]{2},\zeta_3)$\rule[-.3\baselineskip]{0pt}{0.7\baselineskip}} & \parbox{6.5cm}{\centering\rule[-.3\baselineskip]{0pt}{1.2\baselineskip} exists $\alpha\in\Z[\zeta_3]$ with $\frak{p}=(\alpha)$,\\ $\alpha\equiv 1\bmod 6$ \rule[-.3\baselineskip]{0pt}{0.7\baselineskip}} \\\hline
\end{tabular}
\end{center}
\vfill$\text{}$
\end{landscape}

For example, in $\Z[\zeta_3]$ we have
\[31=(1+6\zeta_3)(1+6\zeta_3^2)\]
and each factor is prime in $\Z[\zeta_3]$, and in $\ringofintegersof{L}$ we have
\[1+6\zeta_3^a=-\prod_{b=0}^2(1-\zeta_3^a+\zeta_3^b\sqrt[3]{2})\]
for each $a=1,2$.

Here is another example. Let $p$ be an odd prime and $m$ squarefree, $p\nmid m$. Let $K=\Q(\sqrt{m})$. We have
\[\ringofintegersof{K}/p\ringofintegersof{K}\cong\Z[\sqrt{m}]/p\Z[\sqrt{m}]\]
because $p\neq 2$. Note that this is isomorphic to
\[\Z[\sqrt{m}]/p\Z[\sqrt{m}]\cong \Z[T]/(T^2-m,p)\cong \F_p[T]/(T^2-m).\]
If $\legendre{m}{p}=-1$, then $\F_p[T]/(T^2-m)$ is a field and $(p)$ is maximal.

If $\legendre{m}{p}=1$, then $\F_p[T]/(T^2-m)\cong\F_p[T]/(T-a)(T-b)\cong\F_p\times\F_p$, and $(p)=(p,a-\sqrt{m})(p,b-\sqrt{m})$.

I saw a book that said class field theory was the greatest theory in number theory, and I misunderstood and thought that number theorists only studied stupid things now, because the greatest theory was already completed. But this was my misunderstanding.