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We will use the notation of Legendre symbols:
\[ \left( \frac{m}{p} \right) = \begin{cases}
  0 & \text{if $p \mid m$} \\
  1 & \text{if $\exists a \neq 0 \in \ZZ / p\ZZ$
  with $a^2 \equiv m \pmod{p}$} \\
  -1 & \text{otherwise}
\end{cases} \]
See \courseref[Number Theory]{NT} for some
properties.

\begin{flashcard}[legendre-prime-ideal-facts-thm]
\begin{theorem*}
  Let $K = \QQ(\sqrt{m})$. Let $p \in \ZZ$ be
  prime and suppose $m$ is square-free, with $m
  \neq 0, 1$. Then:
  \cloze{
  \begin{enumerate}[(1)]
    \item $p$ is ramified in $K$, that is $\exists
      P \subset \O_K$ such that $p\O_K = P^2$, if
      and only if $p$ is od and $p \mid m$, or $p$
      is even and $m \equiv 2, 3 \pmod{4}$.
    \item $p$ is split in $K$, that is $\exists
      P_1, P_2 \subset \O_K$ such that $p\O_K =
      P_1 P_2$, if and only if $p$ is odd and
      $\legendre mp = 1$ or $p = 2$ and $m \equiv
      1 \pmod{8}$.
    \item $p$ is inert, that is $p\O_K$ is a
      prime, if and only if $p$ is odd and
      $\legendre{m}{p} = -1$ or $p = 2$ and $m
      \equiv 5 \pmod{8}$.
  \end{enumerate}
  }
\end{theorem*}

\begin{proof}
  \cloze{If $p$ is odd or if $p = 2$ and $m \equiv 2, 3
  \pmod{4}$, then we can apply \nameref{dedekind}
  with $g(x) = x^2 - m$, because $p \nmid [\O_K :
  \ZZ[\sqrt{m}]]$. If $p = 2$ and $m \equiv 1
  \pmod{4}$, then we can apply \nameref{dedekind}
  with $g(x) = x^2 - m + \frac{1 - m}{4}$, which
  is the minimal polynomial of $\frac{1 +
  \sqrt{m}}{2}$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[class-group-defn]
\begin{definition*}[Class group]
  \glssymboldefn{Cl}{Cl}{Cl}
  \cloze{Write $\mathcal{I}$ for the set of fractional
  ideals in $K$, which form an abelian group under
  multiplication. Let $\mathcal{P}$ denote the
  principal fractional ideals, which form a
  subgroup. The class group of $K$ is
  \[ \Cl(K) = \mathcal{I} / \mathcal{P} .\]}
\end{definition*}
\end{flashcard}

\vspace{-1em}
We have seen that for all $I \in \mathcal{I}$,
there exists $a \in \ZZ$ such that $aI \subset
\O_K$, that is $aI$ is an integral ideal. Thus
each class in $\Cl(K)$ contains integral ideals.

Alternatively, $\Cl(K)$ can be defined as
equivalence classes of integral ideals under $I
\sim J$, where $I \sim J$ if and only if $\exists
\alpha \in K$ such that $I = \alpha J$.

\begin{flashcard}[class-number-defn]
\begin{definition*}[Class number]
  \glssymboldefn{clnum}{$h(K)$}{$h(K)$}
  \cloze{The \emph{class number} of $K$ is $h(K) =
  |\Cl(K)|$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
$h(K) = 1$ if and only if $\O_K$ is a PID (which
we also showed before happens if and only if
$\O_K$ is a UFD).

\begin{theorem*}
  For all number fields, $h(K) < \infty$.
\end{theorem*}

\vspace{-1em}
In order to prove this, we need a couple of
results:

\begin{flashcard}[minkowski-bound-thm]
\begin{theorem*}[Minkowski's bound]
  \label{mink_bound_1}
  \cloze{Let $K$ be a \gls{nf}, let $I \subset \O_K$ be
  an ideal. Write $s$ for the number of pairs of
  complex embeddings of $K$. Then $\exists \alpha
  \in I$ such that
  \[ |N(\alpha)| \le \frac{d^1}{d^d} \left(
  \frac{4}{\pi} \right)^s \N(I) \sqrt{\nfdisc (K)}
  .\]
  Then by Stirling's Approximation,
  \[ \frac{d^1}{d^d} = (1 + \sigma(1)) \sqrt{2\pi
  d} e^{-d} .\]}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[minkowski-bound-2-thm]
\begin{corollary*}[Minkowski's bound 2]
  \label{mink_bound_2}
  \cloze{Let $K$ be a \gls{nf}, and let $s$ be the
  number of pairs of complex embeddings of $K$.
  Then every ideal class in
  $\Cl(K)$ contains an integral ideal $I$ with
  \[ N(I) \le \frac{d^1}{d^d} \left( \frac{4}{\pi}
  \right)^s \sqrt{\nfdisc(K)} .\]}
\end{corollary*}

\begin{proof}
  \cloze{Let $I$ be an ideal. Let $J \subset \O_K$ be an
  ideal in the class of $I^{-1}$. We apply the
  \nameref{mink_bound_1} to $J$, so there is
  $\alpha \in J$ such that $N(\alpha) \le \cdots
  \N(J) \sqrt{\nfdisc(K)}$. Since $\alpha \in J$,
  $J \mid \langle \alpha \rangle$, so $\alpha
  J^{-1} \subset \O_K$ is an ideal in the class of
  $I$. Also,
  \[ \N(\alpha J^{-1}) = |N(\alpha)| \N(J)^{-1}
  \le \frac{d^1}{d^d} \left( \frac{4}{\pi}
  \right)^s \sqrt{\nfdisc(K)} .\]}
\end{proof}
\end{flashcard}

This implies $h(K) < \infty$ because of:

\begin{flashcard}[finitely-many-ideals-bounded-norm-lemma]
\begin{lemma*}
  Let $X \in \RR_{> 0}$. Then there are only
  finitely many ideals in $\O_K$ of norm $\le X$.
\end{lemma*}

\begin{proof}
  \cloze{Each ideal of norm $\le X$ is the product of at
  most $\log_2(X)$ primes. The primes in those
  decompositions lie over rational primes $\le
  X$. For each such prime, there at most $d$
  primes of $\O_K$ \glsref[lo]{lying over} it.}
\end{proof}
\end{flashcard}

Computation of $\Cl(K)$:
\begin{enumerate}[(1)]
  \item Calculate $X = \frac{d^1}{d^d} \left(
    \frac{4}{\pi} \right)^s \sqrt{\nfdisc(K)}$.
    For $K = \QQ(\sqrt{m})$, we get:
    \[ X = \begin{cases}
      \frac{\sqrt{m}}{2} & \text{if $m > 1$ and $m
      \equiv 1 \pmod 4$} \\
      \sqrt{m} & \text{if $m > 1$ and $m \equiv 2,
      3 \pmod{4}$} \\
      \frac{2\sqrt{-m}}{\pi} & \text{if $m < 0$
      and $m \equiv 1 \pmod 4$} \\
      \frac{4\sqrt{-m}}{\pi} & \text{if $m < 0$
      and $m \equiv 2, 3 \pmod 4$}
    \end{cases} \]
  \item List all rational primes $\le X$.
  \item Split all of these rational primes in
    $\O_K$, and make a list of all prime ideals
    with norm $\le X$, say $P_1, \ldots, P_k$.
  \item Figure out when $P_1^{m_1} \cdots
    P_k^{m_k}$ is principal for some $m_1, \ldots,
    m_k \in \ZZ$.
\end{enumerate}

\begin{flashcard}[minkowski-bound-3-coro]
\begin{corollary*}[Minkowski bound 3]
  \label{mink_bound_3}
  \cloze{\[ \nfdisc(K) \ge \frac{d^{2d}}{(d^1)^2} \left(
  \frac{\pi}{4} \right)^{2s} .\]
  This follows from $\N(I) \ge 1$ and
  \nameref{mink_bound_2}.}
\end{corollary*}
\end{flashcard}