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\begin{flashcard}[intbasis-iff-tdisc-minimal-thm]
\begin{theorem*}
  Let $K$ be a \gls{nf}. Then $\alpha_1, \ldots,
  \alpha_d \in \O_K$ is integral basis if and only
  if \cloze{$|\tdisc(\alpha_1, \ldots,
  \alpha_d)|$} is
  minimal \cloze{among all $\QQ$-linear indepdendent
  tuples.}
\end{theorem*}

\begin{proof}
  \cloze{Let $\alpha_1, \ldots, \alpha_d$ be such a
  tuple. Let $\beta \in \O_K$. We need to prove
  that $\beta \in M = \alpha_1 \ZZ + \cdots +
  \alpha_d \ZZ$. Then
  \[ \mdisc(M + \beta\ZZ) = |M + \beta \ZZ /
  M|^{-2} \mdisc(M) \implies |M + \beta \ZZ / M| =
  1 ,\]
  so $\beta \in M$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[disc-of-nf]
\begin{definition*}[Discriminant of a number field]
  \glspropdefn{nfdisc}{discriminant}{\gls{nf}}
  \glssymboldefn{nfdisc}{disc}{disc}
  \cloze{The \emph{discriminant} of a \gls{nf} is
  the \gls{tdisc} of any \gls{intbasis}.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Quadratic fields: $K = \QQ(\sqrt{m})$, $m$
  square-free, $m \neq 0$. Two cases:
  \begin{enumerate}[(1)]
    \item $m \equiv 2, 2 \pmod{4}$: $\O_K = \ZZ +
      \sqrt{m} \ZZ$,
      \[ \nfdisc(K) =
      \begin{vmatrix}
        1 & \sqrt{m} \\
        1 & -\sqrt{m}
      \end{vmatrix}^2 = 4m \]
    \item $m \equiv 1 \pmod{4}$: $\O_K = \ZZ +
      \frac{1 + \sqrt{m}}{2} \ZZ$,
      \[ \nfdisc(K) =
      \begin{vmatrix}
        1 & \frac{1 + \sqrt{m}}{2} \\
        1 & \frac{1 - \sqrt{m}}{2}
      \end{vmatrix}^2 = m \]
  \end{enumerate}
\end{example*}

\begin{flashcard}[q-div-disc-beta-in-OK-prop]
\begin{proposition*}
  Let $\alpha_1, \ldots, \alpha_d \in \O_K$ be
  $\QQ$-linearly independent. Then $\exists q \in
  \ZZ_{\ge 0}$ such that \cloze{$q^2 \tdisc(\alpha_1,
  \ldots, \alpha_d)$ and all $\beta \in \O_K$ can
  be written as
  \[ \beta = \frac{a_1 \alpha_1 + \cdots + a_d
  \alpha_d}{q} \]
  with $a_1, \ldots, a_d \in \ZZ$.}
\end{proposition*}

\begin{proof}
  \cloze{Set
  \[ q = \left( \frac{\tdisc(\alpha_1, \ldots,
  \alpha_d)}{\nfdisc(K)} \right)^{1/2} \]
  Then
  \[ |\ub{\O_K / \alpha_1 \ZZ \oplus \cdots \oplus
  \alpha_d \ZZ}_{= M}| = q \]
  $\beta \in \O_K$, $q\beta = 0$ in $M$, so
  $q\beta \in \alpha_1\ZZ \oplus \cdots \oplus
  \alpha_d\ZZ$.}
\end{proof}
\end{flashcard}

\subsubsection*{Unique factorisation of ideals}

Consider $K = \QQ(\sqrt{-5})$, $\O_K =
\ZZ[\sqrt{-5}]$. We have
\[ 2 \cdot 3 = 6 = (1 + \sqrt{-5})(1 - \sqrt{-5}) \]
In order to have unique factorisation, would have
to have these elements split into smaller element.
Say $2 = \pi_1 \pi_2$. $N(2) = 4$, $N(1 +
\sqrt{-5}) = 1 + 5 = 6$. We would need $N(\pi_1) =
\pm 2$. No such $\pi_1, \pi_2$.

\begin{definition*}[Ideal]
  A set $I \subset \O_K$ is an ideal if
  \begin{align*}
    \alpha, \beta \in I
    &\implies \alpha + \beta \in I \\
    \alpha \in I, \beta \in \O_K
    &\implies \alpha \beta \in I
  \end{align*}
\end{definition*}

\begin{example*}
  The principal ideal generated by $\beta \in
  \O_K$ is
  \[ \{\beta \cdot \alpha : \alpha \in \O_K\} =
  \beta\O_K = \langle \beta \rangle = \langle
  \beta \rangle_{\O_K} \]
  Observe that $\langle \beta \rangle = \langle
  \alpha \rangle$ if and only if $\beta = u\alpha$
  for some unit $u \in \O_K^\times$.
\end{example*}

\begin{flashcard}[ideal-product-defn]
\begin{definition*}[Product of ideals]
  Let $I, J \subset \O_K$ be two ideals. We
  define
  \[ IJ = \cloze{\{\alpha_1 \beta_1 + \cdots + \alpha_k
  \beta_k : \alpha_1, \ldots, \alpha_k \in I,
  \beta_1, \ldots, \beta_k \in J\}} .\]
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{itemize}
    \item The set of ideals with this
      multiplication is a semi-group.
    \item $\alpha \mapsto \langle \alpha \rangle$
      is a homomorphism.
  \end{itemize}
\end{remark*}

\begin{definition*}[Prime ideal]
  An ideal $P \subsetneq \O_K$ is a prime ideal if
  the following holds: whenever $\alpha \beta \in
  P$ for some $\alpha, \beta \in \O_K$, then at
  least one of $\alpha, \beta$ is in $P$.
\end{definition*}

\vspace{-1em}
\textbf{Fact:} This is equivalent to $\O_K / P$
being an integral domain (recall that an integral
domain is a commutative, unital ring without
$0$-divisors).

\textbf{Fact:} $\langle a \rangle$ is a prime
ideal $\iff$ $\alpha$ is a prime in $\O_K$.

\begin{flashcard}[unique-factorisation-of-ideals-thm]
\begin{theorem*}
  Let $K$ be a \gls{nf}. Then all non-zero ideals
  in $\O_K$ are a product \cloze{of non-zero prime
  ideals, and this factorisation is unique up to
  the order of the factors.}
\end{theorem*}
\end{flashcard}

\begin{remark*}
  Addition on ideals can be defined as
  \[ I + J = \{\alpha + \beta : \alpha \in I,
  \beta \in J\} \]
  But this does not make the set of ideals a ring.
  Also, $\langle \alpha \rangle + \langle \beta
  \rangle \neq \langle \alpha + \beta \rangle$ in
  general.
\end{remark*}

\begin{flashcard}[ideals-of-OK-are-noetherian]
\begin{lemma*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item All ideals in $\O_K$ are finitely
      generated. That is, they are of the form
      $\beta_1 \O_K + \cdots + \beta_k \O_K$ for
      some $\beta_1, \ldots, \beta_k \in \O_K$.
    \item \cloze{If $I_1 \subset I_2 \subset I_3 \subset
      \cdots$ is a chain of ideals, then there
      exists $k$ such that $I_k = I_{k + 1} = I_{k
      + 2} = \cdots$.}
    \item \cloze{Any collection of ideals contains a
      maximal one with respect to $\subset$.}
  \end{enumerate}
\end{lemma*}

\fcscrap{
\vspace{-1em}
This is called Noetherian property.
}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \cloze{$I \subset \O_K$ is finitely generated
      as a $\ZZ$-module, which is even stronger
      than (1).}
    \item \cloze{$I = \bigcup_{i = 1}^\infty I_j$ is an
      ideal, so $I = \beta_1 \O_K + \cdots +
      \beta_k \O_K$. Then there exists $m$ such
      that $\beta_1, \ldots, \beta_k \in I_m$.
      Then $I = I_m = I_{m + 1} = \cdots$.}
    \item \cloze{Suppose not. Then there is an infinite
      chain of ideals
      \[ I_1 \subsetneq I_2 \subsetneq I_3
      \subsetneq \cdots \]
      contradicting (2).} \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}