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Remarks:
\begin{itemize}
  \item $\O_K$ is not a prime ideal.
  \item $\{0\}$ is not an integral domain (note
    that $\{0\}$ is a ring, with $1 = 0$).
  \item $\langle 0 \rangle \subset R$ is a prime
    ideal if and only if $R$ is an integral domain.
  \item \glsnoundefn{prime}{prime}{primes}
    \textbf{$I \subset \O_K$ is a prime if it is
    a non-zero prime ideal.}
\end{itemize}

\begin{definition*}[Maximal ideal]
  An ideal $I \subsetneq \O_K$ is maximal if the
  only ideals $J$ with $I \subset J \subset \O_K$
  are $I$ and $\O_K$.
\end{definition*}

\vspace{-1em}
\textbf{Fact:} $I$ is maximal if and only if $\O_K / I$ is
a field.

\begin{flashcard}[primes-and-max-ideals-same-in-OK]
\begin{lemma*}
  In $\O_K$, \glspl{prime} and maximal ideals are the
  same.
\end{lemma*}

\begin{proof}
  \cloze{First we prove that $\O_K / I$ is finite for all
  non-zero ideals. Enough to show that the rank of
  $I$ is $d = [K : \QQ]$ as a $\ZZ$-module. Take
  an \gls{intbasis} $\alpha_1, \ldots, \alpha_d
  \in \O_K$. Let $0 \neq \beta \in I$. Then
  $\beta \alpha_1, \ldots, \beta \alpha_d \in I$
  is linearly independent over $\QQ$. Then
  $\rank(I) = d$. Now the lemma follows by the
  fact that finite integral domains are fields.
  Hint: Show that $\O_K / I$ is equal to its field
  of fractions.}
\end{proof}
\end{flashcard}

\begin{flashcard}[finitely-generated-alpha-in-OK]
\begin{lemma*}
  Let $\alpha \in K$. Suppose that there is a
  finitely generated $\O_K$-module $M \subset K$
  such that $\alpha M \subset M$. Then $\alpha \in
  \O_K$.
\end{lemma*}

\fcscrap{
\begin{remark*}
  Integral domains that satisfy this property with
  the field of fractions playing the role of $K$
  are called integrally closed.
\end{remark*}
}

\begin{proof}
  $M$ is also finitely generated as a
  $\ZZ$-module, because $\O_K$ is finitely
  generated as a $\ZZ$-module. Then $\alpha$ is an
  algebraic integer, hence $\alpha \in \O_K$.
\end{proof}
\end{flashcard}

An integral domain satisfying the conclusions of
all 3 lemmas is called a Dedekind domain.

Let $I$ be a non-zero ideal. By the Noetherian
property, there exists a maximal ideal $P$ such
that $P \supset I$. Then $P$ is a \gls{prime}. It would
be great if we had:
\[ I \supset J \iff \exists I_2 \text{ ideal such
that } I I_2 = J .\]
Observe that:
\begin{itemize}
  \item This holds for principal ideals:
    \begin{align*}
      \langle \beta \rangle \subset \langle
      \alpha \rangle
      &\iff \beta \in \langle \alpha \rangle \\
      &\iff \beta = \gamma \alpha
      &&\text{for some $\gamma$} \\
      &\iff \langle \beta \rangle = \langle \gamma
      \rangle \langle \alpha \rangle
    \end{align*}
  \item The $\Leftarrow$ direction is trivial.
    Indeed, if $\alpha \in I$, $\beta \in I_2$,
    then $\alpha \beta \in I$. The collection
    of all possible such $\alpha \beta$ generate
    $J$, so indeed $J \subset I$.
\end{itemize}
If this was true, we could write $I = PI_1$ for
some ideal $I_1$.

\begin{flashcard}[fractional-ideal-defn]
\begin{definition*}[Fractional Ideal]
  \glsnoundefn{fideal}{fractional
  ideal}{fractional ideals}
  \cloze{A \emph{fractional ideal} is a finitely
  generated $\O_K$-submodule of $K$.}
\end{definition*}
\end{flashcard}

\begin{note*}
  We extend the definition of multiplication of
  ideals to get multiplication of \glspl{fideal}.
\end{note*}

\begin{flashcard}[fideal-iff-aI-is-ideal-lemma]
\begin{lemma*}
  If $I \subset K$ is a \gls{fideal}, then
  $\exists a \in \ZZ$ such that $a \cdot I$ is an
  ideal. Conversely, if $I \subset \O_K$ is an
  ideal, then $\alpha \cdot I$ is a \gls{fideal}
  for all $\alpha \in K$.
\end{lemma*}

\begin{proof}
  \cloze{Let $\alpha_1, \ldots, \alpha_k$ generate $I$ as
  an $\O_K$-module. Write them as $\QQ$-linear
  combinations of an \gls{intbasis}. Take $a$ to
  be a common denominator of all the coefficients.
  Then $a \alpha_j \in \O_K$. Hence $aI
  \subset \O_K$. Also, $aI$ is an $\O_K$-module.
  Then $aI$ is an ideal.

  Conversely, if $I$ is an ideal, then it is
  a finitely generated $\O_K$-module, then so is
  $\alpha I$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[prime-ideal-inverse-prop]
\begin{proposition*}
  Let $P$ be a \gls{prime}. Then there exists a
  \gls{fideal} $P'$ such that $PP' = \langle 1
  \rangle$.
\end{proposition*}

\begin{proof}
  \cloze{Let $P' = \{\alpha \in K \st \alpha P \subset
  \O_K\}$. This is an $\O_K$-module. Moreover,
  $\beta P' \subset \O_K$ for any $0 \neq \beta
  \in P$. Then $\beta P'$ is finitely generated as
  a $\ZZ$-module. Then $P'$ is also finitely
  generated, so $P'$ is a \gls{fideal}. Observe
  $P'P \subset \O_K$, hence it is an ideal (note
  that \glspl{fideal} contained in $\O_K$ are
  always ideals). Also by $\O_K \subset P'$, $PP'
  \supset P \O_K = P$. $\O_K \supset P'P \supset
  P$, so $P'P$ is $\O_K$ or $P$. To exclude the
  second possibility, we show that there exists
  $\alpha \in P' \setminus \O_K$. Then we cannot
  have $\alpha P \subset P$, because that would
  imply $\alpha \in \O_K$, by $\O_K$ being
  integrally closed.

  Let $0 \neq \beta \in P$. Let $k$ be the smallest number
  such that there exists $Q_1, \ldots, Q_k$ primes
  with $Q_1, \ldots, Q_k \subset \langle \beta
  \rangle$ (see next lemma for existence of $k$).
  Note that $Q_1, \ldots, Q_k \subset P$. Since
  $P$ is a prime ideal, there exists $j$ with $Q_j \subset
  P$ (we use the fact that $IJ \subset P \implies
  I \subset P$ or $J \subset P$). But $Q_j$ is a
  maximal ideal, so $Q_j = P$. Let $\gamma \in Q_1
  \cdots Q_{j - 1} Q_{j + 1} \cdots Q_k \setminus
  \langle \beta \rangle$. Such a $\gamma$ exists by the
  minimality of $k$. Then $\gamma \notin \langle
  \beta \rangle \implies \frac{\gamma}{\beta}
  \notin \O_K$. Then $P\gamma \in \langle \beta
  \rangle \implies \frac{\gamma}{\beta} P \subset
  \O_K$. So we can take $\alpha =
  \frac{\gamma}{\beta}$.
  }
\end{proof}
\end{flashcard}