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\begin{flashcard}[prime-factorisation-of-ideals-lemma]
\begin{lemma*}
  Let $0 \neq I \subset \O_K$ be an ideal. Then
  there are primes $P_1, \ldots, P_k \subset \O_K$
  such that $I \supset P_1 P_2 \cdots P_k$.
\end{lemma*}

\begin{proof}
  \cloze{Trivial if $I$ is a prime. Suppose that the lemma
  is false. Let $I$ be maximal among the ideals
  for which it fails (since $\O_K$ is Noetherian).
  Then $I$ is not a prime.
  Then there exists $\alpha, \beta \in \O_K
  \setminus I$ such that $\alpha\beta \in I$. Then
  \[ \ub{(I + \langle \alpha \rangle)}_{\supsetneq
  I}\ub{(I + \langle \beta \rangle)}_{\supsetneq I}
  \subset I \]
  By hypothesis, there exists $Q_1, \ldots, Q_l,
  R_1, \ldots, R_m \subset \O_K$ primes such that
  \[ Q_1 \cdots Q_l \subseteq I + \langle \alpha
  \rangle \qquad \text{and} \qquad R_1 \cdots R_m
  \subseteq I + \langle \beta \rangle .\]
  Multiplying these together, we see that the
  lemma holds for $I$ also.}
\end{proof}
\end{flashcard}

\begin{flashcard}[unique-prime-factorisation-of-ideals-thm]
\begin{theorem*}
  Non-zero ideals in $\O_K$ are products of primes
  in a unique way up to the order of the factors.
\end{theorem*}

\begin{proof}
  \cloze{Let $i$ be a non-zero ideal. Let $P_1 \subsetneq
  \O_K$ be an ideal that is maximal among those
  that contain $I$. Then $P_1$ is a maximal ideal,
  hence prime. Let $I_1 = I \cdot P^{-1}$
  ($P^{-1}$ is notation for $P'$ from the
  Proposition about $PP' = \langle 1 \rangle$).
  Observe that $I_1 P = I$ and $I_1 \subset \O_K$
  is an ideal. This is because $I_1 = I \cdot
  P^{-1} \subset PP^{-1} = \langle 1 \rangle =
  \O_K$. Also, $I_1 \supsetneq I$, for otherwise
  we would have $\alpha I \subset I$ for all
  $\alpha \in P^{-1}$, and this would imply $P'
  \subset \O_K$. Keep going with this, and we get
  sequences $P_1, P_2, \ldots$ and $I_1 \subsetneq
  I_2 \subsetneq I_3 \subsetneq \cdots$ such that
  $I_{j - 1} = I_j P_j$. This must terminate, so
  $I_k = \O_K$ for some $k$. Then
  \[ I = P_1 I_1 = P_1 P_2 I_2 = \cdots = P_1 P_2
  \cdots P_k I_k = P_1 \cdots P_k .\]

  We now show that $P_1 \cdots P_k = Q_1 \cdots
  Q_l$ implies $k = l$ and $P_j = Q_{\sigma(j)}$
  for some permutation $\sigma$. It is enough to
  show that $P_1 = Q_j$ for some $j$, because then
  the claim follows by induction on $k + l$.
  Observe that $P_1 \supset P_1 \cdots P_k = Q_1
  \cdots Q_l$. By the argument for the proof of
  the lemma, $P_1$ must be equal to one of the
  $Q_j$'s.}
\end{proof}
\end{flashcard}

\begin{flashcard}[fractional-ideals-group]
\begin{corollary*}
  For all non-zero fractional ideals $I \subset
  K$, there exists $I^{-1} \subseteq K$ a
  fractional ideal such that $II^{-1} = \langle 1
  \rangle$. That is, fractional ideals form a
  group.
\end{corollary*}

\begin{proof}
  \cloze{If $I \subset \O_K$ is an ideal, then $I = P_1
  \cdots P_k$ for some primes. We can use the
  lemma and take: $I^{-1} = P_1^{-1} \cdots
  P_k^{-1}$. In the general case, $I = J_1 \cdots
  J_2^{-1}$, where $J_1, J_2 \subseteq \O_K$. In
  fact we can take $J_2 = \langle a \rangle$ for
  some $a \in \ZZ$. Then use the special case, and
  take $I^{-1} = J^{-1} J_2$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[ideal-subset-iff-factor-coro]
\begin{corollary*}
  Let $0 \neq I, J \subset \O_K$ be ideals. Then
  \[ I \supset J \iff \exists I_2 \subset \O_K
  \text{ such that } II_2 = J .\]
\end{corollary*}

\begin{proof}
  \cloze{Take $I_2 = J \cdot I^{-1}$. We need to show
  that $J \cdot I^{-1} \subseteq \O_K$. Let
  $\alpha \in J \cdot I^{-1}$. Then $\alpha I
  \subset J \subset I$, so by integrally
  closedness, $\alpha \in \O_K$ as needed.}
\end{proof}
\end{flashcard}

\begin{flashcard}[O-K-UFD-iff-PID]
\begin{corollary*}
  $\O_K$ is a UFD if and only if it is a PID.
\end{corollary*}

\begin{proof}
  \cloze{PID $\implies$ UFD holds in general.

  Suppose it is a UFD. All elements $\alpha \in
  \O_K$ are the product of primes. All principal
  ideals are products of principal prime ideals.
  Let $I$ be an ideal, and let $\beta \in I$.
  $\langle \beta \rangle \subset I \implies I \mid
  \langle \beta \rangle$. So all prime factors of
  $I$ are prime factors of $\langle \beta
  \rangle$, hence principal.}
\end{proof}
\end{flashcard}
