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Recall $J \mid I \iff J \supset I$, by a Corollary
from last time.

\begin{flashcard}[gcd-lcm-defn]
\begin{definition*}[$\gcd$ and $\lcm$]
  \cloze{$\gcd(I_1, I_2)$ is the smallest ideal $J$ such
  that $J \mid I_1$, $J \mid I_2$.

  $\lcm(I_1, I_2)$ is the largest ideal $J$ such
  that $I_1, I_2 \mid J$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\textbf{Fact:}
\begin{align*}
  \gcd(I_1, I_2)
  &= I_1 + I_2 = \{\alpha + \beta :
  \alpha \in I, \beta \in J\} \\
  \lcm(I_1, I_2)
  &= I_1 \cap I_2
\end{align*}

\subsubsection*{Norm of ideals}

\begin{flashcard}[norm-of-ideals]
\begin{definition*}[Norm of an ideal]
  \glssymboldefn{idealnorm}{$N(I)$}{$N(I)$}
  \cloze{Let $I \subset \O_K$ be an ideal. Then $N(I) =
  |\O_K / I|$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\textbf{Recall:} If $I \neq \langle 0 \rangle$,
then $\N(I) < \infty$. If $\alpha_1, \ldots,
\alpha_d$ generate $I$ as a free $\ZZ$-module,
then
\[ \N(I) = \left( \frac{\tdisc(\alpha_1, \ldots,
\alpha_d)}{\nfdisc(K)} \right)^{1/2} \]

\begin{flashcard}[multiplicativity-of-norm]
\begin{proposition*}
  Let $I, J \subset \O_K$ be non-zero ideals. Then
  \[ \N(IJ) = \N(I) \cdot \N(J) .\]
\end{proposition*}

\begin{proof}
  \cloze{Enough to prove when $J$ is a prime. This
  special case implies that
  \[ \N(P_1 \cdots P_k) = \N(P_1) \cdots \N(P_k) \]
  for primes $P_1, \ldots, P_k$. Apply this to the
  factorisation of $I, J, IJ$ to deduce the
  general case.

  Now let $J$ be a prime. Observe $\O_K / I \cong
  (\O_K / IJ) / (I / IJ)$. So
  \[ \N(I) = \N(IJ) / |I / IJ| .\]
  So it is enough to show $\N(J) = |I / IJ|$.

  Let $\alpha_1, \ldots, \alpha_{N(J)}$ be
  representatives for the cosets in $\O_K / J$.
  Let $\beta \in I \setminus IJ$.

  \textbf{Claim:} $\beta \alpha_1, \ldots, \beta
  \alpha_{N(J)}$ are representatives for $I / IJ$.

  Proof:
  \begin{enumerate}[(1)]
    \item Show $\forall \gamma \in I$, $\exists
      \alpha_j$ such that $\gamma \equiv \beta
      \alpha_j \pmod{IJ}$. Enough to show that
      $\exists \alpha \in \O_K$ such that $\gamma
      \equiv \beta \alpha \pmod{IJ}$, because
      $\exists \alpha_j \equiv \alpha \pmod{J}$.
      Need to find $\alpha$ such that $\gamma -
      \beta \alpha \in IJ$. This is the same as
      showing that $\gamma \in IJ + \langle \beta
      \rangle$. Note $\langle \beta \rangle = I
      \cdot P_1 \cdots P_k$, where none of the
      $P_j$'s are $J$. Now
      \begin{align*}
        IJ + \langle \beta \rangle
        &= \gcd(IJ, \langle \beta \rangle) \\
        &= I
      \end{align*}
      That is good because $\gamma \in I$.
    \item Want to show $\beta \alpha_i \equiv \beta \alpha_j
      \pmod{IJ}$ implies $i = j$. We have $IJ \mid
      \langle \beta \rangle \langle \alpha_i -
      \alpha_j \rangle$. This is
      \begin{align*}
        IJ
        &\mid I \cdot P_1 \cdots P_k \langle
        \alpha_i - \alpha_j \rangle \\
        \implies IJ
        &\mid P_1 \cdots P_k \langle \alpha_i -
        \alpha_j \rangle \\
        J
        &\mid \langle \alpha_i - \alpha_j \rangle
        \\
        \implies i
        &= j
      \end{align*}
  \end{enumerate}}
\end{proof}
\end{flashcard}

\begin{flashcard}[ideal-norm-equals-field-norm-lemma]
\begin{lemma*}
  Let $\alpha \neq 0 \in \O_K$. Then $N(\langle
  \alpha \rangle) = |N_{K / \QQ}(\alpha)|$.
\end{lemma*}

\begin{proof}
  \cloze{Let $\alpha_1, \ldots, \alpha_d$ be an
  \gls{intbasis}. Then
  \[ \langle \alpha \rangle = \alpha \alpha_1 \ZZ
  \oplus \cdots \oplus \alpha \alpha_d \ZZ .\]
  Now we can calculate:
  \begin{align*}
    \N(\langle \alpha \rangle)^2
    &= \frac{\tdisc(\alpha \alpha_1, \ldots,
    \alpha \alpha_d)}{\tdisc(\alpha_1, \ldots,
    \alpha_d)}
  \end{align*}
  and
  \begin{align*}
    \tdisc(\alpha \alpha_1, \ldots, \alpha
    \alpha_d)
    &= \det(\sigma_i(\alpha \alpha_j))^2 \\
    \sigma_1(\alpha)^2 \cdots \sigma_d(\alpha)^2
    \cdot \det(\sigma_i(\alpha_j))^2 \\
    &= N_{K / \QQ}(\alpha)^2 \nfdisc(K) \qedhere
  \end{align*}}
\end{proof}
\end{flashcard}

\begin{flashcard}[nf-extension-ideal-association]
Let $L / K$ be an extension of \glspl{nf}. Given
an ideal $I \subset \O_K$, we can associate to it
an ideal in $\O_L$:
\cloze{\[ I \cdot \O_L = \{\alpha_1 \beta_1 + \cdots +
\alpha_k \beta_k : \alpha_i \in I, \beta_i \in
\O_L\} \]
This is indeed an ideal in $\O_L$.

It is the smallest ideal that contains $I$.

\textbf{Fact:}
\[ (I_1 \O_L) \cdot (I_2 \O_L) = (I_1 I_2)\O_L \]
}
\end{flashcard}

\begin{flashcard}[nf-extension-ideal-association-reverse]
Given an ideal $I \subset \O_L$, we can associate
to it one in $\O_K$:\cloze{ $I \cap \O_K$. Again this is
an ideal. In general:
\[ (I \cap \O_K)(I_2 \cap \O_K) \neq (I_1 I_2 \cap
\O_K) \]
}
\end{flashcard}

\begin{flashcard}[ideal-association-division-equivalence-lemma]
\begin{lemma*}
  The following are equivalent for $P \subset
  \O_K$ and $Q \subset \O_L$ primes:
  \begin{enumerate}[(1)]
    \item $Q \mid P\O_L$.
    \item $Q \cap \O_K = P$.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1) $\implies$ (2)]
    \item[(1) $\implies$ (2)] \cloze{$Q \supset P\O_L \supset
      P$. So $Q \cap \O_K \supset P$. But $P$ is a
      maximal ideal, so enough to show that $Q
      \cap \O_K \subsetneq \O_K$. And this follows
      by $1 \notin Q$.}
    \item[(2) $\implies$ (1)] \cloze{(2) implies $Q
      \supset P$ and hence $Q \supset P \O_L$
      because $Q$ is an ideal. Then $Q \mid P
      \O_L$.}\qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[lies-over-defn]
\begin{definition*}[Lying above]
  \glsverbdefn{la}{lies above}{lies above}
  \glsverbdefn{lo}{lies over}{lies over}
  \glsverbdefn{lu}{lies under}{lies under}
  \cloze{Let $P \subset \O_K$, $Q \subset \O_L$ be
  primes. If $Q \mid P \O_L$ (or equivalently $Q
  \cap \O_K = P$), we say that $Q$ lies above or
  over $P$, and $P$ lies under or below $Q$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[lies-under-existence-uniqueness-lemma]
\begin{lemma*}
  For all primes $Q \subset \O_L$, there is a
  unique prime in $\O_K$ that \gls{lu} it. For
  all primes $P \subset \O_K$, there is at least
  one in $\O_L$ that \gls{lo} it.
\end{lemma*}

\begin{proof}
  \cloze{
  \phantom{}
  \begin{enumerate}[(i)]
    \item Need to show $Q \cap \O_K$ is a prime.
      Observe that $1 \notin Q \cap \O_K$, so is a
      proper ideal. Since $\O_L / Q$ is finite,
      the image of $\O_K$ ($\O_K / Q \cap \O_K$)
      in it is also finite. Since $\O_K$ is
      infinite, $Q \cap \O_K \neq \langle 0
      \rangle$. Suppose that $\alpha, \beta \in
      \O_K$ and $\alpha\beta \in Q \cap \O_K$.
      Then $\alpha\beta \in Q$, a prime ideal,
      hence $\alpha \in Q$ or $\beta \in Q$. So
      $\alpha \in Q \cap \O_K$ or $\beta \in Q
      \cap \O_K$. So $Q \cap \O_K$ is indeed a
      prime.
    \item We only need to show that $P\O_L$ is a
      proper ideal, because then it has prime
      factors. To that end: $\O_L =
      (P\O_L)(P^{-1}\O_L)$. If $\O_L = P\O_L$,
      then
      \[ \O_L = \O_L (P^{-1} \O_L) = P^{-1}\O_L \]
      so $P^{-1} \subset \O_L$. But we have seen
      that $P^{-1}$ contains elements which are
      not algebraic integers.
  \end{enumerate}
}
\end{proof}
\end{flashcard}
