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\begin{flashcard}[dedekind-pf-2]
\begin{proposition*}
  \prompt{Second step of \nameref{dedekind}:}
  \cloze{$P\O_L / Q_j$ is a factor of $(\O_K / P)[X] /
  \langle \ol{g}_j \rangle$ (``factor'' is another
  way of saying ``quotient of'').}
\end{proposition*}

\fcscrap{
\vspace{-1em}
Two possible factors (since $\ol{g}_j$ is
irreducible, so $(\O_K / P)[X] / \langle \ol{g}_j
\rangle$ is a field): $P\O_L / Q_j \cong \{0\}$,
so $Q_j = P\O_L$, or $P\O_L / Q_j \cong (\O_K /
P)[X] / \langle \ol{g}_j \rangle$, in which case
$Q_j$ is a prime and $\inertdeg(Q_j \mid P) = \deg
g_j$.

For $A \subset R$, we use $\langle A \rangle_R$ to
denote the ideal generated by $A$ in $R$.

\begin{lemma*}
  Let $R_1 \stackrel{\varphi_1}{\to} R_2
  \stackrel{\varphi_2}{\to} R_3$ be surjective
  homomorphisms of rings. Let $A \subset R_1$ such
  that $\langle \varphi_1(A) \rangle_{R_2} =
  \Ker(\varphi_2)$. Then:
  \[ \Ker(\varphi_2 \circ \varphi_1) = \langle A
  \rangle_{R_1} + \Ker(\varphi_1) .\]
\end{lemma*}

\vspace{-1em}
Key point is to show:
\[ \varphi_1(\langle A \rangle_{R_1}) = \langle
\varphi_1(A) \rangle_{R_2} .\]
This uses the surjectivity of $\varphi_1$.
}

\begin{proof}[Proof of Proposition]
  \cloze{First we prove
  \[ (\O_K / \ul{P})[X] / \langle \ol{g}_i \rangle
  \cong \O_K[\alpha] / \langle \ul{P}, g_j(\alpha)
  \rangle \]
  \[ \begin{tikzcd}[ampersand replacement=\&]
    \& (\O_K / \ul{P})[X] \ar[r, "\varphi_2"]
    \& (\O_K / \ul{P})[X] / \langle \ol{g}_j
    \rangle \\
    \O_K[X] \ar[ur, "\varphi_1"] \ar[dr, "\chi_1"]
    \\
    \& \O_K[\alpha] \ar[r, "\chi_2"]
    \& \O_K[\alpha] / \langle \ul{P}, g_j(\alpha)
    \rangle
  \end{tikzcd} \]
  $\varphi_2 \circ \varphi_1$: Let $A = \{g_j\}$.
  Then $\varphi_1(g_j) = \ol{g}_j$ generates
  $\langle \ol{g}_j \rangle = \Ker(\varphi_2)$.
  \[ \Ker(\varphi_2 \circ \varphi_1) = \langle g_j
  \rangle \O_K[X] + P\O_K[X] .\]
  $\chi_2 \circ \chi_1$: Let $A = \ul{P} \cup
  \{g_j\}$. $\chi_1(A) = \ul{P} \cup
  \{g_j(\alpha)\}$ generates $\Ker(\chi_2)$.
  \[ \Ker(\chi_2 \circ \chi_1) = \ul{P} \O_K[X] +
  \langle g_j \rangle \O_K[X] + \langle g \rangle
  \O_K[X] .\]
  Noet $g \equiv g_j \circ h \pmod{\ul{P}}$ (where
  $h$ is the product of the other $g_i$'s).
  \[ \begin{cases}
    g_j h \in \langle g_j \rangle_{\O_K[X]} \\
    g - g_j h \in \ul{P} \O_K[X]
  \end{cases} \implies g \in \ul{P} \O_K[X] +
  \langle g_j \rangle_{\O_K[X]} \]
  So the RHS of the two earlier equations are
  equal, so $\varphi_2 \circ \varphi_1$ and
  $\chi_2 \circ \chi_1$ have the same kernel.

  Observe $Q_j \cap \O_K[\alpha] \supset \langle
  \ul{P}, g_j(\alpha) \rangle_{\O_K[\alpha]}$.
  $\O_K[\alpha] / Q_j \cap \O_K[\alpha]$ is a
  quotient of $\O_K[\alpha] / \langle \ul{P},
  g_j(\alpha) \rangle$. Enough to show that
  \[ \O_L / Q_j \cong \O_K[\alpha] / (Q_j \cap
  \O_K[\alpha]) \]
  $\O_L \stackrel{\varphi}{\to} \O_L / Q_j$,
  $\varphi(\O_K[\alpha]) \cong \O_K[\alpha] / (Q_j
  \cap \O_K[\alpha])$. Enough to show:
  $\O_K[\alpha] + Q_j = \O_L$. Look at $\O_L /
  (\O_K[\alpha] + Q_j)$ in the category of abelian
  groups. This is a quotient of both $\O_L /
  \O_K[\alpha]$ and $\O_L / Q_j$.
  \[ [\O_L : \O_K[\alpha] + Q_j] \mid \gcd(\ub{[\O_L :
  \O_K[\alpha]]}_{p \nmid}, \ub{[\O_L : Q_j]}_{=
  N(Q_j)}) = 1 \]
  where $N(Q_j)$ is a power of $p$ because $Q_j$
  lies above $\ul{P}$ that lies above $p$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[Qi-plus-Qj-equals-OL-prop]
\begin{proposition*}
  \prompt{Third step of \nameref{dedekind}:}
  If $i \neq j$, then $Q_i + Q_j = \O_L$.
\end{proposition*}

\begin{proof}
  \cloze{$\ol{g}_i, \ol{g}_j$ are two distinct
  irreducible polynomials in $(\O_K / \ul{P})
  [X]$, a Euclidean domain. By Euclidean
  algorithm, there exists $\ol{h}_i, \ol{h}_j \in
  (\O_K / \ul{P})[X]$ such that
  \[ \ol{h}_i \ol{g}_i + \ol{h}_j \ol{g}_j = 1 .\]
  Let $h_i, h_j$ be lifts of $h_i$ and $h_j$ in
  $\O_K[X]$.
  \[ h_i g_i + h_j g_j \equiv 1 \pmod{\ul{P}} .\]
  There exists $f \in \ul{P} \O_K[X]$ such that
  \[ \ub{h_i(\alpha) g_i(\alpha)}_{\in Q_i} + \ub{h_j(\alpha)
  g_j(\alpha)}_{\in Q_j} + \ub{f(x)}_{\in \ul{P}} = 1 \]
  So $1 \in Q_i + Q_j$, so $Q_i + Q_j = \O_L$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[proof-of-dedekind]
\begin{proof}[Proof of \nameref{dedekind}]
  Recall: $P\O_L\ supset Q_1^{e_1} \cdots
  Q_r^{e_r}$.

  % TODO
\end{proof}
\end{flashcard}