Give some intuition/motivation for the Frobenius element and decompos…

…ition group

tex/alg-NT/finite-field.tex

@@ -182,6 +182,7 @@ \section{The Galois theory of finite fields}
 	\[ \sigma_p(x) = x^p \]
 	is an automorphism, and moreover fixes $\FF_p$.
 \end{theorem}
+This is called the Frobenius automorphism, and will re-appear later on in \Cref{ch:frobenius-element}.
 \begin{proof}
 	It's a homomorphism since it fixes $1$,
 	respects multiplication,

tex/alg-NT/frobenius.tex

@@ -1,4 +1,4 @@
-\chapter{The Frobenius element}
+\chapter{The Frobenius element}\label{ch:frobenius-element}
 Throughout this chapter $K/\QQ$ is a Galois extension with Galois group $G$,
 $p$ is an \emph{unramified} rational prime in $K$, and $\kp$ is a prime above it.
 Picture:
@@ -12,6 +12,12 @@ \chapter{The Frobenius element}
 	\QQ & \supset & \ZZ & (p) & \FF_p
 \end{tikzcd}
 \end{center}
+
+We recall that the $p$-th power map $\sigma \colon \FF_{p^f} \to \FF_{p^f}$ is an automorphism, and it's called the Frobenius map on $\FF_{p^f}$.
+We can try to extend this map to a $K \to K$ map by $\sigma(x) = x^p$, unfortunately this doesn't make it a field automorphism.
+
+Surprisingly, it is nevertheless possible to extend this to some field automorphism $\sigma \in \Gal(K/\QQ)$.
+
 If $p$ is unramified, then one can show there
 is a unique $\sigma \in \Gal(K/\QQ)$ such that
 $\sigma(\alpha) \equiv \alpha^p \pmod{\kp}$ for every prime $p$.
@@ -24,7 +30,7 @@ \section{Frobenius elements}
 	Assume $K/\QQ$ is Galois with Galois group $G$.
 	Let $p$ be a rational prime unramified in $K$, and $\kp$ a prime above it.
 	There is a \emph{unique} element $\Frob_\kp \in G$
-	with the property that
+	with the property that, for all $\alpha \in \OO_K$,
 	\[ \Frob_\kp(\alpha) \equiv \alpha^{p} \pmod{\kp}. \]
 	It is called the \vocab{Frobenius element} at $\kp$, and has order $f$.
 \end{theorem}

tex/alg-NT/ramification.tex

@@ -262,22 +262,27 @@ \section{(Optional) Decomposition and inertia groups}
 	How are $\Gal\left( (\OO_K/\kp) / \FF_p \right)$
 	and $\Gal(K/\QQ)$ related?
 \end{quote}
-Absurdly enough, there is an explicit answer:
-\textbf{it's just the stabilizer of $\kp$, at least when
-$p$ is unramified}.
+
+First, every $\sigma \in \Gal(K/\QQ)$ induces an automorphism of $\OO_K$, which induces a map
+$\OO_K \to \OO_K/\kp$ by
+\[ \alpha \mapsto \sigma(\alpha) \pmod\kp. \]
+For this to induce a map in $\Gal\left( (\OO_K/\kp) / \FF_p \right)$, it's necessary that $\sigma(\kp) \subseteq \kp$. So, we consider the subset of automorphisms that fixes $\kp$:
 \begin{definition}
 	Let $D_\kp \subseteq \Gal(K/\QQ)$ be the stabilizer of $\kp$, that is
 	\[ D_\kp \defeq \left\{ \sigma \in \Gal(K/\QQ) \mid \sigma\kp = \kp \right\}. \]
 	We say $D_\kp$ is the \vocab{decomposition group} of $\kp$.
 \end{definition}
-Then, every $\sigma \in D_\kp$ induces an automorphism of $\OO_K / \kp$ by
-\[ \alpha \mapsto \sigma(\alpha) \pmod\kp. \]
+Note that this definition is in fact equivalent to the set of $\sigma$ such that $\sigma(\kp) \subseteq \kp$,
+because a field isomorphism fixes the ideal norm $\Norm(\kp)$.
+
 So there's a natural map
 \[ D_\kp \taking\theta \Gal\left( (\OO_K/\kp) / \FF_p \right) \]
 by declaring $\theta(\sigma)$ to just be ``$\sigma \pmod \kp$''.
 The fact that $\sigma \in D_\kp$ (i.e.\ $\sigma$ fixes $\kp$)
 ensures this map is well-defined.
 
+Surprisingly, every element of $\Gal\left( (\OO_K/\kp) / \FF_p \right)$ arises this way from some field automorphism of $K$.
+
 \begin{theorem}[Decomposition group and Galois group]
 	\label{thm:decomposition}
 	Define $\theta$ as above. Then