Some changes (#209)

* Clean up some labels and refs

* Give some intuition/motivation for the Frobenius element and decomposition group

* Fix the proof of Frobenius map restriction

* Update frobenius.tex

---------

Co-authored-by: Evan Chen <[email protected]>

tex/H113/sylow.tex

@@ -1,5 +1,5 @@
 \chapter{Find all groups}
-\label{chapter:sylow}
+\label{ch:sylow}
 The following problem will hopefully never be proposed at the IMO.
 \begin{quote}
 	Let $n$ be a positive integer and let $S = \left\{ 1,\dots,n \right\}$.

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 +1,5 @@
 \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 +13,12 @@ \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 +31,7 @@ \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}
@@ -338,6 +345,8 @@ \section{Frobenius elements behave well with restriction}
 	\Frob_{\kP} \colon L \to L \]
 and want to know how these are related.
 
+Both maps $\Frob_{\kP}$ and $\Frob_{\kp}$ induce the power-of-$p$ map in the corresponding quotient field, hence we would expect them to be naturally the same.
+
 \begin{theorem}
 	[Restrictions of Frobenius elements]
 	Assume $L/\QQ$ and $K/\QQ$ are both Galois.
@@ -346,22 +355,13 @@ \section{Frobenius elements behave well with restriction}
 	i.e.\ for every $\alpha \in K$,
 	\[ \Frob_\kp(\alpha) = \Frob_{\kP}(\alpha). \]
 \end{theorem}
-%\begin{proof}
-%	We know
-%	\[ \Frob_{\kP}(\alpha) \equiv \alpha^p \pmod{\kP}
-%		\quad \forall \alpha \in \OO_L \]
-%	from the definition.
-%	\begin{ques}
-%		Deduce that
-%		\[ \Frob_{\kP}(\alpha) \equiv \alpha^p \pmod{\kp}
-%			\quad \forall \alpha \in \OO_K. \]
-%		(This is weaker than the previous statement in two ways!)
-%	\end{ques}
-%	Thus $\Frob_{\kP}$ restricted to $\OO_K$ satisfies the
-%	characterizing property of $\Frob_\kp$.
-%\end{proof}
 \begin{proof}
-	TODO: Broken proof. Needs repair.
+	First, $K/\QQ$ is normal, so $\Frob_{\kP}$ fixes the image of $K$, that is,
+	$\Frob_{\kP} \restrict{K} \in \Gal(K/\QQ)$ is well-defined.
+
+	We have the natural map $\phi \colon \OO_K \to \OO_L \to \OO_L/\kP$, and the quotient map $q\colon \OO_K \to \OO_K / \kp$. Since $\kP \subseteq \kP$, then $\phi$ factors through $q$ to give a natural field homomorphism $\OO_K / \kp \to \OO_L / \kP$.
+
+	Since a field homomorphism is injective, $\Frob_{\kP}$ induces the power-of-$p$ map on $\OO_L / \kP$, and everything is commutative, the theorem follows.
 \end{proof}
 In short, the point of this section is that
 \begin{moral}

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

tex/frontmatter/digraph.tex

@@ -13,7 +13,7 @@
 \reqhalfcourse 55,45:{Ch 2,6-8}{\hyperref[part:basictop]{Topology}}{}
 \halfcourse 33,45:{Ch 9-15,18}{\hyperref[part:linalg]{Lin Alg}}{}
 \halfcourse 5,35:{Ch 16}{\hyperref[part:groups]{Grp Act}}{}
-\halfcourse 5,24:{Ch 17}{\hyperref[chapter:sylow]{Grp Classif}}{}
+\halfcourse 5,24:{Ch 17}{\hyperref[ch:sylow]{Grp Classif}}{}
 \halfcourse 30,35:{Ch 19-22}{\hyperref[part:repth]{Rep Th}}{}
 \halfcourse 45,43:{Ch 23-25}{\hyperref[part:quantum]{Quantum}}{}
 \halfcourse 64,38:{Ch 26-30}{\hyperref[part:calc]{Calc}}{}

tex/linalg/fourier.tex

@@ -344,7 +344,7 @@ \section{Summary, and another teaser}
 \section{Parseval and friends}
 Here is a fun section in which you get to learn a lot of big names quickly.
 Basically, we can take each of the three results
-from Proposition~\ref{prop:orthonormal},
+from \Cref{prop:orthonormal},
 translate it into the context of our Fourier analysis
 (for which we have an orthonormal basis of the Hilbert space),
 and get a big-name result.

tex/measure/pontryagin.tex

@@ -181,7 +181,7 @@ \section{The orthonormal basis in the compact case}
 \end{itemize}
 
 \section{The Fourier transform of the non-compact case}
-If $G$ is LCA but not compact, then Theorem~\ref{thm:god} becomes false.
+If $G$ is LCA but not compact, then \Cref{thm:god} becomes false.
 On the other hand, it's still possible to define $\wh G$.
 We can then try to write the Fourier coefficients anyways:
 let \[ \wh f(\xi) = \int_G f \cdot \ol{e_\xi} \; d\mu \]