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\title{Math 129 HW 9}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{The Different and its Properties}

Let $L/K$ be an extension of number fields and $S,R$ the associated number rings.
In this section we define the \textit{different} of an additive subgroup $A\subset L$ and explore some of its
properties.

Recall that $L$ is a $K$ vector space equipped with a canonical symmetric nondegenerate bilinear form
\begin{eqnarray*}
	\tr^L_K:L\longrightarrow K.
\end{eqnarray*}
Therefore, for any additive subgroup $A\subset L$ as above, we can define the vector space dual $A^*$ by
\begin{eqnarray*}
	A^*=\{\alpha\in L:\tr^L_K(\alpha A)\subset R\}.
\end{eqnarray*}
We may also extend the definition of the inverse of any fractional ideal to an additive subgroup $A$ by defining
\begin{eqnarray*}
	A^{-1}=\{\alpha\in L:\alpha A\subset S\}.
\end{eqnarray*}
In the case that $A$ is in fact a fractional ideal, both $A^*$ and $A^{-1}$ are also fractional ideals.

\begin{defn}
	Let $A\subset L$ be an additive subgroup.  The \textbf{different} of $A$ is defined as
	\begin{eqnarray*}
		\diff A=(A^*)^{-1}.
	\end{eqnarray*}
\end{defn}
By the above remarks, $\diff A$ is a fractional ideal whenever $A$ is.  We will be especially interested in
the different of $S$ as this encodes important information about the extension $L/K$.  First, we study the
special case $S=R[\alpha]$ for some $\alpha\in S$.

\begin{thm}
	Let $S=R[\alpha]$ for some $\alpha\in S$ and let $f\in R[X]$ be the minimal polynomial of $\alpha$.  
	Then $\diff S$ is the principal ideal $f^{\prime}(\alpha)S$.
\end{thm}

\begin{proof}
	Since $\diff S=(R[\alpha]^*)^{-1}$, we first determine the structure of the dual
	$R[\alpha]^*$.  To do this, we need to know a dual basis of $R[\alpha]$.  We have shown that
	if 
	\begin{eqnarray}
		f(X)&=&a_0+a_1X+\cdots+a_{n-1}X^n\\
		&=&(X-\alpha)(\gamma_0+\gamma_1x+\cdots+\gamma_{n-1}x^{n-1})
	\label{factor}
	\end{eqnarray}
	where $\gamma_i\in S$, then the dual basis of $1,\alpha,\ldots,\alpha^{n-1}$ with respect
	to $\tr^L_K$ is
	\begin{eqnarray*}
		\left\{\frac{\gamma_0}{f^{\prime}(\alpha)},\ldots,\frac{\gamma_{n-1}}{f^{\prime}(\alpha)}\right\}.
	\end{eqnarray*}
	Moreover, it is not difficult to see, by comparing coefficients of $X^i$ in \ref{factor} that
	the $\gamma_i$ are given by the formulae
	\begin{eqnarray*}
		\gamma_{n-i}=\alpha^{i-1}+a_{n-1}\alpha^{i-2}+\cdots+a_{n-i+1}.
	\end{eqnarray*}
	Therefore, 
	\begin{eqnarray*}
		R[\alpha]&=&R+R\alpha+\cdots+R\alpha^{n-1}\\
		&=&R\gamma_0+R\gamma_1+\cdots+R\gamma_{n-1}.
	\end{eqnarray*}
	Thus, we have
	\begin{eqnarray*}
		R[\alpha]^*&=&f^{\prime}(\alpha)^{-1}(R\gamma_0+R\gamma_1+\cdots+R\gamma_{n-1})\\
		&=&f^{\prime}(\alpha)^{-1}R[\alpha].
	\end{eqnarray*}
	It now follows that $\diff R[\alpha]=f^{\prime}(\alpha)S$.
\end{proof}

Recall that when $S=R[\alpha]$ for $\alpha\in S$ we have $\disc S=\N^L_K(\alpha)$.  Notice that from the definition
of the norm of an ideal, we have
\begin{eqnarray}
	\N^L_K(\diff S)&=&\N^L_K(f^{\prime}(\alpha)S)\nonumber\\
	&=&\N^L_K(\alpha)R\nonumber\\
	&=&(\disc S) R.
\label{diff}
\end{eqnarray}
In general, $\diff S$ is the $S$--ideal generated by the set
\begin{eqnarray*}
	\left\{f^{\prime}(\alpha):L=K(\alpha)\quad\text{and}\quad \alpha\in S\right\},
\end{eqnarray*}
though we will not prove this.  We simply note that the above argument can be used to show that
whenever $L=K(\alpha)$ and $\alpha\in S$ then $f^{\prime}(\alpha)\in\diff S$.
The equality \ref{diff} suggests an intimate relationship between the different and the discriminant
in general, even when $S$ can not be written as $R[\alpha]$ for any $\alpha\in S$.  

\begin{thm}\label{Qdiv}
	Let $P$ be a prime of $K$ and $Q$ a prime of $L$ above $P$.  Then $Q$ is ramified over $P$
	if and only if $Q|\diff S$.  Moreover, if $e=e(Q|P)$ then $Q^{e-1}|\diff S$. 
\end{thm}

The different satisfies another important property: namely, that of multiplicitivity in towers.

\begin{thm}\label{tower}
	Let $K\subset L\subset M$ be number field extensions, and $R,S,T$ the associated number rings.
	Further, let $\diff(T|S),\diff(S|R),\diff(T|R)$ denote the differents corresponding to the
	extensions $M/L,L/K,M/K$ respectively.  Then  
	\begin{eqnarray*}
		\diff(T|R)=\diff(T|S)\diff(S|R),
	\end{eqnarray*}
	where we are considering the $S$--ideal  $\diff(S|R)$ as a $T$--ideal.
\end{thm}

In the special case that we are considering the extension of number fields $K/\Q$, we have the following result:
\begin{thm}\label{difdisc}
	\begin{eqnarray*}
		\|\diff R\|=|\disc R|.
	\end{eqnarray*}
\end{thm}

A very useful consequence of this result is
\begin{cor}
	Let $M/L$ be an extension of number fields; $T,S$ the associated number rings.  Then
	\begin{eqnarray*}
		\disc(S)^{[M:L]}|\disc(T).
	\end{eqnarray*}
\end{cor}

\begin{proof}
	Set $K=Q$ in Theorem \ref{tower} to obtain
	\begin{eqnarray*}
		\diff(T)=\diff(T|S)\diff(S).
	\end{eqnarray*}
	Since $\diff(S)$ is an $S$--ideal, we have
	\begin{eqnarray*}
		\|\diff(S)\|=S^{[M:L]}.
	\end{eqnarray*}
	Therefore, since $\|IJ\|=\|I\|\|J\|$ for any ideals $I,J$, we have by Theorem \ref{difdisc}
	\begin{eqnarray*}
		|\disc(T)|&=&\|\diff(T)\|\\
		&=&\|\diff(T|S)\|\|\diff(S)\|\\
		&=&\|\diff(T|S)\| S^{[M:L]},
	\end{eqnarray*}
	which completes the proof.
\end{proof}


\section{Higher Ramification Groups}

Let $L/K$ be a Galois extension of number fields, $S,R$ the associated number rings,
and set $G=\Gal(L/K)$.  Let $P$ be a prime of $K$.  For each prime $Q$ of $L$ over $P$, 
recall that the \textbf{Decomposition} and \textbf{Inertia} groups are given by
\begin{alignat*}{4}
	D&=D(Q|P)\ &=&\left\{\sigma\in G:\sigma Q=Q\right\}\\
	I&=I(Q|P)\ &=&\left\{\sigma\in G: \sigma(\alpha)\equiv\alpha\mod Q\quad\forall\alpha\in S\right\}.\\
\end{alignat*}

From now on, we suppose that $P,Q$ are fixed.
Notice that some power of $Q$ is principal, say $Q^j=(\beta)$.  Then if $\sigma\in I$
we have $\sigma(\beta)\equiv\beta\mod Q$, so that $\sigma(Q)^j\equiv Q^j\equiv 0\mod Q$.
Since $\sigma(Q)$ is a prime lying over $P$ whose $j^{\text{th}}$ power is divisible by 
$Q$, we see that $\sigma(Q)=Q$ so that $\sigma\in D$ and $I\subset D$.

\begin{defn}\label{ramgrp}
	For $m\geq -1$, the \textbf{higher ramification groups} are defined by
	\begin{eqnarray*}
		V_m=\left\{\sigma\in G:\sigma(\alpha)\equiv\alpha\mod Q^{m+1}\quad\forall\alpha\in S\right\}.	
	\end{eqnarray*}
\end{defn}

Notice that $V_{-1}=G$ and $V_0=I$.  We obviously have the chain of subgroups
\begin{eqnarray*}
	G=V_{-1}\supseteq V_0\supseteq V_1\supseteq\cdots,
\end{eqnarray*}
moreover, eeach $V_i$ is a normal sungroup of $D$.  This follows from a short computation:
Let $\tau\in V_i$ and $\sigma\in D$.  Moreover, denote by $v_Q(\beta)$ the exact power of
$Q$ dividing $\beta\in S$.  Then for any $\alpha\in S$ we have
\begin{eqnarray*}
	v_Q(\sigma\tau\sigma^{-1}\alpha-\alpha)&=&V_Q(\sigma(\tau\sigma^{-1}\alpha-\sigma^{-1}\alpha))\\
	&=&V_Q(\tau\sigma^{-1}\alpha-\sigma^{-1}\alpha)
\end{eqnarray*}
since $\sigma$ fixes $Q$.  Therefore, since $\sigma^{-1}S=S$, we see that $V_i\triangleleft D$.  This in fact
shows that $V_i\triangleleft V_{i-1}$, so that we may consider the quotients \
\begin{eqnarray*}
	V_{i-1}/V_i
\end{eqnarray*}
as groups.  Also, suppose that $\sigma\in V_i$ for all $i$.  Then $\sigma(\alpha)-\alpha\equiv 0\mod Q^i$
for all $i$ and all $\alpha\in S$.  But if $\sigma(\alpha)-\alpha\equiv 0\mod Q^i$ for all $i\geq 0$ then
we must have $\sigma(\alpha)-\alpha=0$ (since this element of $S$ is divisible by arbitrarily high powers
of $Q$, its norm must be infinite).  But if $\sigma\alpha=\alpha$ for all $\alpha\in S$ then $\sigma=1$.
This shows that
\begin{eqnarray*}
	\bigcap_{i=1}^{\infty}V_i=\{1\},
\end{eqnarray*}
and since $G$ is a finite group, it follows that $V_i=\{1\}$ for all sufficiently large $i$ so that the
quotient $V_{i-1}/V_i$ is trivial for all sufficiently large $i$.

It can be shown that given $P\in K$ with $v_Q(P)=1$, for each $\sigma\in I$ there exists $\alpha_{\sigma}\in S$,
uniquely determined modulo $Q$ such that $\sigma(P)\equiv\alpha P\mod Q^2$.  Moreover, it is not difficult to 
show that 
\begin{eqnarray*}
	\sigma\tau(P)\equiv\alpha_{\sigma}\alpha_{\tau} P\mod Q^2
\end{eqnarray*}
for any $\sigma,\tau\in I$.  It follows that $\alpha_{\sigma\tau}\equiv\alpha_{\sigma}\alpha_{\tau}\mod Q^2$.
Therefore, we have a homomorphism 
\begin{eqnarray*}
	I\longrightarrow (S/Q)^*
\end{eqnarray*}
given by $\sigma\mapsto \alpha_{\sigma}$.  In fact, the kernal of this homomorphism is $V_1$ so that we have

\begin{thm}\label{first}
	The mapping $V_0/V_1\longrightarrow (S/Q)^*$ given by taking any coset representative
	$\sigma\in V_0$ to $\alpha_{\sigma} \in S$ is an injective group homomorphism. 
\end{thm}

This result has a generalization:

\begin{thm}\label{higher}
	Let $i\geq 2$.  Then there exists an injective group homomorphism
	\begin{eqnarray*}
		V_{i-1}/V_i\longrightarrow S/Q,
	\end{eqnarray*}
	where we are considering $S/Q$ as an additive group.
\end{thm}

There are many nice group theoretic consequences of Theorems \ref{first} and \ref{higher}.
For example, since $Q$ is a prime of $L$, the quotient $S/Q$ is a finite field.  As such,
the multiplicative group $(S/Q)^*$ is cyclic of order $|S/Q|-1$.  Since $V_0/V_1$ may be 
imbedded into $(S/Q)^*$ and any subgroup of a cyclic group is cyclic, we have

\begin{cor}
	The group $V_0/V_1$ is cyclic of order dividing $|S/Q|-1$.
\end{cor}

Again, $S/Q=\F_{p^r}$ where $p=P\cap \Z$.  Therefore, since $V_{i-1}/V_i$ may be viewed as
a subgroup of the abelian additive group $\F_{p^r}$, the structure theorem for finite
abelian groups tells us that $V_{i-1}/V_i$ decomposes as the direct sum of cyclic groups
$C_{d_i}$ where $d_{i}|d_{i+1}$.  Clearly each $d_i$ is a power of $p$, viz. $d_i=p^{j_i}$.
However, the additive group of $\F_{p^r}$ has no elements of order greater than $p$, while
if $C_{p^{j}}$ occurs in the above decomposition for some $j>1$, then we do get an element
of order greater than $p$.  Thus we have

\begin{cor}
	For $i\geq 2$, the group $V_{i-1}/V_i$ is isomorphic to a direct sum of cyclic groups of
	order $p$.
\end{cor}

Continuing along this trajectory, we have

\begin{prop}
	Let $L,K,S,R,Q,P,p,D,I,V_i$ be as above.  Then $V_i$ is the $p$--Sylow subgroup of $I$, D and $I$ are
	both solvable groups, and if $K$ contains a prime which becomes a power of a prime in $L$ then
	$G$ is solvable.
\end{prop}

More can be said if $D/V_1$ is abelian.

\begin{prop}
	Let $D/V_1$ be abelian.  Then the map of Theorem \ref{first} maps $I/V_1$ to a subgroup
	of $(R/P)^*$.
\end{prop}

Therefore, since $|R/P|=\|P\|$, we see that when $D/V_1$ is abelian, $I/V_1$ is cyclic of order dividing
$\|P\|-1$.

Leaving these group--theoretical consequences aside, we have the important Theorem

\begin{thm}[Hilbert's Formula]
	Let $L/K$ be a Galois extension of number fields with corresponding rings of integers $S,R$.
	Let $P$ be a prime of $K$ and $Q$ a prime of $L$ above $P$, and let the $V_i$ be as in
	Definition \ref{ramgrp}.  Theorem \ref{Qdiv} tells us that $Q^{e-1}|\diff S$, where
	$e=e(Q|P)=|V_0-1|$.  In fact, if
	\begin{eqnarray*}
		k=\sum_{m\geq 0}(|V_m|-1),
	\end{eqnarray*}
	then $Q^k$ is the exact power of $Q$ dividing $\diff S$.
\end{thm}

\end{document}