\documentclass[12pt]{article}

\usepackage{amsmath,amsfonts,amsthm}

\theoremstyle{definition}
\newtheorem{defn}{Definition}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}

\renewcommand{\P}{\mathbb{P}1}
\renewcommand{\H}{\mathbb{H}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\F}{\mathbb{F}}
\DeclareMathOperator{\tr}{Tr}
\DeclareMathOperator{\diff}{diff}
\DeclareMathOperator{\disc}{disc}
\DeclareMathOperator{\N}{N}
\DeclareMathOperator{\Gal}{Gal}

\title{Math 129 HW 10}
\author{Bryden Cais}

\begin{document}
\maketitle

\noindent\textbf{Question 2:}
\begin{enumerate}

	\item Since $L/K$ is a Galois extension of degree $p$, for any prime $Q$ in $L$ above $P$, we have
	\begin{eqnarray*}
		e(Q|P)f(Q|P)d(Q|P)=p.
	\end{eqnarray*}
	Since $p$ is a prime in $\Z$, we see that one of $e,f,d$ is $p$ and the other two are $1$.  Therefore,
	$P$ is inert or totally ramified or totally split in $L$.

	\item Let $x=\pi^sy$ with $s\in\N$ and $y\not\in P$.  Suppose that $p\not| s$.  Let $Q$ be a prime above
	$P$ in $L$.  Then since $(\alpha)^p=P^s(y)$, we see that $Q|(\alpha)^p$ and hence $Q|(\alpha)$.  Thus,
	$Q^p|(\alpha)^p$ so that $Q^p|P^s(y)$.  Now since $y\not\in P$, we must have $Q\not|y$, for if
	$y\in Q$ then $y\in R\cap Q=P$, a contradiction.  Therefore, we see that $Q^p|P^s$.  Since $(s,p)=1$,
	we have $Q^p|P$.  But then we must have $Q^p=P$ and $P$ is totally ramified.

	\item Now suppose that $s=0$ so that $x=y\not\in P$.
	\begin{enumerate}
		\item Suppose that $\bar{x}\in R/P^m$ is a $p^{\text{th}}$ power but $\bar{x}\in R/P^{m+1}$ is not.
		\begin{enumerate}
		\item Suppose that $m\leq p-1$ and let $\beta\in R$ be a solution to 
		\begin{eqnarray*}
			u^p\equiv x\mod P^m.
		\end{eqnarray*}
		Now let $w=\beta-\alpha$, where $\alpha\in L$ is a root of $u^p-x=0$.  Since 
		\begin{eqnarray*}
			\N^L_K(w)=\beta^p-x,
		\end{eqnarray*}
		we see that $P^m|\N^L_K(w)$.  Therefore, $\gcd(P,w)$ is a $R$--ideal distinct from $P$ (since
		if $P|w$ then $P^p|\N^L_K(w)$ and $m<p$), that is, $P$ is not prime.  We know that if $P$ splits
		it does so completely.  We will show, however, that $P$ splits completely if and only if $\bar{x}\in R/P^{p+1}$
		is a $p^{\text{th}}$ power, so that in this case, $P$ does not split.  Hence,
		since $P$ is not prime and does not split, it must be totally ramified by part 1.

		\item Now suppose that $m=p$.  As above, let $\beta\in R$ be a solution to $u^p\equiv x\mod P^m$,
		and let $\alpha$ be as before.  For any $z\in P^{-1}-R$, set $w_z=z(\beta-\alpha)$, and observe that 
		$w_z$ is a root of
		\begin{eqnarray}
			f(u)=(u-z\beta)^p+z^px=0
		\label{poly}
		\end{eqnarray}
		Using the binomial theorem to expand the above polynomial, we obtain
		\begin{eqnarray*}
			(u-z\beta)^p+z^px=u^p+\sum_{j=1}^{p-1}\binom{p}{j} u^{p-j} z^j-z^p(\beta^p-x).
		\end{eqnarray*}
		Since $\binom{p}{j}$ is divisible by $p$ and $P^{p-1}=p$, we see that $\binom{p}{j} z^j$
		is integral for $1\leq j\leq p-1$.  Moreover, since $\beta^p-x\equiv 0\mod P^p$, we see that
		$z^p(\beta^p-x)$ is integral.  Therefore, $f(u)\in R[u]$, and as such,
		$w_z\in S$.  Notice that 
		\begin{eqnarray*}
			f^{\prime}(w_z)=pw_z^{p-1}.
		\end{eqnarray*}
		Since $P^{p-1}=p$ and $z\in P^{-1}$, we see that $f^{\prime}(w_z)\not\in P$.  Since the
		different $\diff_{L/K}$ is the ideal generated by $f^{\prime}(\gamma)$ for all $\gamma\in S$,
		we see that $P\not|\diff_{L/K}$ and hence $P\not|\disc_{L/K}$.

		\item In the next problem, we will show that $P$ splits in $L$ if and only if $\bar{x}\in R/P^{p+1}$
		is a $p^{\text{th}}$ power.  Hence, if $m>p$ then $P$ splits completely since $\bar{x}\in R/P^{p+1}$
		is a $p^{\text{th}}$ power, so that by the next question, $\bar{x}\in R/P^{l}$
		is a $p^{\text{th}}$ power for all $l\geq 1$.  This contradicts the assumption that $\bar{x}\in R/P^{m+1}$
		is not a $p^{\text{th}}$ power.  Therefore, $m>p$ never happens.
		\end{enumerate}

		\item We show that if $P$ splits in $L$ then $\bar{x}\in R/P^{m}$
		is a $p^{\text{th}}$ power for all $m\geq 1$ and that if $\bar{x}\in R/P^{p+1}$ is a $p^{\text{th}}$ power,
		then $P$ splits completely.
		First suppose that $P$ splits.  We know that it must split completely, so we write $P=Q_1\cdots Q_p$
		for distinct primes $Q_j$ of $L$.  Moreover, we have $e(Q_j|P)=f(Q_j|P)=1$.  We show that the rings
		$S/Q_1^m$ and $R/P^m$ are equal for all $m\geq 1$.  To do this, we show that
		$Q_1^m\cap R=P^m$ for all $m\geq 1$, from which it follows that $R/P^m$ is a subring of $S/Q_1^m$ for all $m$.
		Since $f(Q_1|P)=1$, we have 
		\begin{eqnarray*}
			|S/Q_1^m|=|S/Q_1|^m=|R/P|^m=|R/P^m|,
		\end{eqnarray*}
		from which the equality follows.
		We proceed by induction.  Certainly for $m=1$ we have $Q_1\cap R=P$.  Now suppose the result holds for $m-1$.
		Then we have
		\begin{eqnarray*}
			P^m\subset Q_1^m\cap R\subset Q_1^{m-1}\cap R=\subset P^{m-1}.
		\end{eqnarray*}
		If we can show that $Q_1^m\cap R\not=P^{m-1}$ then we will be done since $P$ is a prime ideal.
		But if $Q_1^m\cap R=P^{m-1}$, then $Q_1^m|P^{m-1}$ so that $e(Q_1|P)>1$, which is false.
		This gives the claim.  Now since $S/Q_1^m$ and $R/P^m$ are equal, there exists $\beta\in R$ with
		$\alpha\equiv\beta\mod Q_1^m$ for all $m\geq 1$ (where $\alpha$ is as before).  Now take norms to
		obtain
		\begin{eqnarray*}
			\N^L_K(\alpha)\equiv \N^L_K(\beta)\mod\N^L_K(Q_1)^m,
		\end{eqnarray*}
		that is,
		\begin{eqnarray*}
			\beta^p\equiv x\mod P^m,
		\end{eqnarray*}
		as required.

		Now suppose that we have a solution $\beta$ to $u^p\equiv x\mod P^{p+1}$.  As before, for any $z\in P^{-1}-R$,
		set $w=z(\beta-\alpha)$.  Then $w$ is a root of 
		\begin{eqnarray}
			f(u)=(u-z\beta)^p+z^px=0.
		\end{eqnarray}
		as are $w_j=z(\beta-\zeta^{j}\alpha)$ for $j=1,2,\ldots,p-1$ and $\zeta=e^{2\pi i/p}$.
		As before, $f(u)\in R[u]$ so that $w_j\in S$ for each $j$.  Now set
		\begin{eqnarray*}
			Q_j=\gcd(PS,w_jS).
		\end{eqnarray*}
		Certainly $Q_j\not=S$ for $0\leq j\leq p-1$ since $\N^L_K(w_j)=z^p(\beta^p-x)$ is divisible by
		$P$ by hypothesis.  We claim that $Q_1\cdots Q_p=P$.  To see this, notice that from the definition
		of $Q_j$ we have $P\subset Q_j$ for each $j$ so that $Q_1\cdots Q_p\supset P$.  Moreover, since
		we can write
		\begin{eqnarray*}
			Q_j=\gamma_j+\delta_j w_j
		\end{eqnarray*}
		for some $\gamma_j\in PS$ and $\delta_j\in S$, every
		element of $Q_1\cdots Q_p$ is a sum of elements of the form 
		\begin{eqnarray*}
			\prod_{j=0}^{p-1}(\gamma_j+\delta_j z(\beta-\zeta^{j}\alpha))=\gamma+\delta z^p(\beta^p-x),
		\end{eqnarray*}
		where $\gamma\in P$ and $\delta\in S$.  Since $z^p(\beta^p-x)\in P$ by hypothesis, it follows that
		$Q_1\cdots Q_p\subset P$, so that $Q_1\cdots Q_p= P$ as claimed.  Finally, the $Q_j$ are either
		all distinct or all equal by part 1.  But the latter possibility cannot hold since we showed above that
		$P$ does not divide $disc_{L/K}$.  Therefore, $P$ splits completely.

	\end{enumerate}

	\item We can now easily determine the behaviour of $P$ in $L$.  We know that $P$ either totally
	splits or ramifies or it is inert.  Moreover, we know that $P$ splits completely if and only if
	$u^p\equiv x\mod P^{p+1}$ has a solution in $R$.  Also, if $P$ is not totally split
	but the congruence $u^p\equiv x\mod P^{p}$ has a solution in $R$ then $P$ is inert.  If
	The above congruence $u^p\equiv x\mod P^{p}$ has no solution then $P$ is totally ramified.

	\item Suppose that $x\in R^*$.
	\begin{enumerate}

		\item Recall that $\disc_{L/K}$ divides $\disc(\alpha)=\N^L_K(p\alpha^{p-1})=p^px^{p-1}$.
		Since $x\in R^*$, the only primes dividing $\disc_{L/K}$ are those above $p$, that is,
		those primes of $L$ above $P$ (since $P$ is the unique prime in $K$ above $P$).  Therefore,
		the extension $L/K$ is unramified outside $P$.

		\item If $L/\Q$ is Galois, then it is not difficult to see that we in fact have
		\begin{eqnarray*}
			L=MK,
		\end{eqnarray*}
		where $M$ is a degree $p$ Galois extension of $\Q$ (this follows from the Galois correspondence, for example).
		Let $P,\pi,Q$ be primes of $K,M,L$ above $p$.  Now we know that $p$ totally ramifies in $K$ so that
		$e(P|p)=p-1$.  We have shown that $L/K$ is unramified outside of $P$ and that $K/\Q$ is unramified
		outside of $p$.  Consequently, $L/\Q$ is unramified outside of $p$.  If a prime $q\in\Z$ distinct from 
		$p$ ramified in $M$, it would ramify in $L$ so that the only possible ramification in $M/\Q$ occurs
		at $p$.  But \textit{some} prime must ramify, so that $p$ ramifies in $L$.  Moreover, since
		$e(\pi|p)f(\pi|p)d(\pi|p)=p$, we see that $e(\pi|p)=p$ and $p$ totally ramifies in $L$.  Now
		$e(Q|p)=e(Q|P)e(P|p)=e(Q|\pi)e(\pi|p)$ so that $e(Q|p)$ is divisible by $p$ and $p-1$.  Since
		$(p,p-1)=1$, we see that $e(Q|p)=p(p-1)$ and $p$ is totally ramified in $L$ so that $P$ is totally
		ramified in $L$.

		\item If $x=\zeta$, then $L=\Q(\zeta^{1/p})$ is a cyclotomic field and hence Galois.  Therefore,
		$P$ ramifies in $L$ so that $m\leq p-1$.  
	
	\end{enumerate}

\end{enumerate}

\noindent\textbf{Question 3:}

Suppose that there exists $x\in R^*$ that is congruent to a $p^{\text{th}}$ power modulo $P^p$ but is not
a $p^{\text{th}}$ power in $R^*$.  Then $L=K(\sqrt[p]{x})$ is a nontrivial Galois field extension of
degree $p$.  Moreover, we know from question 2 part 3 that $P$ does not ramify in $L$.  By part 5 of question 2, 
$L/K$ is unramified outside of $P$ and is therefore an everywhere unramified Galois extension of degree $p$.
If $K$ has no such extensions, then we colclude that there can be no such $x$ and hence that every $x\in R^*$
that is congruent to a $p^{\text{th}}$ power in $R^*$ modulo $P^p$ is in fact a $p^{\text{th}}$ power in $R^*$.

\end{document}