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\title{Math 129 HW 8}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{}
\begin{enumerate}
	
	\item Marcus \# 33
	\begin{enumerate}

		\item Let $m\equiv 2$ or $3\mod 4$ be a positive squarefree integer.  We know that
		the ring of integers of $K=\Q(\sqrt{m})$ is $\Z[\sqrt{m}]$.  Moreover, since roots of unity
		of $K$ are just $\pm 1$ (since $K\subset\R$), we know that every unit in $K$ is $\pm u^h$
		for some $h\in\Z$ where $u$ is a fundamental unit.  Denote by $\bar{s}$ the image of $s\in K$
		under the generator of $\Gal(K/\Q)$.  Let $b$ be the least positive integer such that
		$mb^2\pm 1=a^2$ for some positive integer $a$.  We know that there must be such a $b$ since every 
		unit $x+y\sqrt{m}$ satisfies $my^2\pm1=x^2$.  Write $u=c+d\sqrt{m}$ and without loss
		of generality, assume that $c,d>0$.  Certainly $c\not=0$ since then the square of $u$
		is rational.  Then we can at any rate choose $c>0$ by multiplying
		$u$ by $-1$ if necessary, and notice that for any unit $z$, $\bar{z}$ is also a unit so that
		$\bar{z}=\pm u^h$ for some $h$ which implies that $z=\pm\bar{u}^h$ whence $\bar{u}$ is 
		also a fundamental unit.  Therefore, if $u=c+d\sqrt{m}$ with $c>0$ and $d<0$ replace $u$
		by $\bar{u}$.  Since $a+b\sqrt{m}$ is a unit, we write $a+b\sqrt{m}=\pm(c+d\sqrt{m})^r$
		for some $r\in\Z$.  Notice first that the $+$ sign must hold above since $a,b,c,d$ are all
		positive.  Moreover, because $a,b,c,d$ are positive, $c+d\sqrt{m},a+b\sqrt{m}$ are both 
		greater than $1$, so that $r\geq 1$.  Since we have assumed $b$ to be minimal, 
		we have $d\geq b$.  If $d\geq b+1$ then $c^2\geq md^2-1\geq mb^2+(2b+1)m-1>mb^2+1\geq a^2$
		since $b\neq 0$, obviously, and $m\equiv 2$ or $3\mod 4$ is at least $2$.  Therefore, if 
		$d>b$ then $c>a$ so that $a+b\sqrt{m}<c+d\sqrt{m}$, which is impossible since $r\geq 1$.
		Therefore we conclude that $b=d$ so that $a^2,c^2$ are equal or differ by $2$ (since
		$mb^2\pm1=a^2,mb^2\pm1=c^2$).  But no two
		squares differ by $2$ so that $a=c$ and $r=1$
		and $a+b\sqrt{m}$ is a fundamental unit.


		\item The argument is virtually identical to before.  We know that
		the ring of integers of $K=\Q(\sqrt{m})$ is $\Z[\frac{1+\sqrt{m}}{2}]$.  As before, the roots of unity
		of $K$ are just $\pm 1$ and every unit in $K$ is $\pm u^h$
		for some $h\in\Z$ where $u$ is a fundamental unit.  Denote by $\bar{s}$ the image of $s\in K$
		under the generator of $\Gal(K/\Q)$.  Let $b$ be the least positive integer such that
		$mb^2\pm 4=a^2$ for some positive integer $a$.  We know that there must be such a $b$ since every 
		unit $(x+y\sqrt{m})/2$ satisfies $my^2\pm4=x^2$.  Write $u=(c+d\sqrt{m})/2$ and without loss
		of generality, assume that $c,d>0$.  The argumant is identical to that above.  
		Since $(a+b\sqrt{m})/2$ is a unit, we write $(a+b\sqrt{m})/2=\pm((c+d\sqrt{m})/2)^r$
		for some $r\in\Z$.  Again, the $+$ sign must hold above since $a,b,c,d$ are all
		positive.  Moreover, because $a,b,c,d$ are positive, $a\equiv b,c\equiv d\mod 2$, and $m>1,d,b\neq 0$, 
		the numbers $(c+d\sqrt{m})/2,(a+b\sqrt{m})/2$ are both 
		greater than $1$, so that $r\geq 1$.  Since we have assumed $b$ to be minimal, 
		we have $d\geq b$. If $d\geq b+1$ then $c^2\geq md^2-4\geq mb^2+(2b+1)m-4>mb^2+4\geq a^2$
		since $b\neq 0$, obviously, and $m\equiv 1\mod 4$ is at least $5$.  Therefore, if 
		$d>b$ then $c>a$ so that $a+b\sqrt{m}<c+d\sqrt{m}$, which is impossible since $r\geq 1$.
		We conclude that $b=d$ so that $a^2,c^2$ are equal or they differ by $8$.  If they differ by $8$ then
		one of $a,c$ is $1$ and the other is $3$.  Since we are supposing that $(a+b\sqrt{m})/2=((c+d\sqrt{m})/2)^r$
		with $r\geq 1$ and $b=d$ we must have $c=1$ and $a=3$, whence $b=d=1$ and $m=5$, and indeed
		$((1+\sqrt{5})/2)^2=(3+\sqrt{5})/2$.  Except in this case, we therefore obtain, as above, 
		$b=d,c=a$ so that $u$ is a fundamental unit.  The procedure in genreal, as the above example indicates,
		is to minimize $b$ first and then minimize $a$.

		\item Applying this procedure to the integers $5,6,7$ we find that fundamental units for
		$\Z[\frac{1+\sqrt{5}}{2}],\Z[\sqrt{6}],\Z[\sqrt{7}]$ are, respectively,
		\begin{eqnarray*}
			u_5&=&\frac{1+\sqrt{5}}{2}\\
			u_6&=&5+2\sqrt{6}\\
			u_7&=&8+3\sqrt{7}.
		\end{eqnarray*}
		
	\end{enumerate}

\end{enumerate}

\section{}
\begin{enumerate}

	\item Marcus \# 43
	\begin{enumerate}
		
		\item Let $\tau$ be the automorphism of $K$ given by complex conjugation, and let $H=\langle\tau\rangle.$
		If $\tau$ is trivial on $K$ then $\# H=1$, otherwise $\tau$ has order $2$ and $\# H=2$.  Clearly
		we have $K\cap \R=K^H$.  The Galois corespondence then gives $[K:K\cap\R]=\# H$ so that
		$K$ has degree $1$ or $2$ over $K\cap\R$.

		\item Let $\sigma:K\cap\R\rightarrow\C$ be an embedding.  Then since $K/Q$ is normal, $\sigma$ extends
		to an automorphism $\tilde{\sigma}:K\rightarrow K$ whose restriction to $K\cap\R$ is $\sigma$.  If
		$\sigma$ is a real embedding, $\sigma(K\cap\R)\subset\R$ so that $\tilde{\sigma}(K\cap\R)\subset K\cap\R$,
		that is, $\sigma:K\cap\R\rightarrow K\cap\R$ is an automorphism of $K\cap\R$.  If $K\cap\R$ has no non--real 
		embeddings in $\C$ then every embedding of $K\cap\R$ is an automorphism of $K\cap\R$ so that $K\cap\R$ is
		a normal extension of $\Q$.  Conversely, is $\sigma$ is not a real embedding, then 
		$\sigma(K\cap\R)\not\subset K\cap\R$ so that $K\cap\R/Q$ is not normal. 

		\item Notice that complex conjugation $\tau$ is in the center of $G=\Gal(K/\Q)$ if and only if $\langle\tau\rangle=H$
		is normal in $G$.  By the Galois correspondence, $H\triangleleft G$ if and only if $K\cap\R/\Q$ is a normal extension
		with Galois group $G/H$.  By part b, $K\cap\R/\Q$ is a normal extension if and only if there are no non--real
		embeddings of $K\cap\R$.  If $K=K\cap\R$ then $\tau=1$ is trivial and hence in the center.  In this case,
		$U=U\cap\R$ since $K$ is real and hence $U/U\cap\R$ is finite.  Now suppose that $K\neq K\cap\R$.  Then
		$\# H=2$.  Let $r,2s$ be the number of real and non--real embeddings of $K$ in $\C$.  Similarly,
		let $u,2v$ be the number of real and non--real embeddings of $K\cap\R$.  Then we know that
		\begin{eqnarray*}
			U&\simeq& C\times\Z^{r+s-1}\\
			U\cap\R&\simeq& \{\pm1\}\times\Z^{u+v-1},
		\end{eqnarray*}
		where $C$ is a cyclic group.  Furthermore, we have the relations $r+2s=[K:\Q]=n$ and $u+2v=[K\cap\R:\Q]=n/2$.
		Since we have
		supposed that $K\neq K\cap\R$, it must be that $K$ has no real embeddings.  Therefore $r=0,s=n/2$.  
		Now if there are no non--real embeddings of $K\cap\R$ then $v=0$ and $u=n/2$.  
		It follows that $U\simeq C\times\Z^{n/2-1}$ and $U\cap\R\simeq \{\pm 1\}\times\Z^{n/2-1}$ so that
		$U/U\cap\R\simeq C/\{\pm1\}=\{\text{Roots of unity in $K$}\}/\{\pm 1\}$ is finite.  Conversely,
		if $U/U\cap\R$ is finite then $r+s=u+v$.  If in fact $U=U\cap\R$ then $r=u,s=v$ and complex conjugation
		is trivial.  Otherwise, we may assume $K\neq K\cap\R$ so that $r=0$ as before, $s=n/2$ and $u+2v=n/2$.  This gives 
		two equations in two unknowns with solution $u=n/2,v=0$, i.e. $K\cap\R$ has no non--real embeddings.
		Thus we conclude that $U/U\cap\R$ is finite if and only if complex conjugation is in the center of $G$.
	\end{enumerate}

	\item Marcus \# 45 Let $K/\Q$ be abelian and let $u\in\mathcal{O}_K$ be a unit.  If $K$ is real
	then obviously every unit is the product of a real unit and a root of unity.  So assume that $K$ is not
	real.  Denote complex conjugation by $s\mapsto\bar{s}$ and put $v=u/\bar{u}$.  Since $K/\Q$ is abelian,
	complex conjugation commutes with every element of $\Gal(K/\Q)$ so that every Galois conjugate
	of $u/\bar{u}$ has absolute value $1$.  We showed on an earlier problem set (and will give the idea again
	in the next problem) that this implies that $v$ is a root of unity in $K$.  Therefore, write
	\begin{eqnarray*}
		u=\bar{u}\zeta.
	\end{eqnarray*}
	Noe either $\zeta^{1/2}\in K$ or it is in a degree $2$ extension of $K$.  We can write
	$u\zeta^{-1/2}=\bar{u}\zeta{1/2}$ so that $u\zeta{-1/2}$ is real.  Thus we can write $u$ as
	a real unit times a root of unity where the factors are in $K$ or a degree $2$ extension
	of $K$.

	\item Marcus \# 46
	\begin{enumerate}

		\item For an odd prime $p$ let $\omega=e^{2\pi i/p}$ and put $R=\Z[\omega]$.  Let $u\in R$ be a unit.
		Consider 
		\begin{eqnarray*}
			v=\frac{u}{\bar{u}}.
		\end{eqnarray*}
		Clearly, $|v|=1$.  Moreover, the Galois conjugates of $v$ are $v^i$ for $1\leq i\leq p-1$ so that
		$|\sigma(v)|=|v^i|=1$ for all $\sigma\in\Gal(\bar{\Q}/\Q)$.  We showed on an earlier problem
		set that this implies that $v$ is a root of unity (recall the argument: $|\sigma(v^n)|=1$ for all 
		$\sigma\in\Gal(\bar{\Q}/\Q)$ and $n$, but $\Z[\omega]$ can only contain finitely many 
		such algebraic integers since they are the roots of monic irreducible polynomials of bounded
		degree and bounded coefficients.  This implies $v^n=v^m$ for some $m\neq n$).  Since $v$ is a root
		of unity in $\Z[\omega]$, we may write $v=\pm\omega^k$ for some $k$.  We now claim that the $+$ sign
		holds.  Suppose not.  Then $u/\bar{u}=-\omega^k$ whence $u^p=-\bar{u^p}$.  Now write
		\begin{eqnarray*}
			u=a_0+a_1\omega+\cdots+a_{p-2}\omega^{p-2}
		\end{eqnarray*}
		with $a_i\in\Z$.
		Since $x\mapsto x^p$ is the familiar Frobenius automorphism, we see that $u^p\equiv a_0+a_1+\cdots+a_{p-2}\mod p$.
		We then also have $\bar{u^p}\equiv a_0+a_1+\cdots+a_{p-2}\mod p$ so that $\bar{u^p}\equiv u^p\mod p$.
		Since we are supposing that $\bar{u^p}\equiv -u^p\mod p$, we must have $u^p\equiv 0\mod p$ since $p$ is odd.
		Since $u$ is a unit, this is a contradiction.  Thus we have $u=\bar{u}\omega^k$.  Now choose $j$ so that \
		$2j\equiv k\mod p$.  Then we have
		\begin{eqnarray*}
			u\omega^{-j}&=&\bar{u}\omega^j\\
			&=&\bar{u\omega^{-j}}
		\end{eqnarray*}
		so that $u\omega^{-j}\in\R$.  Therefore, $u=\omega^k\cdot u\omega^{-k}$ is the product of a root of unity
		in $R$ and a real unit.

		\item We showed in part $1$ of this question that $U/U\cap\R\simeq C/\{\pm1\}$ where $C$ is the group
		of roots of unity in $R$.  In the case $R=\Z[\omega]$, we find that $C=\pm\langle\omega\rangle$,
		so that 
		\begin{eqnarray*}
			U/U\cap\R\simeq\langle\omega\rangle.
		\end{eqnarray*}
		where $U$ is the group of units in $\Z[\omega]$.
		Moreover, $\Z[\omega+\omega^{-1}]=\Z[\omega]\cap\R$ since $\Q(\omega+\omega^{-1})$ is the fixed field
		of complex conjugation.  Therefore, the group of units $U^{\prime}$ in $\Z[\omega+\omega^{-1}]$ is $U\cap\R.$
		Piecing the information together, we see that $U\simeq\langle\omega\rangle\times U^{\prime}$.

	\end{enumerate}

	\item Marcus \# 48
	\begin{enumerate}
		
		\item Let $\omega=e^{2\pi i/m}$ and set $\alpha=e^{\pi i/m}$.  We have
		\begin{eqnarray*}
			1-\omega^k&=&e^{\pi ik/m}\left(e^{-\pi ik/m}-e^{\pi ik/m}\right)\\
			&=&-2i\alpha^k\frac{e^{\pi ik/m}-e^{-\pi ik/m}}{2i}\\
			&=&-2i\alpha^k\sin(k\pi/m)
		\end{eqnarray*}
		for any $k\in\Z$.  Therefore, we conclude that
		\begin{eqnarray*}
			\frac{1-\omega^k}{1-\omega}=\alpha^{k-1}\frac{\sin(k\pi/m)}{\sin(\pi/m)}.
		\end{eqnarray*}

		\item If $k$ is odd then $k-1$ is even to that $\alpha^{k-1}=\omega^{(k-1)/2}$.  If $k$
		is even then we are supposing that $m$ must be odd so that $-\alpha^{k-1}=\alpha^{m+k-1}=\omega{(m+k-1)/2}$
		since $\alpha^m=-1$ and $m+k-1$ is even.  Thus, in any case, as long as $m,k$ are not both even
		$\alpha^k=\pm\omega^h$ for some $h\in\Z$.

		\item If $(k,m)=1$ then certainly $k,m$ are not both even.  Therefore, by parts a and b we have
		\begin{eqnarray*}
			u_k=\frac{\sin(\pi k/m)}{\sin(\pi/m)}=\pm\omega^h\frac{1-\omega^k}{1-\omega}.
		\end{eqnarray*}
		Again, if $(k,m)=1$ then the Galois conjugates of $\omega^k$ are the same as the Galois conjugates
		of $\omega$.  Therefore,
		\begin{eqnarray*}
			\n(u_k)&=&\pm\n(\omega)\n\left(\frac{1-\omega^k}{1-\omega}\right)\\
			&=&1,
		\end{eqnarray*}
		so that $u_k$ is a unit.

	\end{enumerate}

\end{enumerate}

\section{}
\begin{enumerate}

	\item
	\begin{enumerate}

		\item We have $\pi i(a)=Sa\cap K$.  Obviously $a\in Sa\cap K$.  Suppose now that $x\in Sa\cap K$.
		Then $x\in Sa$ and $x\in K$.  Thus we can write $x=s\alpha$ for $s\in S$ and $\alpha\in K$.
		Then $s=x/\alpha\in K$ so that $s\in S\cap K=R$.  Therefore $x\in Ra=a$.  Thus, $Sa\cap K=a$.

		\item No, $\pi$ is not a group homomorphism.  For consider a prime $P$ of $K$ ramified in $L$ (We know
		such a prime exists for any nontrivial extension $L/K$).  Then write $P=QQ_1$ where $Q$ is prime.
		Then $\pi(Q)=P$ and $\pi(Q_1)\supset P$.  But $\pi(QQ_1)=\pi(P)=P$ which is not equal to
		$\pi(Q)\pi(Q_1)=P P^{'}$ where $P^{'}\supset P$ is clearly not all of $R$.  

	\end{enumerate}

	\item
	\begin{enumerate}

		\item Marcus \# 15
		\begin{enumerate}
			
			\item Let $K\subset L\subset M$ be fields and let $Q$ be a prime of $M$.  Let $T,S,R$ denote
			the rings of integers pf $M,L,K$ respectively.
			Then
			\begin{eqnarray*}
				\n_{M/K}(Q)=P^{f(Q|P)}
			\end{eqnarray*}
			by definition.  Letting $Q_L$ be the unique prime in $L$ below $Q$, we have
			$f(Q|P)=f(Q|Q_L)f(Q_L|P)$ which enables us to write
			\begin{eqnarray*}
				\n_{M/K}(Q)&=&(P^{f(Q|Q_L)})^{f(Q_L|P)}\\
				&=&\N(P^{f(Q|Q_L)})\\
				&=&\N\n_{M/L}(P),
			\end{eqnarray*}
			where we have used the multiplicitive extension of $\n$ in the second line.  Now
			extending this result multiplicitively to all ideals $I\subset T$ we
			see that $\n_{M/K}(I)=\N\n_{M/L}(I)$.

			\item Let $\alpha\in L$ be nonzero.  We now show that $\N(\alpha S)$ is the principal ideal in
			$R$ generated by the element $\N(\alpha)$.  Let $M$ be the normal closure of $L/K$.  Then
			by the first part, we have $\n_{M/K}(\alpha T)=\N\n_{M/L}(\alpha T)$.  Since $M/L$ is a normal
			extension, we already know (see Marcus ex. \# 14) that $\n_{M/L}(\alpha T)$ is the $S$ ideal generated 
			by the element $\n_{M/L}(\alpha)=\alpha^{[M:L]}$ since $\alpha\in L$.  Therefore
			\begin{eqnarray*}
				\N\n_{M/L}(\alpha T)&=&\N(\alpha^{[M:L]}S)\\
				&=&\N((\alpha S)^{[M:L]})\\
				&=&\N(\alpha S)^{[M:L]}.
			\end{eqnarray*}
			On the other hand, since $M/K$ is Galois, 
			\begin{eqnarray*}
				\n_{M/K}(\alpha T)&=&\n_{M/K}(\alpha) R\\ 
				&=&\N\n_{M/L}(\alpha) R\\
				&=&\N(\alpha^{[M:L]}) R\\
				&=&\N(\alpha)^{[M:L]} R\\
				&=&(\N(\alpha) R)^{[M:L]},
			\end{eqnarray*}
			so that $\N(\alpha S)^{[M:L]}=(\N(\alpha) R)^{[M:L]}$, from which we conclude that
			$\N(\alpha S)=\N(\alpha) R$.

			\item Now let $K=\Q$, and let $Q$ be a prime in $L$ with $p\in\Z$ the unique primebelow.  
			In this case we have
			\begin{eqnarray*}
				\|Q\|=|S/Q|,
			\end{eqnarray*}
			which is a finite field of characteristic $p$.  Since it is an extension of degree $f(Q|p)$ of
			$\F_p$, we conclude that $\|Q\|=p^{f(Q|P)}$.  On the other hand, we have shown that
			$\n_{L/\Q}(Q)=p^{f(Q|P)}$ is the principal ideal of $\Z$ generated by $p^{f(Q|p)}$.  Putting things
			together, we see that $\n_{L/\Q}(Q)$ is the principal ideal in $\Z$ generated by the number $\|Q\|$.
			Using the fact that $\|IJ\|=\|I\|\|J\|$ for any ideals $I,J$ and that 
			$\n_{L/\Q}(IJ)=\n_{L/\Q}(I)\n_{L/\Q}(J)$, we see that $\n_{L/\Q}(I)$ is the principal ideal in $\Z$
			generated by $\|I\|$ for any ideal $I\subset S$.

		\end{enumerate}

		\item Since $\N$ is multiplicitive, it is enough to show the result when $a=P$ is prime.
		We have $i(P)=SP=Q_1^{e_1}Q_2^{e_2}\cdots Q_r^{e_r}$ so that 
		\begin{eqnarray*}
			\N\circ i(P)&=&\prod_{j=1}^r\N(Q_i)^{e_i}\\
			&=&\prod_{j=1}^r P^{f(Q_i|P)e_i}\\
			&=&P^{\sum_{i=1}^r f(Q_i|P)e_i}\\
			&=&P^{[L:K]}.
		\end{eqnarray*}

		\item Marcus \# 16.
		\begin{enumerate}
			\item Let $\phi:G(S)\rightarrow G(R)$ be the map taking any ideal in a given class $C$
			and sending $C$ to the class of $\N(I)$.  This is well defined because if $[I]=[J]$ then
			$I=\alpha J$ for some $\alpha\in L$ whence $\N(I)=\N(alpha)\N(J)$, that is $[\N(I)]=[\N(J)]$
			since $\N(\alpha)\in K$.  That $\phi$ is a homomorphism follows immediately from the multiplicitivity
			of $\N$.

			\item Let $Q$ be a prime lying over the prime $P$ and let $d_Q,d_P$ be the orders of the classes
			of $Q,P$ in $\Cl(S),\Cl(R)$ respectively.  Then we have $\N(Q)=P^{f(Q|P)}$.  Since $[Q]^{d_Q}=1$
			and $\phi$ is a homomorphism we have $[\phi(Q)^{d_Q}]=[P^{d_Qf(Q|P)}]=[P]^{d_Qf(Q|P)}=1$ so that
			$d_P|d_Qf(Q|P)$.
 		\end{enumerate}

		\item Suppose that $[L:K]$ is prime to $\#\Cl(K)$.  Let $p$ be a prime dividing $\#\Cl(K)$.  Then
		$\Cl(K)$ contains an element of order $p$, say $P_p$.  If $p|f(Q|P_p)$ for every
		prime $Q$ over $P_p$ then the formula $[L:K]=\sum_{i=1}^r e(Q|P)f(Q|P)$ tells us that $p|[L:K]$.
		Thus, if $p\not|[L:K]$ then $p\not|f(Q|P)$ for some $Q$ over $P$.  Therefore, by part b above,
		$p|d_Q$.  Since $d_Q$ is the order of an element of $\Cl(L)$, we see that $d_Q|\#\Cl(L)$ and hence
		$p|\Cl(L)$.  Thus every prime divisor of $\Cl(K)$ divides $\Cl(L)$ when $[L:K]$ and $\#\Cl(K)$
		are relatively prime.  

		\item Marcus \# 17.  Let $K=\Q(\sqrt{-23})$, $L=\Q(\omega)$, where $\omega=e^{2\pi i/23}$.
		Set $R=\mathcal{O}_K$, let $P=(2,\frac{1+\sqrt{-23}}{2})$ be a prime over $2$ (in fact
		$2=P\bar{P}$), and let $Q$ be a prime of $S=\Z[\omega]$ lying over $P$.
		\begin{enumerate}
	
			\item Since $\disc(L)=\pm23^h$ for some integer $h$, we see that $2\not|\disc(L)$.
			We know that $f(Q|2)$ is the order of $2\mod 23$.  This must be a factor of $22$.
			Since $2^{11}=2048=23\cdot 89+1$, we see that $f(Q|2)=f$ (since $L/\Q$ is Galois)
			is $11$.  On the other hand, $f(Q|2)=f(Q|P)f(P|2)$ and since $2=P\bar{P}$ is totally split in
			$K$, we have $f(P|2)=1$ and therefore $f(Q|P)=11$.  Again, $L/K$ is Galois so that
			$e(Q|P)f(Q|P)d=[L:K]=11$.  But $f(Q|P)=11$, which forces $d=e(Q|P)=1$, so that
			$P$ is inert in $L$.

			\item Notice that $\n(\frac{-3+\sqrt{-23}}{2})=2^3=\n(P^3)$.  It is not difficult
			to see that the principal ideals generated by $\frac{-3+\sqrt{-23}}{2},\frac{-3-\sqrt{-23}}{2}$ are
			relatively prime, for if $I$ divides both of them, then $-3\in I$ and $\sqrt{-23}\in I$
			so that $\n(I)|\n(3)=9$ and $\n(I)|\n(\sqrt{-23})=-23$ whence $\n(I)=1$.  Therefore,
			we see that $P^3=(\frac{-3+\sqrt{-23}}{2})$ or $(\frac{-3-\sqrt{-23}}{2})$ and $\bar{P}^3$
			is the other ideal.  At any rate, $P^3$ is principal.  If $P$ were principal, say $P=(\alpha)$
			then we would have $P=(a+b\sqrt{-23})/2$ for some $a,b\in\Z$ of the same parity.
			In this case, we have 
			\begin{eqnarray*}
				\n(P)=\frac{a^2+23b^2}{4}=2,
			\end{eqnarray*}
			so that $a^2+23b^2=8$.  This is clearly impossible.  

			\item By \# 16b, we have $d_P|d_Qf(Q|P)$, where $d_P,d_Q$ are the orders of $P,Q$ respectively
			in their respective class groups.  We know that $d_P=3$ and $f(Q|P)=11$.  Since $(3,11)=1$,
			we then have $3|d_Q$ so that $Q$ is not principal.

			\item We have seen that $2=QQ^{'}$ where $Q,Q^{'}$ are not principal.  Suppose that 
			$2=\alpha\beta$ with $\alpha,\beta\in S$.  Then $Q|(\alpha\beta)$ so that
			$Q|(\alpha)$ or $Q|(\beta)$.  If it happens that $Q$ divides one of $(\alpha),(\beta)$ and
			$Q^{'}$ divides the other, then without loss of generality we write $(\alpha)=Q\mathcal{a}$
			and $(\beta)=Q^{'}\mathcal{b}$ for ideals $\mathcal{a},\mathcal{b}$.  Moreover, since $\alpha\beta=2$
			and $QQ^{'}=2$ we see that $\mathcal{a}\mathcal{b}=1$.  But this implies that $\mathcal{a}=\mathcal{b}=(1)$,
			so that $(\alpha)=Q$.  But $(\alpha)$ is principal and $Q$ is not, a contradiction.  Therefore,
			both $Q,Q^{'}$ divide one of $\alpha,\beta$.  Without loss of generality assume that they divide $\alpha$.
			Then $2|\alpha$ and since $\alpha\beta=2$, we see that $\beta$ is a unit.
			
		\end{enumerate}
	
	\end{enumerate}

	\item

\end{enumerate}

\section{}
\begin{enumerate}

	\item Marcus \# 5.  Let $K,L$ be number fields with $L/K$ normal and $\Gal(L/K)=G$, and let $P$
	be a prime of $K$.  By ``intermediate field'' we mean a field $F\neq K,L$ with $K\subset F\subset L$. 
	\begin{enumerate}
	
		\item Suppose that $P$ is inert in $L$.  In this case, $f=\# G$ and $e=d=1$.  
		Since $[L^D:K]=d$ and $[L:L^I]=e$, we see that $L^D=K$ and $L^I=L$ so that
		$I=\{1\}$ and $D=G$.  Since $D/I$ is cyclic (generated by the Frobenius automorphism),
		we see that $G=D/I$ is cyclic.

		\item Suppose that $P$ is totally ramified in every intermediate field but not in $L$.  If $I\not=\{1\}$ then 
		in particular
		$P$ is totally ramified in $L^I$.  Therefore, since $e(Q^I|P)=1$ where $Q^I$ is any prime of $L^I$ over
		$P$, we see that $[L^I:K]=e(Q^I|P)=1$ whence $L^I=K$ and $I=G$.  Therefore, $[L:L^I]=e(Q|P)=[L:K]$
		so that $P$ is totally ramified in $L$ also.  Therefore, $I=\{1\}$ and $e(Q|P)=[L:L^I]=1$.  Since 
		$e(Q|P)=e(Q|Q^H)e(Q^H|P)$ for any subgroup $H$ of $G$,
		we find that $e(Q^H|P)=1$ for every subgroup.  Thus, if $G$ has any nontrivial subgroups $H$ then 
		$P$ is unramified in the intermediate field $L^H$, contradicting our assumption, so that there can be
		no intermediate fields $F$ such that $L\subset F\subset L$ with $F\neq K,F\neq L$.  It follows that
		the order of $G$ is prime since if $d|\#G$ then we have a subgroup of $G$ of order $d$ and hence 
		an intermediate field.  But if $\#G$ is prime, then $G$ is cyclic of prime order.

		\item Suppose that every intermediate field contains a unique prime over $P$ but $L$ does not.  If $D\not=\{1\}$ 
		then in particular we have a unique prime $Q$ over $P$ in $L^D$.  But $e(Q^D|P)=f(Q^D|P)=1$ so that 
		$[L^D:K]=e(Q^D|P)f(Q^D|P)d=d$ whence
		$L^D=K$ and $D=G$.  Therefore, $[L:L^D]=e(Q|P)f(Q|P)=[L:K]$ where $Q$ is any prime of $L$ over $P$.  But this
		implies that there is a unique prime in $L$ over $P$.  Therefore, $D=\{1\}$ and $e(Q|P)f(Q|P)=[L:L^D]=1$ so that 
		$e(Q|P)=f(Q|P)=1$.  As above,
		by the multiplicitivity of $e,f$ we see that $e(Q^H|P)=f(Q^H|P)=1$ for any subgroup $H$ of $G$.
		Thus, if $G$ has any nontrivial subgroups $H$ then 
		$P$ does not have a unique prime over it in the intermediate field $L^H$, contradicting our assumption, so 
		that there can be no intermediate fields $F$ such that $L\subset F\subset L$ and $F\neq K,F\neq L$.  It
		follows as above that $G$ is cyclic of prime order.

		\item Suppose that $P$ is unramified in every intermediate field $F$ but ramified in $L$.  Since
		$L^I$ is the largest intermediate field in which $P$ is unramified, we see that $L^I$ contains
		every intermediate field of $L/K$.  Moreover, since $P$ is ramified in $L$, $L\neq L^I$ so that
		$I$ is not trivial.  The Galois correspondence then tells us that every nontrivial subgroup $H$ of $G$
		contains $I$.  The group $I$ is then the unique smallest nontrivial subgroup of $G$.  It is smallest 
		because it is contained in every subgroup of $G$ and it is unique because if $J$ were another such group,
		we have $\# J=\# I$ and $I\subset J$ which implies $I=J$.  Moreover, we see that $I\triangleleft G$
		since for any $g\in G$ the group $gIg^{-1}$ has the same size as $I$ and must contain $I$ since
		it is a nontrivial subgroup of $G$, so that $gIg^{-1}=I$ for any $g\in G$.  Now since $I$ has no nontrivial 
		subgroups (if $J$ were a sungroup of $I$, it would be a subgroup of $G$ and would then contain $I$), 
		we see that $I$ must have prime order for otherwise it would have a nontrivial (proper) Sylow subgroup.
		Now let $\# I=p$ and suppose that $q|\#G$ for some prime $q$.  The $G$ contains an element of order $q$ and
		hence a subgroup of order $q$.  Since this subgroup is nontrivial, it has $I$ as a subgroup so that $p|q$
		implying that $p=q$.  Therefore, $G$ has prime power order.  Finally, cince the center of a $p$--group
		is nontrivial, we see that $I$ is contained in the center of $G$.

		\item Suppose that $P$ splits completely in every intermediate field but not in $L$.
		Since $L^D$ is the largest intermediate field in which $P$ splits completely, we see that $L^D$ contains
		every intermediate field of $L/K$.  Moreover, since $P$ does not split completely in $L$, $L\neq L^D$ so that
		$D$ is not trivial.  The Galois correspondence then tells us that every nontrivial subgroup $H$ of $G$
		contains $D$.  The group $D$ is then the unique smallest nontrivial subgroup of $G$.  It is smallest 
		because it is contained in every subgroup of $G$ and it is unique because if $J$ were another such group,
		we have $\# J=\# D$ and $D\subset J$ which implies $D=J$.  Moreover, we see that $D\triangleleft G$
		since for any $g\in G$ the group $gDg^{-1}$ has the same size as $D$ and must contain $D$ since
		it is a nontrivial subgroup of $G$, so that $gDg^{-1}=D$ for any $g\in G$.  Now since $D$ has no nontrivial 
		subgroups (if $J$ were a sungroup of $D$, it would be a subgroup of $G$ and would then contain $D$), 
		we see that $D$ must have prime order for otherwise it would have a nontrivial (proper) Sylow subgroup.
		Now let $\# D=p$ and suppose that $q|\#G$ for some prime $q$.  The $G$ contains an element of order $q$ and
		hence a subgroup of order $q$.  Since this subgroup is nontrivial, it has $D$ as a subgroup so that $p|q$
		implying that $p=q$.  Therefore, $G$ has prime power order.  Finally, cince the center of a $p$--group
		is nontrivial, we see that $D$ is contained in the center of $G$.

		As an example of this, consider $L=\Q(\zeta_5),K=\Q$.  Then $G=\mathbf{C}_4$ and we have a unique
		subfield $F=\Q(\sqrt{5})=L^H$ where $H=\mathbf{C}_2$ is in the center of $G$.  Let $p=19$ (or in fact
		any prime congruent to $-1\mod5$).  Since $(5,19)=1$, $19\not|\disc(L)$.  We have the factorizations
		\begin{eqnarray*}
			x^4+x^3+x^2+x+1&\equiv& (x^2+5x+1)(x^2+15x+1)\mod 19\\
			x^2-5&\equiv&(x-9)(x+9)\mod 19,
		\end{eqnarray*}
		where all of the factors above are irreducible.  We thus have the factorizations
		\begin{eqnarray*}
			19&=&(19,\sqrt{5}-9)(19,\sqrt{5}+9)\quad\text{in $\Q(\sqrt{5})$},\\
			19&=&(19,\zeta_5^2+5\zeta_5+1)(19,\zeta_5^2+15\zeta_5+1)\quad\text{in $\Q(\zeta_5)$}.
		\end{eqnarray*}
		Therefore, $19$ splits completely in every intermediate field but not in $L$.

		\item Suppose that $P$ is inert in every intermediate field but not in $L$.  We distinguish two cases.
		Either $P$ is ramified in $L$ of it splits into at least two distinct primes in $L$.  In the latter
		case, part c applies and we have that $G$ is cyclic of prime order.  In the former case, part d
		applies and we have that $G$ is of prime power order and contains a unique nontrivial normal subgroup
		(in fact in the center of $G$) of prime order.  Since $L^D$ is the largest intermediate field
		in which $P$ splits completely, we see that $L^D=K$ so that $D=G$.  We know that $D/I$ is cyclic; therefore
		$G/I$ is cyclic.  But $I$ is in the center of $G$, and this implies that $G$ is cyclic.  Therefore
		$G$ is cyclic of prime power order.
	\end{enumerate}
\end{enumerate}

\end{document}