\documentclass[12pt]{article}

\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amscd}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}

\theoremstyle{definition}
\newtheorem{defn}{Definition}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}

\renewcommand{\P}{\mathbf{P}1}
\renewcommand{\H}{\mathbf{H}}
\newcommand{\R}{\mathbf{R}}
\newcommand{\C}{\mathbf{C}}
\newcommand{\Z}{\mathbf{Z}}
\newcommand{\Q}{\mathbf{Q}}
\DeclareMathOperator{\ind}{Index}
\DeclareMathOperator{\disc}{disc}

\title{Math 129 HW 6}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{Multiplicities in the Discriminant}

20)     Suppose that the elements $\alpha_{ij}\beta$ for $1\leq i\leq r$, $1\leq j\leq e_i$ and $\beta\,\in\,B_i$
	are linearly dependent over $R/P$.  That is, we have a relation
	\begin{eqnarray}
	    \sum_{i=1}^r\sum_{j=1}^{e_i}\sum_{\beta\in B_i} c_{ij\beta}\alpha_{ij}\beta\equiv 0\mod P,
	\label{relation}
	\end{eqnarray}
	where $c_{ij\beta}\,\in\,R/P$.  From the definition of $\alpha_{ij}$ as an element in the ideal
	$(Q_i^{j-1}-Q_i^j)\cap (\bigcap_{h\neq i}Q_h^{e_h})$, we readily see that $Q_i^k|\alpha_{ij}$
	for all $j>k$ and $Q_i^k|\alpha_{lj}$ for all $l\not=i$ and any $j$ as long as $k\leq e_i$.
	Therefore, considering \ref{relation} modulo $Q_i$ for each $i$ we see that
	\begin{eqnarray*}
	    \sum_{\beta\in B_i}c_{i1\beta}\alpha_{i1}\beta\ \equiv 0\mod Q_i
	\end{eqnarray*}
	for each $i$.  Since $B_i$ is a set of coset representatives for a basis of $S/Q_i$ and
	$\alpha_{i1}\not\equiv0\mod Q_i$, we must have
	$c_{i1\beta}\equiv 0\mod Q_i$ for all $i$ and all $\beta$.  Now suppose that we have $c_{ik\beta}\equiv 0\mod Q_i$
	for all $i$ and $\beta$.  Then consider \ref{relation} module $Q_i^{k+1}$.  We then obtain
	\begin{eqnarray*}
	    \sum_{j=1}^{k+1}\sum_{\beta\in B_i} c_{ij\beta}\alpha_{ij}\beta\ equiv 0\mod Q_i^{k+1}.
	\end{eqnarray*}
	Since $c_{ij\beta}\equiv 0\mod Q_i$ for all $i,\beta$ and $j\leq k$, we then see that
	\begin{eqnarray*}
	    \sum_{\beta\in B_i}c_{i(k+1)\beta}\alpha_{i(k+1)}\beta\ \equiv 0\mod Q_i,
	\end{eqnarray*}
	which implies, as before, that $c_{i(k+1)\beta}\equiv 0\mod Q_i$ for all $i$ and all $\beta$.  By induction, then
	we have that $c_{ij\beta}\equiv 0\mod Q_i$ for all $1\leq i\leq r$, $1\leq j\leq e_i$ and all $\beta\in B_i$.
	Viewing the $c_{ij\beta}$ as elements of $R$ (that is, picking representatives of the $P$ cosets in $R$)
	we see that $c_{ij\beta}\,\in\,(Q_i)$ for each $i$.  Since $c_{ij\beta}\,\in\,R$, we have
	$c_{ij\beta}\,\in\,R\cap Q_i$, that is, $c_{ij\beta}\,\in\,(P)$.  This holds for each $i,j,\beta$, so that
	$c_{ij\beta}=0$ in $R/P$ for all $i,j,\beta$, which shows that the $\alpha_{ij}\beta$ are linearly
	independent over $R/P$.

21)     Suppose that $p||S/G|$ where $G$ is the free abelian group generated by $\alpha_1,\ldots,\alpha_n$.  Then
	$S/G$ contains an element of order $p$, say $\beta$ with $p\beta\equiv 0\mod G$.  That is, $p\beta\,\in\,G$
	so that we have $p\beta=x_1\alpha_1+cdots+x_n\alpha_n$ with $x_n\,\in\,\Z$.  We therefore have
	$x_1\alpha_1+cdots+x_n\alpha_n\equiv 0\mod p$ so that $x_i\equiv 0\mod p$ for all $i$ since the $\alpha_i$
	are independent mod $p$.  We therefore have $x_i=py_i$ for some $y_i\,\in\,\Z$ so that
	$\beta=y_1\alpha_1+\cdots+y_n\alpha_n$ is in $G$ whence $\beta=0$ in $S/G$ and does not have order $p$,
	a contradiction.  Hence, $p\not| |S/G|$.  Since exercise 27 of chapter 2 tells us that
	$\disc(\alpha_1,\ldots,\alpha_n)=|S/G|^2\disc(S)$, we have that
	$\disc(\alpha_1,\ldots,\alpha_n)=m\disc(S)$ with $p\not|m$.

	Now let $\alpha_1,\ldots,\alpha_n$ be the set constructed in exercise 20.  We know that this set is
	independent mod $p$.  From the definition of this set, we readily see that a given prime
	$Q$ over $p$ divides exactly $k=\sum_{j=1}^{r}(e_j-1)f_j$ of the $\alpha_i$.  We have shown that if
	$\sigma\,:\,K\longrightarrow \C$ is any imbedding then $\sigma^{-1}(Q)$ is also a prime over $p$ and
	therefore divides $\alpha_i$ for $k$ values of $i$, so that $Q|\sigma(\alpha_i)$ for $k$ values of $i$.
	Therefore, if $\{\sigma_1,\ldots,\sigma_s\}$ are the imbeddings of $K$ into $\C$, we see that
	for every $j$, we have $Q|\sigma_j(\alpha_i)$ for exactly $k$ values of $i$.  Since 
	$\disc(\alpha_1,\ldots,\alpha_n)$ is the determinant $|\sigma_j(\alpha_i)|$, we see that
	$Q^k$ divides $\disc(\alpha_1,\ldots,\alpha_n)$.  Since the discriminant is in $\Z$, we have
	that $Q^k\cap \Z=p^k$ divides $\disc(\alpha_1,\ldots,\alpha_n)$.  Therefore, since 
	$\disc(\alpha_1,\ldots,\alpha_n)=m\disc(S)$ with $p\not|m$, we see that 
	$p^k|\disc(S)$.

28)	Since $\alpha$ is a root of $x^n+a_{n-1}x^{n-1}+\cdots+a_0$ with $p^r|a_j$ for all $j$, we see that
	\begin{eqnarray*}
	    \alpha^n&=&p^r\left(-\frac{a_{n-1}}{p^r}\alpha^{n-1}-\cdots-\frac{a_0}{p^r}\right)\\
	    &:=&p^r\beta,
	\end{eqnarray*}
	with $\beta\,\in\,R$.  Since $p^{r+1}\not|a_0$, we see that $(\beta)$ and $(p^r)$ are
	relatively prime.

	Now write $(\alpha)=Q_1^{e_1}\cdots Q_t^{e_t}$ uniquely as a product of prime ideals.  Then we have
	$\prod_{i=1}^t Q_i^{ne_i}=(p^r)(\beta)$.  Since $(\beta)$ and $(p^r)$ are relatively prime, for each
	$i$ we have $Q_i^{ne_i }$ dividing $(p^r)$ or $(\beta)$.  Let $i_1,\ldots,i_k$ be those $i$ for which
	$Q_i^{ne_i }$ divides $(p^r)$.  Then we must have $Q_j^{ne_j}$ dividing $(\beta)$ for any $j$ which
	is not one of the $i_l$.  Since in fact $(\alpha^n)=(p^r)(\beta)$, we must have
	\begin{eqnarray*}
	    p^r&=&Q_{i_1}^{ne_{i_1}}\cdots Q_{i_1}^{ne_{i_k}}\\
	    &=&(Q_{i_1}^{e_{i_1}}\cdots Q_{i_1}^{e_{i_k}})^n,
	\end{eqnarray*}
	so that $p^r$ is the $n^{\text{th}}$ power of some ideal in $R$.
	
	Now suppose that $(n,r)=1$.  Since $(p^r)=I^n$ for some ideal $I$, we may decompose both sides
	into their unique prime factorizations to obtain
	\begin{eqnarray*}
	    P_1^{re_1}\cdots P_s^{re_s}\ =\ Q_1^{nf_1}\cdots Q_t^{nf_t} 
	\end{eqnarray*}
	from which we obtain $s=t$ and, renumbering if necessary, $re_i=nf_i$ for each $i$.  Therefore,
	$n|re_i$ and since $(n,r)=1$ we conclude that $n|e_i$ so that we may write $e_i=n\epsilon_i$ and hence
	\begin{eqnarray*}
	    p=(P_1^{\epsilon_1}\cdots P_s^{\epsilon_s})^n.
	\end{eqnarray*}
	Therefore, $p$ is the $n^{\text{th}}$ power of some ideal, $p=Q^n$.  Since the primes $\pi_i$
	above $p$ satisfy 
	\begin{eqnarray}
	    \sum_{i=1}^d f(\pi_i|p)e(\pi_i|p)=n
	\label{formula}
	\end{eqnarray}
	we see that $p$ only has one prime
	above and that $p$ is totally ramified.  That is, $d=1, f=1$ and $e=n$.

	By exercise $21$, we see that $p^k=p^{n-\sum_{i=1}^d (f_i)}=p^{n-1}$ divides $\disc(R).$
	In the case that $(n,r)=m>1$, we may show, just as above, that $p=(Q)^{n/m}$ for some ideal $Q$
	and therefore that the ramification indices of the primes above $p$ are all divisible by
	$n/m$.  Write $e_i=m_i(n/m)$ and use formula \ref{formula} to obtain 
	\begin{eqnarray*}
	    \sum_{i=1}^d f_i \leq \sum_{i=1}^d m_i f_i = \sum_{i=1}^d e_i\frac{m}{n} f_i=m, 
	\end{eqnarray*}
	and hence $n-\sum_{i=1}^d\geq n-m$, from which (by exercise $21$ again) it follows that
	$p^{n-m}|\disc(R)$.

\section{Totally Ramified Primes}

24)     Let $K\subset M\subset L$ be field extensions and suppose that a prime $p\,\in\,K$ is
	totally ramified in $L$.  There is only one prime, say $Q$ above $p$ in $L$.  Since there is
	a unique prime below $Q$ in $M$, we see that there is only one prime above $p$ in $M$, say $\pi$.
	We also know that $f(Q|\pi)f(\pi|p)=f(Q|p)=1$, so that $f(\pi|p)=1$ also.  Therefore,
	since $f(\pi|p)e(\pi|p)=e(\pi|p)=[M:K]$, we conclude that $p$ is totally ramified in $M$.

	Now suppose that $L,L'$ are field extensions of $K$ and that $p$ is totally ramified in $L$ and
	unramified in $L'$.  Let $M=L\cap L'$.  Then we know that $p$ is totally ramified in $M$ since
	$M\subset L$ is a subfield.  Therefore, let $\pi$ be the unique prime above $p$ in $M$.  We then have
	$\pi^m=p$ where $m=[M:K]$.  Now suppose that $\pi$ factors in $L'$ as $\pi=Q_1^{e_1}\cdots Q_s^{e_s}$.
	Then we have the factorization $p=Q_1^{me_1}\cdots Q_s^{me_s}$ in $L'$.  Since $p$ is unramified in
	$L'$, we must have $me_i=1$ for all $i$ and therefore $m=1$ whence $M=K$.

\section{Completely Split Primes}

30)	Let $f$ be any nonconstant polynomial over $\Z$.  We show that $f$ has a root modulo $p$ for infinitely
	many primes $p$.  First assume that $f(0)=1$.  We then have $f(n!)\equiv 1\mod n!$ for any $n$.  
	THerefore, the prime divisors of $f(n!)$ are greater than or equal to $n$, from which we deduce that
	there are infinitely many primes $p$ for which $f$ has a root modulo $p$.  If $f(0)=0$ then obviously
	$f$ has a root modulo every prime $p$ since it has a root in $\Z$, so we may suppose that $f(0)\neq 0$.
	Now there are only finitely many primes dividing $f(0)$, so excluding them from our considerations,
	we see that $if $f $f(0)\neq 1$ then $g(x)=f(xf(0))/f(0)$ is well defined modulo $p$ and satisfies
	$g(0)=1$.  Notice that any root of $g$ yields a root of $f$.  We are therefore reduced to the previous 
	case.

	Now let $K=\Q(\alpha)$ be a number field with $g\,\in\,\Z[X]$ the minimal polynomial of $\alpha$.
	Suppose that $g$ factors modulo $p$ as $\bar{g_1}^{e_1}\cdots\bar{g_r}^{e_r}$.
	From \cite[pg. 79]{marcus} we know that if $p\,\in\,\Z$ does not divide $|S/\Z[\alpha]|$ then
	the prime decomposition of $p$ in $K$ is given by
	\begin{eqnarray*}
	    Q_1^{e_1}\cdots Q_r^{e_r},
	\end{eqnarray*}
	where $Q_i^{e_i}=(p,g_i(\alpha))$ and $f(Q_i|p)=\mathop{deg}(g_i)$.  Since there are infinitely many 
	primes $p$ for which the polynomial $g$ has a root modulo $p$, we see that there are infinitely many
	primes $p$ for which $g$ has a linear factor and hence for which there is a prime $Q$ over $p$ with $f(Q|p)=1$.

	Let $K,L$ be number fields with $K\subset L$.  Denote by $\bar{L}$ the normal closure of $L$ over $K$.
	We know that there are infinitely many primes $P_{\bar{L}}$ of $\bar{L}$ with $f(P_{\bar{L}}|P_{\Z})=1$ where $P_{\Z}$ is 
	the unique prime in $\Z$ below $P_{\bar{L}}$.  Since $\bar{L}/K$ is Galois, we have the relation
	$efd=[\bar{L}:\Q]$ where $e=e(P_{\bar{L}}|P_{\Z})$, $f=f(P_{\bar{L}}|P_{\Z})=1$, and $d$ is the number
	of primes over $P_{\Z}$ in $\bar{L}$.  Since only finitely many primes of $\Z$ are ramified in
	$\bar{L}$, we see that there are infinitely many primes $P_{\bar{L}}$ such that the unique prime $P_{\Z}$
	below splits into $[\bar{L}:\Q]$ factors in $\bar{L}$, i.e. splits completely in $\bar{L}$.  Let $P_{K}$
	be the unique prime below $P_{\bar{L}}$ in $K$.  Then using the relations
	\begin{eqnarray*}
	    f(P_{\bar{L}}|P_{L})f(P_{L}|P_{L})f(P_{K}|P_{\Z})&=&f(P_{\bar{L}}|P_{\Z})=1\\
	    e(P_{\bar{L}}|P_{L})e(P_{L}|P_{K})e(P_{K}|P_{\Z})&=&e(P_{\bar{L}}|P_{\Z})=1,
	\end{eqnarray*}
	we see that $e(P_{L}|P_{K})=f(P_{L}|P_{K})=1$ for infinitely many primes $P_{\bar{L}}$.
	Since there are only finitely many primes in $\bar{L}$ above any given prime in $K$, we see that there must be
	infinitely many primes $P_{K}$ of $K$ which split completely in $L$.

	Let $f$ be a nonconstant monic irreducible polynomial over a number ring $R\subset K$ and 
	let $\alpha$ be a root of $f$.  Set $L=K(\alpha)$.  By Theorem $27$ \cite[pg. 79]{marcus},
	we know that if $f=\bar{g_1}^{e_1}\cdots \bar{g_r}^{e_r}\mod P$ where the $\bar{g_i}$ are
	irreducible modulo $P$ then the prime factorization of $P$ in $L$ is given by
	$P=Q_1^{e_1}\cdots Q_r^{r_r}$, where $Q_i=(P,g_i(\alpha))$ and $f(Q_i|P)=\mathop{deg}(g_i)$.
	We have shown that there are infinitely many primes $P$ of $R$ which split completely in $L$,
	and therefore infinitely many $P$ for which all the $g_i$ have degree one, that is, for which
	$f$ splits into linear factors modulo $P$.

\section{Splitting Primes in Biquadratic Extensions}

6)	Let $m,n$ be distinct, squarefree integers and set $K=\Q(\sqrt{m},\sqrt{n})$.  Then $K$ contains
	three quadratic subfields, $\Q(\sqrt{m}),\Q(\sqrt{k}),\Q(\sqrt{k})$, where $k=mn/(m,n)^2$.
	Let $p$ be a prime in $\Z$.  In any examples that follow, we shall make implicit use of
	the following trivial calculation:
	\begin{eqnarray*}
	    \disc(\sqrt{m}+\sqrt{n})&=&N\left(4(\sqrt{m}+\sqrt{n})((\sqrt{m}+\sqrt{n})^2-(m+n))\right)\\
	    &=&2^{12}m^2n^2(m-n)^2\\
	    &=&\ind(\sqrt{m}+\sqrt{n})\disc(R)\\
	    &=&2^b\ind(\sqrt{m}+\sqrt{n})\frac{m^2n^2}{(m,n)^2},
	\end{eqnarray*}
	where $b=0,4$ or $6$, depending on the residues of $m,n\mod 4$.  In this manner, we
	may determine whether or not a given prime divides $\ind(\sqrt{m}+\sqrt{n})$,
	and therefore whether or not Theorem 27 \cite[pg. 79]{marcus} is applicable.

	Suppose that $p$ is ramified in each of the quadratic subfields.  Then we know that $p$ divides
	the discriminant of each of these fields.  But we know that the discriminant of $\Q(\sqrt{d})$
	is $d$ or $4d$ depending on whether $d\equiv 1 \mod 4$ or $d\not\equiv 1\mod 4$.  Therefore,
	if $p$ is a prime other than $2$, it must divide $m,n,$ and $k$.  But the definition of $k$ shows
	that this is impossible, so that $p=2$.  Observe that $\alpha=\sqrt{m}+\sqrt{n}$ is of degree $4$ over
	$\Q$, so that it therefore generates the extension $K/\Q$.  The minimal polynomial of $\alpha$
	over $\Z$ is easily seen to be
	\begin{eqnarray*}
	    f(X)=(X^2-(m+n))^2-4mn.
	\end{eqnarray*}
	Reducing modulo $2$ and using the fact that $X^2-1\equiv(X-1)^2\mod 2$, we see that $f(X)\equiv X^4\mod 2$
	if $m+n$ is even and $f(X)\equiv (X-1)^4\mod 2$ is $m+n$ is odd.  In any case, the prime $p$ (that is, $2$)
	is totally ramified in $K$.  As an example, we let $m=3$, $n=6$, $k=2$.  Then we have the ideal 
	factorizations
	\begin{eqnarray*}
	     2&=&(2,1+\sqrt{3})^2\\
	     2&=&(2,2+\sqrt{6})^2\\
	     2&=&(\sqrt{2})^2
	\end{eqnarray*}	
	so that $2$ is totally ramified in each quadratic subfield.  We also easily have
	\begin{eqnarray*}
	    2\ =\ (2,\sqrt{2}+\sqrt{3}-1)^4
	\end{eqnarray*}
	in $K$.

	We know that $K$ is just the composite of $\Q(\sqrt{m})$ and $\Q(\sqrt{n})$.  By Theorem $31$ \cite[pg. 107]{marcus},
	if a prime $P$ splits completely in $\Q(\sqrt{m})$ and $\Q(\sqrt{n})$ then it splits completely in
	their composite.  Thus, $P$ must split completely.  As an example, we take $m=2$, $n=15$, and $p=7$.
	Then since $1,2$ are squares modulo $7$ we see that $7$ splits in $\Q(\sqrt{m})$ and $\Q(\sqrt{n})$.
	In fact,
	\begin{align*}
	    7&=(7,\sqrt{2}+3)(7,\sqrt{2}-3) & \text{in $\Q(\sqrt{2})$},\\
	    7&=(7,\sqrt{15}+1)(7,\sqrt{15}-1) &\text{in $\Q(\sqrt{15})$},\\	
	    7&=(7,\sqrt{30}+3)(7,\sqrt{30}-3) &\text{in $\Q(\sqrt{30})$},\\
	    7&=(7,\sqrt{15}+\sqrt{2}-2)(7,\sqrt{15}+\sqrt{2}+2)(7,\sqrt{15}+\sqrt{2}-3)(7,\sqrt{15}+\sqrt{2}+3) &\text{in $K$}.
	\end{align*}


	Notice that $P$ cannot be inert in each of the quadratic subfields.  To see this, we observe that
	$P$ is inert in $\Q(\sqrt{m})$ if and only if $x^2-m$ is irreducible modulo $P$.  That is, if and only if
	\begin{eqnarray*}
	    \left(\frac{m}{P}\right)=-1,
	\end{eqnarray*}
	where $\left(\frac{m}{P}\right)$ is the Jacobi symbol of $m\mod P$.  But since $k=mn/(m,n)^2$, we easily see
	that
	\begin{eqnarray*}
	    \left(\frac{k}{P}\right)\ =\ \left(\frac{m}{P}\right)\left(\frac{n}{P}\right),
	\end{eqnarray*}
	so that if $P$ is inert in $\Q(\sqrt{m})$ and $\Q(\sqrt{n})$, then $m,n$ are not squares modulo $P$ and
	therefore $k$ \textit{is} a square modulo $P$ so that $P$ is \textit{not} inert in $\Q(\sqrt{k})$.

	Let $m=3,n=2$ and $p=3$.  The minimal polynomial of $\sqrt{2}+\sqrt{3}$ is $(X^2-5)^2-24\equiv (X^2+1)^2\mod 3$.
	Since $X^2+1$ is irreducible modulo $3$, we find that the factorization of $3$ in $\Q(\sqrt{2},\sqrt{3})$ is
	\begin{eqnarray*}
	    3=(3,2\sqrt{6})^2.
	\end{eqnarray*}

	Now let $m,n$ be as above but consider $p=5$.  We have \cite[pg. 79]{marcus} that 
	$(X^2-5)^2-24\equiv X^4+1\equiv (X^2+2)(X^2-2)\mod 5$.  Since $(X^2\pm2)$ are irreducible modulo $5$
	we find that $5$ splits as
	\begin{eqnarray*}
	    5=(5,2+2\sqrt{6})(5,3+2\sqrt{6})
	\end{eqnarray*}
	in $\Q(\sqrt{2},\sqrt{3})$.

	Finally, let $m=5,n=6$, and $p=5$.  Then since the minimal polynomial of $\sqrt{5}+\sqrt{6}$ is
	$(X^2-11)^2-120\equiv (X-1)^2(X+1)^2\mod 5$, we see that $5$ factors as
	\begin{eqnarray*}
	    5=(5,\sqrt{5}+\sqrt{6}+1)^2(5,\sqrt{5}+\sqrt{6}-1)^2
	\end{eqnarray*}
	in $\Q(\sqrt{5},\sqrt{6})$.

7)	Let $K=\Q(\sqrt{-1})$ and $L=\Q(\sqrt{-5})$.  We show that $2$ is totally ramified in $K,L$, but not in
	$KL$.  
	It is not difficult to see that we have the factorizations
	\begin{eqnarray*}
	    2&=&(1+\sqrt{-1})^2\quad\text{in $K$ and}\\
	    2&=&(2,1+\sqrt{-5})^2\quad\text{in $L$}.
	\end{eqnarray*}
	On the other hand,
	\begin{eqnarray*}
	    2&=&\left(2,\frac{1}{2}(-1+\sqrt{-1})(1+\sqrt{5})\right)^2,
	\end{eqnarray*}
	as you will check (with tears).  That this is a prime ideal follows from the fact that 
	$\frac{1}{2}(-1+\sqrt{-1})(1+\sqrt{5})$ has norm $4$.  We note that this gives an example
	of a prime ($2$) that totally ramifies in $K$ and $L$ but not in $KL$ and of a prime
	that gives trivial residue field extensions for $K,L$, but not for $KL$.  Of course, both
	of these phenomena are unusual and happen because $2$ divides $\ind(\sqrt{-1}+\sqrt{-5})$.

	Now consider $K=\Q(\sqrt{3}),L=\Q(\sqrt{3})$, and $p=7$.  It is clear that $7$ is inert in $K,L$ since
	$3,5$ are quadratic non residues modulo $7$.  However, $7$ factors in $KL$ as
	\begin{eqnarray*}
	    7=(7,3+2\sqrt{15})(7,2+2\sqrt{15}),
	\end{eqnarray*}
	which is readily checked.  This therefore gives an example where a prime $7$ has unique primes lying over it in
	$K,L$ (namely $7$) but not in $KL$.  It also gives an example of a prime that is inert in $K,L$ but not in $KL$.
	That covers all the examples required in the problem.

\begin{thebibliography}{2}

\bibitem{marcus} Marcus, D.A.  \textit{Number Fields}.  Springer Verlag, New York.  1977.

\end{thebibliography}

\end{document}