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\title{Math 129 HW 7}
\author{Bryden Cais}

\begin{document}
\maketitle

\begin{enumerate}

	\item Suppose that $p\leq \frac{-d_K}{4}$.  We know that 
	\begin{eqnarray*}
	\left\{1,\frac{d_K+\sqrt{d_K}}{2}\right\}
	\end{eqnarray*}
	is an integral basis for $\O$, so that if $\N(x)=p$ for some $x\in\O,$ then we would have
	\begin{eqnarray*}
	    p&=&\N\left(a+b\left(\frac{d_K+\sqrt{d_K}}{2}\right)\right)\\
	    &=&\left(a+\frac{bd_K}{2}\right)^2+b^2\frac{-d_K}{4}.
	\end{eqnarray*}
	Since $p\leq\frac{-d_K}{4}$, it follows that we in fact have equality, $b=\pm 1$, $d_K=-4p$, and
	$a=-2bp$.

	\item Suppose that $\O$ is principal.
	\begin{enumerate}

		\item Suppose that $p<\frac{-d_K}{4}$ splits in $\O$ as $p=Q_1Q_2$ for $Q_1,Q_2$ not necessarily
		distinct primes.  Then we have $Q_1=(x)$ for some $x\in\O$ since $\O$ is principal and therefore
		$\N(x)=p$, contradicting problem $1$.  Therefore $p$ is inert.

		\item Suppose that $-d_K>8$.  Part a then tells us that $2$ is inert since $2<-d_K/4$.  Therefore
		$2$ does not ramify so that $2\not|d_K$ which implies that $d_K=d$ and $d\equiv 1\mod 4$.  Therefore,
		$d\equiv 1\quad\text{or}\quad5\mod 8$.  We know that $\O=\Z[\frac{1+\sqrt{d}}{2}]$, and therefore
		that $2$ splits if $d\equiv 1\mod8$ since the minimal polynomial of $\frac{1+\sqrt{d}}{2}$ is
		\begin{eqnarray*}
			X^2-X+\frac{1-d}{4},
		\end{eqnarray*}
		which splits as $X(X-1)\mod2$.  Now if $-d$ is not prime then there exists some $p<\sqrt{-d}$ dividing
		$d_K$.  This implies that $p$ ramifies in $\O,$ that is $p=Q^2$.  Since $-d>8$ and $-d\equiv 3\mod 8$ 
		by the above, we have $-d>16$ (since we are assuming that $d$ is not prime) and theresore that $\sqrt{-d}<-d_K/4$.
		Then $p<-d_K/4$ and $p=Q^2$, which contradicts part a.  Therefore, $-d$ is prime.

		\item Suppose that $-d_K\geq 28$.  Since we have shown that $d_K=d\equiv 5\mod8$, we in fact have $-d_K>28$
		so that $7<-d_K/4$.  By part a, the primes $3,5,7$ are inert in $K$.  That is, the polynomial
		\begin{eqnarray*}
			X^2-X+\frac{1-d}{4}
		\end{eqnarray*}
		is irreducible modulo $3,5,7$.  From this, we obtain the requirements
		\begin{eqnarray*}
			-d&\equiv& 1\mod 3\\
			-d&\equiv& 2,3\mod 5\\
			-d&\equiv& 1,2,4\mod 7.
		\end{eqnarray*}
		We combine the first congruence with the requirement that $-d\equiv 3\mod 8$ from above to obtain
		\begin{eqnarray*}
			-d&\equiv& 19\mod 24\\
			-d&\equiv& 2,3\mod 5\\
			-d&\equiv& 1,2,4\mod 7.
		\end{eqnarray*}
		The chinese remainder theorem gives a unique solution modulo $2^3\cdot3\cdot5\cdot7=840$ to 
		each of the six systems of congruences above.  A short computation shows that $-d$ is congruent
		modulo $840$ to one of
		\begin{eqnarray*}
			43,67,163,403,547,667.
		\end{eqnarray*}
	\end{enumerate}

	\item Here we shall use the Minkowski bound.  We have that every ideal of $\O$ is equivalent to 
	some ideal $J$ with 
	\begin{eqnarray*}
		\N(J)\leq \lfloor\frac{2}{\pi}\sqrt{|d_K|}\rfloor=\lambda.
	\end{eqnarray*}
	For $d=-1,-2,-3,-7$ we have $d_K=-4,-8,-3,-7$ and therefore that $\lambda=2$, so that $\O$ is
	clearly principal for these values of $d$.  For $d=-11,-19$ we have $\lambda =2$.  Since
	$2$ is inert in $\O$ by $2$b, we see that $2$ cannot be the norm of any ideal and therefore 
	every ideal of $\O$ is equivalent to some ideal of norm $1$; i.e. $\O$ is principal for
	these values of $d$ also.  For $d=-43$, we obtain $\lambda=4$.  Since it is enough
	to check that no ideal has prime norm (since such ideals generate the classgroup), we need
	only consider the possibilities $\N(J)=2,3$.  We rule out the first right away using $2$b
	and since $X^2-X+\frac{1+43}{4}\equiv X^2-X+2\mod 3 $ is irreducible, we see that $3$ is inert
	and rule out the second possibility.  We then conclude that $\O$ is principal.  Similarly, for
	$d=-67$, we need only consider the behavior of $X^2-X+17$ modulo $3,5$.  It is readily seen that
	this polynomial is irreducible in both cases, so that $3,5$ are inert in $\O$ and hence, as before, that
	$\O$ is principal.  Finally, we conquer $d=-163$.  Minkowski gives $\lambda=8$ so that we must check
	the behavior of $X^2-X+41$ modulo $3,5,7$.  We find that this polynomial is irreducible for
	each modulus, and therefore that $3,5,7$ are inert and cannot be the norms of any ideals.  Therefore,
	every $\O$ ideal is equivalent to some ideal of norm $1$; that is, $\O$ is principal.

	\item Let's take $N=909$.  We have shown that if $\O$ is principal and $-d_K>28$ then 
	$-d\equiv 43,67,163,403,547,667\mod 840$.  We have also shown that $\O$ is principal
	for $-d=1,2,3,7,11,19,43,67,163$.  This leaves the values $-d=5,6,403,547,667,883,907$.
	When $-d=5,6$ we have $-d_K=20,24$ respectively so that by $2$c we must have $-d_K=d$,
	which rules out the possibility of principality of $\O$ for these values of $d$.  
	Observe that we have the following factorizations:
	\begin{eqnarray*}
		X^2-X+101&\equiv& (X+4)(X+6)\mod 11\\
		X^2-X+137&\equiv& (X+2)(X+8)\mod 11\\
		X^2-X+167&\equiv& (X+4)(X+6)\mod 11\\
		X^2-X+221&\equiv& X(X-1)\mod 13\\
		X^2-X+227&\equiv& (X+4)(X+8)\mod 13.
	\end{eqnarray*}
	This shows that $11$ is not inert in $\O$ for $-d=403,547,667$, so that by $2$a, since $11<-d/4$,
	we see that $\O$ is not principal for these values of $d$.  Similarly, $13$ is not inert in $\O$
	for $-d=883,907$ and hence $\O$ is not principal for these values of $d$ either.  We have shown
	that the only $-d$ in the range [1,909] for which $\O$ is principal are those listed in problem $3$.
	
	\item We now compute $\Cl(\O)$ for $-d=6,14,21,23$.
	\begin{enumerate}
		\item $-d=6$.
		Using Minkowski, we find that every $\O$ ideal is equivalent to some ideal $J$
		with $\N(J)\leq 3$.  Since $-6\equiv 2\mod 4$, we have $\O=\Z[\sqrt{-6}]$.  Since 
		$2,3$ divide $6$, both primes ramify and we find
		\begin{eqnarray*}
			(2)&=&(2,\sqrt{-6})^2=P_2^2\\
			(3)&=&(3,\sqrt{-6})^2=P_3^2.
		\end{eqnarray*}
		Since $(2,\sqrt{-6})(2,\sqrt{-6})=(\sqrt{-6})$ is principal, we have $[P_2]=[P_3]^{-1}$ where
		$[P]$ denotes, as usual, the image of $P$ in $\Cl(\O)$.  Therefore, $\Cl(\O)$ is 
		generated by $[P_2]$, and since $P_2^2$ is principal, we have $[P_2]^2=[1]$.  Therefore,
		\begin{eqnarray*}
			\Cl(\O)\simeq\Z/2\Z.
		\end{eqnarray*}

		\item $-d=14$.
		Again, by Minkowski, we find that every $\O$ ideal is equivalent to some ideal $J$
		with $\N(J)\leq 4$.  Since $-14\equiv 2\mod 4$, we have $\O=\Z[\sqrt{-6}]$.  We look
		at the polynomial $X^2+14$ modulo $2,3$ to determine the factorizations
		\begin{eqnarray*}
			(2)&=&(2,\sqrt{-14})^2=P_2^2\\
			(3)&=&(3,\sqrt{-14}-1)(3,\sqrt{-14}+1)=P_3\bar{P_3}.
		\end{eqnarray*}
		Notice that $(2+\sqrt{-14})(2-\sqrt{-14})=2\cdot3^2=P_2^2(P_3\bar{P_3})^2$.  Using the fact that
		$(2+\sqrt{-14})$ and $(2-\sqrt{-14})$ are conjugate, we find that up to relabeling we have
		the factorization
		\begin{eqnarray*}
			(2+\sqrt{-14})=P_2P_3^2,
		\end{eqnarray*}
		which shows that $[P_3]^2=[P_2]$.  We already know that $P_2^2=(2)$ is principal, hence
		that $[P_2]^2=[1]$.  Therefore, $\Cl(\O)$ is generated by $[P_3]$, which is of order $4$.
		This shows that 
		\begin{eqnarray*}
			\Cl(\O)\simeq\Z/4\Z.
		\end{eqnarray*}

		\item $-d=21$.
		Minkowski shows that we must only consider the behavior of the primes $2,3,5$ in $\O$.
		From the factorizations of $X^2+21$ modulo $2,3,5$, we find
		\begin{eqnarray*}
			(2)&=&(2,\sqrt{-21}-1)^2=P_2^2\\
			(3)&=&(3,\sqrt{-21})^2=P_3^2\\
			(5)&=&(5,\sqrt{-21}+2)(5,\sqrt{-21}-2)=P_5\bar{P_5}.
		\end{eqnarray*}
		From the fact that $(3+\sqrt{-21})(3-\sqrt{-21})=2\cdot3\cdot5$, we see (again using the fact that
		$(3+\sqrt{-21})$ and $(3-\sqrt{-21})$ are conjuagte), that
		\begin{eqnarray*}
			(3+\sqrt{-21})=P_2P_3P_5,
		\end{eqnarray*}
		up to relabeling and units.  At any rate, we see that $P_2P_3P_5$ is principal so that
		$[P_5]=[P_2P_3]^{-1}$.  Notice that this implies that $[P_2]\neq [P_3]^{\pm 1}$
		since then $P_5$ would be principal, which it clearly is not (by considering norms:
		$5\not=a^2+21 b^2$ for any integers $a,b$).  In summary, we have shown that $\Cl(\O)$
		is generated by $[P_2],[P_3]$ with the relations $[P_2]^2=[P_3]^2=1$, and $[P_2],[P_3]^{\pm1}$
		distinct.  Therfore,
		\begin{eqnarray*}
			\Cl(\O)\simeq \mathbf{V}_4,
		\end{eqnarray*}
		the Klein--four group.

		\item $-d=23$.
		Using Minkowski, we find that every $\O$ ideal is equivalent to some ideal $J$
		with $\N(J)\leq 3$.  In this case, $\O=\Z[\frac{1+\sqrt{-23}}{2}]:=\Z[\alpha]$, and we look
		at the irreducible polynomial
		\begin{eqnarray*}
			X^2-X+\frac{1+23}{4}=X^2-X+6
		\end{eqnarray*}
		modulo $2,3$.  This gives the factorizations
		\begin{eqnarray*}
			(2)&=&(2,\alpha)(2,\alpha-1)=P_2\bar{P_2}\\
			(3)&=&(3,\alpha)(3,\alpha-1)=P_3\bar{P_3}.
		\end{eqnarray*}
		Since $\frac{1+\sqrt{-23}}{2}\frac{1-\sqrt{-23}}{2}=6$ and the two factors are conjugate, we must have
		\begin{eqnarray*}
			\left(\frac{1+\sqrt{-23}}{2}\right)=P_2P_3
		\end{eqnarray*}
		up to relabeling.  Therefore, $\Cl(\O)$ is generated by $[P_2]$.  Now, the factorization
		\begin{eqnarray*}
			\frac{3+\sqrt{-23}}{2}\frac{3-\sqrt{-23}}{2}=2^3
		\end{eqnarray*}
		tells us that 
		\begin{eqnarray*}
			\left(\frac{3+\sqrt{-23}}{2}\right)=P_2^3
		\end{eqnarray*}
		up to relabeling and units (it cannot be that 
		only a single or double power of $P_2$ divides $\frac{3+\sqrt{-23}}{2}$, for then we would have
		$\left(\frac{3+\sqrt{-23}}{2}\right)=P_2^2\bar{P_2}$, up to relabeling, which is impossible
		since $[P_2]^2[\bar{P_2}]=[P_2]$ is not principal).  Therefore, $[P_2]^3=1$ so that
		$[P_2]$ must have order $3$.  It follows that
		\begin{eqnarray*}
			\Cl(\O)\simeq \Z/3\Z.
		\end{eqnarray*}
	\end{enumerate}

	\item Let $d\equiv 1,2\mod 4$ be a positive square free integer greater than $4$, so that
	$\O=\Z[\sqrt{-d}]$, and suppose that $3$ does not divide $|\Cl(\O)|$.  Consider the equation 
	\begin{eqnarray*}
		y^2=x^3-d
	\end{eqnarray*}
	which we write as
	\begin{eqnarray*}
		(y-\sqrt{-d})(y+\sqrt{-d})=x^3.
	\end{eqnarray*}
	Assume that we have some solution with $x,y\in\Z$.
	We show first that $(y-\sqrt{-d})$ and $(y+\sqrt{-d})$ are relatively prime principal ideals.
	For let $P$ be a prime ideal containing $(y-\sqrt{-d})$ and $(y+\sqrt{-d})$.  Then $2\sqrt{-d}\in P$
	and since $P|x^3$ we must have $P|x$ and therefore $x\in P$.  This implies that $\N(P)|\N(2\sqrt{-d})$
	and $\N(P)|\N(x)$, that is, $\N(P)|4d$ and $\N(P)|x^2$ (since $x\in Z$).  Clearly $x$ cannot
	be even since then it would be $0\mod 8$ and $x^3-d\equiv 2,3\mod 4$, neither of which are squares.
	Therefore $x$ is odd.  Furthermore, $(x,d)=1$ since if they had a common prime factor, its square
	would divide $y^2$ so that, since its cube has to divide $x^3$, $d$ would not be square free.
	Thus, $N(P)=1$ and $(y-\sqrt{-d})$ and $(y+\sqrt{-d})$ are relatively prime.  Now let $Q$ be any
	prime ideal factor of $(x)$.  Then $Q^3|(y-\sqrt{-d})(y+\sqrt{-d})$ and therefore
	$Q^3$ divides one of $(y-\sqrt{-d})$ or $(y+\sqrt{-d})$.  Repeating this process for each prime ideal
	factor of $(x)$, we find that $J^3=(y+\sqrt{-d})$ for some ideal $J$.  Since $3\not||\Cl(\O)|$,
	$\Cl(\O)$ contains no elements of order $3$, and therefore $J$ must be principal, say
	$J=(a+b\sqrt{-d})$ with $a,b\in\Z$.  Since the only units of $\Z[\sqrt{-d}]$ are $\pm 1$
	(we showed this on an earlier homework) we have
	\begin{eqnarray*}
		y+\sqrt{-d}=\pm(a+b\sqrt{-d})^3,
	\end{eqnarray*}
	which gives the equations
	\begin{eqnarray*}
		y&=&\pm a(a^2-3db^2)\\
		1&=&\pm b(3a^2-b^2d)
	\end{eqnarray*}
	from which we obviously have $b=\pm 1$ and $d=3a^2\pm 1$.  Notice that $d=5$ has
	$\Cl(\O)\simeq \Z/2\Z$ and therefore satisfies all of our hypotheses on $d$ but that
	$5$ is not of the form $3a^2\pm 1$ for any integer $a$.  Therefore, the equation
	$y^2=x^3-5$ has no integer solutions.

	\item  Let $K=\Q(\sqrt[3]{19})$ and put $R=\O$.  In an earlier Marcus excercise, we
	computed a basis for the ring of integers of a pure cubic extension.  Using such a result,
	we determine that an integral basis for $R$ is
	\begin{eqnarray*}
		\left\{1,\alpha,\frac{1+\alpha+\alpha^2}{3}\right\}
	\end{eqnarray*}
	where $\alpha=\sqrt[3]{19}$.  Using this, it is not difficult to calculate the discriminant
	of $K$.  It is found to be $-3\cdot19^2$.  The Minkowski bound is therefore
	\begin{eqnarray*}
		\lambda=\frac{3!}{3^3} \left(\frac{4}{\pi}\right)\sqrt{3\cdot 19^2}<10.
	\end{eqnarray*}
	Marcus kindly tells us that
	\begin{eqnarray*}
		3R=P^2Q
	\end{eqnarray*}
	for primes $P,Q$.  We now consider how $2,5,7$ split in $R$.  The irreducible
	polynomial of $\alpha$ is clearly 
	\begin{eqnarray*}
		X^3-19&\equiv& X^3-1\equiv (X-1)(X^2+X+1)\mod 2\\
		&\equiv& X^3-4\equiv (X+1)(X^2+4X+1)\mod 5\\
		&\equiv& X^3+2\mod 7,
	\end{eqnarray*}
	where the factors listed above are clearly irreducible.  We therefore see that $7$ is inert in $R$ while
	\begin{eqnarray*}
		2R&=& P_2Q_2\\
		5R&=& P_5Q_5
	\end{eqnarray*}
	where $Q_2,Q_5$ have residual degree $2$.  In particular, we have unique primes of 
	norms $2$ and $5$.

	Now it is not difficult to see that
	\begin{eqnarray*}
		\N(\alpha-3) &=& -8\\
		\N(\alpha+1) &=& 20\\
		\N(\alpha-1) &=& 18\\
		\N(2 \alpha-5) &=& 27\\.
	\end{eqnarray*}
	Using the second equality, we find that $(\alpha+1)=P_5P_2^2$ or $P_5Q_2$ up to units.
	The third equality shows that $(\alpha-1)=P_2P^2$ or $P_2Q^2$ up to units.  In either case,
	since $P^2Q$ is principal and therefore $[Q]=[P]^{-2}$, we find that $[P_2]$
	and hence $[Q_2]$ (since $P_2Q_2=(2)$ is principal), and hence $[P_5]$, are all powers of $[P]$ 
	so that $[P]$ generates the class group.

	Now, using the first equality and the fact that there is a unique prime ideal of norm $2$,
	we have $P_2^3=\alpha-3$ up to units (note that it cannot be the product of the prime ideals
	of norms $2$ and $4$ since this is $P_2Q_2=(2)$), so that, since $P_2$ is not principal, 
	$\Cl(R)$ contains an element of order $3$ and therefore has size divisible by $3$.  
	Now both $3$ and $(2\alpha-5)$ have norm $27$ so that one of the following factorizations holds:
	\begin{eqnarray*}
		2\alpha-5 &=& P^3\\
		2\alpha-5 &=& P Q^2\\
		2\alpha-5 &=& Q^3.
	\end{eqnarray*}
	Therefore, in the first two cases, $P$
	has order $3$, while in the last case it has order $6$.  
	Now suppose that $[P]$ has order $6$.  Then $[Q]=[P]^{-2}=[P]^4$.  Similarly, considering the factorizations
	$(\alpha-1)=P_2P^2$ or $P_2Q^2$, we see that $[P_2]=[P]^4$ in any case.  Finally, considering the factorizations
	$(\alpha+1)=P_5P_2^2$ or $P_5Q_2$, we see that $[P_5]=[P]^4$.  By multiplying these ideals together, since
	each is an even power of $P$,
	we see that no ideal $J$ with $\N(J)\leq 9$ is in the same class as $P^3$.  This is impossible
	by the Minkowski bound 
	since we have assumed that $P^3$ is not principal.  Therefore, there are $3$ ideal classes
	and 
	\begin{eqnarray*}
		\Cl(R)\simeq \Z/3\Z.
	\end{eqnarray*}

	\item Let $K$ be a number field and $I$ an ideal of $R=\O$.  Since the class group $\Cl(R)$
	is finite, some power of $I$ is principal, say $I^mR=\alpha R$ with $\alpha\in R$.  Now
	let $L=K(\sqrt[m]{\alpha})$.  Then we certainly have $I^mS=\alpha S=(\sqrt[m]{\alpha}S)^m$.  It
	follows that $I=\sqrt[m]{\alpha}S$ is principal.  Clearly, $[L:K]|m$ is finite.  Now let
	$I_1,\ldots,I_k$ be representatives of the equivalence classes of $R$ ideals.  We know the
	list is finite since $\#\Cl(R)$ is finite.  We therefore have a finite extension $M$ of
	$K$ with each $I_j=\alpha_j\mathcal{O}_M$ a principal $\mathcal{O}_M$ ideal.  
	Notice that if $J\tilde I_j$ then there exists $\beta\in K$ with $\beta I_j=J$, and
	consequently $J=\beta\alpha_j\mathcal{O}_M$ is also a principal $\mathcal{O}_M$ ideal.
	Therefore, every ideal of $\O$ is principal in $M$.  As an example, we determine a degree
	$4$ extension of $\Q(\sqrt{-21})$ in which every ideal of $\Z[\sqrt{-21}]$ is principal.
	We determined above that $\Cl(\Z[\sqrt{-21}])\simeq\mathbf{V}_4$ is generated
	by $P_2,P_3$ with
	\begin{eqnarray*}
		P_2^2&=&(2,\sqrt{-21}-1)^2=2\\
		P_3^2&=&(3,\sqrt{-21})^2=3.
	\end{eqnarray*}
	Notice that if we find an extension of $\Q(\sqrt{-21}])$ in which $P_2,P_3$ are principal, then 
	every ideal generated by $P_2,P_3$ will also be principal since these generate the class group.
	By the above argument, $P_2,P_3$ are principal in the extension $\Q(\sqrt{-21},\sqrt{2},\sqrt{3})$,
	which is of degree $4$ over $\Q(\sqrt{-21})$.

	Let $\mathcal{A}$ denote the set of algebraic integers.  We show that every finitely generated
	ideal of $\mathcal{A}$ is principal.  Let $I=(a_1,\ldots,a_n)$ be a finitely generated ideal
	of $\mathcal(A)$.  Let $K=\Q(a_1,\ldots,a_n)$.  Then $I$ is an $\O$ ideal.  Therefore, by the 
	previous problem, there exists a finite extension $L$ of $K$ such that $I\mathcal{O}_L$ is
	principal, say $I=(\alpha)$.  But $\mathcal{O}_L\subset \mathcal{A}$ so that $\alpha\in\mathcal{A}$ 
	and $I=\alpha\O$ was principal to begin with.  As an example of an ideal of $\mathcal{A}$ that
	is not finitely generated, consider the ideal $I$ generated by $\sqrt{n}$ for $n=2,3,4,\ldots$.
	Then $I$ cannot be principal since if $I=(\alpha)$ for some $\alpha\in\mathcal{A}$ then 
	$\sqrt{n}\in\alpha\mathcal{A}$ for every $n\geq 2$ so that $\N(\sqrt{n})|\N(\alpha)$
	for each $n$.  But $\N(\sqrt{n})=n$, which implies that $\N(\alpha)$ is unbounded.  This is impossible
	and we see that $I$ is not finitely generated since it is not principal.

\end{enumerate}
\end{document}