%All of my files are now written in Latex2e ....!
\documentclass[11pt]{article}
\usepackage{a4wide}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{latexsym}

\def\Tr{\mathrm{Trace}}
\def\Trace{\mathrm{Trace}}
\def\frak{\mathfrak}
\def\ord{\mathrm{ord}}
\def\ra{\rightarrow}
\def\B{\mathcal B}
\def\N{\mathrm N}
\def\lra{\longrightarrow}
\def\Ker{\mathrm{Ker}}
\def\Disc{\mathrm{Disc}}
\def\O{\mathcal O}
\def\Q{\mathbb Q}
\def\QQ{\mathbb QQ}
\def\C{\mathbb C}
\def\CC{\mathbb C}
\def\Z{\mathbb Z}
\def\ZZ{\mathbb Z}
\def\R{\mathbb R}
\def\W{\mathcal W_N}
\def\k{\kappa}
\def\Remark{\bf Remark\rm. }
\setlength{\parindent}{0mm}
\def\gotha{\mathfrak{a}}
\def\gotho{\mathfrak{o}}
\def\gothp{\mathfrak{p}}
\def\gothq{\mathfrak{q}}

\begin{document}

%%%%%%% Introduction

\newtheorem{theorem}{Theorem}[section]
\newtheorem{df}{Definition}
\newtheorem{lemma}{Lemma}[section]
\newtheorem{cor}{Corollary}
\newtheorem{conj}{Conjecture}

\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Solutions for Problem Sets 1 and 2
\end{center}

\section{Gaussian Integers}

1. Pair each of $1, \dots, p-1$ with its inverse mod $p$. Then only 1 and
$p-1$ are paired with themselves. Thus $1 \cdots (p-1)$ can be rewritten
as $1 \cdot (p-1)$ times the product of pairs $xx^{-1}$, and so is
congruent to $-1$ modulo $p$.

\

2. In fact, every euclidean ring is a principal ideal ring, since any
nonzero element of an ideal of minimal norm generates the ideal. In fact,
we will show that every principal ideal ring $R$ is factorial. To prove
existence of prime factorization, pick $\alpha$ a nonzero element of $R$.
If $\alpha$ generates a prime ideal, then $\alpha$ is irreducible and we
are done. Otherwise, the ideal $(\alpha)$ is contained in a maximal ideal
$(\beta_1)$; write $\alpha = \beta_1 \alpha_1$. If $\alpha_1$ is irreducible,
we are done again. Otherwise, $(\alpha_1)$ is contained in a maximal ideal
$(\beta_2)$, and we may put $\alpha_1 = \beta_2 \alpha_2$. If this
process stops at some point, we obtain a factorization of $\alpha$
into irreducibles. Otherwise, we obtain an increasing sequence
$(\alpha) \subset (\alpha_1) \subseteq \cdots$ of ideals which does not
terminate, contradicting the fact that every principal ideal ring is
noetherian. Thus the process must stop somewhere, and prime factorizations
exist. As for uniqueness, if $\alpha_1 \cdots \alpha_m = \beta_1 \cdots
\beta_m$ is an equality of two products of irreducibles, then 
$\beta_1 \cdots \beta_m$ belongs to the prime ideal $(\alpha_1)$, so
one of the $\beta_i$ must be a multiple of $\alpha_1$, say $\beta_1$.
In fact, $\beta_1$ and $\alpha_1$ must generate the same ideal, since
both generate maximal ideals. Thus we can factor off $\alpha_1$
from both sides and keep going.

\

3. One method is to start with $x$ such that $x^2+1$ is divisible by $p$
(which can be found by exhaustive search, or by taking $x \equiv ((p-1)/2)!
\pmod{p}$), then compute a generator of $(p, x+i)$ by imitating the proof
from lecture that the Gaussian integers form a euclidean ring.
That is, given generators $(\alpha, \beta)$, replace them by
$(\beta, \gamma)$, where $\gamma-\alpha$ is a multiple of $\beta$
and $|\gamma| < |\beta|$. If $\gamma=0$ at some point, then $\beta$ generates
the ideal $(p, x+i)$, and if $\beta = a+bi$, then $a^2+b^2=p$.

\

4. I know, I know, you can do this without the Gaussian integers. So sue me.
Anyway, $x$ and $y$ can't both be odd by mod 4 considerations, so assume
$x$ is odd and $y$ is even. Factor $z^2 = (x+yi)(x-yi)$. Then
the GCD of $x+yi$ and $x-yi$ divides both $2x$ and $2yi$. If $x$ and $y$
have a common divisor, so does $z$, contrary to assumption. Thus the GCD
of $x+yi$ and $x-yi$ also divides $2$, but since $x$ is odd and $y$ is 
even, $x+yi$ is coprime to 2. Thus $x+yi$ and $x-yi$ are relatively prime.
That means, up to units, each one is a perfect square. That is, there exist
$u,v$ such that $x+yi = (u+vi)^2 c$ for some unit $c$.
Since $(u+vi)^2 = u^2-v^2 + 2uv$, $x$ is odd and $y$ is even, $c$ must be
1 or $-1$. If $c = -1$, we can rewrite $-(u+vi)^2$ as $(-v+ui)^2$. So without
loss of generality, assume $c=1$; then $x = u^2-v^2$, $y = 2uv$, and
$z = u^2+v^2$, as desired.

\

5. We use the norm $f(a + b\sqrt{2}) = |a^2-2b^2|$. (Note that the absolute
value is needed in this case!) Clearly this norm is multiplicative. Now
given $\alpha = a+b\sqrt{2}$ and $\beta = c+d\sqrt{2}$

\

6. 
Let $s(n)$ be the number of ways to write $n$ as a sum of two squares
(by which I'll mean the number of solutions of $x^2+y^2=n$ over the 
rational integers); then $s(n)$ is equal to the number of Gaussian integers
of norm $n$. In particular, if $m$ and $n$ are relatively prime,
$\alpha$ runs over the Gaussian integers of norm $m$,
and $\beta$ runs over the Gaussian integers of norm $n$,
then $\alpha \beta$ runs over the divisors of $mn$, but each 
divisor of $mn$ shows up four times. (If a divisor shows up once as
$\alpha \beta$, it shows up as $(\alpha i^m)(\beta i^{-m})$ for $m=0,1,2,3$.)
Thus $4s(mn) = s(m)s(n)$.

Suppose $n = p^e$ for $p$ a prime congruent to 3 modulo 4. Then $p$
is a prime in the Gaussian integers, so if $n = \alpha \overline{\alpha}$,
both of $\alpha$ and $\overline{\alpha}$ are divisible by $p^{e/2}$.
This is impossible if $e$ is odd, so in that case $s(p^e) = 0$; if 
$e$ is even, then the only possible $\alpha$ are $p^{e/2} i^m$ for $m=0,1,2,3$.
Thus in this case $s(p^e) = 4$.

Next, suppose $n = p^e$ for $p$ a prime congruent to 1 modulo 4. Then
$p$ factors as $\alpha \overline{\alpha}$ for some prime $\alpha$.
If $n = \beta \overline{\beta}$, then $\beta$ must be of the form
$\alpha^l \overline{\alpha}^{e-l} i^m$ for $l \in \{0, \dots, e\}$
and $m \in \{0,1,2,3\}$. In this case, $s(p^e) = 4(e+1)$.

Putting it all together, we get for $n = \prod_i p_i^{e_i}$,
\[
s(n) = 4 \prod_{i} \frac{s(p_i^{e_i})}{4} = 4 \prod_i
\begin{cases}
e_i+1 & p_i \equiv 1 \pmod{4} \\
0 & p_i \equiv 3 \pmod{4}, e_i \equiv 1 \pmod{2} \\
1 & p_i \equiv 3 \pmod{4}, e_i \equiv 0 \pmod{2}.  
\end{cases}
\]

\section{Number Fields}

1. We start with the basis $1$ and $\sqrt{D}$, which may or may not be
integral. Note that $\Trace(1) = 1 + 1 = 2$ and $\Trace(\sqrt{D}) = \sqrt{D} + (-\sqrt{D} = 0$. Thus
\[
\Disc(1, \sqrt{D}) = \det 
\begin{pmatrix}
  \Trace(1 \cdot 1) & \Trace(1 \cdot \sqrt{D}) \\
\Trace(\sqrt{D} \cdot 1) & \Trace{\sqrt{D} \cdot \sqrt{D}}
\end{pmatrix}
= \det 
\begin{pmatrix}
  2 & 0 \\ 0 & 2D 
\end{pmatrix}
= 4D.
\]
Recall that if $m$ is
the index, in the ring of integers, of the $\ZZ$-submodule
generated by this basis, then $m^2$ divides the discriminant of the basis.
Since we assumed $D$ is squarefree, either the basis $1, \sqrt{D}$ is integral,
or it generates a $\ZZ$-submodule of the ring of integers of index 2.

First suppose $D \equiv 1 \pmod{4}$. In that case, $(1 + \sqrt{D})/2$
has minimal polynomial $x^2 + x + (1-D)/4$, and so is an integer.
The basis $1, (1+\sqrt{D})/2$ generates a $\ZZ$-submodule between the ring
of integers and $\ZZ + \ZZ\sqrt{D}$, but the index between these is 2
and $\ZZ+\ZZ\sqrt{D}$ has index 2 in $\ZZ+\ZZ(1+\sqrt{D})/2$.
Thus $1, (1+\sqrt{D})/2$ is an integral basis.

Next, suppose $D \not\equiv 1 \pmod{4}$. In this case, we claim
$1, \sqrt{D}$ is already an integral basis; since the index of its
$\ZZ$-span in the ring of integers is at most 2, it suffices to check that
if $a,b$ are not both even, then $(a + b\sqrt{D})/2$ is not an algebraic
integer. In fact, it suffices to check this for $(a,b) = (1,0), (0,1), (1,1)$.
In these cases, the minimal polynomials of these are $x-1/2$,
$x^2-D/4$, and $x^2+x+(1-D)/4$, none of which have integer coefficients.
Thus $1, \sqrt{D}$ is an integral basis. In summary: an integral
basis is given by
\[
\begin{cases}
1, \frac{1+\sqrt{D}}{2} & D \equiv 1 \pmod{4} \\
1, \sqrt{D} & D \not\equiv 1 \pmod{4}.
\end{cases}
\]

\

2. First we compute $\Trace(1) = 3$, $\Trace(\sqrt[3]{2}) = 0$,
$\Trace(\sqrt[3]{4}) = 0$. (Either add up the conjugates in $\CC$ or
write out the matrices by which these act on $\QQ(\sqrt{3})$ as a
$\QQ$-vector space.) Thus 
\[
\Disc(1, \sqrt[3]{2}, \sqrt[3]{2^2}) =
\det 
\begin{pmatrix}
  \Trace(1\cdot 1) & \Trace(1 \cdot \sqrt[3]{2}) & \Trace(1 \cdot
\sqrt[3]{2^2} \\
\Trace(\sqrt[3]{2} \cdot 1) & \Trace(\sqrt[3]{2} \cdot \sqrt[3]{2})
& \Trace(\sqrt[3]{2} \cdot \sqrt[3]{2^2} \\
\Trace(\sqrt[3]{2^2} \cdot 1) & \Trace(\sqrt[3]{2^2} \cdot
\sqrt[3]{2}) & \Trace(\sqrt[3]{2^2} \cdot \sqrt[3]{2^2})
\end{pmatrix} =
\det 
\begin{pmatrix}
  3 & 0 & 0 \\ 0 & 0 & 6 \\ 0 & 6 & 0
\end{pmatrix}
= -2^2 \cdot 3^3.
\]
Thus the index of $\ZZ + \ZZ \sqrt[3]{2} + \ZZ \sqrt[3]{2^2}$ in the
full ring of integers is either 1, 2, 3 or 6.

We next check that the index is not divisible by 2. That is, we check
that if $a,b,c$ are rational
integers, not all even, then $\alpha = (a + b \sqrt[3]{2} +
c \sqrt[3]{2^2})/2$ is not an algebraic integer.
One could do this by computing the minimal polynomials for all
possible choices of $a,b,c \in \{0,1\}$, but in this case there is a shortcut.
If $\alpha$ is an algebraic integer,
so is $\alpha \sqrt[3]{2^2} = b + c \sqrt[3]{2} + (a/2) \sqrt[3]{2^2}$,
as then is $(a/2) \sqrt[3]{2^2}$. But the latter has minimal polynomial
$x^3 - a^3/2$, which has integer coefficients only if $a$ is even.
Thus $(b \sqrt[3]{2} + c \sqrt[3]{2^2})/2$ must also be an integer;
multiplying by $\sqrt[3]{2}$, we find that $b$ must be even, then $c$.

Finally, we check that the index is not divisible by 3. Again, we must
check that if $a,b,c$ are rational integers, not all divisible by 3, then
$\alpha = (a + b \sqrt[3]{2} + c \sqrt[3]{2^2})/3$ is not an algebraic integer.
It suffices to check this for $a,b,c \in \{0,1,2\}$. Again, one can list all
of the minimal polynomials of these elements, but one can also do a bit better.
Note that if we put $\beta = 1 + \sqrt[3]{2}$
and $\gamma = 1 - \sqrt[3]{2} + \sqrt[3]{2^2}$, then
$\beta \gamma = 3$. Now $\alpha \gamma =
(a+2b-2c)/3 + (-a+b+2c)\sqrt[3]{2}/3 + (a-b+c)\sqrt[3]{2^2}/3$
is equal to an element of $\ZZ + \ZZ\sqrt[3]{2} + \ZZ\sqrt[3]{2^2}$ plus
$(a-b+c)(1 + \sqrt[3]{2} + \sqrt[3]{2^2})/3$. This can only be integral
if $a-b+c$ is divisible by 3, since
$1+\sqrt[3]{2} + \sqrt[3]{2^2}$ has minimal polynomial $x^3- 3x^2-3x+1$
whose constant coefficient is not divisible by any cube.

We now know that if $\alpha$ is integral, then $a-b+c$ is divisible by 3.
Next, suppose $c=0$. Then
Then $\alpha(1 + \sqrt[3]{2} + \sqrt[3]{2^2}) =
(a+2b)/3 + (b+a)\sqrt[3]{2}/3 + (b+a) \sqrt[3]{2^2}/3$ is integral,
so $a+2b$ is divisible by 3, which now implies $a=b=0$. 
A similar argument applies if $a=0$, or if $b=0$. The only remaining
cases are $a=c=1, b=2$ or $a=c=2, b=1$. These are equivalent:
in the former, the minimal polynomial of $\alpha$ is
$x^3-x^2-x-1/3$, which is not integral.

We conclude $1, \sqrt[3]{2}, \sqrt[3]{2^2}$ is an integral basis.

\

3. From what we have shown in class, it suffices to check that the
discriminant of the polynomial $P(x)$ equals the discriminant of the
basis $1, x, \dots, x^{n-1}$. Let $r_1, \dots, r_n$ be the roots of
$P(x)$ in $\CC$, which are precisely the conjugates of $x$ in $\CC$.
The discriminant of the basis $1, x, \dots, x^{n-1}$ is
\[
(\det (r_i^{j-1})_{i,j=1}^n)^2 = 
\prod_{i<j} (r_i-r_j)^2
\]
by the properties of the Vandermonde determinant. (See Neukirch, p. 11.
One proof of the Vandermonde formula: view both sides as polynomials in
the $r_i$, note that the determinant vanishes whenever $r_i=r_j$, then
check that the coefficient of some monomial matches up.)
The latter is precisely the discriminant of $P$.

\

4. One example is $P(x) = x^3+2x+2$, which has discriminant $-140$.
The proof that $1, x, x^2$ is an integral basis is similar to the
argument given in problem 2.

\

5. Note $2\cos (2\pi/n) = \zeta_n + \zeta_n^{-1}$ and
$2\sin (2\pi/n) = -i(\zeta_n + \zeta_n^{-1}$; since $i, \zeta_n$ and
$\zeta_n^{-1}$ are algebraic integers, these combinations are too.
From this representation, we also see that the conjugates of
$\cos (2\pi/n)$ are all of the form $\cos (2\pi k/n)$. If 
$\cos (2\pi/n)$ were an algebraic integer, its norm would be an integer;
but the absolute value of the product of the conjugates is less than 1
for $n \geq 5$, contradiction. The argument for $\sin (2\pi/n)$ is similar.

\

6. The root of a polynomial $P$ with algebraic integer coefficients is an
algebraic number; let $K$ be the number field generated by the roots
and coefficients of $P$. The integrality of the root now follows from the
fact that $\gotho_K$ is integrally closed, which we checked in class.

\section{Interlude on Symmetric Polynomials}

1. Let $d$ be a positive integer; we check that the homogeneous symmetric
polynomial
of degree $d$ with integer coefficients can be written by integer
polynomials in $\sigma_1, \dots, \sigma_n$. We order the monomials
$x_1^{e_1} \cdots x_n^{e_n}$ in lexicographic order. That is, if
two monomials have different $e_1$, whichever has the bigger $e_1$
is the bigger of the two. If the $e_1$'s are equal, compare $e_2$'s, and so on.

Now given a homogeneous symmetric polynomial $P$ with integer coefficients,
let $c x_1^{e_1} \cdots x_n^{e_n}$ be the largest monomial in lexicographic
order. In particular, we must have $e_1 \geq \cdots \geq e_n$, otherwise
the monomial with the exponents sorted in order also appears and is larger
in lexicographic order. Then $P - c\sigma_1^{e_1-e_2}\cdots 
\sigma_{n-1}^{e_{n-1}-e_n}\sigma_n^{e_n}$ is symmetric with integer
coefficients, and its largest monomial is strictly smaller than 
that of $P$. Thus this process eventually stops at the zero polynomial,
and $P$ has been expressed as an integer polynomial in the $\sigma_i$.

\

2. We just give the argument for $\alpha+\beta$ here, as the argument
for $\alpha \beta$ is similar. Let $\alpha_1, \dots, \alpha_m$ be
the conjugates of $\alpha$ in $\CC$, and let $\beta_1, \dots, \beta_n$
be the conjugates of $\beta$ in $\CC$. Let $P(x)$ be the polynomial
$\prod_{i=1}^m \prod_{j=1}^n (x - \alpha_i - \beta_j)$. 
The polynomial $\prod_{i=1}m \prod_{j=1}^n (x - t_i - u_j)$
can be viewed as a symmetric polynomial in the $u_j$ with coefficients
in the ring $\ZZ[t_1, \dots, t_m]$, so by the previous problem is
a polynomial in the elementary symmetric functions of the $u_i$. But
the coefficients of this polynomial are themselves symmetric functions
of the $t_i$, so can be written as integer polynomials in the elementary
symmetric functions of the $t_i$. Substituting $t_i = \alpha_i$
and $u_j = \beta_j$, the elementary symmetric functions of
the $t_i$ and $u_j$
become the coefficients of the minimal polynomial of $\alpha$ and $\beta$,
which are integers. Thus $P$ becomes a monic
polynomial with integer coefficients, so its root $\alpha+\beta$ is
an algebraic integer.

\section{Conjugates, Trace and Norm}

1. Let $[\Q(\alpha):\Q]=n$. Let
$$X^n + a_{1} X^{n-1} + \ldots a_n$$
be the minimal polynomial of $\alpha$ over $\Q$. 
Since $\alpha$ is algebraic we may assume that $a_i \in \Z$.
Let $\alpha_i$ be the conjugates of $\alpha$, and let
$\sigma_k$ be the $k^{th}$ symmetric polynomial.
Using the triangle inequality, and
the identity $|\alpha_i| = 1$,
we conclude that
$$\|a_i\| = \|\sigma_k(\alpha_1,\ldots,\alpha_n)\| \le
\binom{n}{k}.$$
Since this bound depends only on $n$, we conclude that there
can only be \emph{finitely many}
algebraic integers of degree $n$ with this property,
since there are only
finitely many polynomials for each $n$ satisfying the above bound.
 On the other hand,
if $\alpha$ has this property, then so does $\alpha^k$ for every $k$.
Thus $\alpha^i = \alpha^j$ for some $i \ne j$ and we conclude 
(since $\alpha \ne 0$) that
$\alpha$ is a root of unity.

\

2.  Let $\alpha$ be an algebraic number of degree $d$.
Suppose that $\Tr(\alpha^n) \in \Z$ for all $n$.
Let $\alpha_i$ for $i=1,\ldots,d$ be the conjugates of
$\alpha$.  Observe
that symmetric polynomials (in $\alpha_i$) may be expressed
in terms of the ``Newton polynomials'', which are symmetric
polynomials given by taking the sum of the $k^{th}$ powers
(i.e. $\sum \alpha^k_i$). This follows
from the formal identity:
$$ \exp \left(- \sum_{k=1}^{\infty} \frac{t^k}{k}
 \sum_{i=1}^d \alpha^k_i \right)
= \exp \left( \sum_{i=1}^d  \log(1-t \alpha_i)\right)
= \prod_{i=1}^d (1 - t \alpha_i)$$
In particular, the minimal polynomial of $\alpha$ has
denominators bounded purely in terms of $d$, and so
$c_d \cdot \alpha$ is an algebraic integer
for some universal constant $c_d$. Applying the
same argument to $\alpha^n$ for all $n$ we conclude
that $c_d \cdot \alpha^n$ is also an algebraic integer
for all $n$.  One may infer  that $\alpha$
itself is algebraic in various ways. One such way is
as follows. Suppose that $(\alpha) = \frak{p}^{a_1}_1
\ldots \frak{p}^{a_n}_n$. We wish to prove that
$a_i \ge 0$ for all $i$. If $a_i < 0$, then for
$c_d \cdot \alpha$ to be integral we must have 
$c_d \in \frak{p}^{-a_i}_i$. Applying the same 
argument to $\alpha^n$ we must have
$c_d \in \frak{p}^{-a_i \cdot n}_i$ for all $n$, which
is a contradiction. Alternatively, note that the 
$\ZZ$-submodule generated by $1, \alpha, \alpha^2, \dots$
is finitely generated, since it's contained in the finitely
generated $\ZZ$-module $c_d^{-1} \gotho$, where $\gotho$
is the ring of integers of $\QQ(\alpha)$. That implies 
$\alpha$ is algebraic by a theorem from lecture.


\section{Ideals and Unique Factorization}

1. Let $R = \Z[\sqrt{-7}]$. Let $\eta = (2,1 + \sqrt{-7})$. 
Then
$$\eta^2 = (4,-6 + 2 \sqrt{-7},2 + 2 \sqrt{-7}) = (4,2 + 2 \sqrt{-7})
= (2)(2,1 + \sqrt{-7}) = (2) \eta. $$
Assuming unique factorization for ideals in
$R$ we conclude that $\eta = (2)$. Yet since
the element
$(1 + \sqrt{-7})/2 \notin R$,  $(1 + \sqrt{-7})$ cannot lie
in $(2)$, and we are done.



\

2(i). Suppose first of all that there exists a solution to
 $x^2 = d \mod p$. Clearly we may take $2x \le p$, and thus
write $x^2 - d = p m$ with $m < p$.
Let 
$\frak{p}_1 = (p,x + \sqrt{d})$ and $\frak{p}_2 = (p,x - \sqrt{d})$.
We observe that
$$\frak{p}_1 \frak{p}_2 = (p^2,p(x \pm \sqrt{d}),pm)$$
since $m < p$, $(m,p) = 1$ and so $(p^2,pm) = (p)$. Thus
$\frak{p}_1 \frak{p}_2 = (p)$. 
Observe that $\frak{p}_1$ is not the unit ideal, since  if
$\frak{p}_1 = (1)$ then by symmetry $\frak{p}_2 = (1)$ and
$(p) = \frak{p}_1 \frak{p}_2 = (1)$. On the other hand, 
$\frak{p}_1 \ne (p)$ since  for $p > 2$,
$(x + \sqrt{d})/p$ is not an algebraic integer (see previous 
assignment).

\

2(ii). In the other direction, suppose that $p \O$ is not prime in $\O$, and that $p$ does not
divide $2d$.
Then we may write $(p) = \frak{a} \frak{b}$ with 
$\frak{a}$ and $\frak{b}$ neither $(p)$ nor the unit ideal.
Write $p = ab$ with $a \in \frak{a}$
and $b \in \frak{b}$. Since $\N(p):=\mathrm{Norm}(p) = p^2$, we must
have
$N(a)N(b) = p^2$.  If $N(a) = 1$, then $a$ is a unit and $\frak{a}$ is
the  unit ideal. Thus $p | N(a)$, and similarly $p | N(b)$. Thus
$N(a) = \pm p$. Write $a = x + y \sqrt{d}$, with $2x, 2y \in \Z$.
Then
$$N(a) = x^2 - d y^2 = \pm p$$
Suppose that $p|y$. Then $p|x^2$, and thus
$p | x$.
This would imply, however, that
$p^2 | N(a) = \pm p$, which is clearly impossible. Thus $p$ does
not divide $y$. In particular,
since $p \ne 2$,
$y$ is invertible modulo $p$, and
$(x y^{-1})^2 \equiv d \mod p$,
which was to be shown.

\

3. Suppose $\gotha$ is an integral ideal which is not principal. Let
$\alpha$ be a nonzero element of $\gotha$ of minimum norm. By 
assumption, there exists $\beta \in \gotha \setminus (\alpha)$. Choose
$\gamma \in \ZZ[\sqrt{-5}]$ to minimize $|\delta|$, where
$\delta = \beta/\alpha - \gamma$; then
in particular, $\Re(\delta) \in [-1/2, 1/2]$ and
$\Im(\delta) \in [-\sqrt{5}/2, \sqrt{5}/2]$. On the other hand, by
assumption, $\delta \alpha = \beta - \gamma \alpha$ is a nonzero element
of $\gotha$, so $|\delta \alpha| \geq |\alpha|$, which is to say
$|\delta| \geq 1$. Since $|\Re(\delta)| \leq 1/2$, it follows that
$|\Im(\delta)| \geq \sqrt{3}/2$. Without loss of generality,
assume $\Im(\delta)>0$; then $\sqrt{3}-\sqrt{5}
< \Im(2\delta - \sqrt{-5}) < 0$. If $c$ is the closest integer to
$\Re(2\delta)$, then put $\epsilon = 2\delta - \sqrt{-5} - c$; we
have 
\[
|\epsilon|^2 \leq \Re(\epsilon)^2 + \Im(\epsilon)^2
\leq \frac{1}{4} + (\sqrt{3}-\sqrt{5})^2 < 1.
\]
If $\epsilon \neq 0$, then $\epsilon \alpha$ is an element of
$\gotha$ of norm less than $|\alpha|$, contradiction. Thus
$\epsilon = 0$ and so $2\beta \in (\alpha)$. Since this holds for
all $\beta$, we conclude $[\gotha:(\alpha)] = 2$ and that $2\gotha \subseteq
(\alpha)$.

Now let $\gothp = (2, 1+\sqrt{-5})$ and note that $\gothp^2 = (2)$.
That means that $(\alpha) | 2\gotha = \gothp^2 \gotha$ and
$\gotha | (\alpha)$. That means that $\gotha$ is either equal to
$(\alpha), \gothp^{-1} (\alpha)$ or $\gothp^{-2} (\alpha)= (\alpha/2)$.
The first and third choices are principal ideals, so we must have
$\gotha = \gothp^{-1} (\alpha)$. Thus $\gotha$ is in the ideal class
of $\gothp$, as desired.

\

4. Let $I$ be an ideal in a Dedekind domain $R$. Let $a \in I$ be a
non zero element. Then $(a)$ is a non-trivial ideal in $R$, and
by the hint in the exercise, $R/a$ is a PID. On the other hand,
since $(a) \subseteq I$, there is a commutative diagram:
$$R \lra R/a \lra R/I.$$
Let $J$ be the kernel of $R/a \ra R/I$. Since $R/a$ is principal,
we may write $J = b + aR$ in $R/a$, with $b \in R$.
Thus $(a,b) = \Ker(R \ra R/I) = I$. (This last claim could be
expanded upon briefly, but it is completely trivial).


\

5. Since the group of  ideals is  generated by  prime ideals, it
suffices to prove the following statement for Dedekind domains: 
for any (non-zero) prime ideal
$\frak{p}$, there exists another ideal $\frak{q}$ such that
$\frak{q}$ represents the same element of the class group of
$\frak{p}$, and $\frak{q}$ is prime to any particular choice of
finitely many ideals:
$\frak{p}_1$, $\frak{p}_2, \ldots,\frak{p}_n$. Since $R$
is a Dedekind domain, $\frak{p}_i + \frak{p}_j = R$ for all
$i,j$. By the
Chinese remainder theorem:
$$R/\frak{p}^2 \frak{p}_1 \frak{p}_2 \ldots \frak{p}_n 
\simeq R/\frak{p}^2 \oplus R/\frak{p}_1 \oplus \ldots \oplus R/\frak{p}_n.$$
In particular, since $\frak{p} \ne \frak{p}^2$, 
there exists
an element $r \in R$ such that $r \in \frak{p}\setminus \frak{p}^2$, and
$r \ne \frak{p}_i$ for any $i$. 

Let us consider the factorization of the ideal $(r)$.
By construction, the ideal $\frak{p}$ occurs exactly once, and
the ideals $\frak{p}_i$ do not occur. Write
$(r) = \frak{p} \frak{q}$.
Then $\frak{q}$ is prime to $\frak{p}$ and $\frak{p}_i$ for all $i$.
On the other hand, $\frak{q}$ is the \emph{inverse} element to $\frak{p}$
in the class group, since their product is principal. One finishes
the proof
by either taking some appropriate power of $\frak{q}$ (since the 
class group is finite), or more appropriately, taking
$N(\frak{q})/\frak{q}$, which is manifestly an integral ideal.
(Or for a third argument, repeat the argument above with
$\frak{p}$ replaced by $\frak{q}$, and $\frak{p}_i$ replaced
by $\frak{p}_i \cup \frak{p}$ to produce an ideal inverse
to $\frak{q}$ in the classgroup, and coprime to $\frak{p}$ and
$\frak{p}_i$.)

\

6. We imitate the proof of Theorem I.3.1 in Neukirch; the key point is that
$K[x]$ is a principal ideal domain. (Every nonzero ideal in $K[x]$ is generated
by an element of minimum degree.) Let $R$ be the integral closure of $K[x]$
in $L$. By Proposition 2.10, every ideal of $R$ is a finitely generated
$K[x]$-module, and hence is a finitely generated ideal. Thus $R$ is noetherian.
Moreover, since $R$ is the integral closure of $K[x]$ in $L$, it is 
automatically integrally closed. (This was essentially a problem on the
first homework; see also section 2 of Chapter 1 of Neukirch.) Now let
$\gothp$ be a nonzero prime ideal of $R$; it remains to show that $\gothp$ must
be maximal. For any $y \in \gothp$ nonzero, there exists a polynomial
$y^n + a_1 y^{n-1} + \cdots + a_n = 0$. Thus $a_n \in \gothp$, so
the intersection $\gothq = \gothp \cap K[x]$ is a nonzero ideal; it is
also prime because $K[x]/\gothq$ injects into $R/\gothp$ and the latter is
an integral domain. Thus $\gothq$ is maximal, so $K[x]/\gothq$ is a field
and $R/\gothp$ is an integral extension of $K[x]/\gothq$. That forces
$R/\gothp$ to be a field: if $z \in R/\gothp$ has minimal polynomial
$z^m + b_1 z^{m-1} + \cdots + b_m$ over $K[x]/gothq$, then $b_m \neq 0$,
so $z$ is invertible. We conclude every nonzero prime ideal of $R$
is maximal, so $R$ is a Dedekind domain.


\end{document}