\documentclass[12pt]{amsart}
\parindent=0pt \parskip=8pt

\usepackage{verbatim}
\usepackage{fullpage}
\usepackage{amssymb}

\input{../tex/isomath}
\input{../tex/mathenv}

\renewcommand{\theequation}{\arabic{problem}.\arabic{equation}}
\renewcommand{\labelenumi}{(\alph{enumi})}
\def\ve{\varepsilon}
\def\SR{S^{-1}R}
\def\ophi{\overline{\phi}}
\def\opsi{\overline{\psi}}
\newcommand{\ann}{\mathrm{ann}}
\newcommand{\Hom}{\mathrm{Hom}}
\newcommand{\OK}{\mathcal{O}_K}
\def\fa{\mathfrak{a}}
\def\fb{\mathfrak{b}}
\def\fc{\mathfrak{c}}
\def\fp{\mathfrak{p}}
\def\Rp{R_{\mathfrak{p}}}
\newcommand*{\pig}[1]{\langle #1 \rangle}

%--------------------------------------------------------------------
\begin{document}
\title{Algebraic Number Theory  $\cdot$ Spring 2004 \\ Assignment 1 }
\author{John Kerl}
\date{\today}
\maketitle

%% ----------------------------------------------------------------
\begin{problem}
Prove or disprove and salvage:  Let $R$ be an integral domain, and let $F$
be a field contained as a subring in $R$.  Suppose $R$ is finitely generated
as an $F$-module.  Then $R$ is a field.

%[First proof, secondhand]

\begin{proof} It suffices to show the existence of inverses for non-zero
elements since $R$ is already commutative with identity.  Since $R$ is an
$F$-module for a field $F$, $R$ is by definition a vector space over $F$.  Let
$a$ be any non-zero element of $R$ (such exists since $R$ has identity).  Then
the map $\tau_a$ given by $r \mapsto ar$ is a linear transformation from $R$ to
$R$.  To see that $\tau_a$ is one-to-one, suppose $ar=as$ for some $r,s \in R$.
Then $a(r-s)=0$ which since $R$ is an integral domain implies $a=0$ (contrary
to hypothesis) or $r=s$.  The image of $\tau_a$ is inside the
finite-dimensional vector space $R$ and so $\tau_a$ is onto as well.  Thus 1,
which is in $R$, has a preimage and so there is a $b$ in $R$ such that $ab =
1$.  \end{proof}

%% ----------------------------------------------------------------
%\begin{proof}[Second proof]
% How to make a field:
% o F a field
% o R = F[x]
% o F[x] / <r(x)>   w/ r(x) irred.

% r in R = sum_{i=1}^n f_i r_i   for some f_i's in F and r_i's in R

% r   = f1 r1 + ... + fn rn

% r   = f11 r1 + ... + f1n rn
% r^2 = f21 r1 + ... + f2n rn
% ...
% r^n = fn1 r1 + ... + fnn rn

% [ r     ]   [ f11 f12 ... f1n ] [ r1 ]
% [ r^2   ]   [ f21 f22 ... f2n ] [ r2 ]
% [ ...   ] = [  .   .       .  ] [ ...]
% [ r^n   ]   [ fn1 fn2 ... fnn ] [ rn ]

% If add 
% [ r^n+1 ] = [ fn1 fn2 ... fnn ] [ r? ]
% then l.d. by rank?

% ----> big problem:  no r_{n+1}

% So (if I could get past that), then r^n = sum fi ri
% hence r^n - sum fi ri = 0
% hence r is a root of a degree-n poly
% F[x] is a UFD so r is a root of a degree-m poly m(x).

% Make isom R ~ F[x] / <m(x)> ??  What if r is 0 or 1 etc.
% Need somehow a *maximal* r -- whatever that means.
%\end{proof}

%% ----------------------------------------------------------------
% Multiplication by r:

%   [ r1 ]   [ f11 f12 ... f1n ] [ r1 ]
% r [ r2 ]   [ f21 f22 ... f2n ] [ r2 ]
%   [ ...] = [  .   .       .  ] [ ...]
%   [ rn ]   [ fn1 fn2 ... fnn ] [ rn ]

% Take c(x) = chpoly of matrix; r is a root of it.
% Maybe c(x) is irred and maybe not.
% r is a root of a factor, m(x).

% Make isom R ~ F[x] / <m(x)> ??  What if r is 0 or 1 etc.
% Need somehow a *maximal* r -- whatever that means.
%\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Show:
\begin{enumerate}
\item If $R$ is a principal ideal domain, then $R$ is a Dedekind domain.
\item If $R$ is a Dedekind unique factorization domain, then $R$ is a
	principal ideal domain.
\end{enumerate}

\begin{proof}

For the first item, let $R$ be a PID.  To see that $R$ is Noetherian, observe
that all ideals of $R$ are principal, hence quite finitely generated.  Prime
ideals of $R$ are maximal in a PID.  Since a PID is a UFD,
$R$ is integrally closed in its quotient field.

\bigskip

For the second item, let $R$ be a Dedekind UFD and let $\fp$ be a prime ideal
of $R$.  Since $R$ is Noetherian, $\fp$ is finitely generated and is equal to
$a_1 R + \ldots + a_n R$ for some $a_i$'s in $R$, not all zero.  Let $a$ be a
non-zero $a_i$.  Since $R$ is a UFD, $a=\prod_{i=1}^n p_i e^i$ for some
irreducible elements $\{p_i\}$ of $R$.  Since $\fp$ is a prime ideal, one of
the $p_i$'s, say $p_{i_0}$, is in $\fp$, and so $\pig{p_{i_0}} \subseteq \fp$.
Since $R$ is Dedekind, prime ideals are maximal implying that $\pig{p_{i_0}} =
\fp$.  Thus any prime ideal of $R$ is principal.

Now let $\fa$ be any ideal of $R$, not necessarily prime.  Since $R$ is
Dedekind, $\fa$ factors into the product of (not necessarily distinct) prime
ideals $\prod_{i=1}^n\fp_i$, where each $\fp_i = \pig{p_i}$ for some
$p_i \in \fp_i$.  Recall that for ideals $A$ and $B$, the product $AB$
is elements of the form $\sum a_i b_i$.  Here, any $a \in \fa$
is $\sum_{j=1}^m q_{1j} \cdots q_{nj}$ where the $q_{ij}$'s are elements of
$\fp_i$, hence of the form $c_{ij}p_i$.  Then
\begin{eqnarray*}
	a &=& \sum_{j=1}^m c_{1j}p_1 \cdots c_{nj}p_n \\
	&=& \sum_{j=1}^m (c_{1j}\cdots c_{nj}) (p_1 \cdots p_n) \\
	&=& (p_1 \cdots p_n) \sum_{j=1}^m (c_{1j}\cdots c_{nj})
\end{eqnarray*}
such that $\fa = \pig{\prod p_i}$.  Since an arbitrary ideal of $R$ is
principal, $R$ is a PID.

\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Let $K$ be an algebraic number field.  Show that the set of all fractional
ideals of $K$ forms a group under multiplication.

\begin{proof}

Let $\OK$ be the integral closure of $\Z$ in $K$.  Recall that a fractional
ideal of $K$ is a finitely generated non-zero $\OK$ submodule of $K$.  If $\fa$
and $\fb$ are fractional ideals of $K$ then we define $\fa \fb = \{
\sum_{i=1}^n a_i b_i : a_i \in \fa, b_i \in \fb, n \in \Z_+\}$.  Let $G$ be the
set of all fractional ideals of $K$.  Note that this is a non-empty set since
it contains at least $K$ itself.

To see that $G$ is closed under multiplication, let $z =\sum_{k=1}^n a_k b_k
\in \fa \fb$.  Since $\fa$ is a finitely generated $\OK$ module, it has an
$\OK$-base $\{a_i\}$.  Likewise $\fb$ has an $\OK$-base $\{b_j\}$.  Then any $a
\in \fa$ is of the form $\sum_{i=1}^l{x_i a_i}$ with coefficients $x_i \in \OK$.
Likewise, any $b \in \fb$ is of the form $\sum_{j=1}^m{y_j b_j}$ with
coefficients $y_j \in \OK$.  Thus
\begin{eqnarray*}
	z &=& \sum_{k=1}^n a_k b_k \\
	&=& \sum_{k=1}^n
		\left(\sum_{i=1}^{l_k}{x_i a_i}\right)
		\left(\sum_{j=1}^{m_k}{y_j b_j}\right)
\end{eqnarray*}
which by commutativity of the field $K$ is a finite sum of terms of the form
$xyab$ with $x,y \in \OK$, $a \in \{a_i\}$ and $b \in \{b_j\}$.  Thus
$\fa \fb$ is generated by the finite set $\{a_i b_j\}$.
Since $K$ is closed as a field, $\fa\fb$ is certainly contained in $K$.
Also, $\fa \fb$ trivially satisfies the module axioms, and so is an
$\OK$-submodule of $K$.  To see that $\fa\fb$ is non-zero, observe that
$\fa$ and $\fb$ are non-zero and hence contain non-zero elements $\alpha$ and
$\beta$ respectively.  Since $\OK$ is an integral domain, $\alpha \beta$ is a
non-zero element of $\fa \fb$.

To see that multiplication in $G$ is associative: 
\begin{eqnarray*}
	\fa\fb &=& \left\{\sum_i a_i b_i \right\} \\
	\fb\fc &=& \left\{\sum_k b_k c_k \right\} \\
	(\fa\fb)\fc &=& \left\{\sum_j \left(\sum_i a_{ij} b_{ij}\right) c_j
		\right\} \\
	&=& \left\{\sum_j \left(\sum_i a_{ij} b_{ij} c_{ij} \right)
		\right\} \textrm{ letting $c_{ij} = c_j$ }\\
	\fa(\fb\fc) &=& \left\{\sum_l a_l \left(\sum_k b_{kl} c_{kl}\right)
		\right\} \\
	&=& \left\{\sum_l \left(\sum_k a_{kl} b_{kl} c_{kl}\right)
		\right\} \textrm{ letting $a_{kl} = a_l$ }
\end{eqnarray*}

Multiplication in $G$ is commutative (i.e. $G$ is in fact an \emph{abelian}
group) since $K$ is commutative:
\begin{eqnarray*}
\fa \fb &=& \left\{ \sum_{i=1}^n a_i b_i : a_i \in \fa, b_i \in \fb, n \in \Z_+
	\right\} \\
\fb \fa &=& \left\{ \sum_{i=1}^n b_i a_i : a_i \in \fa, b_i \in \fb, n \in \Z_+
	\right\} = \fa \fb
\end{eqnarray*}

The identity of $G$ is $\OK$:  Since an $R$-module $M$ is closed under
multiplication by elements of $R$, $\fa \OK \subseteq \fa$.  On the other
hand, $\fa$ is contained in $\fa\OK$ since $\OK$, being a $\Z$-order, must
contain all of $\Z$ including 1.

To see that $G$ contains inverses for all its elements, define, given $\fa \in
G$, $$\fa^{-1} = \{k \in K : k \fa \subseteq \OK\}$$ The remainder takes
several lemmas which you proved in class.  I could copy and paste them in, but
that seems unproductive; I've proved the group axioms that were left sketchy.
\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Let $K$ be an algebraic number field.  Show that every fractional ideal of $K$
may be written (uniquely up to ordering of the factors) as a product of
(possibly negative) integral powers of finitely many prime ideals of $\OK$.

\begin{proof}
The proof is in two steps.  First, any fractional ideal $\fa$ of $K$ may be
written as the quotient of integral ideals $\fb/\fc$.  Second, the prime
factorizations of $\fb$ and $\fc$ may be combined to obtain a factorization for
$\fa$.

First, let $\fa$ be a fractional ideal of $K$.  Since $\fa$ is a finitely
generated $\OK$-submodule of $K$, $\fa = \pig{a_1/b_1, \ldots, a_n/b_n}$ for
$a_i$'s and $b_i$'s in $\OK$.  Let $\mu=\prod_{i=1}^n b_i$.  Let $\fb =
\pig{\mu a_1/b_1, \ldots, \mu a_n / b_n}$ and let $\fc = \pig{\mu}$, whence
$\fc^{-1} = \pig{1/\mu}$.  Thus $\fb$ and $\fc$ are integral ideals of $\OK$.
Write $\fb/\fc$ for $\fb\fc^{-1}$.

To see that $\fa \subseteq \fb/\fc$, let $a = \sum_{i=1}^m r_i a_i/b_i$.  Then
$a = \sum_{i=1}^m \mu r_i a_i/b_i \mu \in \fb/\fc$.  To see that $\fa \supseteq
\fb/\fc$, let $d = \sum_{i=1}^m r_i \mu a_i/b_i \mu \in \fb/\fc$.  Since $\OK$
is an integral domain, $d = \sum_{i=1}^m r_i a_i/b_i \in \fa$.  Thus $\fa =
\fb/\fc$.

Second, since $\OK$ is Dedekind I may write
$$ \fb = \prod_{i=1}^m \fp_i^{n_i}, \; \fc = \prod_{i=1}^m \fp_i^{d_i}$$
for prime ideals $\fp_i$ where for convenience I here permit $n_i$, $d_i = 0$
such that the same $m$ appears in both places.

Since the non-zero fractional ideals of $K$ form an abelian group under
multiplication,
$$ \fc^{-1} = \left(\prod_{i=1}^m \fp_i^{d_i}\right)^{-1}
	= \prod_{i=1}^m \fp_i^{-d_i}
$$
and therefore
$$ \fa = \fb/\fc = \fb\fc^{-1} = \prod_{i=1}^m \fp_i^{n_i-d_i} $$
\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Give an example of an integral domain which is not integrally closed in
its field of quotients.
\begin{answer}
See problem 6.
\end{answer}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Let $F$ be a field and let $R=F[x^2,x^3]$, a subring of the ring of polynomials
$F[x]$.
\begin{enumerate}
\item By using that $F[x^2] \subset R \subset F[x]$, show that $R$ is
	Noetherian and that every non-zero prime ideal of $R$ is maximal.
\item Show that $R$ is not a Dedekind domain.
\end{enumerate}

\begin{proof} Since $F$ is Noetherian and $R$ is a finitely generated
$F$-module, $R$ is a Noetherian $F$-module.  Thus for all $F$-submodules $\fa$
of $R$, there is a finite set $\{a_1,\ldots,a_n\}$ of elements of $\fa$ such
that for all $a \in \fa$ there are $f_1, \ldots, f_n \in F$ such that $a =
\sum_{i=1}^n f_i a_i$.  I need to show that for all $R$-submodules $\fa$ of
$R$, there is a finite set $\{a_1,\ldots,a_n\}$ of elements of $\fa$ such that
for all $a \in \fa$ there are $r_1, \ldots, r_n \in R$ such that $a =
\sum_{i=1}^n r_i a_i$.  Any $R$-submodule of $R$ is also an $F$-submodule of
$R$, and the $f_i$'s are also $r_i$'s.  Therefore $R$ is a Noetherian
$R$-module, and thus a Noetherian ring.

To see that every non-zero prime ideal of $R$ is maximal, I want to say that
every ideal of $R$ is also an ideal of $F[x^2]$, which is a PID.  However, this
doesn't quite follow.
\begin{itemize}
\item $\fp$ is a prime ideal of $F[x^2,x^3]$: given.
\item $\implies \fp$ is an ideal of $F[x^2]$: Non-emptiness and closure
	under addition depend on $\fp$, not on the containing ring;
	the ideal property carries over since $F[x^2]$ is contained in
	$F[x^2,x^3]$.
\item $\implies \fp$ is a \emph{prime} ideal of $F[x^2]$:  For $a, b \in 
	F[x^2,x^3]$, $ab \in \fp \implies a \in \fp$ or $b \in \fp$.
	Since $F[x^2] \subset F[x^2,x^3]$, this holds for $a,b \in F[x^2]$ as
	well.
\item $\implies \fp$ is a maximal ideal of $F[x^2]$: since $F[x^2]$ is a PID.
\item $\implies ? \fp$ is a maximal ideal of $F[x^2,x^3]$: ?? (\ldots
	incomplete).
\end{itemize}

To show that $R=F[x^2,x^3]$ is not a Dedekind domain, since $R$ is Noetherian
and every non-zero prime ideal of $R$ is maximal it suffices to show that $R$
is not integrally closed in its field of quotients, $K$.  For $R$ to differ
from its integral closure in $K$, there must be an element of $K$ which is not
in $R$ but which is integral over $R$.  Let $f(y) = y^2 + x^2 y - x^2 - x^3$, a
monic quadratic in $R[y]$.  Notice that $x \in K$ since $x = x^3 / x^2$ but $x
\not\in R$.  Yet $f(y) = 0$.

\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\bigskip
\begin{problem}
Is the ring $F[x,y]$ of polynomials in two variables over a number field
$F$ a Dedekind domain?  Prove that your answer is correct.

\begin{proof}[Counterexample]
Let $F=\Q$ and let $f(x)g(x) \in \pig{x}$.  Then necessarily $f(x)g(x) = x
h(x)$ for some $h(x) \in \Q[x]$.  Since $\Q$ is a field, $\Q[x]$ is a UFD; $x$
is irreducible since linear.  Thus either $f(x)$ or $g(x)$ must contain $x$ in
its factorization, hence $f(x) \in \pig{x}$ or $g(x) \in \pig{x}$.  Therefore
$\pig{x}$ is prime.  Since $\Q[x,y] \supsetneq \pig{x,y} \supsetneq \pig{x}$
while $\pig{x}$ is prime, $\pig{x}$ is a non-maximal prime ideal and therefore
$\Q[x,y]$ is not Dedekind.
\end{proof}
\end{problem}

%% ----------------------------------------------------------------
\end{document}
