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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 2}}
\author{William Stein}
\date{Tuesday, February 10, 2004}
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\maketitle

{\tt Announcements:} (1) Barry Mazur will talk about his popular math
book ``Imagining Numbers'' at the Harvard COOP tonight at 7pm. (2) I
have office hours Tuesdays and Thursdays 2--3pm in SC 515.
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We will do some serious commutative algebra today, which will provide
a powerful algebraic foundation for understanding the more refined
number-theoretic structures associated to number fields.

In the first section we establish the standard properties of
Noetherian rings and modules, including the Hilbert basis theorem.  We
also observe that finitely generated abelian groups are Noetherian
$\Z$-modules, which fills the gap in our proof of the structure
theorem for finitely generated abelian groups.  After establishing
properties of Noetherian rings, we consider the rings of algebraic
integers and discuss some of their properties.


\section{Noetherian Rings and Modules}
Let~$R$ be a commutative ring with unit element.  We will frequently
work with $R$-modules, which are like vector spaces but over a ring.
More precisely, recall that an {\em $R$-module} is an additive abelian
group $M$ equipped with a map $R\times M \to M$ such that for all~$r,
r'\in R$ and all $m, m'\in M$ we have $(r r')m = r(r' m )$, $(r + r')m
= rm + r' m$, $r(m+m') = rm + rm'$, and $1m=m$.  A {\em submodule}
is a subgroup of $M$ that is preserved by the action of $R$.

\begin{example}
The set of abelian groups are in natural bijection with
$\Z$-modules.
\end{example}

A {\em homomorphism} of $R$-modules $\vphi:M\to N$ is a abelian group
homomorphism such that for any $r\in R$ and $m\in M$ we have
$\vphi(rm) = r\vphi(m)$.  A {\em short exact sequence} of $R$-modules
$$0 \to L \xra{f} M \xra{g} N \to 0$$
is a specific choice of injective homomorphism $f:L\to M$ and 
a surjective homomorphism $g:M\to N$ such that $\im(f) = \ker(g)$.


\begin{definition}[Noetherian] An $R$-module~$M$ is {\em Noetherian} if every
submodule of $M$ is finitely generated.  A ring~$R$ is {\em Noetherian} if~$R$ is
Noetherian as a module over itself, i.e., if every ideal of~$R$ is finitely generated.
\end{definition}

Notice that any submodule $M'$ of $M$ is Noetherian, because if every submodule of $M$
is finitely generated then so is every submodule of $M'$, since submodules of $M'$
are also submodules of $M$.  

\begin{definition}[Ascending chain condition]
An $R$-module $M$ {\em satisfies the ascending chain condition} if
every sequences $M_1\subset M_2 \subset M_3 \subset \cdots$ of
submodules of~$M$ eventually stabilizes, i.e., there is some $n$ such
that $M_n=M_{n+1}=M_{n+2}=\cdots$.
\end{definition}

\begin{proposition}
If $M$ is an $R$-module, then the following are equivalent:
\begin{enumerate}
\item $M$ is Noetherian,
\item $M$ satisfies the ascending chain condition, and
\item Every nonempty set of submodules of $M$ contains at least one
maximal element.
\end{enumerate}
\end{proposition}
\begin{proof}
{\bf $1\implies 2$:} Suppose $M_1\subset M_2\subset \cdots$ is a
sequence of submodules of $M$.  Then $M_\infty=\union_{n=1}^{\infty}
M_n$ is a submodule of $M$.  Since $M$ is Noetherian, there is a
finite set $a_1,\ldots, a_m$ of generators for $M$.  Each $a_i$
must be contained in some $M_j$, so there is an $n$ such that
$a_1,\ldots, a_m\in M_n$.  But then $M_{k}=M_n$ for all $k\geq n$,
which proves that the ascending chain condition holds for~$M$.

\noindent{}{\bf $2\implies 3$:} Suppose 3 were false, so there exists
a nonempty set~$S$ of submodules of~$M$ that does not contain a
maximal element.  We will use~$S$ to construct an infinite ascending
chain of submodules of~$M$ that does not stabilize.  Note that~$S$ is
infinite, otherwise it would contain a maximal element.  Let $M_1$ be
any element of~$S$.  Then there is an $M_2$ in $S$ that contains
$M_1$, otherwise $S$ would contain the maximal element $M_1$.
Continuing inductively in this way we find an $M_3$ in $S$ that
properly contains $M_2$, etc., and we produce an infinite ascending
chain of submodules of $M$, which contradicts the ascending chain
condition.

\noindnet{}{\bf $3\implies 1$:} Suppose 1 is false, so there is a
submodule $M'$ of~$M$ that is not finitely generated.  We will show
that the set~$S$ of all finitely generated submodules of $M'$ does not
have a maximal element, which will be a contradiction.  Suppose~$S$
does have a maximal element~$L$.  Since~$L$ is finitely generated and
$L\subset M'$, and $M'$ is not finitely generated, there is an $a\in
M'$ such that $a\not\in L$.  Then $L'=L+Ra$ is an element of~$S$ that
strictly contains the presumed maximal element~$L$, a contradiction.
\end{proof}

\begin{lemma}\label{lem:noetherianexact}
If 
$$
 0 \to L \xra{f} M \xra{g} N \to 0
$$ 
is a short exact sequence of~$R$-modules, then~$M$ is Noetherian if and 
only if both~$L$ and~$N$ are Noetherian.
\end{lemma}
\begin{proof}
First suppose that~$M$ is Noetherian.  Then~$L$ is a submodule of~$M$, so~$L$
is Noetherian.   If $N'$ is a submodule of~$N$, then the inverse image of $N'$
in~$M$ is a submodule of~$M$, so it is finitely generated, hence its image
$N'$ is finitely generated.  Thus~$N$ is Noetherian as well.

Next assume nothing about~$M$, but suppose that both~$L$ and~$N$ are
Noetherian.  If $M'$ is a submodule of $M$, then $M_0=\vphi(L)\cap M'$
is isomorphic to a submodule of the Noetherian module $L$, so $M_0$ is
generated by finitely many elements $a_1,\ldots, a_n$.  The quotient
$M'/M_0$ is isomorphic (via $g$) to a submodule of the Noetherian
module~$N$, so $M'/M_0$ is generated by finitely many elements
$b_1,\ldots, b_m$. For each $i\leq m$, let $c_i$ be a lift of $b_i$ to
$M'$, modulo $M_0$.  Then the elements $a_1,\ldots, a_n, c_1,\ldots,
c_m$ generate $M'$, for if $x\in M'$, then there is some element $y\in
M_0$ such that $x-y$ is an $R$-linear combination of the $c_i$,
and~$y$ is an $R$-linear combination of the $a_i$.
\end{proof}


\begin{proposition}\label{prop:noethfg}
Suppose~$R$ is a Noetherian ring.  Then an $R$-module~$M$ is 
Noetherian if and only if it is finitely generated. 
\end{proposition}
\begin{proof}
If~$M$ is Noetherian then every submodule of~$M$ is finitely generated
so~$M$ is finitely generated.  Conversely, suppose~$M$ is finitely
generated, say by elements $a_1,\ldots, a_n$.  Then there is a
surjective homomorphism from $R^n=R\oplus \cdots \oplus R$ to~$M$ that
sends $(0,\ldots,0,1,0,\ldots,0)$ ($1$ in $i$th factor) to $a_i$.
Using Lemma~\ref{lem:noetherianexact} and exact sequences of
$R$-modules such as $0\to R\to R\oplus R\to R\to 0$, we see
inductively that $R^n$ is Noetherian.  Again by
Lemma~\ref{lem:noetherianexact}, homomorphic images of Noetherian
modules are Noetherian, so~$M$ is Noetherian.
\end{proof}

\begin{lemma}\label{lem:surjnoetherian}
Suppose $\vphi:R\to S$ is a surjective homomorphism of rings 
and $R$ is Noetherian.   Then $S$ is Noetherian.
\end{lemma}
\begin{proof}
The kernel of $\vphi$ is an ideal $I$ in $R$, and
we have an exact sequence 
$$ 
  0 \to I \to R \to S \to 0
$$
with~$R$ Noetherian.  By Lemma~\ref{lem:noetherianexact}, it follows
that~$S$ is a Noetherian $R$-modules.  Suppose~$J$ is an ideal of~$S$.
Since~$J$ is an $R$-submodule of~$S$, if we view~$J$ as an $R$-module,
then~$J$ is finitely generated.  Since~$R$ acts on~$J$ through~$S$,
the $R$-generators of~$J$ are also $S$-generators of~$J$, so~$J$ 
is finitely generated as an ideal.  Thus~$S$ is Noetherian.
\end{proof}

\begin{theorem}[Hilbert Basis Theorem]
If $R$ is a Noetherian ring and $S$ is finitely generated as a ring
over~$R$, then~$S$ is Noetherian.  In particular, for any~$n$ the
polynomial ring $R[x_1,\ldots, x_n]$ and any of its quotients are
Noetherian.
\end{theorem}
\begin{proof}
Assume first that we have already shown that for any $n$ the
polynomial ring $R[x_1,\ldots, x_n]$ is Noetherian.  Suppose $S$ is
finitely generated as a ring over $R$, so there are generators
$s_1,\ldots, s_n$ for $S$.  Then the map $x_i\mapsto s_i$ extends
uniquely to a surjective homomorphism $\pi: R[x_1,\ldots, x_n] \to S$,
and Lemma~\ref{lem:surjnoetherian} implies that $S$ is Noetherian.

The rings $R[x_1,\ldots, x_n]$ and $(R[x_1,\ldots,x_{n-1}])[x_n]$ are
isomorphic, so it suffices to prove that if~$R$ is Noetherian then
$R[x]$ is also Noetherian.  (Our proof follows \S12.5 of Artin's {\em
Algebra}.)  Thus suppose $I$ is an ideal of $R[x]$ and that~$R$ is
Noetherian.  We will show that $I$ is finitely generated.

Let $A$ be the set of leading coefficients of polynomials in $I$ along
with $0$.  If $a,b\in A$ are nonzero with $a+b\neq 0$, then there are
polynomials $f$ and $g$ in $I$ with leading coefficients $a$ and $b$.
If $\deg(f)\leq \deg(g)$, then $a+b$ is the leading coefficient of
$x^{\deg(g)-\deg(f)}f + g$, so $a+b\in A$.  If $r\in R$ and $a\in A$
with $ra\neq 0$, then $ra$ is the leading coefficient of $rf$, so
$ra\in A$.  Thus $A$ is an ideal in $R$, so since $R$ is Noetherian
there exists $a_1,\ldots, a_n$ that generate $A$ as an ideal.  Since
$A$ is the set of leading coefficients of elements of $I$, and the
$a_j$ are in $I$, we can choose for each $j\leq n$ an element $f_j\in
I$ with leading coefficient $a_j$.  By multipying the $f_j$ by some power
of~$x$, we may assume that the $f_j$ all have the same degree $d$.

Let $S_{<d}$ be the set of elements of~$I$ that have degree strictly less
than~$d$.  This set is closed under addition and under multiplication by
elements of~$R$, so $S_{<d}$ is a module over~$R$.  The module $S_{<d}$ is submodule of
the $R$-module of polynomials of degree less than~$n$, which is
Noetherian because it is generated by $1,x,\ldots, x^{n-1}$.  Thus
$S_{<d}$ is finitely generated, and we may choose generators $h_1,\ldots,
h_m$ for $S_{<d}$.

Suppose $g\in I$ is an arbitrary element.  We will show by induction
on the degree of~$g$ that~$g$ is an $R[x]$-linear combination of
$f_1,\ldots, f_n, h_1,\ldots h_m$.  Thus suppose this statement is
true for all elements of~$I$ of degree less than the degree of~$g$.
If the degree of~$g$ is less than~$d$, then $g\in S_{<d}$, so~$g$ is
in the $R[x]$-ideal generated by $h_1,\ldots, h_m$.  Next suppose
that~$g$ has degree $e\geq d$.  Then the leading coefficient~$b$
of~$g$ lies in the ideal~$A$ of leading coefficients of~$g$, so there
exist $r_i\in R$ such that $b=r_1 a_1 + \cdots + r_n a_n$.  Since
$f_i$ has leading coefficient $a_i$, the difference $g- x^{e-d} r_i
f_i$ has degree less than the degree~$e$ of~$g$.  By induction $g-
x^{e-d} r_i f_i$ is an $R[x]$ linear combination of $f_1,\ldots, f_n,
h_1,\ldots h_m$, so $g$ is also an $R[x]$ linear combination of
$f_1,\ldots, f_n, h_1,\ldots h_m$.  Since each $f_i$ and $h_j$ lies in
$I$, it follows that $I$ is generated by $f_1,\ldots, f_n, h_1,\ldots
h_m$, so $I$ is finitely generated, as required.
\end{proof}

Properties of Noetherian rings and modules will be crucial in the rest
of this course.  We have proved above that Noetherian rings have many
desirable properties.

\subsection{$\Z$ is Noetherian}
The ring $\Z$ of integers is Noetherian because every ideal of $\Z$ is
generated by one element. 
\begin{proposition}\label{prop:zpid}
Every ideal of the ring $\Z$ of integers is principal.
\end{proposition}
\begin{proof}
Suppose~$I$ is a nonzero ideal in~$\Z$.  Let~$d$ the least positive
element of~$I$.  Suppose that $a\in I$ is any nonzero element of~$I$.
Using the division algorithm, write  $a=dq + r$, where~$q$ is an integer
and $0\leq r < d$.  We have $r=a-dq\in I$ and $r<d$, so our assumption
that $d$ is minimal implies that $r=0$, so $a=dq$ is in the ideal generated
by~$d$.   Thus~$I$ is the principal ideal generated by~$d$.
\end{proof}

Proposition~\ref{prop:noethfg} and \ref{prop:zpid} together imply that
any finitely generated abelian group is Noetherian.  This means that
subgroups of finitely generated abelian groups are finitely generated,
which provides the missing step in our proof of the structure theorem
for finitely generated abelian groups.

\section{Rings of Algebraic Integers}
Fix an algebraic closure $\Qbar$ of $\Q$.  For example, $\Qbar$ could
be the subfield of the complex numbers $\C$ generated by all roots in
$\C$ of all polynomials with coefficients in $\Q$.

Much of this course is about algebraic integers.
\begin{definition}[Algebraic Integer]
An element $\alpha\in\Qbar$ is an {\em algebraic integer} if it is a
root of some monic polynomial with coefficients in $\Z$.
\end{definition}

\begin{definition}[Minimal Polynomial]
The {\em minimal polynomial} of $\alpha\in\Q$ is the monic polynomial
$f\in\Q[x]$ of least positive degree such that $f(\alpha)=0$.
\end{definition}
The minimal polynomial of $\alpha$ divides any polynomial~$h$ such
that $h(\alpha)=0$, for the following reason. If $h(\alpha)=0$, use
the division algorithm to write $h=qf + r$, where $0\leq \deg(r) <
\deg(f)$.  We have $r(\alpha) = h(\alpha) - q(\alpha) f(\alpha) = 0$,
so $\alpha$ is a root of~$r$.  However,~$f$ is the polynomial of least
positive degree with root~$\alpha$, so $r=0$.


\begin{lemma}\label{lem:minpolint}
If~$\alpha$ is an algebraic integer, then the minimal polynomial
of~$\alpha$ has coefficients in~$\Z$.
\end{lemma}
\begin{proof}
(From S-D, page 2.) Suppose $f\in\Q[x]$ is the minimal polynomial of
$\alpha$ and $g\in\Z[x]$ is a monic integral polynomial such that
$g(\alpha)=0$.  As mentioned after the definition of minimal
polynomial, we have $g=fh$, for some $h\in\Q[x]$.  If $f\not\in\Z[x]$,
then some prime~$p$ divides the denominator of some coefficient of
$f$.  Let $p^i$ be the largest power of~$p$ that divides any
denominator of~$f$, and likewise let $p^j$ be the largest power of~$p$
that divides any denominator of~$g$.  Then $p^{i+j}g = (p^if)(p^j g)$,
and if we reduce both sides modulo $p$, then the left hand side is $0$
but the right hand side is a product of two nonzero polynomials in
$\F_p[x]$, hence nonzero, a contradiction.
\end{proof}

\begin{proposition}\label{prop:intfg}
An element $\alpha\in\Qbar$ is integral if and only if $\Z[\alpha]$ is
finitely generated as a $\Z$-module.
\end{proposition}
\begin{proof}
Suppose~$\alpha$ is integral and let $f\in\Z[x]$ be the monic minimal polynomial
of~$\alpha$ (that $f\in\Z[x]$ is Lemma~\ref{lem:minpolint}).  Then $\Z[\alpha]$
is generated by $1,\alpha,\alpha^2,\ldots,\alpha^{d-1}$, where~$d$ is
the degree of~$f$.   Conversely, suppose $\alpha\in\Qbar$ is such that
$\Z[\alpha]$ is finitely generated, say by elements 
$f_1(\alpha), \ldots, f_n(\alpha)$.  Let~$d$ be any integer bigger
than the degree of any~$f_i$.  Then there exist integers $a_i$ such
that $\alpha^d = \sum a_i f_i(\alpha)$, hence~$\alpha$ satisfies
the monic polynomial $x^d - \sum a_i f_i(x) \in \Z[x]$, so~$\alpha$
is integral.
\end{proof}

The rational number $\alpha=1/2$ is not integral.  Note that
$G=\Z[1/2]$ is not a finitely generated $\Z$-module, since $G$ is infinite
and $G/2G=0$.

\begin{proposition}
The set $\Zbar$ of all algebraic integers is a ring, i.e., the sum and
product of two algebraic integers is again an algebraic integer.
\end{proposition}
\begin{proof}
Suppose $\alpha, \beta\in \Z$, and let $m, n$ be the degrees of the
minimal polynomials of $\alpha, \beta$, respectively.  Then
$1,\alpha,\ldots,\alpha^{m-1}$ span $\Z[\alpha]$ and
$1,\beta,\ldots,\beta^{n-1}$ span $\Z[\beta]$ as $\Z$-module.  Thus
the elements $\alpha^i\beta^j$ for $i \leq m, j\leq n$ span
$\Z[\alpha, \beta]$.  Since $\Z[\alpha + \beta]$ is a submodule of the
finitely-generated module $\Z[\alpha, \beta]$, it is finitely
generated, so $\alpha+\beta$ is integral.  Likewise, $\Z[\alpha\beta]$
is a submodule of $\Z[\alpha, \beta]$, so it is also finitely
generated and $\alpha\beta$ is integral.
\end{proof}

Recall that a {\em number field} is a subfield~$K$ of $\Qbar$ such
that the degree $[K:\Q] := \dim_\Q(K)$ is finite.
\begin{definition}[Ring of Integers]
The {\em ring of integers} of a number field~$K$ is the ring 
$$
  \O_K = K \cap \Zbar = \{x \in K : \text{ $x$ is integral}\}.
$$
\end{definition}


\begin{example}
The field $\Q$ of rational numbers is a number field of degree $1$,
and the ring of integers of $\Q$ is $\Z$.
The field $K=\Q(i)$ of Gaussian integers has degree $2$ and $\O_K = \Z[i]$.
The field $K=\Q(\sqrt{5})$ has ring of integers $\O_K = \Z[(1+\sqrt{5})/2]$.
According to MAGMA, the ring of integers of $K=\Q(\sqrt[3]{9})$ is
$\Z[\sqrt[3]{3}]$, where $\sqrt[3]{3}=\frac{1}{3}(\sqrt[3]{9})^2$.
\end{example}

\begin{lemma}
Let $\O_K$ be the ring of integers of a number field.  Then 
$\O_K\cap \Q = \Z$ and $\Q\O_K = K$.
\end{lemma}
\begin{proof}
If $\alpha\in \O_K\cap\Q$, write $\alpha=a/b$ in lowest terms.  Since
$a/b$ is integral, by Proposition~\ref{prop:intfg}, the ring $\Z[a/b]$ is
finitely generated as a $\Z$-module.  If $b\neq 0$, then $G=\Z[a/b]$
is not even finitely generated as an abelian group, since~$G$ is
torsion free and $G/bG=0$.  Thus $\O_K\cap \Q=\Z$.

To prove that $\Q\O_K=K$, suppose $\alpha\in K$, and let
$f(x)\in\Q[x]$ be the minimal monic polynomial of~$\alpha$.  For any
positive integer~$d$, the minimal monic polynomial of $d\alpha$ is
$d^{\deg(f)}f(x/d)$, i.e., the polynomial obtained from $f(x)$ by
multiplying the coefficient of $x^{\deg(f)}$ by~$1$, multiplying the
coefficient of $x^{\deg(f)-1}$ by~$d$, multiplying the coefficient of
$x^{\deg(f)-2}$ by $d^2$, etc.  If~$d$ is the least common multiple of
the denominators of the coefficients of~$f$, then the minimal monic
polynomial of $d\alpha$ has integer coefficients, so $d\alpha$ is
integral and $d\alpha\in \O_K$.  This proves that $\Q\O_K = K$.
\end{proof}


Next time we will prove the following proposition:
\begin{proposition}
The ring of integers $\O_K$ of a number field is Noetherian.
\end{proposition}
We will also develop some basic properties of norms, traces, and
discriminants, and give more properties of rings of integers in the
general context of Dedekind domains.

\end{document}