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\begin{document}

\chapter*{September 17:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Number Fields
\item 
Examples
\item 
Algebraic integers
\end{itemize}

\section{Number Fields}

The simplest field of characteristic~0 is of course
the field~$\Q$ of rational numbers.  Perhaps the next
object of study should be finite extensions of this field;
in fact this turns out to be an extraordinarily fertile
area of study, full of many open problems.  The rest of
the course will be concerned with such fields.

\begin{defin}
A \defn{number field} is a finite extension of $\Q$.
\end{defin}

A later exercise will show that every number field is generated
by a single element, i.e., is a root field of an irreducible
polynomial.  If $F = \Q(\alpha)$ is a number field then the
minimal polynomial $\m_\alpha(x)$ factors as a product of
distinct linear factors in its splitting field.  If one desires,
one can take the roots to be complex roots, so that the
splitting field, and the number field, lie inside the
complex numbers.  We then let
\begin{itemize}
\item
$r$ be the number of real roots of $\m_\alpha$, and
\item
$s$ be the number of pairs of complex conjugate roots.
\end{itemize}
Thus
$$r + 2s = n := \degg(f).$$
The number of embeddings of $F$ into~$\C$ is clearly~$n$, i.e., there
is an embedding for each choice of root.  

It is largely a matter of taste as to whether one thinks of number
fields as lying explicitly inside the complex numbers, but there
is no loss of generality in doing this, and we will do so when it
is convenient.

\section{Examples}

The simplest number field is of course the field
of rational numbers.

The fields
$$\Q(i), \quad \Q(\sqrt{-3}) = \Q(\omega), \quad \Q(\sqrt{5})$$
are all examples of quadratic number fields, and by earlier
remarks we know that any such field is of the form
$\Q(\sqrt{d})$ where $d$ is a rational number that is not a perfect square.

A quadratic field $\Q(\sqrt{d})$ is said to be \defn{real} if $d > 0$
and \defn{imaginary} if $d < 0$.  These correspond to 
$r = 2, s = 0$ and $r = 0, s = 1$ respectively.
Any quadratic field is a galois
extension, and we will denote the nontrivial automorphism, which
takes $\sqrt{d}$ to $-\sqrt{d}$ by $x \mapsto x'$.

The fields
$$
\Q(\sqrt[3]{2}), \quad \Q[x]/(x^3+x^2-2x-1$$
are ``cubic'' number fields of degree~3 over~$\Q$.
In this case $r = s = 1$ or $r = 3, s = 1$ according to whether
a defining polynomial has 1 or 3 real roots.  A cubic field
might or might not be galois.

A ``cyclotomic'' field $\Q(e^{2\pi i/n})$ obtained by
adjoining a primitive $n$-th root of unity.
Cyclotomic fields turn out to be galois extensions, since $\Q(e^{2 \pi i/n})$
is the splitting field of~$x^n-1$.  Later we will learn that the
``cyclotomic'' polynomial
$$f_n(x) := \prod_{\gcd(k,n)=1} (x - e^{2 \pi i k/n})$$
is irreducible and is therefore the minimal polynomial of
$\zeta_n := e^{2 \pi i/n}$.
Thus the $[\Q(e^{2 \pi i/n}:\Q] = \phi(n)$, where $\phi(n)$ is the
Euler-phi function, i.e., the number of positive integers less than~$n$
that are relatively prime to~$n$.
By the basic lifting result, this means that for every such integer~$k$
there is a unique automorphism $\sigma_k$ such that
$$\sigma_k(\zeta_n) = \zeta_n^k.$$
From the obvious fact that $\sigma_k \sigma_l = \sigma_{kl}$ we see
that the galois group $G(\Q(\zeta_n)/\Q) $ is isomorphic to the
group multiplicative group $(\Z/n\Z)^*$.

A deep theorem due to Kronecker and Weber asserts that any galois
extension of $\Q$ whose galois group is abelian
is contained in a cyclotomic extension; this is one of the important
theorems of ``class field theory.''

\section{Algebraic integers}

To actually do algebraic number theory in a number field we need to go
beyond field theory and have ``integers'' $\Z_F$ in a number field~$F$,
just as to do usual number theory one works in the integers $\Z$ 
inside~$\Q$.

\begin{defin}
Let $F$ be a number field.  Define 
$$\Z_F = \left\{ \alpha\in F\ :\
\m_\alpha(x) \in \Z[x] \right\}.$$
We call $\Z_F$ the ring of \defn{algebraic integers} in $F$.
\end{defin}

One of our first tasks is to show that $\Z_F$ is in fact a ring,
i.e., that the sum and product of algebraic integers are algebraic
integers.  It is convenient to first develop a useful criterion
for integrality.

\begin{theorem}
Let $F$ be a number field and $\alpha\in F$. 
Then the following are equivalent. 
\begin{enumerate}
\item $\alpha$ is an algebraic integer, i.e., $\m_\alpha(x) \in \Z_F$.
\item The ring $\Z[\alpha]$ has a finitely generated additive group.
\item $\alpha$ is contained in a subring with a finitely generated additive group.
\item There exists a finitely generated additive group $L$ inside~$F$
such that $\alpha L \subset L$.
\end{enumerate}
\end{theorem}

\begin{proof}
If $\alpha$ is an algebraic integer then $\alpha^n$ can be
expressed, using the minimal polynomial, as an integral linear
combination of $1, \alpha, \alpha^2, \cdots, \alpha^{n-1}$.
By a simple induction argument, all higher powers of~$\alpha$
are in the span of those~$n$ elements, and the additive
group of $\Z[\alpha$ is generated by those elements.  This
proves that (1) implies~(2).

$(2) \Rightarrow (3)$ and $(3) \Rightarrow (4)$ are obvious.
Suppose that $L$ is a finitely generated group inside~$F$ with
the property that $\alpha L \subset L$.  Choose a basis
(generating set)
$\beta_1, \ldots, \beta_n$ for~$L$. 
We have:  $$ \alpha
\beta_i = \sum_j a_{ij} \beta_j. $$ We can write this in matrix
notation as $$ \alpha\left[ \matrix{ \beta_1 \cr \vdots \cr \beta_n
\cr} \right] =
    \left[ \matrix{ a_{11} & \cdots & a_{1n} \cr
	\vdots &  & \vdots \cr a_{n1} & \cdots & a_{nn} \cr }\right]
   \left[ \matrix{ \beta_1 \cr \vdots \cr \beta_n \cr }\right]. $$ Let
$A=(a_{ij})$. It is now clear that $\alpha$ is an eigenvalue of this
matrix, i.e.  $\det(\alpha I - A) = 0$. This implies that the minimal
polynomial for $\alpha$ divides the characteristic polynomial of $A$.
Since the latter is a monic polynomial over $\Z$, Gauss' lemma implies
that $\m_\alpha(x) \in \Z[x]$, and $\alpha$ is an algebraic
integer.\end{proof}

\begin{example}
We determine the algebraic integers in an arbitrary quadratic field.
Let $F = \Q(\sqrt d)$, where we may assume that $d$ is a squarefree
integer. An arbitrary element of $F$ can be written $z = a+b{\sqrt d}$.
Its minimal polynomial is then 
$$ \m_z(x) = x^2 - 2ax + (a^2-db^2). $$ 
The integrality of the linear term implies that $a$ has the form
$a = {A\over2}$, where $A\in\Z$.  
The squarefreeness of~$d$ implies that $b$ also has at worst a 2 in
the denominator, i.e.,
$b = {B\over2}$, for $B\in\Z$. 
In order for $a^2-db^2$ to be an integer it follows that $A$ and $B$
have the same parity.
We summarize these observations as follows.

\begin{theorem}
Let $d\in\Z$ be a squarefree integer, and $F = \Q(\sqrt d)$. Then if $d
\equiv 2,3\ {\rm (mod\ 4)}$, $$\Z_F = \Z[d].$$ If, on the other hand,
$d \equiv 1,4\ {\rm (mod\ 4)}$, $$\Z_F = \Z[{1+\sqrt d \over 2}] =
\{\frac{A+B\sqrt{d}}{2} : A \equiv B \bmod 2 \}.$$
\end{theorem}
\end{example}

\end{document}



