\documentclass[12pt]{article}
\usepackage{amsfonts, amsthm, amsmath}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parskip}{0pt}
\setlength{\parindent}{20pt}

\def\CC{\mathbb{C}}
\def\FF{\mathbb{F}}
\def\PP{\mathbb{P}}
\def\QQ{\mathbb{Q}}
\def\RR{\mathbb{R}}
\def\ZZ{\mathbb{Z}}

\def\head#1{\medskip \noindent \textbf{#1}.}
\def\fixme#1{\textbf{FIXME! #1}}

\begin{document}

\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Lectures 2-3: Algebraic Numbers and Algebraic Integers \\
August 29 and 31, 2001
\end{center}


\head{Addendum to Monday's lecture}
To clarify the confusing point from the end of last time: we had a 
prime $p \equiv 1 \pmod{4}$ and an integer $a$ such that $a^2+1 \equiv 0
\pmod{p}$. We took $c+di = \gcd(a+i, p)$, noticed that $c^2+d^2 =
|c+di|^2$ divides $p^2$ and wanted to know that $c^2+d^2$ was equal to $p$,
not $p^2$. Two ways to see this:
\begin{enumerate}
\item (What I said in class.) Note that $|c+di|^2$ also divides
$|a+i|^2$. If the latter is not divisible by $p^2$, great. If it is,
repeat with $a$ replaced by $a+p$, since $(a+p)^2+1 = (a^2+1)+p^2+2ap$
cannot also be divisible by $p^2$.
\item
The fact that $c+di$ divides $p$ is the same as saying that
$p/(c+di) = p(c-di)/(c^2+d^2)$ belongs to $\ZZ[i]$. If $c^2+d^2=p^2$, then
this means $(c-di)/p$ is in $\ZZ[i]$, so $c/p$ and $d/p$ are integers. But then
$p$ divides $c+di$ and $c+di$ divides $a+i$, so $p$ divides $a+i$,
which it doesn't.
\end{enumerate}
Strictly speaking, I should have also checked that $c^2+d^2$ is not equal to 1,
but I'll leave that to you. (Hint: for all $\alpha$ in the ideal generated
by $a+i$ and $p$, $|\alpha|^2$ is a multiple of $p$.)

\head{Reference} Neukirch, Section 1.2. (Warning: the book works in somewhat
greater generality than I will work in class. But you should read it
anyway.)

\head{Jargon watch} A \emph{principal ideal domain} is an integral domain
(ring without zero divisors) in which each ideal is generated by some element.
Examples: $\ZZ$, $K[x]$ for any field $K$, and (from Lecture~1) $\ZZ[i]$.
Nonexample: $K[x,y]$, since the ideal $(x,y)$ has no single generator.

The \emph{discriminant} of a monic polynomial $P(x)$
with roots $r_1, \dots, r_n$ is defined as $\prod_{i<j} (r_i-r_j)^2$.
This expression can be computed as a polynomial in the coefficients of $P$
(this follows from the symmetric polynomials theorem on the problem set).
It can also be computed by a built-in Magma function, as follows.
(The first line creates a ring $R$ of polynomials over the rationals with $x$
as the variable. After that, Magma knows to interpret $x^3+x+1$ as an element
of $R$.)
\begin{verbatim}
> R<x> := PolynomialRing(RationalField());
> Discriminant(x^3+x+1);
-31
\end{verbatim}

\head{Outline of lectures}
\begin{enumerate}
\item Define algebraic number, number field (and integral field extension).
\item Define algebraic integer, ring of integers (and integral closure).
\item Examples: roots of unity, $\cos (2\pi/n)$, $(1+\sqrt{-3})/2$,
$\sqrt{2}+\sqrt{3}$. Also a couple of nonexamples: $e, \pi$.
\item
As an aside, mention Gelfond-Schneider theorem: if $\alpha, \beta$ are
algebraic numbers and
$\beta \notin \QQ$, then $\alpha^\beta$ is not algebraic.
\item Explain linear algebraic proof that sum, product of algebraic
numbers/integers is algebraic. Allude to symmetric functions proof (on
homework).
\item Discuss minimal polynomial. Note that an algebraic number is integral
if and only if its minimal polynomial is integral. Compute some examples.
\item Define trace, norm, trace pairing. Show that the trace pairing is
nondegenerate for separable extensions. Define integral basis (absolute and relative).
\item Reinterpret trace and norm for Galois extensions.
\item Examples: quadratic fields, cyclotomic fields, the maximal real
subfield of a cyclotomic field.
\item Define discriminant of a number field.
\end{enumerate}

\end{document}