integral basis of quadratic fieldIntegralBasisOfQuadraticField2008-04-08 12:01:392008-04-11 14:33:00Derivationintegral basis% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
\PMlinkescapeword{order}
Let $m$ be a squarefree integer $\neq 1$. All numbers of the quadratic field $\mathbb{Q}(\sqrt{m})$ may be written in the form
\begin{align}
\alpha = \frac{j+k\sqrt{m}}{l},
\end{align}
where $j,\,k,\,l$ are integers with\, $\gcd(j,\,k,\,l) = 1$\, and\, $l > 0$.\, Then $\alpha$ (and its algebraic conjugate \,$\alpha' = \frac{j-k\sqrt{m}}{l}$) satisfy the equation
\begin{align}
x^2+px+q = 0,
\end{align}
where
\begin{align}
p = -\frac{2j}{l}, \quad q = \frac{j^2-k^2m}{l^2}.
\end{align}
We will find out when the number (1) is an algebraic integer, i.e. when the coefficients $p$ and $q$ are rational integers.
Naturally, $p$ and $q$ are integers always when\, $l = 1$.\, We suppose now that\, $l > 1$.\, The latter of the equations (3) says that $q$ can be integer only when
$$(\gcd(j,\,l))^2 = \gcd(j^2,\,l^2) \mid k^2m$$
(see divisibility in rings).\, Because\, $\gcd(j,\,k,\,l) = 1$,\, we have by Euclid's lemma that\, $\gcd(j,\,l) \mid m$.\, Since $m$ is squarefree, we infer that
\begin{align}
\gcd(j,\,l) = 1.
\end{align}
In order that also $p$ were an integer, the former of the equations (3) implies that\, $l = 2$.
So, by the latter of the equations (3),\, $4 \mid j^2-k^2m$, i.e.
\begin{align}
k^2m \equiv j^2 \pmod{4}.
\end{align}
Since by (4),\, $\gcd(j,\,2) = 1$,\, the integer $j$ has to be odd.\, In order that (5) would be valid, also $k$ must be odd.\, Therefore,\, $j^2 \equiv 1 \pmod{4}$\, and\, $k^2 \equiv 1 \pmod{4}$,\, and thus (5) changes to
\begin{align}
m \equiv 1 \pmod{4}.
\end{align}
If we conversely assume (6) and that $j,\,k$ are odd and\, $l = 2$, then (5) is true, $p,\,q$ are integers and accordingly (1) is an algebraic integer.
We have now obtained the following result:
\begin{itemize}
\item When\, $m \not\equiv 1 \pmod{4}$,\, the integers of the field $\mathbb{Q}(\sqrt{m})$ are
$$a+b\sqrt{m}$$
where $a,\,b$ are arbitrary rational integers;
\item when\, $m \equiv 1 \pmod{4}$,\, in \PMlinkescapetext{addition} to the numbers $a+b\sqrt{m}$, also the numbers
$$\frac{j+k\sqrt{m}}{2},$$
with $j,\,k$ arbitrary odd integers, are integers of the field.
\end{itemize}
Then, it may be easily inferred the
\textbf{Theorem.}\, If we denote
%\begin{align*}
\[
\omega :=
\begin{cases}
& \frac{1+\sqrt{m}}{2} \quad \mbox{when } m \equiv 1\pmod{4},\\
& \sqrt{m} \quad \mbox{ when } m \not\equiv 1\pmod{4},
\end{cases}
\]
%\end{align*}
then any integer of the quadratic field $\mathbb{Q}(\sqrt{m})$ may be expressed in the form
$$a+b\omega,$$
where $a$ and $b$ are uniquely determined rational integers.\, Conversely, every number of this form is an integer of the field.\, One says that 1 and $\omega$ form an integral basis of the field.
\begin{thebibliography}{9}
\bibitem{K.V.} {\sc K. V\"ais\"al\"a}: {\em Lukuteorian ja korkeamman algebran alkeet}.\, Tiedekirjasto No. 17.\quad Kustannusosakeyhti\"o Otava, Helsinki (1950).
\end{thebibliography}