integral basis of quadratic field

Let $m$ be a squarefree integer $\ne 1$. All numbers of the quadratic field $\mathbb{Q}(\sqrt{m})$ may be written in the form

$\alpha ={\displaystyle \frac{j+k\sqrt{m}}{l}},$ (1)

where $j,k,l$ are integers with  $\gcd (j,k,l)=1$  and  $l>0$.  Then $\alpha$ (and its algebraic conjugate  ${\alpha }^{\prime }=\frac{j-k\sqrt{m}}{l}$) satisfies the equation

$x^2+px+q=\mathrm{\hspace{0.33em}0},$ (2)

where

$p=-{\displaystyle \frac{2j}{l}},q={\displaystyle \frac{j^2-k^{2}m}{l^2}}.$ (3)

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

$${(\gcd (j,l))}^2=\gcd (j^2,l^2)\mid k^{2}m$$

(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

$\gcd (j,l)=1.$ (4)

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),  $4\mid j^2-k^{2}m$, i.e.

$k^{2}m\equiv j^2(mod4).$ (5)

Since by (4),  $\gcd (j,\mathrm{\hspace{0.17em}2})=1$,  the integer $j$ has to be odd.  In order that (5) would be valid, also $k$ must be odd.  Therefore,  $j^2\equiv 1(mod4)$  and  $k^2\equiv 1(mod4)$,  and thus (5) changes to

$m\equiv 1(mod4).$ (6)

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:

• When  $m\not\equiv 1(mod4)$,  the integers of the field $\mathbb{Q}(\sqrt{m})$ are

  $$a+b\sqrt{m}$$

  where $a,b$ are arbitrary rational integers;

• when  $m\equiv 1(mod4)$,  in 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.

Then, it may be easily inferred the

Theorem.  If we denote

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.

Title	integral basis of quadratic field
Canonical name	IntegralBasisOfQuadraticField
Date of creation	2014-02-27 10:24:31
Last modified on	2014-02-27 10:24:31
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Derivation
Classification	msc 11R04
Synonym	canonical basis of quadratic field
Synonym	quadratic integers
Related topic	PropertiesOfQuadraticEquation
Related topic	Gcd
Related topic	ExamplesOfRingOfIntegersOfANumberField
Related topic	SomethingRelatedToFundamentalUnits
Related topic	CanonicalBasis