integral basis of quadratic field

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

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