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[parent] integral basis of quadratic field (Derivation)

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

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

where $ j,\,k,\,l$ are integers with $ \gcd(j,\,k,\,l) = 1$ and $ l > 0$. Then $ \alpha$ (and its algebraic conjugate $ \alpha' = \frac{j-k\sqrt{m}}{l}$) satisfy the equation
$\displaystyle x^2+px+q = 0,$ (2)

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

$\displaystyle (\gcd(j,\,l))^2 = \gcd(j^2,\,l^2) \mid k^2m$
(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
$\displaystyle \gcd(j,\,l) = 1.$ (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^2m$, i.e.

$\displaystyle k^2m \equiv j^2 \pmod{4}.$ (5)

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
$\displaystyle m \equiv 1 \pmod{4}.$ (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 \pmod{4}$, the integers of the field $ \mathbb{Q}(\sqrt{m})$ are
    $\displaystyle a+b\sqrt{m}$
    where $ a,\,b$ are arbitrary rational integers;
  • when $ m \equiv 1 \pmod{4}$, in addition to the numbers $ a+b\sqrt{m}$, also the numbers
    $\displaystyle \frac{j+k\sqrt{m}}{2},$
    with $ j,\,k$ arbitrary odd integers, are integers of the field.

Then, it may be easily inferred the

Theorem. If we denote

\begin{displaymath} \omega := \begin{cases} & \frac{1+\sqrt{m}}{2} \quad \mbox{w... ...sqrt{m} \quad \mbox{ when } m \not\equiv 1\pmod{4}, \end{cases}\end{displaymath}

then any integer of the quadratic field $ \mathbb{Q}(\sqrt{m})$ may be expressed in the form

$\displaystyle a+b\omega,$
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.

Bibliography

1
K. V¨AISÄLÄ: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17.    Kustannusosakeyhtiö Otava, Helsinki (1950).



"integral basis of quadratic field" is owned by pahio. [ full author list (2) ]
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See Also: properties of quadratic equation, gcd, examples of ring of integers of a number field, table of some fundamental units, canonical basis

Other names:  canonical basis of quadratic field, quadratic integers

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Cross-references: integral basis, odd integers, field, odd, implies, Euclid's lemma, divisibility in rings, rational, coefficients, algebraic integer, equation, algebraic conjugate, quadratic field, numbers, integer, squarefree

This is version 7 of integral basis of quadratic field, born on 2008-04-08, modified 2008-04-11.
Object id is 10490, canonical name is IntegralBasisOfQuadraticField.
Accessed 407 times total.

Classification:
AMS MSC: 11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
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