integral basis of quadratic field
Naturally, and are integers always when . We suppose now that . The latter of the equations (3) says that can be integer only when
(see divisibility in rings). Because , we have by Euclid’s lemma that . Since is squarefree, we infer that
In order that also were an integer, the former of the equations (3) implies that .
So, by the latter of the equations (3), , i.e.
Since by (4), , the integer has to be odd. In order that (5) would be valid, also must be odd. Therefore, and , and thus (5) changes to
If we conversely assume (6) and that are odd and , then (5) is true, are integers and accordingly (1) is an algebraic integer.
We have now obtained the following result:
When , the integers of the field are
where are arbitrary rational integers;
when , in to the numbers , also the numbers
with 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 may be expressed in the form
where and are uniquely determined rational integers. Conversely, every number of this form is an integer of the field. One says that 1 and form an integral basis of the field.
- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).
|Title||integral basis of quadratic field|
|Date of creation||2014-02-27 10:24:31|
|Last modified on||2014-02-27 10:24:31|
|Last modified by||pahio (2872)|
|Synonym||canonical basis of quadratic field|