basis of ideal in algebraic number field

Theorem.  Let ${𝒪}_K$ be the maximal order of the algebraic number field $K$ of degree $n$.  Every ideal $𝔞$ of ${𝒪}_K$ has a basis, i.e. there are in $𝔞$ the linearly independent numbers  ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$  such that the numbers

$$m_1{\alpha }_1+m_2{\alpha }_2+\mathrm{\dots }+m_n{\alpha }_n,$$

where the $m_i$’s run all rational integers, form precisely all numbers of $𝔞$.  One has also

$$𝔞=({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n),$$

i.e. the basis of the ideal can be taken for the system of generators of the ideal.

Since  $\{{\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n\}$  is a basis of the field extension $K/\mathbb{Q}$, any element of $𝔞$ is uniquely expressible in the form $m_1{\alpha }_1+m_2{\alpha }_2+\mathrm{\dots }+m_n{\alpha }_n$.

It may be proven that all bases of an ideal $𝔞$ have the same discriminant $\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)$, which is an integer; it is called the discriminant of the ideal.  The discriminant of the ideal has the minimality property, that if ${\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n$ are some elements of $𝔞$, then

$$\mathrm{\Delta }({\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n)\geqq \mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)\mathit{\hspace{1em}}\text{or}\mathit{\hspace{1em}}\mathrm{\Delta }({\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n)=0$$

But if  $\mathrm{\Delta }({\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n)=\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)$,  then also the ${\beta }_i$’s form a basis of the ideal $𝔞$.

Example.  The integers of the quadratic field $\mathbb{Q}(\sqrt{2})$ are $l+m\sqrt{2}$ with  $l,m\in \mathbb{Z}$.  Determine a basis $\{{\alpha }_1,{\alpha }_2\}$  and the discriminant of the ideala)  $(6-6\sqrt{2},\mathrm{\hspace{0.17em}9}+6\sqrt{2})$,  b) $(1-2\sqrt{2})$.

a) The ideal may be seen to be the principal ideal $(3)$, since the both generators are of the form  $(l+m\sqrt{2})\cdot 3$  and on the other side,  $3=0\cdot (6-6\sqrt{2})+(3-2\sqrt{2})(9+6\sqrt{2})$.  Accordingly, any element of the ideal are of the form

$$(m_1+m_2\sqrt{2})\cdot 3=m_1\cdot 3+m_2\cdot 3\sqrt{2}$$

where $m_1$ and $m_2$ are rational integers.  Thus we can infer that

$${\alpha }_1=3,{\alpha }_2=3\sqrt{2}$$

is a basis of the ideal concerned.  So its discriminant is

b) All elements of the ideal $(1-2\sqrt{2})$ have the form

$\alpha =(a+b\sqrt{2})(1-2\sqrt{2})=(a-4b)+(b-2a)\sqrt{2}\mathit{\hspace{1em}}\text{with }a,b\in \mathbb{Z}.$(1)

Especially the rational integers of the ideal satisfy  $b-2a=0$,  when  $b=2a$  and thus  $\alpha =a-4\cdot 2a=-7a$.  This means that in the presentation  $\alpha =m_1{\alpha }_1+m_2{\alpha }_2$  we can assume ${\alpha }_1$ to be $7$.  Now the rational portion $a-4b$ in the form (1) of $\alpha$ should be splitted into two parts so that the first would be always divisible by 7 and the second by $b-2a$, i.e.  $a-4b=7m_1+(b-2a)x$; this equation may be written also as

$$(2x+1)a-(x+4)b=7m_1.$$

By experimenting, one finds the simplest value  $x=3$,  another would be  $x=10$.  The first of these yields

$$\alpha =7(a-b)+(b-2a)(3+\sqrt{2})=m_1\cdot 7+m_2(3+\sqrt{2}),$$

i.e. we have the basis

$${\alpha }_1=7,{\alpha }_2=3+\sqrt{2}.$$

The second alternative  $x=10$  similarly would give

$${\alpha }_1=7,{\alpha }_2=10+\sqrt{2}.$$

For both alternatives,  $\mathrm{\Delta }({\alpha }_1,{\alpha }_2)=392$.