basis of ideal in algebraic number field

Let $\mathcal{O}_{K}$ be the maximal order of the algebraic number field $K$ of degree $n$.  Every ideal $\mathfrak{a}$ of $\mathcal{O}_{K}$ has a basis, i.e. there are in $\mathfrak{a}$ the linearly independent numbers  $\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n}$  such that the numbers

 $m_{1}\alpha_{1}+m_{2}\alpha_{2}+\ldots+m_{n}\alpha_{n},$

where the $m_{i}$’s run all rational integers, form precisely all numbers of $\mathfrak{a}$.  One has also

 $\mathfrak{a}=(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n}),$

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

Since  $\{\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n}\}$  is a basis of the field extension $K/\mathbb{Q}$, any element of $\mathfrak{a}$ is uniquely expressible in the form $m_{1}\alpha_{1}+m_{2}\alpha_{2}+\ldots+m_{n}\alpha_{n}$.

It may be proven that all bases of an ideal $\mathfrak{a}$ have the same discriminant $\Delta(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n})$, which is an integer; it is called the discriminant of the ideal.  The discriminant of the ideal has the minimality property, that if $\beta_{1},\,\beta_{2},\,\ldots,\,\beta_{n}$ are some elements of $\mathfrak{a}$, then

 $\Delta(\beta_{1},\,\beta_{2},\,\ldots,\,\beta_{n})\geqq\Delta(\alpha_{1},\,% \alpha_{2},\,\ldots,\,\alpha_{n})\quad\mbox{or}\quad\Delta(\beta_{1},\,\beta_{% 2},\,\ldots,\,\beta_{n})=0$

But if  $\Delta(\beta_{1},\,\beta_{2},\,\ldots,\,\beta_{n})=\Delta(\alpha_{1},\,\alpha_% {2},\,\ldots,\,\alpha_{n})$,  then also the $\beta_{i}$’s form a basis of the ideal $\mathfrak{a}$.

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 ideal a)  $(6\!-\!6\sqrt{2},\,9\!+\!6\sqrt{2})$,  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

 $\alpha_{1}=3,\quad\alpha_{2}=3\sqrt{2}$

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

 $\Delta(\alpha_{1},\,\alpha_{2})=\left|\begin{matrix}3&3\sqrt{2}\\ 3&-3\sqrt{2}\end{matrix}\right|^{2}=648.$

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

 $\displaystyle\alpha=(a+b\sqrt{2})(1-2\sqrt{2})=(a-4b)+(b-2a)\sqrt{2}\quad\mbox% {with\, }a,\,b\in\mathbb{Z}.$ (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

 $\alpha_{1}=7,\quad\alpha_{2}=3+\sqrt{2}.$

The second alternative  $x=10$  similarly would give

 $\alpha_{1}=7,\quad\alpha_{2}=10+\sqrt{2}.$

For both alternatives,  $\Delta(\alpha_{1},\,\alpha_{2})=392$.

 Title basis of ideal in algebraic number field Canonical name BasisOfIdealInAlgebraicNumberField Date of creation 2013-03-22 17:51:15 Last modified on 2013-03-22 17:51:15 Owner pahio (2872) Last modified by pahio (2872) Numerical id 9 Author pahio (2872) Entry type Theorem Classification msc 12F05 Classification msc 11R04 Classification msc 06B10 Synonym basis of ideal in number field Related topic IntegralBasis Related topic IdealNorm Related topic AlgebraicNumberTheory Defines basis of ideal Defines ideal basis Defines discriminant of the ideal