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# basis of ideal in algebraic number field

Theorem. 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, $\Delta(\alpha_{1},\,\alpha_{2})=392$.

## Mathematics Subject Classification

12F05*no label found*11R04

*no label found*06B10

*no label found*

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