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 $(l+m\sqrt{2})\cdot3$ 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 $$(m_1+m_2\sqrt{2})\cdot3 = m_1\cdot3+m_2\cdot3\sqrt{2}$$ 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

with

(1)