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Revision difference : basis of ideal in algebraic number field |
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| \textbf{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 {\em basis}, i.e. there are in $\mathfrak{a}$ the linearly independent numbers\, $\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n$\, such that the numbers |
\textbf{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 {\em basis}, i.e. there are in $\mathfrak{a}$ the linearly independent numbers\, $\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n$\, such that the numbers |
| $$m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n,$$ |
$$m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n,$$ |
| where the $m_i$'s run all rational integers, form precisely all numbers of $\mathfrak{a}$.\, One has also |
where the $m_i$'s run all rational integers, form precisely all numbers of $\mathfrak{a}$.\, One has also |
| $$\mathfrak{a} = (\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n),$$ |
$$\mathfrak{a} = (\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n),$$ |
| i.e. the basis of the ideal can be taken for the system of generators of the ideal.\\ |
i.e. the basis of the ideal can be taken for the system of generators of the ideal.\\ |
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| 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 $m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n$. |
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 $m_1\alpha_1+m_2\alpha_2+\ldots+m_n\alpha_n$. |
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| 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 {\em 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 |
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 {\em 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 |
| $$\Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n) \geqq \Delta(\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n) |
$$\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$$ |
\quad \mbox{or} \quad \Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n) = 0$$ |
| 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 $\mathfrak{a}$.\\ |
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 $\mathfrak{a}$. |
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| \textbf{Example.}\, The integers of the quadratic field $\mathbb{Q}(\sqrt{2})$ are $l+m\sqrt{2}$ with\, $l,\,m \in \mathbb{Z}$.\, Determine the basis and the discriminant of the ideal\, $(6-6\sqrt{2},\,9+6\sqrt{2})$. |
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| This 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,\, $0\cdot(6-6\sqrt{2})+(3-2\sqrt{2})(9+6\sqrt{2}) = 3$.\, Accordingly, any element of the ideal are of the form |
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| $$(m_1+m_2\sqrt{2})\cdot3 = m_1\cdot3+m_2\cdot3\sqrt{2}$$ |
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| where $m_1$ and $m_2$ are rational integers.\, Thus we can infer that |
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| $$\alpha_1 = 3, \quad \alpha_2 = 3\sqrt{2}$$ |
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| is a basis of the ideal concerned.\, So its discriminant is |
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| $$\Delta(\alpha_1,\,\alpha_2) = |
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| \left|\begin{matrix} |
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| 3 & 3\sqrt{2}\\ |
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| 3 & -3\sqrt{2} |
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| \end{matrix}\right|^2 = 648.$$ |
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