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Version 21 |
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\textbf{Theorem:} \, {\em Every ideal of a Dedekind domain can be generated by two of its elements.}
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Theorem: {\em Every ideal of a Dedekind domain can be generated by two of its elements.}
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| {\em Proof}. Let $\mathfrak{a}$ be an arbitrary ideal of a Dedekind domain $R$. Let $\mathfrak{b}$ be such an ideal of $R$ that $\mathfrak{ab}$ is a principal ideal $(\beta)$. The lemma to which this entry is attached gives also an element $\gamma$ and an ideal $\mathfrak{c}$ of $R$ such that $\mathfrak{ac} = (\gamma)$ and $\mathfrak{b+c} = R$. Then we have |
{\em Proof}. Let $\mathfrak{a}$ be an arbitrary ideal of a Dedekind domain $R$. Let $\mathfrak{b}$ be such an ideal of $R$ that $\mathfrak{ab}$ is a principal ideal $(\beta)$. The lemma to which this entry is attached gives also an element $\gamma$ and an ideal $\mathfrak{c}$ of $R$ such that $\mathfrak{ac} = (\gamma)$ and $\mathfrak{b+c} = R$. Then we have |
| $$\mathfrak{a} = \gcd(\mathfrak{ab}, \mathfrak{ac}) = |
$$\mathfrak{a} = \gcd(\mathfrak{ab}, \mathfrak{ac}) = |
| \gcd((\beta), (\gamma)) = (\beta, \gamma)$$ |
\gcd((\beta), (\gamma)) = (\beta, \gamma)$$ |
| because $\gcd(\mathfrak{b}, \mathfrak{c}) = \mathfrak{b+c} = R = (1)$. |
because $\gcd(\mathfrak{b}, \mathfrak{c}) = \mathfrak{b+c} = R = (1)$. |
| An open question is whether or not the two-generator property can be generalized to the invertible ideals of Pr\"ufer domains (and Pr\"ufer rings). |
An open question is whether or not the two-generator property can be generalized to the invertible ideals of Pr\"ufer domains (and Pr\"ufer rings). |
