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Revision difference : two-generator property
Version 29 Version 28
\begin{thmplain} \begin{thmplain}
\,\,Every ideal of a Dedekind domain can be generated by two of its elements. \,\,Every ideal of a Dedekind domain can be generated by two of its elements.
\end{thmplain} \end{thmplain}
{\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).
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem{EM}{\sc Eben Matlis:} {\em The two-generator problem for ideals}. \,-- The Michigan Mathematical Journal \textbf{17} $N^o$ 3 (1970). \bibitem{EM}{\sc Eben Matlis:} {\em The two-generator-problem for ideals}. \,-- The Michigan Mathematical Journal \textbf{17} $N^o$ 3 (1970).
\end{thebibliography} \end{thebibliography}