Say $R$ is a domain such that for any nonzero ideal $I$ of $R$ and any $a \in I$, we have $I = \langle a,b\rangle$ for some $b \in I$. Is $R$ a Dedekind domain? If not, what additional assumptions do I need to force it to be one?


Having every ideal be generated by two elements is not enough. A nice counterexample is $R = \mathbb{Z}[\sqrt{-3}]$. Since $R$ is isomorphic to $\mathbb{Z}^2$ as a $\mathbb{Z}$-module, every ideal in $R$ is a free $\mathbb{Z}$-module of rank at most $2$, hence certainly can be generated by two elements. On the other hand, $R$ is not integrally closed, since $\frac{1+\sqrt{-3}}{2}$ is integral over $R$.

However, the condition in the body of your question is sufficient (even when you impose the condition $a \neq 0$: as written you are characterizing principal ideal domains!): a domain is a Dedekind domain if and only if for every ideal $I$ and every nonzero $a \in I$, there is $b \in I$ such that $I = \langle a,b \rangle$. See e.g. $\S$ 20.5 of my commutative algebra notes. This property is sometimes called "$(1+\epsilon)$-generation of ideals".

  • $\begingroup$ Whoops, I realize my question title is different from the body text of the question and this may be misleading... $\endgroup$ – Sam Hopkins Dec 3 '13 at 18:32
  • $\begingroup$ Nice answer as usual. At the end of the proof of Thm 20.12, I think you mean $I=\mathfrak p I+bR_{\mathfrak p}$ before applying Nakayama. $\endgroup$ – Cantlog Dec 4 '13 at 21:56

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