3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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".

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.