In a Dedekind domain every ideal is either principal or generated by two elements.

Question

Prove that in a Dedekind domain every ideal is either principal or generated by two elements.

Help me some hints.

Thanks a lot!

Answer 1

Hints:

Let $R$ be a Dedekind domain, and let $I\subseteq R$ be any ideal. Factor $I$ as $\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_n^{e_n}$. Take ANY $\alpha\in I-\{0\}$ and factor $(\alpha)$ as $\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_n^{f_n}\mathfrak{q}_1^{g_1}\cdots\mathfrak{q}_m^{g_m}$. We know that $f_i\geqslant e_i$ since $(\alpha)\subseteq I$ so that $I\mid (\alpha)$. Now explain why there exists some $\beta\in R$ such that $v_{\mathfrak{p}_i}(\beta)=e_i$ and $v_{\mathfrak{q}_i}(\beta)=0$. Explain then why

$$(\alpha,\beta)=\text{gcd}((\alpha),(\beta))=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_n^{e_n}=I$$

This, in fact, shows that for any $I$ and any $\alpha\in I-\{0\}$ you can always find a complementary generator.

Answer 2

It is enough to show that R/(x) is a PIR for each nonzero nonunit x. If you can use the structure theorem for rings in which every ideal is a product of prime ideals (called general ZPI-rings), this is immediate, since any zero-dimensional such ring is a PIR. (A general ZPI-ring is a finite direct product of Dedekind domains and PIR's with one prime.)

Alternatively, it is not hard to directly prove that R/(x) is a PIR. (This is a much easier special case of the theorem). Being a zero-dimensional semi-quasi-local general ZPI-ring, it is a finite direct product of (necessarily quasi-local) connected such rings. These factors are PIRs since each principal ideal is a power of the maximal ideal, so R/(x) is a PIR.