# Can a Dedekind domain have a power basis over a ring that isn't a Dedekind domain?

This aim of this question is to determine whether there exists a proof (or some counterexample) to the following statement : "If $$R$$ is a subring of Dedekind domain $$S$$, such that $$S$$ has a power basis as an $$R$$-module, then $$R$$ is itself a Dedekind domain".

The context is the following: I have recently been working on formalizing (using Lean) the Dedekind-Kummer theorem, which can be stated as follows [Proposition 8.3, Algebraic number theory, Neukirch] :

Let $$\mathcal{O}$$ be a Dedekind domain with field of fractions $$K$$ and $$L$$ a finite separable extension of $$K$$, and set $$\mathcal{O'}$$ to be the integral closure of $$\mathcal{O}$$ in $$L$$. If we pick a prime ideal $$\mathfrak{p}$$ of $$\mathcal{O}$$ and $$\theta \in \mathcal{O}'$$ such that the conductor ideal $$\mathcal{C}$$ of $$\mathcal{O}[\theta]$$ is coprime with $$\mathfrak{p}$$ and $$L=K(\theta)$$, then the prime factorisations of $$\mathfrak{p} \mathcal{O'}$$ and $$\overline{f}$$ have the same shape (in other words, there is a bijection between the sets of prime factors of $$\mathfrak{p} \mathcal{O'}$$ and $$\overline{f}$$ that preserves multiplicities), where $$f$$ is the minimal polynomial of $$\theta$$ over $$K$$ and $$\overline{f}$$ is the reduction mod $$\mathfrak{p}$$ of $$f$$. ​​​​​​

So far the case when $$\mathcal{O'} = \mathcal{O}[\theta]$$ has been fully formalized, but the formalization does not use the assumption that $$\mathcal{O}$$ is a Dedekind domain. The aim of this question is thus to determine whether or not this is actually a generalization of the original statement (at least for this particular case).​​​​​​

If $$R\subseteq S$$ with $$S$$ Dedekind and free as an $$R$$-module, then $$R$$ is Dedekind because every $$R$$-ideal $$I$$ is projective (hence invertible, if non-zero). For $$I\otimes_RS$$ $$\cong$$ $$IS$$ is projective over $$S$$, hence over $$R$$. But $$I\otimes_RS$$ $$\cong$$ $$I^{\oplus n}$$ - or, more generally, $$I\otimes_RS$$ $$\cong$$ $$\bigoplus_{\alpha\in A}I$$ when $$S$$ $$\cong$$ $$\bigoplus_{\alpha\in A}R$$. So $$I$$, a direct summand of $$I\otimes_RS$$, is also projective over $$R$$.
Edit In fact, if $$S$$ is integral over $$R$$ (for example, finitely generated as a module), it suffices that $$S$$ is flat over $$R$$. For then we have $$I\otimes_RS$$ $$\cong$$ $$IS$$, which is flat over $$S$$. But $$S/R$$ has lying-over, so that $$\mathfrak{m}S$$ $$\ne$$ $$S$$ for every maximal ideal $$\mathfrak{m}$$ of $$R$$. Thus $$S$$ is faithfully flat over $$R$$. This implies that $$I$$ is flat over $$R$$. So $$R$$ is Prüfer (finitely generated ideals are flat), and hence all non-zero finitely generated ideals are invertible. And $$R$$ must be Noetherian (if $$I_1\subseteq I_2\subseteq\cdots$$ are ideals of $$R$$, we have $$I_nS$$ $$=$$ $$I_{n+1}S$$ for some $$n$$; but $$I_nS\cap R$$ $$=$$ $$I_n$$ by faithful flatness, and likewise for $$I_{n+1}$$).
• Just as a note, since $R$ is a domain, flat + finitely generated actually implies projective (mathoverflow.net/questions/33522/…). Oct 21, 2022 at 9:12
• Thanks, that avoids the use of Prüfer domains. Another point is that, formally, fields do not count as Dedekind domains. So the statement should read: if $R$ $\subseteq$ $S$, with $S$ Dedekind and faithfully flat over $R$, then $R$ is either a field or a Dedekind domain. Oct 21, 2022 at 14:43
• Note that, conversely, if $R$ is Dedekind and $S\supseteq R$ a domain (or just torsion-free over $R$), then $S$ is flat over $R$. Oct 23, 2022 at 15:43