Overrings of Dedekind domains as localizations

Question

I am taking an independent study where I organize and present weekly material on algebraic number theory to my professor and receive feedback. Next week I am going to cover some miscellaneous topics, including localization. This note is alluring in that it is short and sweet and uses technology I have just gotten to at this point. I started going through the argument though and I am stuck at the very beginning on the equation labelled $(\star)$.

I wish to prove that if $R$ is Dedekind and $R\subset A\subset K$ where $K$ is the fraction field of $R$, then $$A=\bigcap_{{\frak p}A\ne A}R_{\frak p} \tag{$\star$}$$

[Edit: with the additional hypothesis that $R$'s fractional ideal classes have finite order.]

where $R_{\frak p}$ is the localization $(R\setminus{\frak p})^{-1}R$ which I define via $S^{-1}R=\min\{T\subseteq K\mid S\subseteq T^\times\}$. I can prove that $S^{-1}R=$"$R[S^{-1}]$" so all elements of $S^{-1}R$ can be written with a denominator in $S$ and conversely any fraction with a denominator in $S$ is in $S^{-1}R$. I can also argue that every $x\in K$ (and hence in $A$) can be written in the form $a/b$ where $a,b\in R$ and $a\not\in(b)$, or even more strongly where $a,b$ share no common factor (which in terms of ideals means $(a,b)\triangleleft R$ is not divisible by any nontrivial principal ideal, if that's worth anything).

To show the LHS is contained in the RHS, I must show ${\frak p}A\ne A\Rightarrow A\subseteq R_{\frak p}$, which means ${\frak p}A<A$ (strictly) implies all $x\in A$ can be written as $a/b$ with $a\in R$, $b\in R\setminus\frak p$. I am not sure how to achieve this though. I am also having trouble with the other inclusion. My first idea was to say $a/b$ is in the intersection, factor $bR={\frak q}_1\cdots{\frak q}_e$, argue ${\frak q}_iA\ne A$ was not possible since then, combined with $b\in{\frak p}_i$, it would follow $a/b\not\in R_{{\frak q}_{\large i}}$, a contradiction, hence ${\frak q}_iA=A$ for each $i$ and then we have the resulting equality $bA=bRA={\frak q}_1\cdots{\frak q}_eA=A\ni 1$ hence $b^{-1}\in A$.

However I am not able to prove $b\in{\frak q}_i\Rightarrow a/b\not\in R_{{\frak q}_{\large i}}$: I need to show that if $a/b$ in simplest terms has $b\in\frak p$ then it cannot be rewritten into another form that has a denominator not in $\frak p$. I would generate ideals at this point and consider exponents of $\frak p$ in factorizations, but I'm not certain $a\not\in\frak p$. (It is possible for both numerator and denominator to be in a prime ideal but still in simplest form: take the ratio of two generators of a nonprincipal ideal for example.)

It'd also be nice to know if $(\star)$ does not hold for non-Dedekind domains, and if so what's a domain and an overring in which it fails (and why).

Going back to the original claim of the note, it'd be nice also to know of simple Dedekind domains that have fractional ideal classes of infinite order. I did a rudimentary google search but the only direct mention involved ideas like Picard groups and elliptic curves, which I haven't covered yet.

Answer 1

You are right that the statement in question is not obvious. However it does hold for all Dedekind domains (with no additional hypotheses on the class group). A detailed study of overrings of Dedekind domains occurs in $\S$ 23 of my commutative algebra notes. However, the discussion here is slightly incomplete: the proof of Theorem 23.3 -- which is in fact what you are asking about -- is deferred to the section on Prüfer domains, which is not yet finished.

Let me give a simpler proof of this fact as a deduction from things that are already in the notes. First we need that the maximal ideals of the overring $A$ are precisely the maximal ideals $\mathfrak{p}$ of $R$ such that $\mathfrak{p} A \subsetneq A$. This is Proposition 23.2. Then we use the fact that for any integral domain $A$ with fraction field $K$, if we intersect the valuation rings of $K$ containing $A$ then we get the integral closure of $A$ (Theorem 17.15). In particular we get $A$ back again iff $A$ is integrally closed, which it is in this case. Also, by Krull-Akizuki every valuation ring containing a Dedekind domain is a DVR, and the DVR overrings of a Dedekind domain correspond to the maximal ideals of the domain. This answers your question.

My notes go on to treat the theorem that is proved in David Krumm's note (a solution to a homework exercise in a graduate course he took from me a few years ago; he has since gotten his PhD) in a more leisurely manner than David's one-page solution: see in particular Theorem 23.5.

Answer 2

Here is a counterexample when $R$ is not a Dedekind domain. Consider

$$k[t^2,t^3] \subseteq k[t] \subseteq k(t)= \operatorname{Frac} k[t^2,t^3].$$

The ring $ R := k[t^2,t^3]$ cannot possibly be a Dedekind domain because it is not integrally closed. Now we claim for every prime $\mathfrak{p}$ of $R$ that $\mathfrak{p}k[t] \neq k[t]$. If we had $\mathfrak{p}k[t] = k[t]$ for some prime $\mathfrak{p}$ then localizing we get $\mathfrak{p}_{\mathfrak{p}} k[t]_{\mathfrak{p}} = k[t]_{\mathfrak{p}}$. Here we are thinking of $\mathfrak{p}$ and $k[t]$ as $R$-modules.

Now the point is that $k[t]$ being a f.g. $R$-module implies $k[t]_{\mathfrak{p}}$ is a f.g. $R_\mathfrak{p}$-module and so Nakayama's Lemma implies that $k[t]_{\mathfrak{p}}=0$, contradiction. Thus we see that $$\bigcap_{\mathfrak{p}k[t] \neq k[t]} R_{\mathfrak{p}} = \bigcap_{\mathfrak{p} \in \operatorname{Spec} R}R_{\mathfrak{p}} = R$$ where the last equality comes from the fact that a domain is equal to the intersection of the localizations at all its primes.

So in the case that $R$ is not a Dedekind domain, the equality is not true.

Answer 3

I argue that both conditions are equivalent to the statement that no fractional ideal contained in $A$ has a negative power of $\mathfrak{p}$ in its prime factorization.

...I do wish that people would explain downvotes. This is the only full proof given of the facts in question, and makes no use of heavy technology.

---

$A\subset R_\mathfrak{p} \implies \mathfrak{p} A\neq A$:

Suppose that $\mathfrak{p}A=A$. Then there is some fractional ideal $I\subset A$ with $R\subset\mathfrak{p} I$, so $\operatorname{ord}_\mathfrak{p} I < 0$. But any fractional ideal $J\subset R_\mathfrak{p}$ has $\operatorname{ord}_\mathfrak{p} J \geq 0$, so it follows that $R\not\subset R_\mathfrak{p}$.

---

$\mathfrak{p} A \neq A \implies A\subset R_\mathfrak{p}$:

If $A\not\subset R_\mathfrak{p}$, then there is some fractional ideal $I\subset A$ with $I\not\subset R_\mathfrak{p}$, which is equivalent to the statement that $\operatorname{ord}_\mathfrak{p} I < 0$. In other words, the unique factorization of $I$ into prime ideals contains a negative power of $\mathfrak{p}$.

By clearing denominators, we may further assume that the prime factorization of $I$ contains no negative powers at all, except for a single negative power of $\mathfrak{p}$. Then $\mathfrak{p}I \subset R$. Note that we cannot have $\mathfrak{p} I = \mathfrak{p}$, because the left-hand side is not divisible by $\mathfrak{p}$.

So there is some ideal $J \subset \mathfrak{p}I\subset R$ with $J \not\subset\mathfrak{p}$. Since $\mathfrak{p}$ is a maximal ideal, we have $\mathfrak{p} + J = R$, therefore $\mathfrak{p} A = \mathfrak{p} A + JA = A$.