Localising Dedekind domains

Question

I'm wondering if the following is true:

Let $A\subset B$ be two Dedekind domains with $B$ integral over $A$. Let $Q$ be a non-zero prime ideal in $B$ and $P=Q\cap A$. Then the localisation of $B$ at $A-P$ and $B-Q$ are the same (in Frac($B$) say).

Thoughts: It is sufficient to prove that $\forall t\in B-Q, \exists b\in B \text{ s.t. } bt\in A-P $. But I can't do it.

Background: If $A$ and $B$ also fit a standard setup in algebraic number theory (i.e. fraction fields also form finite separable extension) then we know the different satisfies $D_{S^{-1}B/S^{-1}A}=S^{-1}D_{B/A}$ $(\star)$ where $S$ is a multiplicative subset of $A$. But then in the standard proof (e.g. Serre "Local Fields") that taking the different 'commutes' with completion, it starts with "wlog replace $A$ by its localisation at $A-P$" quoting the localisation result $(\star)$ above to justify the "wlog". I think it takes $S$ to be $A-P$ to give $D_{B_Q/A_P}=(D_{B/A})_P$. I would understand this if what I asked was true.

Thanks!

Answer 1

Now, despite the fact that in general $B_Q \neq B_P$, you can still justify "the without loss of generality" in the argument about the different. Let $Q$ be a prime of $B$. Let $P = Q \cap A$, let $w$ be the absolute value of $L$ induced by $Q$, and let $v$ be the absolute value of $K$ induced by $P$. The restriction of $w$ to $K$ will be a scaling of $v$, and you can regard the completion $K_v$ as being contained in $L_w$.

The big theorem about the local/global different says that $D_{B/A} B_w = D_{B_w/A_v}$. To prove this, Serre is saying that "without loss of generality," he can just prove that $D_{B_P/A_P}B_w = D_{B_w/A_v}$. So to justify the without loss of generality, you need to argue that $D_{B/A} B_w=D_{B_P/A_P}B_w $.

In general, if $J$ is an ideal of $B$, then $J_P$ is an ideal of $B_P$, $J_Q$ is an ideal of $B_Q$, and $JB_w$ is an ideal of $B_w$. If $n$ is the value of $J$ at the prime ideal $Q$ of $B$, then $n$ is also the value of $J_P$ at the prime $QB_P$ of $B_P$, also $n$ is the value of $J_Q$ at the prime $QB_Q$ of $B_Q$, and finally $n$ is also the value of $JB_w$ at the sole prime $QB_w$ of $B_w$. It follows from this principle that $$JB_w = J_PB_w = J_Q B_w$$ Localizing/completing an ideal at a prime doesn't change its value at that prime.

Now from the result $S^{-1}D_{B/A} = D_{S^{-1}B/S^{-1}A}$, you get $(D_{B/A})_P = D_{B_P/A_P}$, and so $$D_{B_P/A_P}B_w = (D_{B/A})_PB_w = D_{B/A}B_w$$

Answer 2

Usually what you claim is not true. If $K \subseteq L$ is a finite separable extension of fields, $K$ is the quotient field of $A$, and $B$ is the integral closure of $A$ in $L$, then for a prime $Q$ of $B$ lying over a prime $P$ of $A$, you have $(B \setminus Q)^{-1}B = (A \setminus P)^{-1}B$ if and only if $Q$ is the only prime of $B$ lying over $P$.

Proof: Write $B_Q$ and $B_P$ instead of $(B \setminus Q)^{-1}B$ and $(A \setminus P)^{-1}B$. You always have $B_P \subseteq B_Q$. In general, $PB$ is equal to $Q_1^{e_1} \cdots Q_g^{e_g}, e_i > 0$, where $Q_1, ... , Q_g$ are all the primes of $B$ lying over $P$. The ring $B_P$ has only finitely many nonzero prime ideals (i.e. maximal ideals): $Q_1B_P, ... , Q_g B_P$. And if you localize $B_P$ at the maximal ideal $Q_iB_P$, you just get the ring $B_{Q_i}$. Now you can recover any integral domain by intersecting its localization at all the maximal ideals: $$B_P = \bigcap\limits_{i=1}^g B_{Q_i}$$ Now for distinct primes $Q, Q'$ of $B$, the localizations $B_Q, B_{Q'}$ are incomparable: neither ring is a subset of the other. The proof follows.