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I have the following situation:

Let $B \subseteq B'$ be a ring extension such that $\text{Quot}(B) = \text{Quot}(B') =: K$ and $\text{dim}(B) = \text{dim}(B') = 1$ where $B'$ is a Dedekind domain. Let $\mathcal{F} = \{ x \in K \mid xB' \subseteq B \}$ be the conductor of $B$ in $B'$.

Now I want to show that every prime ideal $P$ in $B$ coprime to $\mathcal{F}$, i.e. $P + \mathcal{F} = B$, is invertible.

We have already proven that those prime ideals $P$ coprime to the conductor satisfy that the localization $B_{P}$ is a discrete valuation ring. $(*)$

The source states that the desired proposition follows from $(*)$.

My thoughts:

From this post we have:

Proposition: $M$ as a fractional ideal is invertible if and only if $MB_{P}$ is a principal fractional ideal for every maximal ideal $P$ of $B$.

In my situation we have $B_{P}$ is a DVR, hence a local PID und therefore $PB_{P}$ is a principal ideal and clearly invertible as a fractional ideal of $B_{P}$. From $\text{dim}(B) = 1$ we have that every prime ideal $\neq 0$ is maximal, but I don't know how to go on and use the fact that $P + \mathcal{F} = B$.

I'd appreciate any kind of help and input.

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gisma
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  • You must be have already used that $P + \mathcal{F} = B$ since you proved that $B_P$ is a DVR. (Now I hope you can understand my previous comment better.) – user26857 Nov 12 '14 at 16:03
  • The source is part of some lecture notes from my professor and therefore I wasn't sure if I'm allowed to put that here; besides it's written in german. And yes, $P + \mathcal{F} = B$ was used to show that $B_P$ is a DVR. Again, thank you very much! – gisma Nov 12 '14 at 16:26

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Let's use the invertibility criterion you mentioned (changing the notation):

$I$ is invertible iff $IB_Q$ is a principal fractional ideal for every maximal ideal $Q$ of $B$.

Since $\dim B=1$ there is no maximal ideal containing $P$ excepting $P$ itself (of course, the trivial case $P=(0)$ can be excepted), so $PB_Q=B_Q$ for all $Q\ne P$. Is it clear now?

user26857
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  • I thought about this before but I kind of scrapped that idea for no reason. Of course this is the answer, it's clear now! Thank you very much. – gisma Nov 12 '14 at 16:22