# $\mathcal{O}/(\mathcal{O}\cap I\mathcal{O}_P)\cong \mathcal{O}_P/I\mathcal{O}_p$ if $\mathcal{O}$ is a noetherian one dimensional domain.

Let $\mathcal{O}$ be a Noetherian, one dimensional domain and $P\subset \mathcal{O}$ be a prime ideal. I want to prove that if $0\neq I\subseteq P$ we have $\mathcal{O}/(\mathcal{O}\cap I\mathcal{O}_P) \cong \mathcal{O}_P/ I\mathcal{O}_P$.

It's clearly enough to prove that the map $\mathcal{O}\rightarrow \mathcal{O}_P/ I\mathcal{O}_P$ is surjective, but I can't prove that.

This is used in the proof of propostion 12.3 in Neukirch Algebraic Number Theory.

• Feb 24, 2017 at 7:08
• Would the Noetherian tag be appropriate? Feb 24, 2017 at 20:35

Since $\mathcal{O}$ is a $1$-dimensional domain and $I\neq 0$, $\mathcal{O}/I$ is $0$-dimensional. Since it is Noetherian, that means $\mathcal{O}/I$ is the product of its localizations at each of its prime ideals. In particular, the map from $\mathcal{O}/I$ to any localization of $\mathcal{O}/I$ is surjective (since it is just the projection onto a factor of a product). Thus the map $\mathcal{O}/I\to \mathcal{O}_P/I\mathcal{O}_P$ is surjective, which gives the isomorphism you want.
• Hello Eric, can you please explain why $O/I$ being Noetherian implies that $O/I$ is the product of its localizations at each of its prime ideals? Thanks! Jan 17, 2019 at 12:19
Here is the argument that I suspect Neukirch intended. In the first part of the proof of Prop 12.3 in Neukirch's book, he states that if $$P \supseteq I$$, then $$P$$ is the only prime ideal containing $$\widetilde{I}_P$$ where $$\widetilde{I}_P = (\mathcal{O} \cap I\mathcal{O}_P)$$. He doesn't offer a proof, but a simple one is given in Proof of a stronger form of Chinese remainder theorem (12.3) in Neukirch
That means that $$\mathcal{O}/\widetilde{I}_P$$ has a single maximal ideal, $$P/\widetilde{I}_P$$ and so everything not in that ideal is a unit. Therefore, if $$x/s \in \mathcal{O}_P$$ with $$s \notin P$$, and writing modulo $$\widetilde{I}_P$$ with an overbar, $$\overline{s}$$ has an inverse $$\overline{s}' \in \mathcal{O}/\widetilde{I}_P$$ so that $$xs' \in \mathcal{O}$$ maps to $$\overline{x}/\overline{s}$$, showing surjection.