Let $\mathcal{O}$ be a one-dimensional noetherian integral domain (or more specifically, an order in an algebraic number field). Let $\mathfrak{a} \not= 0$ be an ideal of $\mathcal{O}$. Why does $\mathfrak{p} \supseteq \mathfrak{a}$ imply that $\mathfrak{p}$ is the only prime containing $\mathcal{O} \cap \mathfrak{a}\mathcal{O}_{\mathfrak{p}}$?

Then $\mathfrak{p}^{e} \supseteq \mathfrak{a}^{e}$. If $\mathfrak{q}$ is another prime containing $\mathfrak{a}^{ec}$ ($e$/$c$ are extension/contraction to/from $\mathcal{O}_{\mathfrak{p}}$), then $\mathfrak{q}^{e} \supseteq \mathfrak{a}^{ece} = \mathfrak{a}^{e}$. But I assume this latter fact comes from the fact that $\mathfrak{q} \not\subseteq \mathfrak{p}$, hence $\mathfrak{q}^{e} = \mathcal{O}_{\mathfrak{p}}$.

Note that this is from the proof of Proposition 12.3 in Chapter 1 of Neukirch's Algebraic Number Theory.


1 Answer 1


Let $\mathfrak a=\mathfrak q_1\cap\cdots\cap\mathfrak q_n$ be a reduced primary decomposition, and $\mathfrak p_i=\sqrt{\mathfrak q_i}$. Then $\mathfrak p_i$ are all maximal ideals, and one of them equals $\mathfrak p$, say $\mathfrak p_1$. Then $\mathfrak aR_{\mathfrak p}\cap R=\mathfrak q_1$, and we are done.

  • $\begingroup$ I see why now. Earlier in the proof, it is shown that $\mathfrak{a}$ is a finite intersection of "certain" ideals. They just aren't ever referred to as primary, but this is exactly what they are. $\endgroup$
    – Jacob Bond
    Commented Oct 28, 2015 at 17:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .