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Let $\mathcal O$ be an order in a number field and ${\mathfrak a} \neq 0$ an ideal in $\mathcal O$. Then the theorem shows $${\mathcal O}/{\mathfrak a} = \bigoplus_{{\mathfrak p} \supseteq {\mathfrak a}} {\mathcal O}_{\mathfrak p}/ {\mathfrak a}{\mathcal O}_{\mathfrak p}.$$ In the proof we set ${{\tilde{\mathfrak a}}_{\mathfrak p}} = {\mathcal O} \cap {\mathfrak a}{\mathcal O}_{\mathfrak p}$ and it is claimed that using the bijective correspondence between primes and primes after localizing it follows : if ${\mathfrak p} \supseteq {\mathfrak a}$ then $\mathfrak p$ is the only prime ideal containing ${{\tilde{\mathfrak a}}_{\mathfrak p}}$.

The correspondence works only among the primes and I don't see why it is directly applicable to ${{\tilde{\mathfrak a}}_{\mathfrak p}}$.

My approach was : if ${\mathfrak p} \neq {\mathfrak q} \supseteq {\mathfrak a}$ with ${\mathfrak q} \supseteq {{\tilde{\mathfrak a}}_{\mathfrak p}}$ then we can get a contradiction if ${{\tilde{\mathfrak a}}_{\mathfrak p}} {\mathcal O}_{\mathfrak q}$ contain a unit of ${\mathcal O}_{\mathfrak q}$. However, ${\mathfrak a} \subseteq {\mathfrak p} \cap {\mathfrak q}$ and I don't see how a unit can be constructed.

Another way could be to take the multiplicative set $S = {\mathcal O} \setminus ({\mathfrak p} \cup {\mathfrak q})$ and relate ${\mathcal O}_{\mathfrak q}$ to ${\mathcal O}_S$. Here again the same problem occurs.

RobPratt
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Siddhartha
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  • You are correct the 1-1 correspondence does not apply. Instead, you need to use general facts about primary ideals in commutative rings. This is an old question: https://math.stackexchange.com/questions/471378/j-mathcalo-cong-bigoplus-mathfrakp-p-mathcalo-mathfrakp-for-on/471613#471613 – John M Jul 25 '22 at 02:46

2 Answers2

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I can't see how the above answer shows that $\mathfrak{q} \not\supseteq\tilde{\mathfrak{a}}_\mathfrak{p}$ for $\mathfrak{q}$ a prime not equal to $\mathfrak{p}$, which is the assertion in question.

Here's another solution. The ring $\mathcal{O}_\mathfrak{p}$ has only two prime ideals, $(0)$ and $\mathfrak{p}\mathcal{O}_\mathfrak{p}$, and since $\mathfrak{a}\neq (0)$ the ring $\mathcal{O}_\mathfrak{p}/\mathfrak{a}\mathcal{O}_\mathfrak{p}$ has only one prime, namely $\mathfrak{p}\mathcal{O}_\mathfrak{p}$. Hence any element of $\mathfrak{p}\mathcal{O}_\mathfrak{p}$ is nilpotent in $\mathcal{O}_\mathfrak{p}/\mathfrak{a}\mathcal{O}_\mathfrak{p}$.

Since $\mathfrak{p}$ is maximal in $\mathcal{O}$ there exists $d\in \mathfrak{p}\setminus\mathfrak{q}$: the image of $d$ in $\mathcal{O}_\mathfrak{p}/\mathfrak{a}\mathcal{O}_\mathfrak{p}$ is then nilpotent, giving $$ \frac{d^n}{1} = \frac{a}{s} $$ for some $a\in \mathfrak{a}$ and $s\not\in\mathfrak{p}$, and $n>0$, and $d^n \in \tilde{\mathfrak{a}}_\mathfrak{p}\setminus \mathfrak{q}$ (since $\mathfrak{q}$ is prime).

Iohll
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  • Here we are not using Noetherian property of $\mathcal{o}$ , but using the fact it has dim 1 to show $\frac{\mathcal{O}{\mathcal{p}}}{\mathcal{a} \mathcal{O}{\mathcal{p}}}$ has only one prime ideal. Am I right? – NumDio Aug 12 '23 at 10:51
  • It has been a long time since I thought about algebraic number theory, and my commutative algebra is pretty rusty, but that was my idea here! – Iohll Aug 12 '23 at 18:45
  • @Sushant In a local zero-dimensional ring every element of the maximal ideal is nilpotent. In fact, the nilradical equals the maximal ideal, so the noetherianity is not necessary. – user26857 Aug 12 '23 at 20:06
  • Agree. Actually my doubt was where do we need Noetherian property in proving the strong Chinese remainder because an order is is a Noetherian and has krull dim 1. I realised later that it will give us finiteness of no. of prime ideals that lie above $\mathcal{a}$. Thanks. – NumDio Aug 13 '23 at 14:45
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An order has Krull dimension 1, and in particular non-zero prime ideals are maximal. It follows that $\mathcal{O}_{\mathfrak{p}}$ has exactly two prime ideals, the maximal ideal $\bar{\mathfrak{p}}$ corresponding to $\mathfrak{p}$ and $(0)$. Since $\mathfrak{a}\neq 0$, we have $(0)\not\supseteq \bar{\mathfrak{a}}$. Therefore the only possibility is that $\bar{\mathfrak{a}}\subseteq \bar{\mathfrak{p}}$. And this inclusion holds because we had $\mathfrak{a}\subseteq \mathfrak{p}$.

Pedro
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    I know this is an old question, but I'm still confused. Let $\mathfrak{q}\ne 0$ be a prime ideal that contains $\tilde{\mathfrak{a}}{\mathfrak{p}}$. I agree that we must have $\mathfrak{q}=\mathfrak{p}$ if $\mathfrak{q}$ does not vanish in $\mathcal{O}{\mathfrak{p}}$. But is it not possible that $\mathfrak{q}\not\subseteq\mathfrak{p}$? In this case $\mathfrak{q}\mathcal{O}{\mathfrak{p}}=\mathcal{O}{\mathfrak{p}}$. – Lorenzo Andreaus Sep 03 '21 at 02:51
  • @LorenzoAndreaus - See discussion here: https://math.stackexchange.com/questions/471378/j-mathcalo-cong-bigoplus-mathfrakp-p-mathcalo-mathfrakp-for-on/471613#471613 - or the new answer given by lohll – John M Jul 25 '22 at 02:46