Proof of a stronger form of Chinese remainder theorem (12.3) in Neukirch Let $\mathcal O$ be an order in a number field and ${\mathfrak a} \neq 0$ an ideal in $\mathcal O$. Then the theorem shows ${\mathcal O}/{\mathfrak a} = \oplus_{{\mathfrak p} \supseteq {\mathfrak a}} {\mathcal O}_{\mathfrak p}/ {\mathfrak a}{\mathcal O}_{\mathfrak p}$. 
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.
 A: 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 $\mathscr{O}$ 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).
A: 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 $\mathfrak{a}\subseteq \mathfrak{p}$.
