$J(\mathcal{O})\cong\bigoplus_{\mathfrak{p}} P(\mathcal{O}_\mathfrak{p})$ for one-dimensional Noetherian domains (from Neukirch) I'm having trouble in understanding Neukirch's proof of the proposition in the title ((12.6), p. 75 of his Algebraic Number Theory book). He uses the following CRT-like proposition:

If $\mathfrak{a}\neq 0$ is an ideal of $\mathcal{O}$, then $\mathcal{O}/\mathfrak{a}\cong
\bigoplus_\mathfrak{p} \mathcal{O}_\mathfrak{p}/\mathfrak{a}\mathcal{O}_\mathfrak{p}$.

To prove the surjectivity of the map $\mathfrak{a}\mapsto (\mathfrak{a}_\mathfrak{p})$ he sets $\mathfrak{a}=\bigcap_\mathfrak{p} a_\mathfrak{p}\mathcal{O}_\mathfrak{p}$ for a given choice of the $a_\mathfrak{p}$'s in $K^*$ (almost all equal to 1) and proves that $\mathfrak{a}$ is a fractional ideal.
Then, in order to prove that $\mathfrak{a}\mathcal{O}_\mathfrak{p}=a_\mathfrak{p}\mathcal{O}_\mathfrak{p}$, he fixes a prime ideal $\mathfrak{p}$ and chooses $c\in\mathcal{O}\setminus\{0\}$ such that $ca_\mathfrak p^{-1}a_\mathfrak q\in\mathcal{O}_\mathfrak{q}$ for all $\mathfrak{q}$. He applies the stated proposition to find an $a\equiv c\pmod{\mathfrak{p}}$ such that $a\in ca_\mathfrak p^{-1}a_\mathfrak q\mathcal{O}_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$.
(1) To which ideal is he applying the proposition?
(2) He then deduces that $ac^{-1}$ is a unit in $\mathcal{O}_\mathfrak{p}$ but I can't see why ($a\equiv c\pmod{\mathfrak{p}}$ is not sufficient).
PS: In the proof of the CRT-like proposition the author says that the fact that $\mathfrak{p}$ is the only prime ideal containing $\mathcal{O}\cap\mathfrak{a}\mathcal{O}_\mathfrak{p}$ follows directly from the standard correspondence for prime ideals in a localization, but I was able to deduce it only by appealing to the primary decomposition. Am I missing something?
 A: For your question (1), if I'm understanding Neukirch's proof correctly, he is applying his CRT to the ideal $c a_\mathfrak{p}^{-1} \mathfrak{a}$, which I believe is an integral ideal of $\mathcal{O}$.
Thus, we can choose $a \in \mathcal{O}$ such that
$$a \equiv c \pmod{ca_\mathfrak{p}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{p}}$$
and
$$a \equiv 0 \pmod{ca_\mathfrak{p}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{q}}$$
for $\mathfrak{q} \neq \mathfrak{p}$.
Now consider 2 cases: either $a_\mathfrak{p}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{p} = \mathcal{O}_\mathfrak{p}$ or $a_\mathfrak{p}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{p} \subseteq \mathfrak{p}\mathcal{O}_\mathfrak{p}$.  In the former case, we get that $\mathfrak{a}\mathcal{O}_\mathfrak{p} = a_\mathfrak{p}\mathcal{O}_\mathfrak{p}$, and we are done.
In the latter case, we have $a \equiv c\pmod{\mathfrak{p}\mathcal{O}_\mathfrak{p}}$, but since $a$ and $c$ are in $\mathcal{O}$, we have that $a \equiv c\pmod{\mathfrak{p}}$, as claimed by Neukirch.
I think this answers your question (1).
I'm not sure about your comment (2). Perhaps you can elaborate on why you think it is sufficient just to know that $a \equiv c\pmod{\mathfrak{p}}$.  I'd be curious to know.
Later addition
With respect to the OP's PS comment about the "strong CRT" theorem Prop 12.3 on pg. 74:
As you say, Neukirch claimed that if $\mathfrak{p} \supseteq \mathfrak{a}$ then $\mathfrak{p}$ is the only prime that contains $\mathcal{O} \cap \mathfrak{a}\mathcal{O}_\mathfrak{p}$, and justifies this by appealing to the usual local prime correspondence theorem (Prop 11.1 on pg 65).  I also was not able to make this argument work, and, just as the OP did, had to appeal to more general facts about primary ideals from commutative algebra.
A: I also encountered this problem just before and tried to deduce the theorem from the answer by @JohnM. I still feel that there might be some gaps in the end. Now, $a\in ca_{\mathfrak{p}}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$. Then
$$ac^{-1}a_{\mathfrak{p}}\in\bigcap_{\mathfrak{q}\neq\mathfrak{p}}\mathfrak{a}\mathcal{O}_\mathfrak{q}.$$
Then $$a_{\mathfrak{p}}\mathcal{O}_\mathcal{\mathfrak{p}}\subseteq\left(\bigcap_{\mathfrak{q}\neq\mathfrak{p}}\mathfrak{a}\mathcal{O}_\mathfrak{q}\right)\mathcal{O}_\mathfrak{p}.$$
We know that $\bigcap_{\text{all } \mathfrak{q}}\mathcal{O}_\mathfrak{q}=\mathcal{O}$. However, the intersection is missing the prime $\mathfrak{p}$. So eventurally the ring should be larger. I got stuck from here. In fact, in the second case, $a_\mathfrak{p}^{-1}\mathfrak{a}\mathcal{O}_\mathfrak{p} \subseteq \mathfrak{p}\mathcal{O}_\mathfrak{p}\subsetneq\mathcal{O}_\mathfrak{p}$. One should get a contradiction.
I don't think Neukirch meant this. He meant to find an element $a$ such that $ac^{-1}\in\mathcal{O}_\mathfrak{p}$ is a unit and $a\in ca_{\mathfrak{p}}^{-1}a_{\mathfrak{q}}\mathcal{O}_\mathfrak{q}$ for all $\mathfrak{p}\neq \mathfrak{q}$. In this case, $ac^{-1}a_{\mathfrak{p}}\in a_{\mathfrak{q}}\mathcal{O}_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$, and apparently, $ac^{-1}a_{\mathfrak{p}}\in a_{\mathfrak{p}}\mathcal{O}_\mathfrak{p}$. Then we can take the intersection to conclude.
This is what I did. We are proving the statement for a fixed prime $\mathfrak{p}$. Say $a_{\mathfrak{p}}=r/s$. As $(a_\mathfrak{q}\mathcal{O}_\mathfrak{q})_\mathfrak{q}$ in the direct sum. So we can simply assume almost all $a_\mathfrak{q}=1$. We only get $a_{\mathfrak{q}_1}=r_1/s_1,...,a_{\mathfrak{q}_m}=r_m/s_m$ are not $1$ and all these $\mathfrak{q}_i$ are distinct from $\mathfrak{p}$. Now, take $c:=rs_1...s_m$. Let's first try to multiply by $c$. Let the ideal $\mathfrak{b}:=(ca_{\mathfrak{p}}^{-1}a_{\mathfrak{q}_1}...a_{\mathfrak{q}_m})\subseteq\mathcal{O}$. Now apply the Chinese remainder theorem to $\mathcal{O}/\mathfrak{b}\to\oplus\mathcal{O}_{\mathfrak{q}}/\mathfrak{b}\mathcal{O}_{\mathfrak{q}}$. What we want is
$$a\equiv 1 \mod \mathfrak{b}\mathcal{O}_{\mathfrak{p}}$$
and for all $\mathfrak{q}\neq\mathfrak{p}$
$$a\equiv 0 \mod \mathfrak{b}\mathcal{O}_{\mathfrak{q}}.$$
Now, $a\in ca_{\mathfrak{p}}^{-1}a_{\mathfrak{q}_1}...a_{\mathfrak{q}_m}\mathcal{O}_{\mathfrak{q}}\subseteq a_{\mathfrak{p}}^{-1}a_{\mathfrak{q}}\mathcal{O}_{\mathfrak{q}}$ for all $\mathfrak{q}\neq\mathfrak{p}$ because by definition, $c$ times any amount of these $a_{\mathfrak{q_i}}$'s is always integral. And for $\mathfrak{p}$, $a\in\mathcal{O}_\mathfrak{p}$ is a unit if $a\equiv1$ modulo the maximal ideal $\mathfrak{p}\mathcal{O}_\mathfrak{p}$. So we would actually like to have $\mathfrak{b}\subseteq\mathfrak{p}$. Our first attempt might not be the case. However, this can be easily resolved by multiplying by a nonzerozero element from $\mathfrak{p}$. Then we get $aa_{\mathfrak{p}}$ satisfying the desired property.
