Ideal in the ring of integers with prescribed quotient Let $K/\Bbb Q$ be a finite extension and $N>1$ an integer. What are some conditions on $N$ such that we can find an ideal $I \subset \mathcal O_K$ such that $\mathcal O_K / I \cong \Bbb Z / N \Bbb Z$ ? We can assume that $K / \Bbb Q$ is Galois if it makes the answer easier.
It is sufficient to assume that $N = p^r$ for some prime number $p$. Let $P$ a prime of $K$ above $p$. If its inertia index $f(P/p)$ is $1$, then $\mathcal O_K / P^r$ has cardinality $p^r$, but I don't see why the natural map
$$\Bbb Z / p^r \Bbb Z \to \mathcal O_K / P^r$$
should be injective (hence an isomorphism).
Thank you!
 A: I can provide a general proof. 

Let $L/K$ be a finite extension of number fields, and $J \subset O_K$ be an ideal. Assume that for any prime $\mathfrak p$ dividing $J$ in $O_K$, there is a prime $P \subset O_L$ above $\mathfrak p$ such that $e(P/\mathfrak p) = f(P/\mathfrak p)=1$ (e.g. $\mathfrak p$ is totally split in $L$).
Then there is an ideal $I \subset O_L$ such that $O_L/I  \cong  O_K/J$ as rings (via the natural map $O_K \to O_L/I$).

Proof : fix a prime $P = P(\mathfrak p) \subset O_L$ above each prime divisor $\mathfrak p \mid J$, such that $e(P/\mathfrak p) = f(P/\mathfrak p)=1$, as in the hypothesis.
Let $I$ be the product ideal 
$$I := \prod_{\mathfrak p \mid J} P(\mathfrak p)^{v_{\mathfrak p}(J)} \subset O_L.$$
We have 
$$|O_L / I| = \prod_{\mathfrak p \mid J} |O_L / P(\mathfrak p)|^{v_{\mathfrak p}(J)}
= \prod_{\mathfrak p \mid J} |O_K / \mathfrak p|^{v_{\mathfrak p}(J) f(P(\mathfrak p)/\mathfrak p)} = 
\prod_{\mathfrak p \mid J} |O_K / \mathfrak p|^{v_{\mathfrak p}(J)}
= |O_K/J|
$$
Therefore, if I show that the natural morphism $\phi : O_K/J \to O_L/I$ is injective, it will be an isomorphism.
The kernel of $O_K \to O_L/I$ is
$$O_K \cap I = \{x \in O_K  : \forall \mathfrak p \mid J, 
v_{P(\mathfrak p)}(x) \geq v_{\mathfrak p}(J)\}$$
and
$v_{P(\mathfrak p)}\vert_{K} = e(P(\mathfrak p) / \mathfrak p) v_{\mathfrak p} = v_{\mathfrak p}$. Therefore $O_K \cap I = J$, which means that $\phi$ is injective.
