Quotient of ring of integers of a number field by powers of a prime ideal

Question

Let $K$ be a number field, $\mathcal O_K$ its ring of integers and suppose $\mathfrak p$ is a prime ideal of $\mathcal O_K$ with inertia degree 1 i.e. $\mathcal O_K/\mathfrak p \cong \mathbb Z/p\mathbb Z$. Is is true that for any $r \in \mathbb N_{\geq 1}$, $\mathcal O_K/\mathfrak p^r \cong \mathbb Z/p^r \mathbb Z$?

Answer 1

You also need that $\mathfrak{p}$ is unramified. For example, if $K = \mathbf{Q}(i)$ and $\mathfrak{p} = (1+i) \subset \mathbf{Z}[i] = \mathcal{O}_K$, then $(2) = \mathfrak{p}^2$ and $\mathcal{O}_K/\mathfrak{p}^2$ has characteristic $2$, not $4$.

But this is the only other condition that is required. Use the following steps:

1. There is an element $y \in \mathcal{O}_K$ such that $y \in \mathfrak{p}$ but not in $\mathfrak{p}^2$.
2. There is an isomorphism of groups $\mathbf{F}_p = \mathcal{O}/\mathfrak{p} \rightarrow \mathfrak{p}^{n-1}/\mathfrak{p}^n$ by sending $x \mapsto x y^{n-1}$.
3. The order of $\mathcal{O}/\mathfrak{p}^{n}$ is $p^n$.
4. (The one step that requires unramified): $p^{n-1}$ is non-zero in $\mathcal{O}_K/\mathfrak{p}^{n}$, because there is a factorization: $$p^{n-1} = \mathfrak{p}^{n-1} \times \cdots $$ and hence $p^{n-1} \notin \mathfrak{p}^n$.
5. Any ring of order $p^n$ were $p^{n-1} \ne 0$ is isomorphic to $\mathbf{Z}/p^n \mathbf{Z}$.