Two meanings of "residue field" I've been conflating two senses of the notion of "residue field" and I realized I don't know why they are equal.
The first sense: If $K$ is a number field and $\mathfrak{p}$ is a prime of $K$ lying over the rational prime $p$, then the residue field is $\mathcal{O}_K/\mathfrak{p}$.
The second sense: If $(K, v)$ is a non-archimedean field with ring of integers $R = \{x \in K : |x|_v \le 1\}$ and maximal ideal $\mathfrak{m} = \{x \in K : |x|_v < 1\}$ then the residue field is $R/\mathfrak{m}$.
These two senses seem to coincide when, in the second definition, we take $K$ to be a number field and $v$ to be the valuation $\operatorname{ord}_{\mathfrak{p}}$ attached to the prime ideal $\mathfrak{p}$. Is it then true that $\mathcal{O}_K/\mathfrak{p} \cong R/\mathfrak{m}$?
I'm able to prove it for $K = \mathbb{Q}$, because here, it's easy to see that $R = \mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $(p)$, and $\mathfrak{m} = pR$, so that $$R/\mathfrak{m} = \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} \cong \mathbb{Z}/p\mathbb{Z} = \mathcal{O}_{\mathbb{Q}}/(p).$$
But the general case eludes me. Any insights or references would be appreciated, thanks!
 A: Looks like it could be helpful to give the full explanation here and help get this question closed up. The claim is if we take $v=\operatorname{ord}_\mathfrak{p}$ and $R=\{x\in K:|x|_v\le 1\}$ then $R=(\mathcal O_K)_\mathfrak{p}$, from which one deduces that $R/\mathfrak m=(\mathcal O_K)_\mathfrak{p}/\mathfrak p(\mathcal O_K)_\mathfrak{p}\simeq\mathcal O_K/\mathfrak{p}$. Let us recall one way of introducing the valuation $v_\mathfrak p$ before relating it to the definition you are working with:

If $D$ is a DVR with maximal ideal $\mathfrak m$, then $D$ is a Dedekind domain and $\mathfrak m$ is the unique nonzero prime ideal, so every fractional ideal has the form $\mathfrak m^n$ for some $n\in\mathbb{Z}$, and we have a valuation $v:\operatorname{Frac}(D)\to\mathbb{Z}\cup\{\infty\}$ by writing $v(x)=n$ when $xD=\mathfrak m^n$. One can see that $D=\{x\in \operatorname{Frac}(D)\mid v(x)\ge0\}$.

Now we relate this to your definition:

Lemma: If $x\in K^\times$ satisfies $x\mathcal O_K=\prod_\mathfrak{p}\mathfrak p^{n_\mathfrak{p}}$ for $n_\mathfrak{p}\in\mathbb Z$, then $n_\mathfrak p=v_\mathfrak{p}(x)$ where $v_\mathfrak{p}$ is the valuation coming from the fact that $(\mathcal O_K)_\mathfrak{p}$ is a DVR and $K=\operatorname{Frac}((\mathcal O_K)_\mathfrak{p})$.


Proof: Fix some prime $\mathfrak q$. Noticing that $\mathfrak p(\mathcal O_K)_\mathfrak{q}=(\mathcal O_K)_\mathfrak{q}$ for $\mathfrak p\neq\mathfrak q$, we have $$x(\mathcal O_K)_\mathfrak{q}=(\prod_\mathfrak{p}\mathfrak{p}^{n_\mathfrak{p}})(\mathcal O_K)_\mathfrak{q}=\mathfrak{q}^{n_\mathfrak{q}}(\mathcal O_K)_\mathfrak{q}=(\mathfrak{q}(\mathcal O_K)_\mathfrak{q})^{n_\mathfrak{q}}$$ which gives the result.

Combining the lemma with the prior paragraph one gets $R=\{x\in K\mid v_\mathfrak{p}(x)\ge0\}=(\mathcal O_K)_\mathfrak{p}$ as claimed.
