[proofplan] We construct the completion $F_\mathfrak{p}$ by completing $F$ at the $\mathfrak{p}$-adic absolute value and show it is a finite extension of $\mathbb{Q}_p$ of degree $e_\mathfrak{p} f_\mathfrak{p}$. The strategy has three parts: (1) embed $\mathbb{Q}_p$ into $F_\mathfrak{p}$ as the closure of $\mathbb{Q}$, (2) show $[F_\mathfrak{p} : \mathbb{Q}_p] = e_\mathfrak{p} f_\mathfrak{p}$ by proving that the natural map $F \otimes_\mathbb{Q} \mathbb{Q}_p \to F_\mathfrak{p}$ restricts to an isomorphism on the $\mathfrak{p}$-component, and (3) identify the residue field as $\mathbb{F}_{p^{f_\mathfrak{p}}}$. [/proofplan] [step:Define the $\mathfrak{p}$-adic absolute value on $F$ and identify $\mathbb{Q}_p \subset F_\mathfrak{p}$] The prime $\mathfrak{p}$ of $\mathcal{O}_F$ defines a discrete valuation $v_\mathfrak{p}: F^\times \to \mathbb{Z}$ by $v_\mathfrak{p}(x) = \max\{n \geq 0 : x \in \mathfrak{p}^n\}$ for $x \in \mathcal{O}_F \setminus \{0\}$, extended to $F^\times$ by $v_\mathfrak{p}(x/y) = v_\mathfrak{p}(x) - v_\mathfrak{p}(y)$. The associated absolute value is \begin{align*} |x|_\mathfrak{p} := p^{-v_\mathfrak{p}(x)/e_\mathfrak{p}} \quad \text{for } x \in F^\times, \qquad |0|_\mathfrak{p} := 0, \end{align*} where $e_\mathfrak{p} = v_\mathfrak{p}(p)$ is the ramification index (so that the restriction to $\mathbb{Q}$ satisfies $|p|_\mathfrak{p} = p^{-v_\mathfrak{p}(p)/e_\mathfrak{p}} = p^{-1} = |p|_p$). The restriction of $|\cdot|_\mathfrak{p}$ to $\mathbb{Q}$ equals $|\cdot|_p$: for $a/b \in \mathbb{Q}^\times$ with $a, b \in \mathbb{Z}$, $v_\mathfrak{p}(a/b) = v_\mathfrak{p}(a) - v_\mathfrak{p}(b) = e_\mathfrak{p}(v_p(a) - v_p(b))$, so $|a/b|_\mathfrak{p} = p^{-(v_p(a) - v_p(b))} = |a/b|_p$. The completion $F_\mathfrak{p}$ is a complete valued field containing $F$ as a dense subfield. The closure of $\mathbb{Q}$ inside $F_\mathfrak{p}$ is a complete field with the $p$-adic absolute value, hence isomorphic to $\mathbb{Q}_p$ by [Existence and Uniqueness of Completions](/theorems/???). We identify this closure with $\mathbb{Q}_p$, so $\mathbb{Q}_p \subset F_\mathfrak{p}$. [guided] The normalisation of the absolute value deserves attention. There are two natural choices: $|x|_\mathfrak{p} = c^{-v_\mathfrak{p}(x)}$ for some $c > 1$, or the normalisation that restricts to the standard $p$-adic absolute value on $\mathbb{Q}$. We choose the latter, which requires $|p|_\mathfrak{p} = p^{-1}$. Since $v_\mathfrak{p}(p) = e_\mathfrak{p}$, this forces $c^{-e_\mathfrak{p}} = p^{-1}$, giving $c = p^{1/e_\mathfrak{p}}$, hence $|x|_\mathfrak{p} = p^{-v_\mathfrak{p}(x)/e_\mathfrak{p}}$. With this normalisation, the restriction to $\mathbb{Q}$ is exactly $|\cdot|_p$, so the closure of $\mathbb{Q}$ in $F_\mathfrak{p}$ is $\mathbb{Q}_p$. Note that the value group $|F^\times|_\mathfrak{p} = \{p^{-k/e_\mathfrak{p}} : k \in \mathbb{Z}\}$ is a discrete subgroup of $\mathbb{R}_{>0}$, confirming that $F_\mathfrak{p}$ is a discretely valued field. [/guided] [/step] [step:Show $[F_\mathfrak{p} : \mathbb{Q}_p] = e_\mathfrak{p} f_\mathfrak{p}$ by constructing a basis] Since $F/\mathbb{Q}$ is a number field of degree $[F:\mathbb{Q}] = d$, we may write $F = \mathbb{Q}(\theta)$ for some $\theta \in \mathcal{O}_F$ (a primitive element). Let $g(x) = \operatorname{min}_\mathbb{Q}(\theta) \in \mathbb{Z}[x]$ be the minimal polynomial, which is monic of degree $d$. The polynomial $g$ factors over $\mathbb{Q}_p$: since $\mathbb{Q}_p$ is complete, [Hensel's Lemma](/theorems/???) and its generalisations allow us to lift factorisations from $\mathbb{F}_p[x]$. More precisely, the factorisation of $g$ over $\mathbb{Q}_p$ corresponds to the decomposition of primes above $p$ in $\mathcal{O}_F$. Write \begin{align*} g(x) = g_1(x) g_2(x) \cdots g_r(x) \quad \text{in } \mathbb{Q}_p[x], \end{align*} where each $g_i$ is irreducible over $\mathbb{Q}_p$ and corresponds to a prime $\mathfrak{p}_i$ of $\mathcal{O}_F$ above $p$. The factor $g_\mathfrak{p}$ corresponding to $\mathfrak{p}$ gives $F_\mathfrak{p} \cong \mathbb{Q}_p[x]/(g_\mathfrak{p})$ as topological fields: the natural map $F = \mathbb{Q}(\theta) \hookrightarrow F_\mathfrak{p}$ sends $\theta$ to a root of $g_\mathfrak{p}$ (since $g(\theta) = 0$ and $\theta$ has $\mathfrak{p}$-adic valuation determined by $g_\mathfrak{p}$), and by density of $F$ in $F_\mathfrak{p}$, the image of $\mathbb{Q}_p[\theta]$ is dense in $F_\mathfrak{p}$. Since $g_\mathfrak{p}$ is irreducible over $\mathbb{Q}_p$ and $\theta$ is a root, $[F_\mathfrak{p} : \mathbb{Q}_p] = \deg g_\mathfrak{p}$. The degree $\deg g_\mathfrak{p} = e_\mathfrak{p} f_\mathfrak{p}$: this follows from the fundamental identity for the local extension $F_\mathfrak{p}/\mathbb{Q}_p$, where the ramification index is $e_\mathfrak{p} = v_\mathfrak{p}(p)$ and the inertia degree is $f_\mathfrak{p} = [\mathcal{O}_{F_\mathfrak{p}}/\mathfrak{m}_{F_\mathfrak{p}} : \mathbb{F}_p]$. [guided] The key idea is that completing $F$ at $\mathfrak{p}$ is the same as taking the $\mathfrak{p}$-adic factor of the tensor product $F \otimes_\mathbb{Q} \mathbb{Q}_p$. More concretely: $F = \mathbb{Q}[x]/(g)$, so $F \otimes_\mathbb{Q} \mathbb{Q}_p \cong \mathbb{Q}_p[x]/(g) \cong \prod_i \mathbb{Q}_p[x]/(g_i)$, where the last isomorphism is the Chinese Remainder Theorem applied to the coprime factorisation $g = g_1 \cdots g_r$ over $\mathbb{Q}_p$. Each factor $\mathbb{Q}_p[x]/(g_i)$ is a finite extension of $\mathbb{Q}_p$ corresponding to a completion of $F$ at one of the primes above $p$. The factor corresponding to $\mathfrak{p}$ gives $F_\mathfrak{p} \cong \mathbb{Q}_p[x]/(g_\mathfrak{p})$, which is a field of degree $\deg g_\mathfrak{p}$ over $\mathbb{Q}_p$. The fundamental identity $[F_\mathfrak{p} : \mathbb{Q}_p] = e_\mathfrak{p} f_\mathfrak{p}$ relates this degree to the ramification index and inertia degree. In the local setting, $e_\mathfrak{p}$ measures how much $p$ ramifies (i.e., $v_\mathfrak{p}(p) = e_\mathfrak{p}$) and $f_\mathfrak{p}$ measures the residue field growth (i.e., $[\mathcal{O}_{F_\mathfrak{p}}/\mathfrak{m}_{F_\mathfrak{p}} : \mathbb{F}_p] = f_\mathfrak{p}$). [/guided] [/step] [step:Identify the residue field of $F_\mathfrak{p}$] The valuation ring of $F_\mathfrak{p}$ is $\mathcal{O}_{F_\mathfrak{p}} = \{x \in F_\mathfrak{p} : |x|_\mathfrak{p} \leq 1\}$, with maximal ideal $\mathfrak{m}_{F_\mathfrak{p}} = \{x \in F_\mathfrak{p} : |x|_\mathfrak{p} < 1\}$. The residue field $k_{F_\mathfrak{p}} := \mathcal{O}_{F_\mathfrak{p}} / \mathfrak{m}_{F_\mathfrak{p}}$ is determined as follows. Since $F$ is dense in $F_\mathfrak{p}$, the image of $\mathcal{O}_F \cap F$ under the residue map is dense in $k_{F_\mathfrak{p}}$ (in the discrete topology on $k_{F_\mathfrak{p}}$, this means surjective). The localisation $(\mathcal{O}_F)_\mathfrak{p}$ maps surjectively onto $k_{F_\mathfrak{p}}$ with kernel $\mathfrak{p} (\mathcal{O}_F)_\mathfrak{p}$, giving \begin{align*} k_{F_\mathfrak{p}} \cong (\mathcal{O}_F)_\mathfrak{p} / \mathfrak{p} (\mathcal{O}_F)_\mathfrak{p} \cong \mathcal{O}_F / \mathfrak{p}. \end{align*} By definition of the inertia degree, $[\mathcal{O}_F / \mathfrak{p} : \mathbb{F}_p] = f_\mathfrak{p}$, so $\mathcal{O}_F / \mathfrak{p} \cong \mathbb{F}_{p^{f_\mathfrak{p}}}$ (a finite field of order $p^{f_\mathfrak{p}}$, since every finite extension of $\mathbb{F}_p$ is of this form). Therefore \begin{align*} k_{F_\mathfrak{p}} \cong \mathbb{F}_{p^{f_\mathfrak{p}}}. \end{align*} [/step] [step:Conclude that $F_\mathfrak{p}$ is a local field] We have shown: - $F_\mathfrak{p}$ is a finite extension of $\mathbb{Q}_p$ of degree $[F_\mathfrak{p} : \mathbb{Q}_p] = e_\mathfrak{p} f_\mathfrak{p}$. - $F_\mathfrak{p}$ is complete (as a completion). - The residue field of $F_\mathfrak{p}$ is $\mathbb{F}_{p^{f_\mathfrak{p}}}$, which is finite. Since $F_\mathfrak{p}$ is a complete discretely valued field with finite residue field, it is a local field. By the [Classification of Local Fields](/theorems/???), it falls into case (1): a finite extension of $\mathbb{Q}_p$ (since $\operatorname{char} F_\mathfrak{p} = 0$). [/step]