[proofplan] A local field $K$ is, by definition, a non-discrete, locally compact, totally disconnected topological field. We treat the two cases according to $\operatorname{char} K$. In characteristic 0, we show $K$ contains $\mathbb{Q}_p$ as a closed subfield for some prime $p$, and then $K/\mathbb{Q}_p$ is a finite extension because local compactness forces finite-dimensionality. In characteristic $p > 0$, the residue field $k$ is a finite field $\mathbb{F}_q$ with $q = p^f$, and we construct an isomorphism $K \cong \mathbb{F}_q((T))$ by choosing a uniformiser and using the coefficient representatives from $k$. [/proofplan] [step:Establish that $K$ carries a non-trivial discrete valuation with finite residue field] Since $K$ is locally compact and totally disconnected, the valuation ring $\mathcal{O}_K = \{x \in K : |x| \leq 1\}$ is a compact open subring. The maximal ideal $\mathfrak{m}_K = \{x \in K : |x| < 1\}$ is the unique maximal ideal of $\mathcal{O}_K$. The residue field $k_K := \mathcal{O}_K / \mathfrak{m}_K$ is finite: the quotient map $\mathcal{O}_K \to k_K$ is continuous and surjective, $\mathcal{O}_K$ is compact, and $k_K$ carries the discrete topology (since $\mathfrak{m}_K$ is open in $\mathcal{O}_K$). A compact discrete space is finite, so $|k_K| = q < \infty$ for some prime power $q = p^f$. The absolute value on $K$ is discrete: the image $|K^\times|$ is a discrete subgroup of $\mathbb{R}_{>0}$. To see this, note that $\mathcal{O}_K^\times = \{x : |x| = 1\}$ is compact and open in $K^\times$, so the quotient $K^\times / \mathcal{O}_K^\times$ is discrete. The map $K^\times / \mathcal{O}_K^\times \to \mathbb{R}_{>0}$ given by $x\mathcal{O}_K^\times \mapsto |x|$ is injective, so $|K^\times|$ is a discrete subgroup of $\mathbb{R}_{>0}$, hence cyclic. Choose a uniformiser $\pi \in \mathfrak{m}_K$ with $|\pi| = \max\{|x| : x \in \mathfrak{m}_K\}$, i.e., a generator of the maximal ideal. Then $|K^\times| = \{|\pi|^n : n \in \mathbb{Z}\}$. [guided] The starting point is to extract the valuation-theoretic structure from the topological hypotheses. Local compactness and total disconnectedness imply that $K$ is a non-archimedean valued field (the topology is induced by a non-archimedean absolute value — this follows from the classification of locally compact fields, but we take it as part of the definition of a local field). The valuation ring $\mathcal{O}_K$ is the closed unit ball, which is compact (a closed subset of a locally compact space is compact when it is also open, which $\mathcal{O}_K$ is). The residue field $k_K = \mathcal{O}_K / \mathfrak{m}_K$ inherits the quotient topology, which is discrete since $\mathfrak{m}_K$ is open. Since $k_K$ is the continuous image of the compact set $\mathcal{O}_K$ under the quotient map, it is compact. A compact discrete space is finite. The discreteness of $|K^\times|$ also follows from compactness: $\mathcal{O}_K^\times$ is an open subgroup of $K^\times$, and the quotient group $K^\times/\mathcal{O}_K^\times \cong \mathbb{Z}$ (generated by a uniformiser $\pi$) parametrises the value group. [/guided] [/step] [step:Case 1 -- Characteristic 0: show $K$ is a finite extension of $\mathbb{Q}_p$] Assume $\operatorname{char} K = 0$. The prime subfield of $K$ is $\mathbb{Q}$. The restriction of $|\cdot|$ to $\mathbb{Q}$ is a non-trivial absolute value (non-trivial because $K$ is non-discrete). Since $k_K$ has characteristic $p$ (as a finite field of order $q = p^f$) and the natural map $\mathbb{Z} \to \mathcal{O}_K \to k_K$ has kernel $p\mathbb{Z}$ (the integers mapping to $0$ in $k_K$ are exactly those divisible by $p$), we have $|p| < 1$ in $K$. Moreover, for any prime $\ell \neq p$, the image of $\ell$ in $k_K$ is nonzero (since $k_K$ has characteristic $p$), so $|\ell| = 1$. By [Ostrowski's Theorem](/theorems/???), the restriction of $|\cdot|$ to $\mathbb{Q}$ is equivalent to $|\cdot|_p$. The closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ inside $K$ (with respect to the topology induced by $|\cdot|$) is a complete valued field with absolute value equivalent to $|\cdot|_p$. By the uniqueness of completions ([Existence and Uniqueness of Completions](/theorems/???)), $\overline{\mathbb{Q}} \cong \mathbb{Q}_p$ as topological fields. Identify $\overline{\mathbb{Q}}$ with $\mathbb{Q}_p$, so $\mathbb{Q}_p \subset K$. Since $K$ is locally compact and $\mathbb{Q}_p \subset K$ is a closed subfield, $K$ is a topological vector space over $\mathbb{Q}_p$. If $[K : \mathbb{Q}_p]$ were infinite, then $K$ would contain subspaces of arbitrarily large finite dimension, and the unit ball $\mathcal{O}_K$ could not be compact (since the closed unit ball in an infinite-dimensional normed space over a locally compact field is not compact). Therefore $[K : \mathbb{Q}_p] = n < \infty$, and $K$ is a finite extension of $\mathbb{Q}_p$. [guided] The argument proceeds in three stages: identify the restriction to $\mathbb{Q}$, recognise the closure of $\mathbb{Q}$ as $\mathbb{Q}_p$, and use local compactness to force finite degree. **Stage 1.** The residue characteristic $p$ pins down the restriction: $|p| < 1$ (since $p$ maps to $0$ in $k_K$), and $|\ell| = 1$ for primes $\ell \neq p$ (since $\ell$ maps to a nonzero element of $k_K$, hence $\ell \in \mathcal{O}_K^\times$). By Ostrowski's Theorem, the restriction is equivalent to $|\cdot|_p$. **Stage 2.** The closure of $\mathbb{Q}$ in $K$ is a complete field with a $p$-adic absolute value, hence isomorphic to $\mathbb{Q}_p$ by uniqueness of completions. **Stage 3.** Local compactness forces $[K:\mathbb{Q}_p] < \infty$. The key fact is that a locally compact topological vector space over a non-discrete locally compact field is finite-dimensional. This can be seen from the compactness of $\mathcal{O}_K$: if $[K:\mathbb{Q}_p]$ were infinite, we could find elements $e_1, e_2, \ldots \in \mathcal{O}_K$ whose images in $\mathcal{O}_K / \pi\mathcal{O}_K$ are $\mathbb{F}_p$-linearly independent, contradicting finiteness of the residue field (which has only $q = p^f$ elements). [/guided] [/step] [step:Case 2 -- Characteristic $p > 0$: construct an isomorphism $K \cong \mathbb{F}_q((T))$] Assume $\operatorname{char} K = p > 0$. The residue field $k_K \cong \mathbb{F}_q$ with $q = p^f$. **Lift the residue field.** Since $k_K = \mathbb{F}_q$ is a finite field of characteristic $p$ and $\operatorname{char} K = p$, the Frobenius endomorphism $x \mapsto x^q$ acts on $\mathcal{O}_K$. The roots of $x^q - x = 0$ in $\mathcal{O}_K$ form a subfield $S \subset \mathcal{O}_K$ isomorphic to $\mathbb{F}_q$: each element of $k_K = \mathbb{F}_q$ satisfies $\bar{x}^q = \bar{x}$, so by [Hensel's Lemma](/theorems/???) (applied as the Simple Root Lifting corollary to $x^q - x$, whose roots in $k_K$ are all simple since the derivative is $qx^{q-1} - 1 = -1 \neq 0$ in characteristic $p$), each element of $k_K$ lifts uniquely to a root of $x^q - x$ in $\mathcal{O}_K$. These lifts form a field $S \cong \mathbb{F}_q$. **Choose a uniformiser.** Let $\pi \in K$ be a uniformiser, i.e., $|\pi|$ generates the value group $|K^\times|$ and $\mathfrak{m}_K = \pi \mathcal{O}_K$. **Construct the isomorphism.** Every element $a \in K^\times$ can be written uniquely as $a = \sum_{i=n}^{\infty} a_i \pi^i$ with $a_i \in S$, $a_n \neq 0$, and $n = v(a) \in \mathbb{Z}$. This follows from iterative approximation: given $a$ with $v(a) = n$, there is a unique $a_n \in S \setminus \{0\}$ with $|a - a_n \pi^n| < |\pi^n|$ (since $a \pi^{-n} \in \mathcal{O}_K^\times$ and its residue class determines $a_n$). Then $v(a - a_n \pi^n) \geq n + 1$, and we repeat. The resulting series converges since $|a_i \pi^i| = |\pi|^i \to 0$. The map \begin{align*} \varphi: K &\to \mathbb{F}_q((T)) \\ \sum_{i=n}^{\infty} a_i \pi^i &\mapsto \sum_{i=n}^{\infty} \bar{a}_i T^i, \end{align*} where $\bar{a}_i$ is the image of $a_i \in S$ under the identification $S \cong \mathbb{F}_q$, is a field isomorphism. It is a homeomorphism when $\mathbb{F}_q((T))$ carries the $T$-adic topology, since both sides have the same valuation structure. [/step] [step:Conclude the classification] Combining both cases: - If $\operatorname{char} K = 0$, then $K$ is a finite extension of $\mathbb{Q}_p$ for the unique prime $p$ determined by the residue characteristic. This is a characteristic 0 local field. - If $\operatorname{char} K = p > 0$, then $K \cong \mathbb{F}_q((T))$ where $q = |k_K| = p^f$. This is a characteristic $p$ local field. Every local field falls into exactly one of these two classes, and the isomorphism is one of topological fields. [/step]