[proofplan] We prove uniqueness, then existence, then completeness. Uniqueness follows from the equivalence of norms on finite-dimensional spaces over a complete field: any two extending absolute values are $K$-norms on $L$, hence equivalent, and since both restrict to $|\cdot|$ on $K$, they must be equal. For existence, we verify that $|\alpha|_L := |N_{L/K}(\alpha)|^{1/n}$ defines a non-archimedean absolute value. Non-degeneracy and multiplicativity follow from properties of the norm map. The strong triangle inequality is established via the [Criterion for the Strong Triangle Inequality](/theorems/???): we show $|\alpha|_L \leq 1$ implies $|\alpha + 1|_L \leq 1$ by identifying $\{\alpha \in L : |\alpha|_L \leq 1\}$ with the integral closure of $\mathcal{O}_K$ in $L$, which is a ring (hence closed under addition). Completeness follows from the equivalence of norms theorem. [/proofplan]

[step:Prove uniqueness of the extending absolute value]Suppose $|\cdot|_L$ and $|\cdot|_L'$ are two absolute values on $L$ that both restrict to $|\cdot|$ on $K$. View $L$ as an $n$-dimensional $K$-vector space. Both $|\cdot|_L$ and $|\cdot|_L'$ are norms on $L$ over $K$: each satisfies $|\alpha \beta|_L = |\alpha|_L |\beta|_L$, hence in particular $|c \alpha|_L = |c| \cdot |\alpha|_L$ for $c \in K$ (using $|c|_L = |c|$), and similarly for $|\cdot|_L'$. Since $K$ is a complete non-archimedean valued field, the theorem on [equivalence of norms on finite-dimensional spaces](/theorems/???) applies: any two norms on the finite-dimensional $K$-vector space $L$ are equivalent. In particular, there exist constants $A, B > 0$ with \begin{align*} A |\alpha|_L \leq |\alpha|_L' \leq B |\alpha|_L \quad \text{for all } \alpha \in L. \end{align*} Since both $|\cdot|_L$ and $|\cdot|_L'$ are absolute values (not just norms), the equivalence can be upgraded to equality. Choose any $a \in K$ with $|a| \neq 0, 1$ (which exists since $|\cdot|$ is non-trivial). For any $\alpha \in L \setminus \{0\}$ and any $m \in \mathbb{Z}$: \begin{align*} |\alpha|_L' = |a^m \alpha|_L' / |a|^m \leq B |a^m \alpha|_L / |a|^m = B |\alpha|_L, \end{align*} and similarly $|\alpha|_L' \geq A |\alpha|_L$. Applying the same bound to $\alpha^k$ for $k \geq 1$: \begin{align*} |\alpha|_L'^k = |\alpha^k|_L' \leq B |\alpha^k|_L = B |\alpha|_L^k. \end{align*} Taking $k$-th roots: $|\alpha|_L' \leq B^{1/k} |\alpha|_L$. Letting $k \to \infty$ gives $|\alpha|_L' \leq |\alpha|_L$. The reverse inequality follows by symmetry. Therefore $|\alpha|_L' = |\alpha|_L$ for all $\alpha \in L$.[/step]

[guided]The uniqueness argument has two parts. First, we use the theorem that all norms on a finite-dimensional vector space over a complete non-archimedean field are equivalent. This gives $A |\alpha|_L \leq |\alpha|_L' \leq B |\alpha|_L$. But equivalence of norms is weaker than equality: it allows a constant factor. The upgrade from equivalence to equality uses the multiplicative structure of absolute values (which norms do not have in general). The key trick: if $|\alpha|_L' \leq B |\alpha|_L$, then replacing $\alpha$ by $\alpha^k$ gives $|\alpha|_L'^k \leq B |\alpha|_L^k$ (by multiplicativity of both absolute values), hence $|\alpha|_L' \leq B^{1/k} |\alpha|_L$. Since $B^{1/k} \to 1$ as $k \to \infty$, we get $|\alpha|_L' \leq |\alpha|_L$. The reverse is symmetric. This "$k$-th root trick" works because absolute values are multiplicative: $|x^k| = |x|^k$. For general norms (which only satisfy $\|cx\| = |c|\|x\|$ for scalars $c \in K$, not for arbitrary multiplication in $L$), this argument fails, and two equivalent norms need not be equal.[/guided]

[step:Verify non-degeneracy and multiplicativity of $|\alpha|_L = |N_{L/K}(\alpha)|^{1/n}$] Define $|\cdot|_L: L \to \mathbb{R}_{\geq 0}$ by $|\alpha|_L := |N_{L/K}(\alpha)|^{1/n}$. **Non-degeneracy.** $|\alpha|_L = 0$ iff $|N_{L/K}(\alpha)| = 0$ iff $N_{L/K}(\alpha) = 0$ (by non-degeneracy of $|\cdot|$ on $K$) iff $\alpha = 0$ (since the field norm $N_{L/K}(\alpha) = 0$ iff $\alpha$ is not invertible in $L$, which for a field means $\alpha = 0$). **Multiplicativity.** For $\alpha, \beta \in L$: \begin{align*} |\alpha \beta|_L = |N_{L/K}(\alpha\beta)|^{1/n} = |N_{L/K}(\alpha) \cdot N_{L/K}(\beta)|^{1/n} = (|N_{L/K}(\alpha)| \cdot |N_{L/K}(\beta)|)^{1/n} = |\alpha|_L \cdot |\beta|_L, \end{align*} where $N_{L/K}(\alpha\beta) = N_{L/K}(\alpha) N_{L/K}(\beta)$ is the multiplicativity of the field norm (which follows from $N_{L/K}(\alpha) = \det(m_\alpha)$ where $m_\alpha: L \to L$ is multiplication by $\alpha$, and $\det(m_{\alpha\beta}) = \det(m_\alpha \circ m_\beta) = \det(m_\alpha)\det(m_\beta)$). **Extension of $|\cdot|$.** For $a \in K$, multiplication by $a$ on $L$ is the scalar map $m_a = a \cdot \operatorname{Id}_L$, so $N_{L/K}(a) = \det(a \cdot \operatorname{Id}_L) = a^n$. Therefore $|a|_L = |a^n|^{1/n} = (|a|^n)^{1/n} = |a|$. [/step]

[step:Identify $\mathcal{O}_L := \{\alpha \in L : |\alpha|_L \leq 1\}$ with the integral closure of $\mathcal{O}_K$ in $L$]Define $\mathcal{O}_L := \{\alpha \in L : |\alpha|_L \leq 1\} = \{\alpha \in L : |N_{L/K}(\alpha)| \leq 1\}$. We show $\mathcal{O}_L$ equals the integral closure $\widetilde{\mathcal{O}_K}$ of $\mathcal{O}_K$ in $L$. **$\mathcal{O}_L \subseteq \widetilde{\mathcal{O}_K}$:** Let $\alpha \in L$ with $|\alpha|_L \leq 1$. Let $f(x) = x^d + a_{d-1}x^{d-1} + \cdots + a_0 \in K[x]$ be the minimal polynomial of $\alpha$ over $K$, where $d = [K(\alpha):K]$. The norm satisfies \begin{align*} N_{L/K}(\alpha) = ((-1)^d a_0)^{[L:K(\alpha)]}, \end{align*} since $N_{L/K}(\alpha) = N_{K(\alpha)/K}(\alpha)^{[L:K(\alpha)]}$ and $N_{K(\alpha)/K}(\alpha) = (-1)^d a_0$ (the constant term of the minimal polynomial, up to sign). Setting $e = [L:K(\alpha)]$: \begin{align*} |N_{L/K}(\alpha)| = |a_0|^e \leq 1, \end{align*} so $|a_0| \leq 1$, i.e., $a_0 \in \mathcal{O}_K$. Since $f$ is irreducible over $K$ and $K$ is complete, the [Irreducibility and Coefficient Bounds](/theorems/???) theorem gives $|a_i| \leq \max(|a_0|, |a_d|) = \max(|a_0|, 1) = 1$ for all $i$ (using $a_d = 1$). Therefore $f \in \mathcal{O}_K[x]$, and $\alpha$ is integral over $\mathcal{O}_K$. **$\widetilde{\mathcal{O}_K} \subseteq \mathcal{O}_L$:** Let $\alpha \in L$ be integral over $\mathcal{O}_K$. The minimal polynomial $f$ of $\alpha$ over $K$ divides the monic polynomial over $\mathcal{O}_K$ witnessing integrality (both are monic, and the minimal polynomial divides every polynomial in $K[x]$ with $\alpha$ as a root). After dividing, $f$ itself has coefficients in $\mathcal{O}_K$ (by Gauss's lemma for the valuation ring, or directly: if $f | g$ with $g \in \mathcal{O}_K[x]$ monic, and $f$ is monic, then the quotient $g/f$ is monic, and all coefficients of $f$ lie in $\mathcal{O}_K$ since $\mathcal{O}_K$ is integrally closed — but we give a more direct argument). Since $\alpha$ is integral over $\mathcal{O}_K$, there is a monic $g \in \mathcal{O}_K[x]$ with $g(\alpha) = 0$. Every conjugate $\sigma(\alpha)$ of $\alpha$ over $K$ (i.e., every root of $f$ in an algebraic closure) also satisfies $g(\sigma(\alpha)) = 0$ (since $g \in K[x]$ and $\sigma$ permutes roots of $f$). Therefore every conjugate of $\alpha$ is a root of $g \in \mathcal{O}_K[x]$, hence integral over $\mathcal{O}_K$. Since $\mathcal{O}_K$ is [integrally closed](/theorems/???) in $K$, and $N_{K(\alpha)/K}(\alpha) = (-1)^d a_0$ is an element of $K$ that is integral over $\mathcal{O}_K$ (being a product of integral elements), we conclude $N_{K(\alpha)/K}(\alpha) \in \mathcal{O}_K$. Therefore $|N_{L/K}(\alpha)| = |N_{K(\alpha)/K}(\alpha)|^e \leq 1$, giving $|\alpha|_L \leq 1$.[/step]

[guided]This is the central step. We need to identify the "unit ball" of $|\cdot|_L$ with a ring (the integral closure) in order to apply the [Criterion for the Strong Triangle Inequality](/theorems/???). **Forward direction ($\mathcal{O}_L \subset \widetilde{\mathcal{O}_K}$):** Given $|\alpha|_L \leq 1$, we need to show all coefficients of the minimal polynomial $f$ of $\alpha$ lie in $\mathcal{O}_K$. The condition $|\alpha|_L \leq 1$ gives $|N_{L/K}(\alpha)| \leq 1$, which pins down the constant term: $|a_0| \leq 1$. The leading coefficient is $a_d = 1$. The [Irreducibility and Coefficient Bounds](/theorems/???) theorem then bounds all intermediate coefficients: $|a_i| \leq \max(|a_0|, 1) = 1$. This is where irreducibility of $f$ and completeness of $K$ are consumed — the coefficient bound theorem requires both. **Backward direction ($\widetilde{\mathcal{O}_K} \subset \mathcal{O}_L$):** Given $\alpha$ integral over $\mathcal{O}_K$, we need $|N_{L/K}(\alpha)| \leq 1$. The norm is (up to sign and a power) the product of $\alpha$'s conjugates over $K$. Each conjugate is also integral over $\mathcal{O}_K$ (as a root of the same minimal polynomial, which divides the monic integral polynomial). The product of integral elements is integral (by the [Integral Closure Is a Ring](/theorems/???) theorem), and an element of $K$ that is integral over $\mathcal{O}_K$ lies in $\mathcal{O}_K$ (since $\mathcal{O}_K$ is [integrally closed](/theorems/???) in $K$). So $N_{L/K}(\alpha) \in \mathcal{O}_K$, giving $|N_{L/K}(\alpha)| \leq 1$.[/guided]

[step:Verify the strong triangle inequality using the criterion] By the [Criterion for the Strong Triangle Inequality](/theorems/???), $|\cdot|_L$ is a non-archimedean absolute value if and only if $|\alpha|_L \leq 1$ implies $|\alpha + 1|_L \leq 1$ for all $\alpha \in L$. (The criterion requires non-degeneracy and multiplicativity, which we verified above.) We check this condition. Suppose $|\alpha|_L \leq 1$. By the identification in the previous step, $\alpha \in \widetilde{\mathcal{O}_K}$ (the integral closure of $\mathcal{O}_K$ in $L$). Since the integral closure is a subring of $L$ (by the [Integral Closure Is a Ring](/theorems/???) theorem), it contains $1$ (as $1 \in \mathcal{O}_K \subset \widetilde{\mathcal{O}_K}$) and is closed under addition. Therefore $\alpha + 1 \in \widetilde{\mathcal{O}_K}$, which means $|\alpha + 1|_L \leq 1$. The criterion is satisfied, so $|\cdot|_L$ is a non-archimedean absolute value on $L$ extending $|\cdot|$ on $K$. [/step]

[step:Conclude completeness of $(L, |\cdot|_L)$] Since $|\cdot|_L$ is a norm on the $n$-dimensional $K$-vector space $L$ (with $K$ complete and non-archimedean), the [equivalence of norms theorem](/theorems/???) implies that $L$ is complete under $|\cdot|_L$. Explicitly: $|\cdot|_L$ is equivalent to the max norm with respect to any $K$-basis of $L$, and the max norm is complete since $K$ is complete. [/step]