[proofplan] We show that for any $\sigma \in \operatorname{Aut}(L/K)$, the map $\alpha \mapsto |\sigma(\alpha)|_L$ is an absolute value on $L$ extending $|\cdot|$ on $K$. By the uniqueness of the extending absolute value (proved in the [Unique Extension](/theorems/???) theorem), it must equal $|\cdot|_L$. [/proofplan] [step:Show $\alpha \mapsto |\sigma(\alpha)|_L$ is an absolute value on $L$ extending $|\cdot|$ on $K$] Define $\|\cdot\|: L \to \mathbb{R}_{\geq 0}$ by $\|\alpha\| := |\sigma(\alpha)|_L$. **Non-degeneracy.** $\|\alpha\| = 0$ iff $|\sigma(\alpha)|_L = 0$ iff $\sigma(\alpha) = 0$ (by non-degeneracy of $|\cdot|_L$) iff $\alpha = 0$ (since $\sigma$ is a field automorphism, hence injective, with $\sigma(0) = 0$). **Multiplicativity.** For $\alpha, \beta \in L$: \begin{align*} \|\alpha\beta\| = |\sigma(\alpha\beta)|_L = |\sigma(\alpha) \cdot \sigma(\beta)|_L = |\sigma(\alpha)|_L \cdot |\sigma(\beta)|_L = \|\alpha\| \cdot \|\beta\|, \end{align*} using that $\sigma$ is a ring homomorphism ($\sigma(\alpha\beta) = \sigma(\alpha)\sigma(\beta)$) and multiplicativity of $|\cdot|_L$. **Strong triangle inequality.** For $\alpha, \beta \in L$: \begin{align*} \|\alpha + \beta\| = |\sigma(\alpha + \beta)|_L = |\sigma(\alpha) + \sigma(\beta)|_L \leq \max(|\sigma(\alpha)|_L, |\sigma(\beta)|_L) = \max(\|\alpha\|, \|\beta\|), \end{align*} using that $\sigma$ is additive and $|\cdot|_L$ is non-archimedean. **Extension of $|\cdot|$.** For $a \in K$, since $\sigma \in \operatorname{Aut}(L/K)$ fixes $K$ pointwise, $\sigma(a) = a$, so $\|a\| = |\sigma(a)|_L = |a|_L = |a|$ (the last equality because $|\cdot|_L$ extends $|\cdot|$ on $K$). Therefore $\|\cdot\|$ is a non-archimedean absolute value on $L$ that restricts to $|\cdot|$ on $K$. [/step] [step:Apply uniqueness to conclude $|\sigma(\alpha)|_L = |\alpha|_L$] By the [Unique Extension of Absolute Values to Finite Extensions](/theorems/???), there is exactly one absolute value on $L$ extending $|\cdot|$ on $K$, namely $|\cdot|_L$. Since $\|\cdot\|$ defined by $\|\alpha\| = |\sigma(\alpha)|_L$ is an absolute value on $L$ extending $|\cdot|$ on $K$ (verified in the previous step), uniqueness forces $\|\cdot\| = |\cdot|_L$, i.e., \begin{align*} |\sigma(\alpha)|_L = |\alpha|_L \quad \text{for all } \alpha \in L. \end{align*} [guided] The proof is a clean application of the uniqueness part of the extension theorem. The only work is verifying that composing $|\cdot|_L$ with a $K$-automorphism $\sigma$ produces another absolute value on $L$ that still extends $|\cdot|$ on $K$. The verification is straightforward: $\sigma$ preserves the field operations (additivity gives the triangle inequality, multiplicativity gives multiplicativity of the composed map), and $\sigma$ fixes $K$ pointwise (so the restriction to $K$ is unchanged). Once both $|\cdot|_L$ and $|\sigma(\cdot)|_L$ are shown to be absolute values on $L$ extending $|\cdot|$ on $K$, the uniqueness theorem forces them to be equal. Note that this result fails if $K$ is not complete: over an incomplete field, the absolute value on an extension need not be unique, and different automorphisms can genuinely change the absolute value. For example, over $\mathbb{Q}$ with $|\cdot|_5$, the field $\mathbb{Q}(\sqrt{2})$ has two distinct extensions of $|\cdot|_5$ (corresponding to the two primes of $\mathbb{Z}[\sqrt{2}]$ above 5), and the nontrivial automorphism $\sqrt{2} \mapsto -\sqrt{2}$ swaps them. [/guided] [/step]