[proofplan] We construct the inertia field $T$ as the unique unramified extension of $K$ whose residue field equals $k_L$, then embed $T$ into $L$ using the classification of unramified extensions. To prove maximality, we show that any unramified subextension $T'/K$ satisfies $T' \subseteq T$ by forming the composite $T'T$, applying the [Stability of Unramified Extensions](/theorems/???), and comparing degrees. [/proofplan] [step:Construct $T$ as the unique unramified extension of $K$ with residue field $k_L$] The residue field $k_L$ is a finite extension of $k_K$. By the [Existence and Uniqueness of Unramified Extensions](/theorems/???), there exists a unique (up to $K$-isomorphism) unramified extension $T/K$ with $k_T \cong k_L$. In particular, $[T:K] = f_{T/K} = [k_T : k_K] = [k_L : k_K] = f_{L/K}$. [guided] We need to find a maximal unramified subextension of $L/K$. The residue field extension $k_L/k_K$ is a finite extension of finite fields, and the classification theorem for unramified extensions says: for every finite extension $\ell/k_K$, there exists a unique unramified extension $E/K$ (inside a fixed algebraic closure $\bar{K}$) with $k_E = \ell$. Applying this to $\ell = k_L$, we obtain an unramified extension $T/K$ with $k_T = k_L$ and $[T:K] = [k_L:k_K] = f_{L/K}$. At this stage, $T$ lives inside $\bar{K}$ but is not yet known to be a subfield of $L$. The next step embeds it. [/guided] [/step] [step:Embed $T$ into $L$ as a subfield] The identity map $k_T = k_L \to k_L$ is a $k_K$-isomorphism. By the [Bijection on Hom-Sets](/theorems/???), $K$-embeddings $T \hookrightarrow \bar{K}$ correspond bijectively to $k_K$-embeddings $k_T \hookrightarrow \bar{k}_K$. Since $k_T = k_L$ and $k_L \subseteq k_{\bar{K}}$, the inclusion $k_L \hookrightarrow k_L$ lifts to a $K$-embedding $\iota: T \hookrightarrow L$. We identify $T$ with $\iota(T) \subseteq L$. Then $T/K$ is an unramified subextension of $L/K$ with $[T:K] = f_{L/K}$. [guided] Why must $T$ embed into $L$ specifically, not just into $\bar{K}$? The key is the rigidity of unramified extensions: a $K$-embedding of $T$ is determined by what it does to the residue field. The identity map $k_T = k_L \to k_L$ is an embedding of $k_T$ into $k_L$, and $k_L$ is the residue field of $L$. By the [Bijection on Hom-Sets](/theorems/???), which establishes a bijection between $\operatorname{Hom}_K(T, \bar{K})$ and $\operatorname{Hom}_{k_K}(k_T, \bar{k}_K)$, this residue-level embedding lifts to a $K$-embedding $\iota: T \hookrightarrow L$. After identifying $T$ with $\iota(T)$, we have $K \subseteq T \subseteq L$ with $T/K$ unramified and $[T:K] = f_{L/K}$. [/guided] [/step] [step:Prove $T$ is the unique maximal unramified subextension] Let $T'/K$ be any unramified subextension of $L/K$. We must show $T' \subseteq T$. Form the composite $T'T$ inside $L$. By the [Stability of Unramified Extensions](/theorems/???), $T'T/K$ is unramified (as the composite of two unramified extensions of $K$). Since $T'T \subseteq L$, we have $[T'T:K] \leq [L:K]$, and therefore \begin{align*} f_{T'T/K} = [T'T:K] \leq [L:K]. \end{align*} The residue field satisfies $k_{T'T} \supseteq k_{T'} \cup k_T = k_T = k_L$, and since $k_{T'T} \subseteq k_L$ (as $T'T \subseteq L$), we get $k_{T'T} = k_L$. Therefore $f_{T'T/K} = [k_L : k_K] = f_{L/K} = [T:K]$. Since $T \subseteq T'T$ and $[T'T:K] = [T:K]$, we conclude $T'T = T$, which forces $T' \subseteq T$. [guided] Suppose $T'/K$ is another unramified subextension of $L/K$. We want to show $T' \subseteq T$, establishing that $T$ is the unique maximal one. Consider the composite $T'T$ inside $L$. Both $T'/K$ and $T/K$ are unramified, so by the [Stability of Unramified Extensions](/theorems/???), part (3), $T'T/K$ is unramified. An unramified extension satisfies $[T'T:K] = f_{T'T/K} = [k_{T'T} : k_K]$. Now $k_{T'T}$ contains both $k_{T'} \supseteq k_K$ and $k_T = k_L$, so $k_{T'T} \supseteq k_L$. On the other hand, $T'T \subseteq L$ implies $k_{T'T} \subseteq k_L$. Therefore $k_{T'T} = k_L$, and $[T'T:K] = [k_L:k_K] = f_{L/K} = [T:K]$. Since $T \subseteq T'T$ and both have the same degree over $K$, we must have $T'T = T$. But $T' \subseteq T'T = T$, so $T' \subseteq T$. This shows $T$ is the unique maximal unramified subextension of $L/K$. [/guided] [/step]