[proofplan] We prove each part by reducing to the residue field level and applying the classification of unramified extensions. For part (1), we express $LM = M(\alpha)$ where $\alpha$ is the Hensel lift of a primitive element for the residue extension, then show the residue degree accounts for the full degree. Parts (2) and (3) follow from the multiplicativity of residue degrees and ramification indices in towers, combined with part (1). [/proofplan] [step:Show $LM/M$ is unramified by lifting the primitive element] Since $L/K$ is unramified, the [Existence and Uniqueness of Unramified Extensions](/theorems/???) gives $L = K(\alpha)$ where $\alpha \in \mathcal{O}_L$ is the Hensel lift of a primitive element $\bar{\alpha}$ for $k_L/k_K$. Let $f \in \mathcal{O}_K[x]$ be the minimal polynomial of $\alpha$ over $K$, so $\bar{f} \in k_K[x]$ is the minimal polynomial of $\bar{\alpha}$ over $k_K$ and $\deg f = \deg \bar{f} = [k_L : k_K] = [L:K]$. The composite field is $LM = M(\alpha)$. Let $\bar{g} \in k_M[x]$ be the minimal polynomial of $\bar{\alpha}$ over $k_M$. Since $\bar{\alpha}$ is a root of $\bar{f} \in k_K[x] \subseteq k_M[x]$, the polynomial $\bar{g}$ divides $\bar{f}$ in $k_M[x]$, and in particular $\bar{g}$ is separable (since $\bar{f}$ is separable, being the minimal polynomial of a generator of an extension of finite fields). Write $\bar{f} = \bar{g}\bar{h}$ in $k_M[x]$ with $\gcd(\bar{g}, \bar{h}) = 1$. By [Hensel's Lemma](/theorems/???), the factorisation $\bar{f} = \bar{g}\bar{h}$ lifts to a factorisation $f = gh$ in $\mathcal{O}_M[x]$ where $g$ is monic with $\deg g = \deg \bar{g}$ and $g \equiv \bar{g} \pmod{\mathfrak{m}_M}$. Since $g(\alpha) \equiv \bar{g}(\bar{\alpha}) = 0 \pmod{\mathfrak{m}_M}$ and $g$ divides $f$ in $\mathcal{O}_M[x]$ with $f(\alpha) = 0$, the polynomial $g$ is the minimal polynomial of $\alpha$ over $M$ (its degree divides $\deg f$ and $\alpha$ is a root). Therefore $[LM:M] = \deg g = \deg \bar{g} = [k_M(\bar{\alpha}) : k_M] = [k_{LM} : k_M]$. Since the residue degree equals the extension degree, the fundamental inequality $[LM:M] \geq e_{LM/M} \cdot f_{LM/M}$ forces $e_{LM/M} = 1$ and $f_{LM/M} = [LM:M]$, so $LM/M$ is unramified. [guided] The idea is to track $\alpha$ — the Hensel lift of the primitive element of the residue extension $k_L/k_K$ — as we adjoin it to $M$ instead of $K$. Since $L/K$ is unramified, the [Existence and Uniqueness of Unramified Extensions](/theorems/???) provides a concrete description: $L = K(\alpha)$ where $\alpha \in \mathcal{O}_L$ satisfies a monic polynomial $f \in \mathcal{O}_K[x]$ whose reduction $\bar{f} \in k_K[x]$ is the minimal polynomial of $\bar{\alpha}$ over $k_K$. The key property is $\deg f = \deg \bar{f} = [k_L : k_K] = [L:K]$. Now consider $LM = M(\alpha)$. The element $\bar{\alpha}$ already lives in $k_L \subseteq k_{LM}$, and its minimal polynomial $\bar{g}$ over $k_M$ divides $\bar{f}$ in $k_M[x]$. Since $\bar{f}$ is separable (it is the minimal polynomial of a generator of an extension of finite fields, and all finite field extensions are separable), $\bar{g}$ is separable too, and $\gcd(\bar{g}, \bar{f}/\bar{g}) = 1$. Write $\bar{f} = \bar{g}\bar{h}$ with $\gcd(\bar{g}, \bar{h}) = 1$ in $k_M[x]$. We apply [Hensel's Lemma](/theorems/???) to the polynomial $f \in \mathcal{O}_M[x]$ (viewing $\mathcal{O}_K \subseteq \mathcal{O}_M$) and the coprime factorisation $\bar{f} = \bar{g}\bar{h}$. Hensel's Lemma requires the base ring to be complete with respect to a non-archimedean absolute value — $\mathcal{O}_M$ satisfies this since $M$ is a local field — and the factors to be coprime modulo $\mathfrak{m}_M$, which we have verified. The conclusion is a lift: $f = gh$ in $\mathcal{O}_M[x]$ with $g$ monic, $\deg g = \deg \bar{g}$, and $g \equiv \bar{g} \pmod{\mathfrak{m}_M}$. Since $f(\alpha) = 0$, we have $g(\alpha) h(\alpha) = 0$ in $\mathcal{O}_M[\alpha]$. Reducing modulo $\mathfrak{m}_{LM}$: $\bar{g}(\bar{\alpha}) = 0$ and $\bar{h}(\bar{\alpha}) \neq 0$ (since $\bar{g}$ is the minimal polynomial of $\bar{\alpha}$ over $k_M$ and $\gcd(\bar{g}, \bar{h}) = 1$). So $h(\alpha)$ is a unit in $\mathcal{O}_{LM}$, hence $g(\alpha) = 0$, and $g$ is the minimal polynomial of $\alpha$ over $M$ (as it is monic and irreducible, being the Hensel lift of an irreducible polynomial). Therefore $[LM:M] = \deg g = \deg \bar{g} = [k_M(\bar{\alpha}) : k_M] = [k_{LM} : k_M]$. The fundamental inequality gives $[LM:M] \geq e_{LM/M} \cdot f_{LM/M} = e_{LM/M} \cdot [k_{LM} : k_M]$. Since $[LM:M] = [k_{LM}:k_M]$, this forces $e_{LM/M} = 1$, so $LM/M$ is unramified. [/guided] [/step] [step:Deduce that any subextension of $L/K$ is unramified] Let $K \subseteq E \subseteq L$ with $L/K$ unramified. By the [Multiplicativity in Towers](/theorems/???), $[L:K] = [L:E][E:K]$ and $f_{L/K} = f_{L/E} \cdot f_{E/K}$. Since $L/K$ is unramified, $[L:K] = f_{L/K}$, so \begin{align*} [L:E][E:K] = f_{L/E} \cdot f_{E/K}. \end{align*} The fundamental inequality gives $[L:E] \geq f_{L/E}$ and $[E:K] \geq f_{E/K}$. The product of these two inequalities yields $[L:E][E:K] \geq f_{L/E} \cdot f_{E/K}$. Since equality holds, both inequalities must be equalities. In particular $[E:K] = f_{E/K}$, so $e_{E/K} = 1$ and $E/K$ is unramified. [/step] [step:Deduce that the composite of two unramified extensions is unramified] Suppose $L/K$ and $M/K$ are both unramified. By part (1), $LM/M$ is unramified, so $e_{LM/M} = 1$ and $[LM:M] = f_{LM/M}$. Since $M/K$ is unramified, $e_{M/K} = 1$ and $[M:K] = f_{M/K}$. By the [Multiplicativity in Towers](/theorems/???), \begin{align*} e_{LM/K} = e_{LM/M} \cdot e_{M/K} = 1 \cdot 1 = 1, \end{align*} and therefore $[LM:K] = f_{LM/K}$, so $LM/K$ is unramified. [/step]