[proofplan] Since $L/K$ is finite, we write $L = K(\alpha_1, \ldots, \alpha_n)$ and build a tower $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ where $K_i = K_{i-1}(\alpha_i)$. Every $\sigma \in \operatorname{Hom}_K(L, E)$ restricts to a $K$-homomorphism $K_i \to E$ at each layer, and the restriction to $K_{i-1}$ determines the restriction to $K_i$ up to a choice of image $\sigma(\alpha_i)$ among the roots of the minimal polynomial $P_{\alpha_i, K_{i-1}}$ in $E$. This gives at most $\deg P_{\alpha_i, K_{i-1}} = [K_i : K_{i-1}]$ choices per layer. Multiplying over all layers and applying the Tower Law yields the bound $|\operatorname{Hom}_K(L, E)| \le \prod_{i=1}^{n} [K_i : K_{i-1}] = [L : K]$. [/proofplan]

[step:Construct a tower of simple extensions from the generators of $L/K$] Since $L/K$ is a finite extension, the [Characterisation of Finite Extensions](/theorems/1252) (part 2) provides elements $\alpha_1, \ldots, \alpha_n \in L$, each algebraic over $K$, such that $L = K(\alpha_1, \ldots, \alpha_n)$. Define the intermediate fields \begin{align*} K_0 := K, \quad K_i := K_{i-1}(\alpha_i) \quad \text{for } i \in \{1, \ldots, n\}. \end{align*} Then $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ is a tower of simple extensions. Each $\alpha_i$ is algebraic over $K$ and hence also algebraic over the larger field $K_{i-1}$, so each extension $K_i / K_{i-1}$ is a finite simple algebraic extension. [/step]

[step:Count the extensions at each layer using the Roots-Homomorphisms Correspondence]We prove by induction on $i$ that the number of $K$-homomorphisms from $K_i$ into $E$ satisfies \begin{align*} |\operatorname{Hom}_K(K_i, E)| \le \prod_{j=1}^{i} [K_j : K_{j-1}]. \end{align*} **Base case ($i = 0$).** The field $K_0 = K$ admits exactly one $K$-homomorphism into $E$, namely the inclusion $\iota: K \hookrightarrow E$. The empty product $\prod_{j=1}^{0} [K_j : K_{j-1}] = 1$, so the bound holds. **Inductive step.** Assume the bound holds for $K_{i-1}$, so there are at most $\prod_{j=1}^{i-1} [K_j : K_{j-1}]$ elements of $\operatorname{Hom}_K(K_{i-1}, E)$. Fix any $\tau \in \operatorname{Hom}_K(K_{i-1}, E)$ and consider the set of $K$-homomorphisms $\sigma: K_i \to E$ whose restriction to $K_{i-1}$ equals $\tau$: \begin{align*} S_\tau := \{\sigma \in \operatorname{Hom}_K(K_i, E) : \sigma|_{K_{i-1}} = \tau\}. \end{align*} Every $\sigma \in S_\tau$ is a $K_{i-1}$-homomorphism from $K_i = K_{i-1}(\alpha_i)$ into $E$ in the following sense: $\sigma$ restricts to $\tau$ on $K_{i-1}$, and $\tau$ is an injective ring homomorphism from $K_{i-1}$ into $E$. Through the identification of $K_{i-1}$ with $\tau(K_{i-1})$, each $\sigma \in S_\tau$ corresponds to a $\tau(K_{i-1})$-homomorphism from $\tau(K_{i-1})(\sigma(\alpha_i))$ into $E$. More precisely, $\sigma(\alpha_i)$ must be a root of $\tau_*(P_{\alpha_i, K_{i-1}})$ in $E$, where $\tau_*(P_{\alpha_i, K_{i-1}})$ denotes the polynomial obtained by applying $\tau$ to each coefficient of the minimal polynomial $P_{\alpha_i, K_{i-1}} \in K_{i-1}[t]$. Since $\tau$ is injective, $\tau_*(P_{\alpha_i, K_{i-1}})$ has the same degree as $P_{\alpha_i, K_{i-1}}$. Moreover, $\sigma$ is completely determined by its value $\sigma(\alpha_i)$, because every element of $K_i = K_{i-1}(\alpha_i)$ is a polynomial expression in $\alpha_i$ with coefficients in $K_{i-1}$, and $\sigma$ acts on such an expression by applying $\tau$ to the coefficients and replacing $\alpha_i$ by $\sigma(\alpha_i)$. By the [Roots-Homomorphisms Correspondence](/theorems/1256) applied to the simple extension $K_i = K_{i-1}(\alpha_i)$ over $K_{i-1}$ (via the identification through $\tau$), the number of such extensions is at most the number of roots of $\tau_*(P_{\alpha_i, K_{i-1}})$ in $E$, which is at most $\deg P_{\alpha_i, K_{i-1}}$. By the [Structure of Simple Algebraic Extensions](/theorems/1251) (part 2), $\deg P_{\alpha_i, K_{i-1}} = [K_i : K_{i-1}]$. Therefore \begin{align*} |S_\tau| \le [K_i : K_{i-1}]. \end{align*} Since every $\sigma \in \operatorname{Hom}_K(K_i, E)$ belongs to $S_\tau$ for a unique $\tau := \sigma|_{K_{i-1}} \in \operatorname{Hom}_K(K_{i-1}, E)$, the sets $\{S_\tau\}_{\tau \in \operatorname{Hom}_K(K_{i-1}, E)}$ partition $\operatorname{Hom}_K(K_i, E)$. Summing the cardinalities: \begin{align*} |\operatorname{Hom}_K(K_i, E)| &= \sum_{\tau \in \operatorname{Hom}_K(K_{i-1}, E)} |S_\tau| \le \sum_{\tau \in \operatorname{Hom}_K(K_{i-1}, E)} [K_i : K_{i-1}] \\ &= |\operatorname{Hom}_K(K_{i-1}, E)| \cdot [K_i : K_{i-1}] \le \prod_{j=1}^{i-1} [K_j : K_{j-1}] \cdot [K_i : K_{i-1}] = \prod_{j=1}^{i} [K_j : K_{j-1}], \end{align*} where the second inequality uses the inductive hypothesis. This completes the induction.[/step]

[guided]The goal is to count the elements of $\operatorname{Hom}_K(K_i, E)$ by decomposing the problem: first choose where $K_{i-1}$ goes (i.e., choose $\tau$), then count how many ways to extend $\tau$ to $K_i = K_{i-1}(\alpha_i)$. **Why does restriction to $K_{i-1}$ give a partition?** Every $K$-homomorphism $\sigma: K_i \to E$ restricts to a $K$-homomorphism $\tau := \sigma|_{K_{i-1}}: K_{i-1} \to E$, since $K \subset K_{i-1} \subset K_i$ and $\sigma$ fixes $K$. Different extensions $\sigma, \sigma'$ with the same restriction $\tau$ differ only in where they send $\alpha_i$. Conversely, two extensions with different restrictions cannot agree on all of $K_i$. This yields a clean partition $\operatorname{Hom}_K(K_i, E) = \bigsqcup_\tau S_\tau$. **How does the Roots-Homomorphisms Correspondence apply?** Fix $\tau \in \operatorname{Hom}_K(K_{i-1}, E)$. Any $\sigma \in S_\tau$ is determined by $\sigma(\alpha_i)$, because every element of $K_i$ has the form $g(\alpha_i)$ for some $g \in K_{i-1}[t]$ (by the [Structure of Simple Algebraic Extensions](/theorems/1251), part 1), and $\sigma(g(\alpha_i)) = \tau_*(g)(\sigma(\alpha_i))$ where $\tau_*(g)$ applies $\tau$ to each coefficient of $g$. The value $\sigma(\alpha_i)$ must satisfy $\tau_*(P_{\alpha_i, K_{i-1}})(\sigma(\alpha_i)) = 0$: this follows from $P_{\alpha_i, K_{i-1}}(\alpha_i) = 0$ and the fact that $\sigma$ sends this equation to $\tau_*(P_{\alpha_i, K_{i-1}})(\sigma(\alpha_i)) = 0$. Conversely, for each root $\beta$ of $\tau_*(P_{\alpha_i, K_{i-1}})$ in $E$, the assignment $g(\alpha_i) \mapsto \tau_*(g)(\beta)$ defines a well-defined $K$-homomorphism $K_i \to E$ extending $\tau$: well-definedness follows because if $g(\alpha_i) = h(\alpha_i)$, then $P_{\alpha_i, K_{i-1}} \mid (g - h)$ in $K_{i-1}[t]$, so $\tau_*(P_{\alpha_i, K_{i-1}}) \mid \tau_*(g - h)$ in $\tau(K_{i-1})[t]$ (since $\tau$ is injective and hence preserves divisibility), so $\tau_*(g)(\beta) = \tau_*(h)(\beta)$. This establishes a bijection between $S_\tau$ and the set of roots of $\tau_*(P_{\alpha_i, K_{i-1}})$ in $E$. Since a nonzero polynomial of degree $d$ over a field has at most $d$ roots, and $\deg \tau_*(P_{\alpha_i, K_{i-1}}) = \deg P_{\alpha_i, K_{i-1}} = [K_i : K_{i-1}]$ (the degree is preserved because $\tau$ is injective), we obtain $|S_\tau| \le [K_i : K_{i-1}]$. **Why is the inequality sometimes strict?** The bound $|S_\tau| \le [K_i : K_{i-1}]$ is strict precisely when the polynomial $\tau_*(P_{\alpha_i, K_{i-1}})$ fails to split completely in $E$, i.e., when some roots of the minimal polynomial are "missing" from $E$. This is the mechanism by which the total count $|\operatorname{Hom}_K(L, E)|$ can fall strictly below $[L : K]$. In the special case where $E$ is algebraically closed, every polynomial splits completely, so all inequalities become equalities.[/guided]

[step:Multiply the layer-by-layer bounds and apply the Tower Law to conclude] Setting $i = n$ in the inductive bound yields \begin{align*} |\operatorname{Hom}_K(L, E)| = |\operatorname{Hom}_K(K_n, E)| \le \prod_{j=1}^{n} [K_j : K_{j-1}]. \end{align*} We evaluate the product on the right by applying the [Tower Law](/theorems/1248) iteratively. For any $i \in \{1, \ldots, n\}$, the Tower Law applied to $K \subset K_{i-1} \subset K_i$ (noting that each $K_j / K_{j-1}$ is a finite extension of finite extensions) gives \begin{align*} [K_i : K] = [K_i : K_{i-1}] \cdot [K_{i-1} : K]. \end{align*} Applying this telescopically from $i = 1$ to $i = n$: \begin{align*} \prod_{j=1}^{n} [K_j : K_{j-1}] = [K_n : K] = [L : K]. \end{align*} Combining with the inductive bound: \begin{align*} |\operatorname{Hom}_K(L, E)| \le [L : K]. \end{align*} [/step]