[proofplan] We proceed by strong induction on $n := [L : K]$. When $n = 1$, the only $K$-homomorphism $L \to \bar{K}$ is the inclusion, and equality holds. For $n > 1$, choose $\alpha \in L \setminus K$ and set $F := K(\alpha)$, so that $K \subsetneq F \subset L$ (with $F = L$ allowed when $L/K$ is itself a simple extension). Each $K$-homomorphism $\sigma \colon F \to \bar{K}$ extends to at most $[L : F]$ homomorphisms $L \to \bar{K}$ by applying the inductive hypothesis to $L/F$ (after transporting the base field via $\sigma$). Meanwhile, the [Roots-Homomorphisms Correspondence](/theorems/1256) gives $|\operatorname{Hom}_K(F, \bar{K})| = |\operatorname{Hom}_K(K(\alpha), \bar{K})| \le [F : K]$, with equality if and only if $P_\alpha$ is separable over $K$. Multiplying yields $|\operatorname{Hom}_K(L, \bar{K})| \le [L : F] \cdot [F : K] = [L : K]$, and equality holds if and only if both $P_\alpha$ is separable over $K$ and $L/F$ is separable — which is equivalent to $L/K$ being separable. [/proofplan]

[step:Handle the base case $[L : K] = 1$] Suppose $[L : K] = 1$, so that $L = K$. A $K$-homomorphism $\sigma \colon K \to \bar{K}$ is a field homomorphism that fixes $K$ pointwise, so $\sigma$ must be the inclusion map $K \hookrightarrow \bar{K}$. Therefore \begin{align*} |\operatorname{Hom}_K(K, \bar{K})| = 1 = [K : K]. \end{align*} Equality holds, and the extension $K/K$ is separable (every element of $K$ has minimal polynomial $t - a$ for some $a \in K$, which has degree $1$ and no repeated roots). This establishes both the inequality and the equality characterisation for the base case. [/step]

[step:Choose a primitive element and set up the inductive tower $K \subset F \subset L$] Suppose $[L : K] = n > 1$ and that the theorem holds for all finite extensions of degree strictly less than $n$. Since $[L : K] > 1$, we have $L \neq K$, so there exists $\alpha \in L \setminus K$. Set $F := K(\alpha)$. Since $\alpha \in L$ is algebraic over $K$ (because $L/K$ is a finite extension, hence algebraic), $F = K(\alpha)$ is a finite extension of $K$ with \begin{align*} [F : K] = \deg P_\alpha \ge 2, \end{align*} where $P_\alpha \in K[t]$ denotes the minimal polynomial of $\alpha$ over $K$. The degree is at least $2$ because $\alpha \notin K$ (if $\deg P_\alpha = 1$, then $P_\alpha = t - a$ for some $a \in K$, forcing $\alpha = a \in K$, a contradiction). By the [Tower Law](/theorems/1248), $[L : K] = [L : F] \cdot [F : K]$. Since $[F : K] \ge 2$, we have $[L : F] = [L : K] / [F : K] \le n/2 < n$. In particular, $[L : F] < n$, so the inductive hypothesis applies to the extension $L/F$. [/step]

[step:Count extensions of each $K$-homomorphism $\sigma \colon F \to \bar{K}$ to $L \to \bar{K}$]Every $K$-homomorphism $\tau \colon L \to \bar{K}$ restricts to a $K$-homomorphism $\tau|_F \colon F \to \bar{K}$. This defines a map \begin{align*} \operatorname{Hom}_K(L, \bar{K}) &\to \operatorname{Hom}_K(F, \bar{K}) \\ \tau &\mapsto \tau|_F \end{align*} that partitions $\operatorname{Hom}_K(L, \bar{K})$ into fibres: for each $\sigma \in \operatorname{Hom}_K(F, \bar{K})$, the fibre over $\sigma$ is \begin{align*} \{\tau \in \operatorname{Hom}_K(L, \bar{K}) : \tau|_F = \sigma\}. \end{align*} Counting by fibres gives \begin{align*} |\operatorname{Hom}_K(L, \bar{K})| = \sum_{\sigma \in \operatorname{Hom}_K(F, \bar{K})} |\{\tau \in \operatorname{Hom}_K(L, \bar{K}) : \tau|_F = \sigma\}|. \end{align*} Fix $\sigma \in \operatorname{Hom}_K(F, \bar{K})$. An extension $\tau$ of $\sigma$ to $L$ is a field homomorphism $\tau \colon L \to \bar{K}$ with $\tau|_F = \sigma$. In other words, $\tau$ is an $F$-homomorphism from $L$ to $\bar{K}$ when we view $\bar{K}$ as an $F$-algebra via $\sigma$: the $F$-algebra structure on $\bar{K}$ is defined by $a \cdot z := \sigma(a) z$ for $a \in F$, $z \in \bar{K}$. We denote the resulting set of $F$-homomorphisms by $\operatorname{Hom}_F(L, \bar{K}_\sigma)$, where the subscript $\sigma$ records that the $F$-algebra structure on $\bar{K}$ is via $\sigma$. Since $\sigma \colon F \to \bar{K}$ is a field embedding, $\sigma(F)$ is a subfield of $\bar{K}$ isomorphic to $F$. The field $\bar{K}$ is algebraically closed, and hence serves as an algebraic closure of $\sigma(F)$. Applying the inductive hypothesis to the extension $L/F$ (which has degree $[L : F] < n$), but viewed through the lens of $\sigma$: \begin{align*} |\operatorname{Hom}_F(L, \bar{K}_\sigma)| \le [L : F], \end{align*} with equality if and only if $L/F$ is separable.[/step]

[guided]The central idea of this step is to decompose the counting problem into two parts: first choose a homomorphism on the smaller field $F$, then count how many ways it extends to $L$. Why can we apply the inductive hypothesis? The extensions of $\sigma$ to $L$ are exactly the $F$-homomorphisms $L \to \bar{K}_\sigma$. The inductive hypothesis applies to extensions of degree less than $n$, and $[L : F] < n$ since $[F : K] \ge 2$. There is a subtlety in applying the induction: the inductive hypothesis for a finite extension $M/E$ states that $|\operatorname{Hom}_E(M, \bar{E})| \le [M : E]$ with equality iff $M/E$ is separable, where $\bar{E}$ is an algebraic closure of $E$. We apply this to $M = L$ and $E = F$. We need an algebraic closure of $F$. Since $\sigma \colon F \hookrightarrow \bar{K}$ is a field embedding and $\bar{K}$ is algebraically closed, the field $\bar{K}$ contains an algebraic closure of $\sigma(F)$. More precisely, the algebraic closure of $\sigma(F)$ inside $\bar{K}$ is isomorphic to $\bar{F}$ (an algebraic closure of $F$). Since we only need an algebraically closed field containing $\sigma(F)$, and any $F$-homomorphism $L \to \bar{K}_\sigma$ lands in $\bar{K}$, the algebraically closed ambient field $\bar{K}$ serves the role of $\bar{F}$. Why does the equality condition transfer correctly? The inductive hypothesis says equality holds iff $L/F$ is separable. This condition depends only on the extension $L/F$ and is invariant under the field isomorphism $\sigma \colon F \xrightarrow{\sim} \sigma(F)$ — the minimal polynomial of any $\beta \in L$ over $F$ maps to the minimal polynomial of $\tau(\beta)$ over $\sigma(F)$, and separability (having distinct roots) is preserved by field isomorphisms.[/guided]

[step:Count the $K$-homomorphisms $F \to \bar{K}$ via the Roots-Homomorphisms Correspondence] Since $F = K(\alpha)$ is a simple algebraic extension, the [Roots-Homomorphisms Correspondence](/theorems/1256) gives a bijection \begin{align*} \operatorname{Hom}_K(K(\alpha), \bar{K}) &\longleftrightarrow \operatorname{Root}_{P_\alpha}(\bar{K}) \\ \sigma &\longmapsto \sigma(\alpha), \end{align*} so $|\operatorname{Hom}_K(F, \bar{K})| = |\operatorname{Root}_{P_\alpha}(\bar{K})|$. Since $\bar{K}$ is algebraically closed, $P_\alpha$ splits completely in $\bar{K}[t]$. Writing $d := \deg P_\alpha = [F : K]$, the polynomial $P_\alpha$ has exactly $d$ roots in $\bar{K}$ counted with multiplicity. The number of distinct roots satisfies \begin{align*} |\operatorname{Root}_{P_\alpha}(\bar{K})| \le d = [F : K], \end{align*} with equality if and only if all $d$ roots are distinct, i.e., $P_\alpha$ has no repeated roots, i.e., $P_\alpha$ is separable. Therefore \begin{align*} |\operatorname{Hom}_K(F, \bar{K})| \le [F : K], \end{align*} with equality if and only if $P_\alpha$ is separable over $K$. [/step]

[step:Multiply the counts and establish the inequality $|\operatorname{Hom}_K(L, \bar{K})| \le [L : K]$] From the fibre-counting formula in Step 3 and the bound on each fibre: \begin{align*} |\operatorname{Hom}_K(L, \bar{K})| &= \sum_{\sigma \in \operatorname{Hom}_K(F, \bar{K})} |\operatorname{Hom}_F(L, \bar{K}_\sigma)| \\ &\le \sum_{\sigma \in \operatorname{Hom}_K(F, \bar{K})} [L : F] \\ &= |\operatorname{Hom}_K(F, \bar{K})| \cdot [L : F] \\ &\le [F : K] \cdot [L : F] \\ &= [L : K], \end{align*} where the last equality is the [Tower Law](/theorems/1248). This establishes the inequality $|\operatorname{Hom}_K(L, \bar{K})| \le [L : K]$.[/step]

[guided]We used two separate bounds, and it is important to track where each one enters. The first inequality, $|\operatorname{Hom}_F(L, \bar{K}_\sigma)| \le [L : F]$, comes from the inductive hypothesis applied to $L/F$. This bound is the same for every $\sigma$, since the fibre count depends on the extension $L/F$ and not on the particular embedding $\sigma$ of the base field. (More precisely, the bound $[L : F]$ is a numerical invariant of $L/F$, and the number of extensions of $\sigma$ to $L$ is at most $[L : F]$ regardless of which $\sigma$ we pick.) The second inequality, $|\operatorname{Hom}_K(F, \bar{K})| \le [F : K]$, comes from the Roots-Homomorphisms Correspondence. The product $[F : K] \cdot [L : F] = [L : K]$ is the Tower Law. Note that we do not need the fibre sizes to be identical for different $\sigma$ — we only need the uniform upper bound $[L : F]$. However, for the equality characterisation in the next step, we will need to know that equality holds in each fibre simultaneously.[/guided]

[step:Characterise when equality holds: $|\operatorname{Hom}_K(L, \bar{K})| = [L : K]$ if and only if $L/K$ is separable] Inspecting the chain of inequalities in the previous step, equality $|\operatorname{Hom}_K(L, \bar{K})| = [L : K]$ holds if and only if both of the following conditions are satisfied: **(i)** $|\operatorname{Hom}_F(L, \bar{K}_\sigma)| = [L : F]$ for every $\sigma \in \operatorname{Hom}_K(F, \bar{K})$, and **(ii)** $|\operatorname{Hom}_K(F, \bar{K})| = [F : K]$. By the inductive hypothesis, condition (i) holds if and only if $L/F$ is separable. By the Roots-Homomorphisms Correspondence and the fact that $\bar{K}$ is algebraically closed, condition (ii) holds if and only if $P_\alpha$ is separable over $K$. We now show that conditions (i) and (ii) together are equivalent to $L/K$ being separable. **($\Leftarrow$)** Suppose $L/K$ is separable. Then every element of $L$ has a separable minimal polynomial over $K$. In particular, $\alpha \in L$ has separable minimal polynomial $P_\alpha$ over $K$, so condition (ii) holds. For condition (i): every element $\beta \in L$ has separable minimal polynomial $P_{\beta, K}$ over $K$. The minimal polynomial $P_{\beta, F}$ of $\beta$ over $F = K(\alpha)$ divides $P_{\beta, K}$ in $\bar{K}[t]$ (since $P_{\beta, K}(\beta) = 0$ and $P_{\beta, K} \in K[t] \subset F[t]$, the minimal polynomial $P_{\beta, F}$ divides $P_{\beta, K}$ in $F[t]$). Since $P_{\beta, K}$ has no repeated roots (being separable), its factor $P_{\beta, F}$ also has no repeated roots. Therefore every element of $L$ has separable minimal polynomial over $F$, which means $L/F$ is separable, giving condition (i). **($\Rightarrow$)** Suppose conditions (i) and (ii) hold. We must show that $L/K$ is separable, i.e., that every element of $L$ has separable minimal polynomial over $K$. Let $\beta \in L$ be arbitrary. Consider the tower $K \subset K(\beta) \subset L$. By condition (ii), $P_\alpha$ is separable over $K$. By condition (i), $L/F = L/K(\alpha)$ is separable, so in particular every element of $L$ has separable minimal polynomial over $K(\alpha)$. We use the following observation: $\beta$ has separable minimal polynomial over $K$ if and only if $|\operatorname{Hom}_K(K(\beta), \bar{K})| = [K(\beta) : K]$. By the Roots-Homomorphisms Correspondence, this is equivalent to $P_{\beta, K}$ having $\deg P_{\beta, K}$ distinct roots in $\bar{K}$, which holds precisely when $P_{\beta, K}$ is separable. Every $K$-homomorphism $K(\beta) \to \bar{K}$ extends to a $K$-homomorphism $L \to \bar{K}$ (since $\bar{K}$ is algebraically closed: choose an algebraic closure, and extend step by step using the existence of roots). The restriction map $\operatorname{Hom}_K(L, \bar{K}) \to \operatorname{Hom}_K(K(\beta), \bar{K})$ is surjective. Applying our theorem (now proved as an inequality) to $K(\beta)/K$, and using the established equality $|\operatorname{Hom}_K(L, \bar{K})| = [L : K]$, we count by fibres over $K(\beta)$: \begin{align*} [L : K] = |\operatorname{Hom}_K(L, \bar{K})| &= \sum_{\rho \in \operatorname{Hom}_K(K(\beta), \bar{K})} |\{\tau \in \operatorname{Hom}_K(L, \bar{K}) : \tau|_{K(\beta)} = \rho\}| \\ &\le |\operatorname{Hom}_K(K(\beta), \bar{K})| \cdot [L : K(\beta)] \\ &\le [K(\beta) : K] \cdot [L : K(\beta)] \\ &= [L : K]. \end{align*} Since the leftmost and rightmost expressions are equal, both inequalities must be equalities. In particular, $|\operatorname{Hom}_K(K(\beta), \bar{K})| = [K(\beta) : K]$, which by the Roots-Homomorphisms Correspondence means $P_{\beta, K}$ has $\deg P_{\beta, K}$ distinct roots, i.e., $P_{\beta, K}$ is separable. Since $\beta \in L$ was arbitrary, $L/K$ is separable.[/step]

[guided]The equality characterisation is the most delicate part of the proof. Let us carefully unpack why the two conditions (i) and (ii) capture separability of $L/K$. **Why does separability of $L/K$ imply (i) and (ii)?** Condition (ii) is immediate: $\alpha \in L$, and separability of $L/K$ means $P_\alpha$ is separable over $K$, which gives equality in the root count. For condition (i), we need $L/F$ to be separable. The key fact is that if $\beta \in L$ has separable minimal polynomial $P_{\beta,K}$ over $K$, then its minimal polynomial $P_{\beta,F}$ over $F \supset K$ also has no repeated roots. Why? Because $P_{\beta,F}$ divides $P_{\beta,K}$ in $F[t]$ (since $P_{\beta,K} \in K[t] \subset F[t]$ and $P_{\beta,K}(\beta) = 0$, the minimal polynomial over the larger field divides it). The roots of $P_{\beta,F}$ are a subset of the roots of $P_{\beta,K}$, and since $P_{\beta,K}$ has all distinct roots, so does $P_{\beta,F}$. **Why does (i) and (ii) imply separability of $L/K$?** This is the harder direction. We cannot simply say "separability of $L/F$ and separability of $P_\alpha$ imply separability of $L/K$" without proof — we need to verify that every element $\beta \in L$ has separable minimal polynomial over $K$, not just over $F$. The argument uses the counting theorem itself as a tool. We know $|\operatorname{Hom}_K(L, \bar{K})| = [L:K]$ (since we have equality). For any $\beta \in L$, we consider the tower $K \subset K(\beta) \subset L$ and count $\operatorname{Hom}_K(L, \bar{K})$ by fibres over $\operatorname{Hom}_K(K(\beta), \bar{K})$. Each fibre has at most $[L : K(\beta)]$ elements (by the inequality, which we have already established), and there are at most $[K(\beta) : K]$ choices for $\rho \in \operatorname{Hom}_K(K(\beta), \bar{K})$. The product $[K(\beta):K] \cdot [L:K(\beta)] = [L:K]$ by the Tower Law. Since the total count equals $[L:K]$, both inequalities must be tight. In particular, $|\operatorname{Hom}_K(K(\beta), \bar{K})| = [K(\beta):K]$, which means $P_{\beta,K}$ has $\deg P_{\beta,K}$ distinct roots — i.e., $P_{\beta,K}$ is separable. Why is the restriction map $\operatorname{Hom}_K(L, \bar{K}) \to \operatorname{Hom}_K(K(\beta), \bar{K})$ surjective? Given $\rho \colon K(\beta) \to \bar{K}$, we need to extend $\rho$ to all of $L$. The extension $L / K(\beta)$ is finite (as a sub-extension of $L/K$). Since $\bar{K}$ is algebraically closed, it contains roots for every polynomial over $\rho(K(\beta))$. By iterated application of the Roots-Homomorphisms Correspondence — adjoin one generator at a time and choose a root of its minimal polynomial in $\bar{K}$ — we extend $\rho$ to a $K$-homomorphism $L \to \bar{K}$. The fibre over $\rho$ is nonempty, so the restriction map is surjective. This is a beautiful instance of a "squeezing" argument: knowing the exact total count and having an upper bound on each summand forces every summand to achieve its maximum. The technique appears frequently in combinatorial and algebraic counting problems.[/guided]