[proofplan] We prove the cycle $(2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1) \Rightarrow (2)$. The implication $(2) \Rightarrow (3)$ is immediate from the definition of separable extension. For $(3) \Rightarrow (4)$, we show that if $P_{\alpha_i}$ is separable over $K$, then its minimal polynomial over the larger field $K(\alpha_1, \ldots, \alpha_{i-1})$ — being an irreducible factor of $P_{\alpha_i}$ — inherits separability because its roots form a subset of the roots of $P_{\alpha_i}$, which are all distinct. The implication $(4) \Rightarrow (1)$ iterates the [Roots-Homomorphisms Correspondence](/theorems/1256) along the tower $K \subset K(\alpha_1) \subset \cdots \subset K(\alpha_1, \ldots, \alpha_n) = L$: separability at each step forces equality at each level, yielding $|\operatorname{Hom}_K(L, \bar{K})| = [L:K]$. Finally, $(1) \Rightarrow (2)$ reverses this tower analysis: if equality $|\operatorname{Hom}_K(L, E)| = [L:K]$ holds globally, then the multiplicative structure of the counting argument forces equality at every intermediate step, which by the Roots-Homomorphisms Correspondence requires every minimal polynomial to have the maximum number of distinct roots, i.e., to be separable. [/proofplan]

[step:$(2) \Rightarrow (3)$: Every set of generators of a separable extension has separable minimal polynomials]Assume $L/K$ is separable, meaning that every $\alpha \in L$ has separable minimal polynomial $P_\alpha$ over $K$. Since $L/K$ is a finite extension, we may write $L = K(\alpha_1, \ldots, \alpha_n)$ for some finite generating set $\alpha_1, \ldots, \alpha_n \in L$ (for instance, a $K$-basis of $L$). By the definition of separability of the extension, each $\alpha_i \in L$ has separable minimal polynomial $P_{\alpha_i}$ over $K$. This is precisely condition (3).[/step]

[guided]This implication is direct from the definitions, but it is worth pausing to see exactly what is being said. Condition (2) states that $L/K$ is separable, which means: *every* element of $L$ has a minimal polynomial over $K$ with no repeated roots. Condition (3) asks for something that appears weaker — it only requires *some* generating set $\{\alpha_1, \ldots, \alpha_n\}$ of $L$ over $K$ such that each generator has a separable minimal polynomial over $K$. Since $L/K$ is finite, $L$ is finitely generated over $K$ as a field extension: we can take any $K$-basis of $L$, say $\{v_1, \ldots, v_m\}$ where $m = [L:K]$, and then $L = K(v_1, \ldots, v_m)$. (One may also use the Primitive Element Theorem when it applies to get a single generator, but it is not needed here.) Each $v_i \in L$, so by condition (2), the minimal polynomial $P_{v_i}$ of $v_i$ over $K$ is separable. Setting $\alpha_i := v_i$ and $n := m$ gives condition (3). The key observation for later: the converse direction — from (3) back to (2) — is where the real work lies. Condition (3) only requires *some* generators to have separable minimal polynomials over $K$, whereas (2) requires *all* elements of $L$ to be separable over $K$. The equivalence of these two conditions is established by the full cycle of implications.[/guided]

[step:$(3) \Rightarrow (4)$: Separability over $K$ implies separability over intermediate fields]Assume $L = K(\alpha_1, \ldots, \alpha_n)$ where each $P_{\alpha_i}$ is separable over $K$. We must show that for each $i \in \{1, \ldots, n\}$, the minimal polynomial of $\alpha_i$ over $K_i := K(\alpha_1, \ldots, \alpha_{i-1})$ is separable (where $K_0 := K$). Fix $i \in \{1, \ldots, n\}$. Let $Q_i \in K_i[t]$ denote the minimal polynomial of $\alpha_i$ over $K_i$. Since $P_{\alpha_i} \in K[t] \subset K_i[t]$ and $P_{\alpha_i}(\alpha_i) = 0$, the minimal polynomial $Q_i$ divides $P_{\alpha_i}$ in $K_i[t]$. That is, there exists $R \in K_i[t]$ such that \begin{align*} P_{\alpha_i}(t) = Q_i(t) \cdot R(t). \end{align*} Let $\bar{K}$ be an algebraic closure of $L$. Since $P_{\alpha_i}$ is separable over $K$, its $\deg P_{\alpha_i}$ roots in $\bar{K}$ are all distinct. Every root of $Q_i$ in $\bar{K}$ is also a root of $P_{\alpha_i}$, and the factorisation $P_{\alpha_i} = Q_i \cdot R$ in $K_i[t] \subset \bar{K}[t]$ shows that no root of $Q_i$ can appear with multiplicity greater than its multiplicity in $P_{\alpha_i}$, which is $1$. Therefore $Q_i$ has $\deg Q_i$ distinct roots in $\bar{K}$, i.e., $Q_i$ is separable.[/step]

[guided]The idea behind this step is straightforward but deserves careful attention. We are given that $P_{\alpha_i}$ has no repeated roots (separability over $K$), and we need to show that $Q_i$ — the minimal polynomial of $\alpha_i$ over the *larger* base field $K_i = K(\alpha_1, \ldots, \alpha_{i-1})$ — also has no repeated roots. Why should enlarging the base field preserve separability? When we pass from $K$ to $K_i$, the polynomial $P_{\alpha_i}$ may factor further (since $K_i[t]$ contains more potential factors than $K[t]$). The minimal polynomial $Q_i$ of $\alpha_i$ over $K_i$ is an irreducible factor of $P_{\alpha_i}$ in $K_i[t]$, so $\deg Q_i \le \deg P_{\alpha_i}$, with equality only when $P_{\alpha_i}$ remains irreducible over $K_i$. The point is that factoring a polynomial into irreducible pieces over a larger field *partitions* its roots (in an algebraic closure) among the factors. Formally: since $P_{\alpha_i} = Q_i \cdot R$ in $K_i[t]$, and this factorisation also holds in $\bar{K}[t]$ (where $\bar{K}$ is an algebraic closure containing $L$), we can write \begin{align*} P_{\alpha_i}(t) = Q_i(t) \cdot R(t) \end{align*} in $\bar{K}[t]$. Now in $\bar{K}[t]$, we can factor completely into linear factors. Each root of $P_{\alpha_i}$ appears with multiplicity exactly $1$ (by separability of $P_{\alpha_i}$ over $K$). Each root of $Q_i$ in $\bar{K}$ is also a root of $P_{\alpha_i}$, so it has multiplicity $1$ as a root of $P_{\alpha_i}$. Since $Q_i$ divides $P_{\alpha_i}$, the multiplicity of any root $\beta$ in $Q_i$ is at most the multiplicity of $\beta$ in $P_{\alpha_i}$, which is $1$. Therefore $Q_i$ has no repeated roots, i.e., $Q_i$ is separable over $K_i$. This argument shows a general principle: **an irreducible factor of a separable polynomial is separable**. The roots of the factor form a subset of the roots of the original polynomial, and a subset of a set of distinct elements consists of distinct elements.[/guided]

[step:$(4) \Rightarrow (1)$: Iterate the Roots-Homomorphisms Correspondence along the tower to count $[L:K]$ embeddings] Assume $L = K(\alpha_1, \ldots, \alpha_n)$ where, for each $i$, the minimal polynomial of $\alpha_i$ over $K_{i-1} := K(\alpha_1, \ldots, \alpha_{i-1})$ is separable (with $K_0 := K$). Set $K_i := K(\alpha_1, \ldots, \alpha_i)$ for $i = 0, 1, \ldots, n$, so that $K_0 = K$ and $K_n = L$, giving a tower \begin{align*} K = K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_n = L. \end{align*} Let $\bar{K}$ denote an algebraic closure of $K$ containing $L$. We set $E := \bar{K}$ and count $|\operatorname{Hom}_K(L, E)|$. Each $K$-homomorphism $\sigma \colon L \to \bar{K}$ restricts to a $K$-homomorphism $\sigma|_{K_{i-1}} \colon K_{i-1} \to \bar{K}$ for each $i$. Since $K_i = K_{i-1}(\alpha_i)$ is a simple extension, the [Roots-Homomorphisms Correspondence](/theorems/1256) applied to the extension $K_i / K_{i-1}$ (viewed relative to a fixed embedding of $K_{i-1}$) tells us that the number of extensions of any fixed $K$-homomorphism $\tau \colon K_{i-1} \to \bar{K}$ to a $K$-homomorphism $K_i \to \bar{K}$ equals \begin{align*} |\{\beta \in \bar{K} : \tau(Q_i)(\beta) = 0\}|, \end{align*} where $Q_i \in K_{i-1}[t]$ is the minimal polynomial of $\alpha_i$ over $K_{i-1}$, and $\tau(Q_i) \in \bar{K}[t]$ is the polynomial obtained by applying $\tau$ to the coefficients of $Q_i$. Since $\tau \colon K_{i-1} \to \bar{K}$ is a field homomorphism, it preserves the property of having distinct roots: $Q_i$ is separable over $K_{i-1}$ (by hypothesis (4)), and $\tau(Q_i)$ has the same degree as $Q_i$ (since $\tau$ is injective on $K_{i-1}$ and preserves the leading coefficient). The roots of $\tau(Q_i)$ in $\bar{K}$ are distinct because $\tau$ induces an isomorphism $K_{i-1} \xrightarrow{\sim} \tau(K_{i-1})$, and separability is preserved under isomorphism: if $Q_i$ has $\deg Q_i$ distinct roots in an algebraic closure of $K_{i-1}$, then $\tau(Q_i)$ has $\deg Q_i$ distinct roots in an algebraic closure of $\tau(K_{i-1})$, which is contained in $\bar{K}$ (since $\bar{K}$ is algebraically closed). Therefore \begin{align*} |\{\beta \in \bar{K} : \tau(Q_i)(\beta) = 0\}| = \deg Q_i = [K_i : K_{i-1}]. \end{align*} Iterating over $i = 1, 2, \ldots, n$: starting from the unique $K$-homomorphism $K_0 = K \hookrightarrow \bar{K}$ (the inclusion), at each step every existing homomorphism $\tau \colon K_{i-1} \to \bar{K}$ extends in exactly $[K_i : K_{i-1}]$ ways to $K_i$. The total count is \begin{align*} |\operatorname{Hom}_K(L, \bar{K})| = \prod_{i=1}^{n} [K_i : K_{i-1}] = [K_n : K_0] = [L : K], \end{align*} where the last equality is the Tower Law applied iteratively. Taking $E := \bar{K}$ gives condition (1).[/step]

[guided]This is the heart of the counting argument, and it proceeds by iterating a single principle — the [Roots-Homomorphisms Correspondence](/theorems/1256) — along a tower of simple extensions. **Setting up the tower.** We decompose $L/K$ into a chain of simple extensions by adjoining one generator at a time: \begin{align*} K = K_0 \subset K_1 = K(\alpha_1) \subset K_2 = K(\alpha_1, \alpha_2) \subset \cdots \subset K_n = K(\alpha_1, \ldots, \alpha_n) = L. \end{align*} At each level, $K_i = K_{i-1}(\alpha_i)$ is a simple extension of $K_{i-1}$, and by hypothesis (4), the minimal polynomial $Q_i$ of $\alpha_i$ over $K_{i-1}$ is separable. **The iteration.** We count $|\operatorname{Hom}_K(L, \bar{K})|$ by building up homomorphisms one step at a time. *Base case ($i = 0$):* There is exactly one $K$-homomorphism from $K_0 = K$ to $\bar{K}$, namely the inclusion $\iota \colon K \hookrightarrow \bar{K}$. *Step $i$:* Given any $K$-homomorphism $\tau \colon K_{i-1} \to \bar{K}$, we ask: in how many ways can $\tau$ be extended to a $K$-homomorphism $\tilde{\tau} \colon K_i = K_{i-1}(\alpha_i) \to \bar{K}$? The [Roots-Homomorphisms Correspondence](/theorems/1256) provides the answer: extensions of $\tau$ to $K_i$ are in bijection with roots of $\tau(Q_i)$ in $\bar{K}$, via the map $\tilde{\tau} \mapsto \tilde{\tau}(\alpha_i)$. The polynomial $\tau(Q_i)$ is obtained by applying $\tau$ to each coefficient of $Q_i$. Since $\tau$ is a field homomorphism (hence injective), $\tau(Q_i)$ has the same degree as $Q_i$. Moreover, $Q_i$ is separable over $K_{i-1}$ by hypothesis, meaning $Q_i$ has $\deg Q_i$ distinct roots in any algebraic closure of $K_{i-1}$. The isomorphism $\tau \colon K_{i-1} \xrightarrow{\sim} \tau(K_{i-1})$ carries this splitting to $\tau(Q_i)$: concretely, if $\beta_1, \ldots, \beta_d$ are the distinct roots of $Q_i$ in some algebraic closure $\overline{K_{i-1}}$, then $\tau(\beta_1), \ldots, \tau(\beta_d)$ are distinct roots of $\tau(Q_i)$ (here $\tau$ extends to $\overline{K_{i-1}}$). Since $\bar{K}$ is algebraically closed and contains $\tau(K_{i-1})$, the polynomial $\tau(Q_i)$ has all $d = \deg Q_i$ roots in $\bar{K}$, and they are distinct. Therefore $\tau$ extends in exactly $\deg Q_i = [K_i : K_{i-1}]$ ways to $K_i$. **Multiplying through the tower.** At each stage $i$, every homomorphism $\tau \colon K_{i-1} \to \bar{K}$ gives rise to exactly $[K_i : K_{i-1}]$ extensions to $K_i$. Starting from the single inclusion $K \hookrightarrow \bar{K}$ and iterating: \begin{align*} |\operatorname{Hom}_K(L, \bar{K})| &= \prod_{i=1}^{n} [K_i : K_{i-1}] = [K_n : K_0] = [L : K], \end{align*} where the product telescopes by the Tower Law: $[L:K] = [K_n : K_{n-1}] \cdots [K_1 : K_0]$. This achieves equality $|\operatorname{Hom}_K(L, \bar{K})| = [L:K]$, which is condition (1) with $E = \bar{K}$. **Why separability is consumed here.** The argument would fail at step $i$ if $Q_i$ were inseparable: in that case, $\tau(Q_i)$ would have fewer than $\deg Q_i$ roots in $\bar{K}$ (due to repeated roots), and $\tau$ would extend in strictly fewer than $[K_i : K_{i-1}]$ ways. The product would then be strictly less than $[L:K]$, giving only the inequality $|\operatorname{Hom}_K(L, \bar{K})| < [L:K]$.[/guided]

[step:$(1) \Rightarrow (2)$: Full embedding count forces every element to be separable]Assume there exists a field $E \supseteq K$ such that $|\operatorname{Hom}_K(L, E)| = [L:K]$. We must show that every $\alpha \in L$ has separable minimal polynomial over $K$. Fix $\alpha \in L$ and let $P_\alpha \in K[t]$ denote its minimal polynomial over $K$. Consider the tower $K \subset K(\alpha) \subset L$ and the corresponding restriction map \begin{align*} \operatorname{Hom}_K(L, E) &\to \operatorname{Hom}_K(K(\alpha), E) \\ \sigma &\mapsto \sigma|_{K(\alpha)}. \end{align*} [claim:The fibre-counting inequality forces $|\operatorname{Hom}_K(K(\alpha), E)| = [K(\alpha) : K]$] The equality $|\operatorname{Hom}_K(L, E)| = [L:K]$ implies $|\operatorname{Hom}_K(K(\alpha), E)| = [K(\alpha) : K]$. [/claim] [proof] For each $\tau \in \operatorname{Hom}_K(K(\alpha), E)$, let $F_\tau := \{\sigma \in \operatorname{Hom}_K(L, E) : \sigma|_{K(\alpha)} = \tau\}$ denote the fibre over $\tau$. Each $\sigma \in \operatorname{Hom}_K(L, E)$ lies in exactly one fibre, so \begin{align*} |\operatorname{Hom}_K(L, E)| = \sum_{\tau \in \operatorname{Hom}_K(K(\alpha), E)} |F_\tau|. \end{align*} By the [Bound on $K$-Homomorphisms](/theorems/1257) applied to the extension $L / K(\alpha)$ (viewed from the base $\tau(K(\alpha))$, noting that each element of $F_\tau$ is a $\tau(K(\alpha))$-homomorphism from $\tau$-twisted $L$ into $E$), we have $|F_\tau| \le [L : K(\alpha)]$ for each $\tau$. Therefore \begin{align*} [L:K] = |\operatorname{Hom}_K(L, E)| = \sum_{\tau} |F_\tau| \le |\operatorname{Hom}_K(K(\alpha), E)| \cdot [L : K(\alpha)]. \end{align*} By the [Roots-Homomorphisms Correspondence](/theorems/1256), $|\operatorname{Hom}_K(K(\alpha), E)| = |\operatorname{Root}_{P_\alpha}(E)| \le \deg P_\alpha = [K(\alpha) : K]$. Substituting: \begin{align*} [L:K] \le [K(\alpha) : K] \cdot [L : K(\alpha)] = [L:K], \end{align*} where the last equality is the Tower Law. Therefore equality holds throughout. [/proof] Since equality holds in the chain \begin{align*} [L:K] \le |\operatorname{Hom}_K(K(\alpha), E)| \cdot [L : K(\alpha)] = [K(\alpha):K] \cdot [L:K(\alpha)] = [L:K], \end{align*} and $|\operatorname{Hom}_K(K(\alpha), E)| \le [K(\alpha) : K]$, we deduce \begin{align*} |\operatorname{Hom}_K(K(\alpha), E)| = [K(\alpha) : K] = \deg P_\alpha. \end{align*} By the [Roots-Homomorphisms Correspondence](/theorems/1256), this means $|\operatorname{Root}_{P_\alpha}(E)| = \deg P_\alpha$, so $P_\alpha$ has $\deg P_\alpha$ distinct roots in $E$. A polynomial with as many distinct roots as its degree has no repeated roots, hence $P_\alpha$ is separable. Since $\alpha \in L$ was arbitrary, every element of $L$ has separable minimal polynomial over $K$, i.e., $L/K$ is separable.[/step]

[guided]This is the most subtle direction. We are given the *global* information that $|\operatorname{Hom}_K(L, E)| = [L:K]$, and we must deduce the *local* information that every single element of $L$ is separable over $K$. The strategy is to show that the global equality forces equality at every level of any tower decomposition. **Fixing an arbitrary element.** Let $\alpha \in L$ be arbitrary. We consider the tower $K \subset K(\alpha) \subset L$ and analyse how the $[L:K]$ homomorphisms distribute across this tower. **The fibre-counting argument.** Each $\sigma \in \operatorname{Hom}_K(L, E)$ restricts to some $\tau := \sigma|_{K(\alpha)} \in \operatorname{Hom}_K(K(\alpha), E)$. Grouping the homomorphisms by their restriction, we obtain the partition \begin{align*} \operatorname{Hom}_K(L, E) = \bigsqcup_{\tau \in \operatorname{Hom}_K(K(\alpha), E)} F_\tau, \end{align*} where $F_\tau = \{\sigma \in \operatorname{Hom}_K(L, E) : \sigma|_{K(\alpha)} = \tau\}$ is the fibre over $\tau$. Taking cardinalities: \begin{align*} |\operatorname{Hom}_K(L, E)| = \sum_{\tau \in \operatorname{Hom}_K(K(\alpha), E)} |F_\tau|. \end{align*} **Bounding each fibre.** For a fixed $\tau \in \operatorname{Hom}_K(K(\alpha), E)$, the elements of $F_\tau$ are precisely the extensions of $\tau$ from $K(\alpha)$ to $L$. Since $\tau \colon K(\alpha) \hookrightarrow E$ is a field embedding, these extensions correspond to $\tau(K(\alpha))$-homomorphisms from (a copy of) $L$ into $E$ that extend $\tau$. By the [Bound on $K$-Homomorphisms](/theorems/1257) applied to $L / K(\alpha)$, the number of such extensions is at most $[L : K(\alpha)]$. So $|F_\tau| \le [L : K(\alpha)]$ for every $\tau$. **Forcing equality.** Now we combine the pieces. We have two inequalities: \begin{align*} |\operatorname{Hom}_K(K(\alpha), E)| &\le \deg P_\alpha = [K(\alpha) : K] \quad \text{(by the Roots-Homomorphisms Correspondence)}, \\ |F_\tau| &\le [L : K(\alpha)] \quad \text{for each } \tau. \end{align*} Multiplying the first by the maximum of the second: \begin{align*} [L:K] = |\operatorname{Hom}_K(L, E)| = \sum_\tau |F_\tau| \le |\operatorname{Hom}_K(K(\alpha), E)| \cdot [L : K(\alpha)] \le [K(\alpha):K] \cdot [L : K(\alpha)]. \end{align*} The Tower Law gives $[K(\alpha):K] \cdot [L:K(\alpha)] = [L:K]$. So the left and right sides are both $[L:K]$, and equality must hold throughout. In particular, $|\operatorname{Hom}_K(K(\alpha), E)| = [K(\alpha) : K] = \deg P_\alpha$. By the [Roots-Homomorphisms Correspondence](/theorems/1256), this means $P_\alpha$ has exactly $\deg P_\alpha$ roots in $E$, all of which are distinct (since a polynomial of degree $d$ has at most $d$ roots in a field). Therefore $P_\alpha$ is separable. **Why "for all $\alpha$" follows.** The element $\alpha \in L$ was arbitrary. The argument above applies to any choice of $\alpha$, producing a tower $K \subset K(\alpha) \subset L$ and forcing $P_\alpha$ to be separable. Therefore every element of $L$ has separable minimal polynomial over $K$, which is the definition of $L/K$ being separable. **What the argument reveals.** The key mechanism is that the global equality $|\operatorname{Hom}_K(L, E)| = [L:K]$ is *rigid*: it leaves no room for any factor in the tower to contribute fewer than the maximum number of extensions. If even a single element $\alpha$ had an inseparable minimal polynomial, the factor $|\operatorname{Hom}_K(K(\alpha), E)|$ would be strictly less than $[K(\alpha):K]$, and the product would fall below $[L:K]$.[/guided]