[proofplan] The strategy is to extract a transcendence basis from a generating set. Given a finite generating set of $K/k$, the existence of a maximal algebraically independent subset follows from a finite descent argument: starting with all generators, repeatedly remove any element algebraic over the rest. The remaining set is then a transcendence basis $\{x_1, \ldots, x_n\}$, and $K_0 := k(x_1, \ldots, x_n)$ is purely transcendental by construction. Each remaining generator is algebraic over $K_0$ by maximality, and finiteness of $K/K_0$ follows because $K$ is generated over $K_0$ by finitely many algebraic elements. Separability in characteristic zero is automatic because every algebraic extension of a characteristic-zero field is separable. [/proofplan] [step:Fix a finite generating set of $K/k$] By hypothesis, $K$ is finitely generated over $k$ as a field. Choose generators \begin{align*} K = k(x_1, \ldots, x_m) \end{align*} for some $m \geq 1$. Since $K/k$ is transcendental, the set $\{x_1, \ldots, x_m\}$ contains at least one element transcendental over $k$ (otherwise every $x_i$ would be algebraic over $k$, making $K = k(x_1, \ldots, x_m)$ algebraic over $k$, contradicting the transcendental hypothesis). [/step] [step:Extract a maximal algebraically independent subset] We claim there is a non-empty subset $S \subset \{x_1, \ldots, x_m\}$ that is algebraically independent over $k$ and maximal among such subsets, in the sense that no proper superset $S \subsetneq S' \subset \{x_1, \ldots, x_m\}$ is algebraically independent. Existence of $S$: Among all algebraically independent subsets of $\{x_1, \ldots, x_m\}$ over $k$ (the empty set is algebraically independent), the collection is finite (there are only $2^m$ subsets in total), so a subset of maximal cardinality exists. Any such subset is maximal in the sense above. By the previous step, at least one $x_i$ is transcendental over $k$, so the singleton $\{x_i\}$ is algebraically independent and $|S| \geq 1$. By reindexing, we may assume $S = \{x_1, \ldots, x_n\}$ for some $1 \leq n \leq m$. Set \begin{align*} K_0 := k(x_1, \ldots, x_n) \subset K. \end{align*} By construction, $\{x_1, \ldots, x_n\}$ is algebraically independent over $k$, so the field extension $K_0/k$ is purely transcendental. [/step] [step:Show that every $x_j$ with $j > n$ is algebraic over $K_0$] Fix $j \in \{n+1, \ldots, m\}$. By maximality of $\{x_1, \ldots, x_n\}$, the set $\{x_1, \ldots, x_n, x_j\}$ is algebraically dependent over $k$: there exists a non-zero polynomial $P \in k[Y_1, \ldots, Y_n, Y_{n+1}]$ such that \begin{align*} P(x_1, \ldots, x_n, x_j) = 0. \end{align*} Write $P$ as a polynomial in $Y_{n+1}$ with coefficients in $k[Y_1, \ldots, Y_n]$: \begin{align*} P(Y_1, \ldots, Y_n, Y_{n+1}) = \sum_{r=0}^{d} c_r(Y_1, \ldots, Y_n) Y_{n+1}^r, \end{align*} where $d = \deg_{Y_{n+1}} P \geq 0$ and at least one $c_r$ is non-zero. If $d = 0$, then $P \in k[Y_1, \ldots, Y_n]$ is non-zero and $P(x_1, \ldots, x_n) = 0$, contradicting the algebraic independence of $\{x_1, \ldots, x_n\}$. Hence $d \geq 1$. Substituting and viewing $P(x_1, \ldots, x_n, Y_{n+1})$ as a polynomial in $Y_{n+1}$ with coefficients in $K_0$: \begin{align*} \tilde P(Y_{n+1}) := \sum_{r=0}^{d} c_r(x_1, \ldots, x_n)\, Y_{n+1}^r \in K_0[Y_{n+1}]. \end{align*} The leading coefficient $c_d(x_1, \ldots, x_n) \in K_0$ is non-zero (else $\tilde P$ would have lower degree, but the largest-$r$ non-zero $c_r$ contributes a non-zero leading coefficient because $\{x_1, \ldots, x_n\}$ is algebraically independent over $k$, so $c_r(x_1, \ldots, x_n) = 0$ would force $c_r = 0$ as a polynomial). Hence $\tilde P$ is a non-zero polynomial in $K_0[Y_{n+1}]$ of degree $d \geq 1$, and $\tilde P(x_j) = P(x_1, \ldots, x_n, x_j) = 0$. Therefore $x_j$ is algebraic over $K_0$. [guided] The argument in this step is a standard manoeuvre: convert algebraic dependence over the ground field $k$ into algebraicity over the intermediate field $K_0$. Let us spell out why each detail is needed. Algebraic dependence of $\{x_1, \ldots, x_n, x_j\}$ over $k$ means: there is a non-zero polynomial relation $P(x_1, \ldots, x_n, x_j) = 0$ with $P \in k[Y_1, \ldots, Y_n, Y_{n+1}]$ non-zero. Why must such a $P$ exist? Because $\{x_1, \ldots, x_n\}$ was chosen maximal among algebraically independent subsets — adjoining $x_j$ destroys algebraic independence, so a non-trivial polynomial relation appears. We separate the variable $x_j$ from the others by writing $P$ as a polynomial in $Y_{n+1}$ with polynomial coefficients in $Y_1, \ldots, Y_n$: \begin{align*} P = c_0(Y_1, \ldots, Y_n) + c_1(Y_1, \ldots, Y_n) Y_{n+1} + \cdots + c_d(Y_1, \ldots, Y_n) Y_{n+1}^d. \end{align*} This decomposition is just regrouping monomials. The non-trivial point: when we substitute $Y_i \to x_i$ for $i \leq n$, do the coefficients $c_r(x_1, \ldots, x_n)$ remain non-zero? In general, polynomial substitution can collapse a non-zero polynomial to zero. But here, $\{x_1, \ldots, x_n\}$ is algebraically independent over $k$, so the substitution map \begin{align*} k[Y_1, \ldots, Y_n] &\hookrightarrow K_0 \\ c(Y_1, \ldots, Y_n) &\mapsto c(x_1, \ldots, x_n) \end{align*} is injective. Hence $c_r \neq 0$ as a polynomial implies $c_r(x_1, \ldots, x_n) \neq 0$ in $K_0$. This is exactly the role of algebraic independence: it gives us a faithful copy of $k[Y_1, \ldots, Y_n]$ inside $K_0$. Now consider the resulting polynomial in $Y_{n+1}$ with coefficients in $K_0$: \begin{align*} \tilde P(Y_{n+1}) := P(x_1, \ldots, x_n, Y_{n+1}) \in K_0[Y_{n+1}]. \end{align*} We claim $\tilde P \neq 0$. If $\tilde P$ were the zero polynomial, then every coefficient $c_r(x_1, \ldots, x_n)$ would be zero in $K_0$, hence each $c_r = 0$ as a polynomial in $k[Y_1, \ldots, Y_n]$ by the injectivity above, hence $P = 0$ — contradicting non-vanishing of $P$. We further need $\deg_{Y_{n+1}} \tilde P \geq 1$. If $d = 0$, then $P = c_0(Y_1, \ldots, Y_n)$ depends only on $Y_1, \ldots, Y_n$, so $P(x_1, \ldots, x_n, x_j) = c_0(x_1, \ldots, x_n) = 0$ would force $c_0 = 0$ by algebraic independence — but $P \neq 0$, contradiction. So $d \geq 1$, and $\tilde P$ is a non-zero polynomial of positive degree over $K_0$ that vanishes at $x_j$. By definition, $x_j$ is algebraic over $K_0$. [/guided] [/step] [step:Conclude that $K/K_0$ is finite] By Step 3, each $x_j$ for $j \in \{n+1, \ldots, m\}$ is algebraic over $K_0$. The field $K = k(x_1, \ldots, x_m) = K_0(x_{n+1}, \ldots, x_m)$ is therefore obtained from $K_0$ by adjoining finitely many algebraic elements. By a standard result on finite field extensions (a finitely generated algebraic extension is finite), $K/K_0$ is a finite extension: \begin{align*} [K : K_0] < \infty. \end{align*} Indeed, building up step by step, $[K_0(x_{n+1}) : K_0]$ is finite (equal to the degree of the minimal polynomial of $x_{n+1}$ over $K_0$), and inductively \begin{align*} [K_0(x_{n+1}, \ldots, x_{n+s}) : K_0(x_{n+1}, \ldots, x_{n+s-1})] < \infty \end{align*} for each $s$, so the total degree $[K : K_0] = \prod_{s=1}^{m-n} [K_0(x_{n+1}, \ldots, x_{n+s}) : K_0(x_{n+1}, \ldots, x_{n+s-1})]$ is finite by the multiplicativity of degree in field-extension towers. [/step] [step:Conclude separability when $\operatorname{char}(k) = 0$] Suppose $\operatorname{char}(k) = 0$. Every algebraic field extension of a characteristic-zero field is separable: the minimal polynomial of any algebraic element has distinct roots in an algebraic closure, equivalently $\gcd(f, \partial f / \partial Y) = 1$, which holds because $f$ being irreducible and $\partial f / \partial Y \neq 0$ (the polynomial $f$ is not a polynomial in $Y^p$ for $p = \operatorname{char} k$, since there is no such $p$) forces $\partial f / \partial Y$ to be coprime to $f$. Applied to the finite (hence algebraic) extension $K/K_0$, we conclude that $K/K_0$ is separable. [/step]