Finitely generated extensions of a perfect field are separably generated

Running through the finite list ${\alpha }_1,\dots ,{\alpha }_n$ and adjoining an element exactly when it is transcendental over the field generated by the elements already adjoined produces, after finitely many steps, an algebraically independent $S\subseteq K$ over which every ${\alpha }_i$ is algebraic; then $K$ is algebraic over $k(S)$ and finitely generated over it, hence finite over $k(S)$ by [F3], so $S$ is a transcendence basis and $[K:k(S){]}_i$ is defined by [F7]. The set of values $[K:k(T){]}_i$, taken over the finite, algebraically independent $T\subseteq K$ with $K\mathrm{/}k(T)$ finite, is a nonempty set of natural numbers and therefore has a least element; fix $T=\{x_1,\dots ,x_d\}$ attaining it, and note that each such $T$ is a transcendence basis of $K\mathrm{/}k$ by [F2]. If $\mathrm{char}k=0$, then $k(T)$ also has characteristic zero, so is perfect by [F4], and $K\mathrm{/}k(T)$ is separable by [F5], so $T$ is already a separating transcendence basis. Henceforth assume $\mathrm{char}k=p>0$. Then Frobenius is surjective on $k$ by [F4], and $k$-linearly independent elements $a_1,\dots ,a_m\in K$ have $k$-linearly independent $p$-th powers: from ${\sum }_i{c}_i{a}_i^p=0$ with $c_i\in k$ and $b_i^p=c_i$ one gets ${\sum }_i(b_i{a}_i{)}^p=0$, hence ${\sum }_i{b}_i{a}_i=0$ because $x\mapsto x^p$ is additive by [F17] and injective on the field $K$, so all $b_i$, hence all $c_i$, vanish.

Suppose $K\mathrm{/}k(T)$ is not separable. By [F6] there is $x_{d+1}:=x\in K$ that is not separable over $K^{\mathrm{\prime }}:=k(x_1,\dots ,x_d)$, and $x$ is algebraic over $K^{\mathrm{\prime }}$ by [F2] since $K\mathrm{/}{K}^{\mathrm{\prime }}$ is algebraic. Choose $F\in k[X_1,\dots ,X_{d+1}]$ nonzero of least total degree with $F(x_1,\dots ,x_{d+1})=0$. Then $F$ is irreducible: a factorisation $F=GH$ with $G,H$ nonconstant satisfies $\mathrm{deg}F=\mathrm{deg}G+\mathrm{deg}H$, and either $G$ or $H$ vanishes at $(x_1,\dots ,x_{d+1})$ with strictly smaller total degree, contradicting minimality.

Suppose every monomial exponent of $F$ were divisible by $p$. Since $k$ is perfect, write each coefficient ${\lambda }_{\alpha }=b_{\alpha }^p$ using [F4]. Then $F=G^p$ for the nonzero polynomial $G={\sum }_{\alpha }{b}_{\alpha }{X}^{\alpha \mathrm{/}p}$, by Frobenius additivity [F17]. Evaluation gives $G(x_1,\dots ,x_{d+1}{)}^p=0$ in the field $K$, hence $G(x_1,\dots ,x_{d+1})=0$, contradicting minimality because $\mathrm{deg}G=\mathrm{deg}F\mathrm{/}p<\mathrm{deg}F$. Thus some variable $X_j$ occurs with an exponent not divisible by $p$; fix such $j$.

Set $L:=k(x_i:1\le i\le d+1,i\mathrm{\ne }j)$ and $T^{\mathrm{\prime }\mathrm{\prime }}:=\{x_i:i\mathrm{\ne }j\}$. Write $F={\sum }_{\nu =0}^e{C}_{\nu }(X_i:i\mathrm{\ne }j)X_j^{\nu }$ with $e\ge 1$ and $C_e\mathrm{\ne }0$. The total degree of $C_e$ is strictly less than that of $F$, so $C_e(x_i:i\mathrm{\ne }j)\mathrm{\ne }0$ by the minimality in step 2.1. Thus $P(T):=F(x_i:i\mathrm{\ne }j;T)$ is a nonconstant polynomial in $L[T]$ vanishing at $x_j$, which proves $x_j$ algebraic over $L$. This makes $T^{\mathrm{\prime }\mathrm{\prime }}$ a transcendence basis of $K\mathrm{/}k$: if $T^{\mathrm{\prime }\mathrm{\prime }}$ were dependent, a maximal independent subset $S⊊T^{\mathrm{\prime }\mathrm{\prime }}$ would be a transcendence basis of $L$ over $k$, and since $L(x_j)=k(x_1,\dots ,x_{d+1})$ is algebraic over $L$ by the preceding coefficient argument, the same finite set $S$ would be a transcendence basis of $k(x_1,\dots ,x_{d+1})$ over $k$ with $\mathrm{\mid }S\mathrm{\mid }<d=\mathrm{\mid }\{x_1,\dots ,x_d\}\mathrm{\mid }$, contradicting that $\{x_1,\dots ,x_d\}$ is a transcendence basis of that field and that all transcendence bases of a finitely generated extension have the same cardinality [F15]; so $T^{\mathrm{\prime }\mathrm{\prime }}$ is independent, $K\mathrm{/}L$ is algebraic, and $L$ is a rational function field over $k$ in the variables $x_i$, $i\mathrm{\ne }j$. Consequently $P(T):=F(x_i:i\mathrm{\ne }j;T)\in L[T]$ is the image of the polynomial $F\in k[X_i:i\mathrm{\ne }j][X_j]$, which has positive $X_j$-degree and is irreducible in $k[X_i:i\mathrm{\ne }j][X_j]=k[X_1,\dots ,X_{d+1}]$; its coefficients are primitive, since a nonunit common factor would factor the irreducible $F$ of positive $X_j$-degree. Thus Gauss' lemma [F12] makes its image irreducible in $L[T]$. Since some $X_j$-exponent of $F$ is not divisible by $p$, the same exponent occurs in $P$, so $P^{\mathrm{\prime }}\mathrm{\ne }0$ and $\gcd (P,P^{\mathrm{\prime }})=1$ because $P$ is irreducible and $\mathrm{deg}{P}^{\mathrm{\prime }}<\mathrm{deg}P$; thus $P$ is separable by [F13]. As $P(x_j)=0$ and $P$ is irreducible, $P$ is a nonzero scalar multiple of the monic minimal polynomial of $x_j$ over $L$ by [F11], so $x_j$ is separable over $L$ and $L(x_j)\mathrm{/}L$ is separable by [F14].

By [F8] and [F9] the inseparable degree is multiplicative, $[E:F{]}_i=[E:M{]}_i[M:F{]}_i$ for finite $F\subseteq M\subseteq E$; and $[L(x_j):L{]}_i=1$ by [F10]. With $K^{\mathrm{\prime }}=k(T)$ and $L(x_j)=K^{\mathrm{\prime }}(x_{d+1})$ we get $[K:L{]}_i=[K:L(x_j){]}_i\cdot [L(x_j):L{]}_i=[K:L(x_j){]}_i$ and $[K:K^{\mathrm{\prime }}{]}_i=[K:L(x_j){]}_i\cdot [L(x_j):K^{\mathrm{\prime }}{]}_i$, while $[L(x_j):K^{\mathrm{\prime }}{]}_i>1$ by [F10] because $K^{\mathrm{\prime }}(x_{d+1})\mathrm{/}{K}^{\mathrm{\prime }}$ is not separable. Therefore $[K:L{]}_i<[K:K^{\mathrm{\prime }}{]}_i$, and $T^{\mathrm{\prime }\mathrm{\prime }}$ is a transcendence basis of $K\mathrm{/}k$ by step 4.1 with a strictly smaller inseparable degree than $T$, contradicting the minimality of step 1.1. Hence $K\mathrm{/}k(T)$ is separable and $T$ is a separating transcendence basis, which proves the theorem; moreover, by [F16] the residual finite separable extension is simple, so $K=k(T)(\gamma )$ for a single element $\gamma$ separable over $k(T)$.