Integral closure in a purely inseparable rational envelope is finite

The elements $y_i:=x_i^{1\mathrm{/}q}$ are algebraically independent over $K^{\mathrm{\prime }}$. First there is an exponent $N\ge 0$ with $K^{\mathrm{\prime }{p}^N}\subseteq K$: by [L1] each element of $K^{\mathrm{\prime }}$ has a $p$-power in $K$, and by [L7] the finite extension $K^{\mathrm{\prime }}$ is spanned over $K$ by finitely many elements ${\gamma }_1,\dots ,{\gamma }_r$, so choosing $N$ as the maximum of the finitely many exponents $n_j$ with ${\gamma }_j^{p^{n_j}}\in K$ gives ${\gamma }_j^{p^N}\in K$ for every $j$; since the $p^N$-th power map is additive and multiplicative by [L2] and $K^{p^N}\subseteq K$, every element of $K^{\mathrm{\prime }{p}^N}=K^{p^N}({\gamma }_1^{p^N},\dots ,{\gamma }_r^{p^N})$ lies in $K$. Now let $P={\sum }_{\alpha }{c}_{\alpha }{Y}^{\alpha }\in K^{\mathrm{\prime }}[Y_1,\dots ,Y_d]$ satisfy $P(y_1,\dots ,y_d)=0$; raising this identity to the $p^{N}q$-th power and using additivity and multiplicativity of the Frobenius iterates [L2] together with $y_i^{p^{N}q}=(y_i^q{)}^{p^N}=x_i^{p^N}$ gives $0={\sum }_{\alpha }{c}_{\alpha }^{p^{N}q}(x_1^{p^N}{)}^{{\alpha }_1}\cdots (x_d^{p^N}{)}^{{\alpha }_d}$ with every coefficient $c_{\alpha }^{p^{N}q}$ in $K$. If a nonzero $R\in K[T_1,\dots ,T_d]$ satisfied $R(x_1^{p^N},\dots ,x_d^{p^N})=0$, then the nonzero polynomial $R(T_1^{p^N},\dots ,T_d^{p^N})$ would vanish at $x_1,\dots ,x_d$, contradicting algebraic independence of the $x_i$ over $K$; so the displayed relation forces $c_{\alpha }^{p^{N}q}=0$ for every $\alpha$, and then $c_{\alpha }=0$ because Frobenius is injective by [L2]. Hence $P=0$ and the evaluation map is injective by [L3]; it is surjective onto the $K^{\mathrm{\prime }}$-subalgebra $T:=K^{\mathrm{\prime }}[y_1,\dots ,y_d]\subseteq E$ generated by the $y_i$. Hence $T$ is isomorphic to the polynomial ring $K^{\mathrm{\prime }}[Y_1,\dots ,Y_d]$ over the field $K^{\mathrm{\prime }}$, so $T$ is an integrally closed domain by [L5], and $\mathrm{Frac}(T)=E$, because $E$ is the smallest field containing $K^{\mathrm{\prime }}$ and the $y_i$ by [L4] while $\mathrm{Frac}(T)$ is the smallest field containing $T$ and equals the field of fractions of $K^{\mathrm{\prime }}[Y_1,\dots ,Y_d]$ under the isomorphism.

$T$ is integral over $R$ and contains it: $R=K[x_1,\dots ,x_d]=K[y_1^q,\dots ,y_d^q]\subseteq K^{\mathrm{\prime }}[y_1,\dots ,y_d]=T$, every $y_i$ is a root of the monic polynomial $Z^q-x_i\in R[Z]$, and every element of the field $K^{\mathrm{\prime }}$ is algebraic, hence integral, over the field $K\subseteq R$. By [L6] the elements of $E$ integral over $R$ form a subring of $E$ containing $R$, containing $K^{\mathrm{\prime }}$ and containing every $y_i$; being a subring, it contains the subring $T$ these elements generate, so $T\subseteq \bar{R}$ for the integral closure $\bar{R}$ of $R$ in $E$. Conversely every $z\in E$ integral over $R$ is integral over $T$, since the monic equation for $z$ over $R$ has coefficients in $R\subseteq T$; as $T$ is integrally closed with fraction field $E$ by step 1.1, such $z$ lies in $T$. Hence $\bar{R}=T$.

The closure $T$ is a finite $R$-module. By [L7] fix a finite $K$-basis ${\beta }_1,\dots ,{\beta }_m$ of $K^{\mathrm{\prime }}$. Every element of $T=K^{\mathrm{\prime }}[y_1,\dots ,y_d]$ is a finite sum of terms $c{y}_1^{a_1}\cdots y_d^{a_d}$ with $c\in K^{\mathrm{\prime }}$ and $a_i\in \mathbb{N}$, and writing $c={\sum }_j{\lambda }_j{\beta }_j$ with ${\lambda }_j\in K$ and reducing exponents modulo $q$ via $y_i^{a_i}=y_i^{q\lfloor a_i\mathrm{/}q\rfloor }{y}_i^{a_i<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}=x_i^{\lfloor a_i\mathrm{/}q\rfloor }{y}_i^{a_i<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}$ shows that the finite list ${\beta }_j{y}_1^{a_1}\cdots y_d^{a_d}$ with $1\le j\le m$ and $0\le a_i<q$ generates $T$ as an $R$-module, using [L3] for the uniqueness of the expressions of elements of the polynomial ring $T$. Hence $T$ is a finitely generated $R$-module, and $R$ is Noetherian by [L8].

Let $K(x_1,\dots ,x_d)\subseteq L\subseteq E$ be an intermediate field. For $z\in L$ the monic polynomial equations over $R$ satisfied by $z$ are the same whether $z$ is regarded in $L$ or in $E$, so the integral closure of $R$ in $L$ is $\bar{R}\cap L=T\cap L$ by step 2.1. Now $T\cap L$ is an $R$-submodule of the finitely generated $R$-module $T$, and $R$ is Noetherian by [L8], so $T\cap L$ is a finitely generated, hence finite, $R$-module by [L8]. In particular, taking $L=E$ recovers the assertion that the closure of $R$ in $E$ is $T=K^{\mathrm{\prime }}[x_1^{1\mathrm{/}q},\dots ,x_d^{1\mathrm{/}q}]$, a polynomial ring over $K^{\mathrm{\prime }}$.