The Stacks project

Proof. Since $k/k_ J$ is finite, the ring extension $k_ J[[x_1^ p, \ldots , x_ d^ p]] \subset k[[x_1, \ldots , x_ d]]$ is finite by Algebra, Lemma 10.97.7. This implies that $A_ J \to A$ is finite.

Let us check properties (1), (2), (3) of Lemma 15.46.4. Proof of (1). For $a \in A$ we see that $a^ p \in A_ J$. Hence $K^ p \subset K_ J$. Proof of (2). Suppose that $f/g^ p \in K$, $f, g \in A$, $g \not= 0$ is contained in $K_ J$ for every choice of $J$. Fix $J$ for the moment. Since $f/g^ p \in K_ J$ we can write $f/g^ p = a/b^ p$ with $a \in A_ J$ and $b \in A$ nonzero. Hence $b^ p f \in A_ J$. For any $A_ J$-derivation $D : A \to A$ we see that $0 = D(b^ pf) = b^ p D(f)$ hence $D(f) = 0$ as $A$ is a domain. Taking $D = \partial _{x_ i}$ and $D = \partial _{y_ j}$ we conclude that $f \in k[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ d^ p]$. Applying a $k_ J$-derivation $\theta : k \to k$ we similarly conclude that all coefficients of $f$ are in $k_ J$, i.e., $f \in A_ J$. Since it is clear that $A^ p = \bigcap \nolimits _ J A_ J$ where $J$ ranges over all subfields as in the lemma we conclude $f \in A^ p$ as desired. Proof of (3). This is clear because $K_{J \cup J'} \subset K_ J \cap K_{J'}$. $\square$