The Stacks project

Proof. Write $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$ and $A' = k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p]$. We also denote $K'$ the fraction field of $A'$. The ring extension $k'[[x_1^ p, \ldots , x_ d^ p]] \subset k[[x_1, \ldots , x_ d]]$ is finite by Algebra, Lemma 10.97.7 which implies that $A' \to A$ is finite. For $f \in A$ we see that $f^ p \in A'$. Hence $K^ p \subset K'$. Any element of $K'$ can be written as $a/b^ p$ with $a \in A'$ and $b \in A$ nonzero. Suppose that $f/g^ p \in K$, $f, g \in A$, $g \not= 0$ is contained in $K'$ for every choice of $k'$. Fix a choice of $k'$ for the moment. By the above we see $f/g^ p = a/b^ p$ for some $a \in A'$ and some nonzero $b \in A$. Hence $b^ p f \in A'$. For any $A'$-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 $k'$-derivation $\theta : k \to k$ we similarly conclude that all coefficients of $f$ are in $k'$, i.e., $f \in A'$. Since it is clear that $A^ p = \bigcap \nolimits _{k'} A'$ where $k'$ ranges over all subfields as in the lemma we win. $\square$