Purely inseparable subgroup descent by Frobenius power ideals

On an affine open $U=\mathrm{Spec}A$ of $G$, let $J\subset A{\otimes }_{k}K$ be the ideal of $H^{\mathrm{\prime }}\cap U_K$. If $j={\sum }_i{a}_i\otimes {\lambda }_i\in J$, then $j^q={\sum }_i{a}_i^q{\lambda }_i^q$ lies in the copy of $A$ inside $A{\otimes }_{k}K$. Let $I_U\subset A$ be the ideal generated by all these $j^q$. Its scalar extension is the ideal $J^{[q]}$ generated by $q$-th powers of sections of $J$. These ideals are the inverse images of the ideal of the Frobenius twist $H^{\mathrm{\prime }(r)}$ under $F_{G_K\mathrm{/}K}^r$, by [F1]. Hence they agree on overlaps after scalar extension; faithfulness of scalar extension shows that the ideals $I_U$ agree on overlaps before extension. They define a quasi-coherent ideal on $G$ and a closed $k$-subscheme $H$ with $H_K=F^{-1}(H^{\mathrm{\prime }(r)})$.

By [F1] the inverse image of a subgroup under relative Frobenius is a subgroup; if $H^{\mathrm{\prime }}$ is normal, its twist and this inverse image are normal. These group and conjugation factorization identities descend to $H$ over $k$: their defining ideal sections are zero after tensoring by the faithful field extension $K$, so are zero already. Thus $H$ is a subgroup and is normal whenever $H^{\mathrm{\prime }}$ is normal.

Since $J^{[q]}\subset J$, $H^{\mathrm{\prime }}$ is a closed subscheme of $H_K$. Every $j\in J$ has $j^q\in J^{[q]}$, so $J$ and $J^{[q]}$ have the same radical. The ideal $J\mathrm{/}{J}^{[q]}$ of $H^{\mathrm{\prime }}$ in $H_K$ is nilpotent: it is finite over the Noetherian coordinate ring, is generated by nilpotent elements, and a product of sufficiently many generators must contain a sufficiently large power of one of them. A finite affine cover supplies one uniform exponent. If $H^{\mathrm{\prime }}$ is affine, [F2] makes its nilpotent thickening $H_K$ affine, and then makes $H$ affine by inseparable descent. If $H^{\mathrm{\prime }}$ is connected, so is $H_K$ because the thickening has the same underlying space; $H_K\to H$ is a homeomorphism by the scalar-extension calculation in [F2], so $H$ is connected. This is the power-ideal descent used in the arbitrary-field structure reduction; it retains nonsmooth thickening rather than replacing it with a smooth reduced subgroup.