High relative Frobenius has smooth scheme-theoretic image

On an affine open with coordinate ring $A$, the comorphism of relative Frobenius is $A{\otimes }_{k,F_k^n}k\to A$, $a\otimes c\mapsto c{a}^{p^n}$. Its image is the subalgebra $k{A}^{p^n}$. If $A$ is generated by $a_1,\dots ,a_r$ over $k$, it is generated as a module over that image by the finitely many monomials $a_1^{j_1}\cdots a_r^{j_r}$ with $0\le j_i<p^n$: reduce higher exponents using $a_i^{p^n}\in k{A}^{p^n}$. Thus the relative Frobenius is finite. Its underlying topological map is a universal homeomorphism: absolute Frobenius fixes prime ideals and is radicial, and the scalar Frobenius base change has the same properties; the relative map is the induced factor. Compatibility of powers with tensor products shows that relative Frobenius commutes with multiplication, inversion, and the identity, hence is a group homomorphism.

Scheme-theoretic images commute with flat field extension: on the affine charts just used their ideals are the kernels of the displayed algebra maps, and tensoring by a field extension preserves kernels. Choose an algebraic closure $\bar{k}$ of $k$ using [F3]. The finite-type scheme $G_{\bar{k}}$ has a finite affine cover. On each chart its nilradical has a bounded nilpotence exponent, so one common $p^n$ kills every nilpotent section in this finite cover. Since $\bar{k}$ is perfect, $\bar{k}{A}^{p^n}=A^{p^n}$ on each chart after scalar extension. If an element of this image is nilpotent, write it as $a^{p^n}$; then $a$ is nilpotent, so $a^{p^n}=0$. The image rings are therefore reduced. By image compatibility, $H_n$ is geometrically reduced for this $n$ and every larger $n$.

The image is a subgroup scheme. On affine charts the map from its coordinate ring into the source coordinate ring is injective; the corresponding product map is also injective because tensor products over $k$ preserve injectivity. The identities defining the group laws on $G$ therefore force multiplication and inversion on $G^{(n)}$ to restrict to its scheme-theoretic image. The identity is in that image. Over $\bar{k}$ this subgroup is reduced. Every irreducible component has a nonempty regular, hence locally standard smooth, open by [F1], and this is smooth under [F2]; in particular there is a smooth $\bar{k}$-rational point. Translating that point to the identity and then translating the identity to any other $\bar{k}$-rational point shows that every closed point of $H_{n,\bar{k}}$ is smooth. The nonsmooth locus is closed by [F1]; if nonempty it would contain a closed point, since a nonzero finite-type algebra over an algebraically closed field has a maximal ideal with that field as residue field. Thus it is empty. Smoothness over $\bar{k}$ gives geometrically regular affine chart rings by the definition in [F2], and geometric regularity descends to the chart rings of $H_n$ by [F2]. These finite-type $k$-algebras are finitely presented, since polynomial rings over $k$ are Noetherian, and are $k$-flat, since modules over a field are vector spaces. Thus all three conditions in the smoothness definition [F2] hold, so $H_n$ is smooth over $k$.

Relative Frobenius is a homeomorphism on underlying spaces by step 1.1 and factors through $H_n$, whose underlying space is the same as $G^{(n)}$ because the map is onto. Hence connectedness of $G$ implies connectedness of $H_n$. The map $G\to H_n$ is finite by the same finite-module calculation as in step 1.1, now with codomain the image algebra. Its fibre over the identity, which is its scheme-theoretic kernel, is finite. AC is used for the algebraic closure and the regularity/smoothness suppliers. The whole twist need not be smooth: for example the twist of ${\mu }_p$ over a perfect field is again ${\mu }_p$, whereas its first Frobenius image is the identity.