Finite-type algebraic group monomorphisms are closed immersions

Let $I$ be the scheme-theoretic image, defined on target affine charts by the kernel of restriction to the source. This is a coherent ideal since the chart rings are Noetherian. It commutes with field extension by [F3]. Products of schematically dominant maps over a field remain schematically dominant: on product affine target charts use injectivity of the coordinate maps and exactness of tensoring; the equality of product global sections with the tensor product follows by applying the finite affine-cover equalizer twice, as in [F3]. The group identities of $f$ then force multiplication, inversion, and identity on $H$ to restrict to $I$: their defining ideal sections vanish after pullback to $G\times G$ or $G$, and schematic dominance detects that vanishing. Thus $I$ is a closed subgroup scheme.

Over an algebraic closure, the image on closed points is a subgroup $S$ of $I(\bar{k})$ and is constructible dense by [F1]. It therefore contains a dense open of the reduced $I$, dense on every irreducible component. For any $y\in I(\bar{k})$, that open and its translate by $y$ intersect, because both are dense opens. A closed point in their intersection gives $y=u{v}^{-1}$ with $u,v\in S$. Hence $S=I(\bar{k})$. The complement of the topological image, if nonempty after scalar extension, would have a closed point: the image is constructible, so a nonempty complement contains a locally closed finite-type subset. Thus $f$ is onto $I$ topologically, and its image in $H$ is closed. This conclusion descends to $k$ by surjectivity of the scalar-extension projections.

Suppose now that the scheme kernel is trivial. For every test scheme, two points with equal image differ by a kernel point, so $f$ is a monomorphism. Over a geometric point in $I$, translation by any source point identifies its fibre with the kernel; by step 2.1 such a source point exists. Thus each geometric fibre is a single reduced point, and $f:G\to I$ is quasi-finite. Apply [F1] and replace the finite factor by the scheme-theoretic closure of the open $G$ in it. We obtain $G\subset \bar{G}$ open and schematically dense, with $\bar{G}\to I$ finite. The boundary image is closed and avoids all generic points of $I$: in a finite morphism the points above a generic target component are generic points of the components dominating it, and every such point lies in the dense open $G$. Hence $f$ is finite over a dense open of $I$.

Over $\bar{k}$, translations of that open by $I(\bar{k})$ cover $I$: any closed point can be translated from a fixed closed point in the open. Each such translation lifts to a source translation by step 2.1, so $f_{\bar{k}}$ is finite on every translated open and hence finite globally. The open immersion $G_{\bar{k}}\subset {\bar{G}}_{\bar{k}}$ is proper by [F2], since its source is finite over $I_{\bar{k}}$ and its target separated over that base. Its image is therefore closed and also schematically dense, so the open image is the whole finite factor. The boundary of $G\subset \bar{G}$ consequently becomes empty after faithful scalar extension and is empty already. Hence $f:G\to I$ is finite over $k$.

A finite monomorphism is a closed immersion. On an affine target chart its finite fibre algebra $D$ over a residue field has $D\otimes D\cong D$ by the diagonal condition, so its dimension is zero or one. In the nonempty case the unit map from that residue field is an isomorphism. The finite cokernel of the target-ring map therefore has zero reduction at every prime and vanishes by [F2]. Thus the ring map is onto on each affine chart. Applied to $f$, this proves the asserted closed immersion into $I$ and hence into $H$. AC is inherited from [F1]–[F3].