A smooth geometrically integral algebraic group has an ample line bundle

Choose a nonempty affine open $U_0\subset G$ and let $D$ be the effective Cartier boundary supplied by [F1]. Products in the group show that multiplication $m:G\times G\to G$ is smooth: the isomorphism $(g,h)\mapsto (gh,h)$ changes it into the smooth first projection. Put $d=\dim G$. Choose by [F2] a nonempty affine open $U\subset G$ with an étale map $U\to {\mathbf{A}}_k^d$. The map is dominant because an étale map is open. If $B$ is its coordinate ring and $A=k[t_1,\dots ,t_d]$, then $B{\otimes }_{A}k(t_1,\dots ,t_d)$ is a finite-dimensional algebra over this function field: it is finite type and étale of dimension zero. Its finitely many algebra generators satisfy monic polynomial equations over that field. Clear the finitely many coefficient denominators, obtaining a nonempty principal open $V\subset {\mathbf{A}}^d$ such that $W=U{\times }_{{\mathbf{A}}^d}V\to V$ is finite étale. It is surjective after further shrinking to its nonempty image. Its degree is a positive constant $r$, because it is finite flat over the integral $V$ and [F2] makes the finite module locally free. Denote this map by $\pi$ and the open immersion into $G$ by $j:W\to G$.

Pull $D$ back by $(g,w)\mapsto gj(w)$ on $G\times W$ and let $T$ be its image under the finite étale map $1\times \pi :G\times W\to G\times V$. This image is closed. Every component of the pulled-back divisor has codimension one, since multiplication is smooth; a finite locally free map between these equidimensional smooth varieties preserves the dimension of each component, so every component of $T$ has codimension one. Take the sum of these prime divisors with coefficient one; by [F3] it is an effective Cartier divisor $E$ with support $T$. For every geometric $v\in V$, its support in the fibre $G\times \{v\}$ is ${\bigcup }_{w\in {\pi }^{-1}(v)}Dj(w{)}^{-1}$, by the definition of the image under a finite map. This finite union of proper closed subsets does not equal the geometrically integral $G$. Consequently restriction of $E$ to that fibre is an effective Cartier divisor (its local equation is nonzero in the integral fibre), with precisely that support.

Apply [F3] to $\mathcal{O}(E)$ on $G\times V$, obtaining an invertible sheaf $L$ on $G$ and an isomorphism $\mathcal{O}(E)\cong {\mathrm{pr}}_G^{\ast }L$. Extend to an algebraic closure $\bar{k}$ using [F4]. For each $v\in V(\bar{k})$, the canonical section of the effective divisor $E_v$ therefore gives a section of $L_{\bar{k}}$, with nonvanishing locus $G_{\bar{k}}\setminus {\bigcup }_{w\in {\pi }^{-1}(v)}{D}_{\bar{k}}j(w{)}^{-1}$. This is affine: it is the intersection of the finitely many affine opens $(G\setminus D)j(w{)}^{-1}$ in the separated scheme $G_{\bar{k}}$. Such intersections are affine by the closed-diagonal argument.

Fix $g\in G(\bar{k})$. The open subset $W^{\mathrm{\prime }}\subset W_{\bar{k}}$ of $w$ for which $gj(w)\mathrm{\notin }{D}_{\bar{k}}$ is nonempty: $j(W)$ is a nonempty open of the irreducible group, and intersects $g^{-1}(G\setminus D)$. Its complement has dimension at most $d-1$, so its image under the finite map $\pi$ is closed of dimension at most $d-1$ in the $d$-dimensional $V_{\bar{k}}$. A rational point $v$ outside that image exists by [F4]. Its entire fibre lies in $W^{\mathrm{\prime }}$, so $g\mathrm{\notin }{E}_v$. Thus the affine section nonvanishing loci of step 3.1 contain every closed point of $G_{\bar{k}}$ and hence cover it: a nonempty closed complement would contain a closed point by [F4]. By [F5], $L_{\bar{k}}$ is ample, and [F4] then gives ampleness of $L$. For $d=0$, geometric integrality and the rational identity imply $G=\mathrm{Spec}k$ and ${\mathcal{O}}_G$ is ample directly; this also covers the zero-divisor-boundary case. AC is carried through [F1]–[F4].