Connected finite-type groups are geometrically connected

In an algebraic closure $\bar{k}$, the connected component of the identity in $G_{\bar{k}}$ is nonempty and open and closed: a Noetherian space has finitely many connected components. Its characteristic idempotent $e\in \mathrm{\Gamma }(G_{\bar{k}},\mathcal{O})$ belongs to $\mathrm{\Gamma }(G,\mathcal{O}){\otimes }_k\bar{k}$ by [F1]. It is already defined over a separable closure $k_s$. Indeed $\bar{k}\mathrm{/}{k}_s$ is purely inseparable; its finitely many coefficient elements belong to a finite extension with some $p^r$-powers in $k_s$. Since an idempotent satisfies $e^{p^r}=e$, the expansion of that power puts it in $\mathrm{\Gamma }(G,\mathcal{O}){\otimes }_k{k}_s$. In characteristic zero the assertion is immediate. Choose a finite Galois $K\mathrm{/}k$ inside $k_s$ containing its coefficients. Each Galois automorphism fixes the identity and therefore preserves this unique geometric connected component and its idempotent. Writing $e$ in finitely many linearly independent coefficients from $\mathrm{\Gamma }(G,\mathcal{O})$ shows its $K$-coefficients are Galois fixed and lie in $k$ by [F1]. Hence the idempotent descends to $G$, which is connected, and must be $1$. Thus $G_{\bar{k}}$ is connected.

If a finite-type group is geometrically reduced, over $\bar{k}$ its reduced group has a smooth point by [F2]. Translate that point to the identity and then to every rational closed point. All closed points are smooth. The nonsmooth locus is closed and, if nonempty, has a closed point by [F2], so it is empty. By geometric-regularity descent in [F2], the group is smooth over $k$. This proof does not assume affineness. For the reduction of an arbitrary group over $\bar{k}$, multiplication and inverse restrict to it: the pullback of a nilpotent ideal vanishes on a reduced source, and products of reduced finite-type schemes over the perfect field are reduced. Thus its reduction is again a group to which the same argument applies.

If $G$ is smooth and connected, its geometric base extension is smooth, reduced, and connected by step 1.1. At a smooth point the local ring is a regular domain, so two different irreducible components cannot meet. The finitely many components are open and closed, and connectedness leaves one; hence the geometric scheme is integral. This proves geometric integrality. AC is inherited from [F1]–[F2].