Every algebraic group has a largest smooth connected affine normal subgroup

The trivial subgroup belongs to the specified class. Its members have nonnegative integer dimensions bounded by $\dim G$, so choose $N$ of largest dimension. For any other member $H$, [F1] gives a member $NH$ containing $N$ and $H$. Maximal dimension forces $\dim NH=\dim N$. Since $NH$ is geometrically integral by [F1], a proper closed subset has smaller dimension; hence $N=NH$ as subsets. Both schemes are smooth and reduced, so their closed-immersion ideal has zero radical and is zero; equality is scheme theoretic. Therefore $H\subset N$. This proves that $N$ is largest and makes it unique.

Form $q:G\to Q=G\mathrm{/}N$ by [F2]. If $Q$ had a nontrivial smooth connected affine closed normal subgroup $A$, its scheme preimage $P=G{\times }_{Q}A$ would be a closed normal subgroup with exact sequence $1\to N\to P\to A\to 1$, since the projection is the base change of $q$. By [F2], $P$ is affine, smooth, and connected. The largest-subgroup assertion then gives $P\subset N$, while its surjection onto $A$ and $P\mathrm{/}N=A$ would force $A$ trivial. This contradiction proves the quotient assertion. AC is inherited from [F1]–[F2].