Rational maps from smooth varieties to abelian varieties extend

First suppose $k$ is algebraically closed. By [F1] and the proper-target part of [F2], the complement of the maximal domain of $f$ contains no codimension-one point. By [F3] and the group-target part of [F2], that complement would be a union of prime divisors if it were nonempty. These statements force it to be empty.

Thus the rational map is a morphism on all of $X$. Any two extensions agree on a dense open; their equalizer is closed since $A$ is separated, and the integral reduced $X$ admits no nonzero ideal vanishing on a dense open. They therefore agree everywhere.

For general $k$, extend to an algebraic closure $\bar{k}$ using [F4]. Smoothness survives scalar extension. The finitely many irreducible components of $X_{\bar{k}}$ are disjoint, because regular local rings are domains, so each is a smooth integral variety. The base extension of the original dense domain is schematically dense, by injectivity of scalar extension on the original affine coordinate rings, and meets each such component densely. The preceding argument on each component therefore gives morphisms which glue to $g:X_{\bar{k}}\to A_{\bar{k}}$ extending $f_{\bar{k}}$. It is defined over a finite extension $K\mathrm{/}k$ inside $\bar{k}$: cover its source by finitely many affine opens lying in inverse images of original affine target opens; the ideals defining these opens inside the finitely many original affine source charts, the coordinate images defining the morphisms, and the finitely many overlap relations all involve finitely many algebraic coefficients. Take $K$ containing those coefficients. The resulting maps glue to $g_K:X_K\to A_K$; their restrictions agree with $f_K$ on its domain, since equality is reflected by the faithful extension $K\to \bar{k}$.

The two base extensions of $g_K$ to $K{\otimes }_{k}K$ agree on the base extension of the original dense domain of $f$. That open is schematically dense even over this possibly nonreduced tensor algebra: on each original affine chart, restriction from its integral coordinate ring to the rational-function field is injective, and tensoring by the $k$-vector space $K{\otimes }_{k}K$ preserves injectivity. Therefore a section of the ideal of their closed equalizer which vanishes on that open is zero. Separatedness of $A$ makes that equalizer closed, so the two morphisms agree everywhere. Apply [F4] to descend $g_K$ to a morphism $X\to A$ extending $f$. The uniqueness argument of step 2.1 works over $k$ as well. AC is carried from the algebraic closure and regularity suppliers.