Finite equivalence relations have saturated affine neighbourhoods around affine-contained orbits

The saturation $s(t^{-1}(X\setminus V))$ is closed, since $s$ is finite, and is a union of entire orbits by composition. Let $V^{\mathrm{\prime }}$ be its complement. It is the largest saturated open contained in $V$ and contains $E$. Write $V=\mathrm{Spec}A$. The closed subset $V\setminus V^{\mathrm{\prime }}$ is defined by an ideal $I$. No prime of any point of the finite orbit $E$ contains $I$, so [F2] gives $f\in I$ nonzero at every point of $E$. Hence $V_f\subset V^{\mathrm{\prime }}$ and contains $E$.

On the saturated $V^{\mathrm{\prime }}$, the restricted relation is still finite locally free. Its norm $N={\mathrm{Norm}}_s(t^{\ast }f)$ is a regular function on $V^{\mathrm{\prime }}$. Its nonvanishing locus consists exactly of the points all of whose relation targets lie in $V_f$: on a residue-field fibre the determinant is nonzero exactly when multiplication by $t^{\ast }f$ is invertible in its finite algebra, equivalently when that element vanishes at none of the fibre's points. Thus this locus is saturated and contains $E$. It is contained in $V_f$, because the identity arrow is one of those targets. Therefore it equals the principal nonvanishing locus of the restriction $N{\mathrm{\mid }}_{V_f}$ in the affine scheme $V_f$, and is affine. It is the required $W$. The norm invariance in [F1] also verifies saturation scheme theoretically. AC is inherited from [F1]–[F2].