Proposition 32.11.2. Let f : X \to S be a morphism of schemes. Assume X is affine and that f is surjective and universally closed1. Then S is affine.
A scheme admitting a surjective integral map from an affine scheme is affine.
Proof. By Morphisms, Lemma 29.41.11 the scheme S is separated. Then by Morphisms, Lemma 29.11.11 we find that f is affine. Whereupon by Morphisms, Lemma 29.44.7 we see that f is integral.
By the preceding paragraph, we may assume f : X \to S is surjective and integral, X is affine, and S is separated. Since f is surjective and X is quasi-compact we also deduce that S is quasi-compact.
By Lemma 32.7.3 we can write X = \mathop{\mathrm{lim}}\nolimits _ i X_ i with X_ i \to S finite. By Lemma 32.4.13 we see that for i sufficiently large the scheme X_ i is affine. Moreover, since X \to S factors through each X_ i we see that X_ i \to S is surjective. Hence we conclude that S is affine by Lemma 32.11.1. \square
Comments (9)
Comment #3025 by Brian Lawrence on
Comment #6678 by Xuande Liu on
Comment #6679 by Johan on
Comment #6889 by Johan on
Comment #6997 by Laurent Moret-Bailly on
Comment #6998 by Johan on
Comment #7000 by Laurent Moret-Bailly on
Comment #7001 by Laurent Moret-Bailly on
Comment #7002 by Johan on