Another argument I really like. We first present two auxiliary lemmata.
Lemma 1. Let be an injective ring homomorphism such that the associated map is closed. Then .
Proof. Since is injective, the map is dominant. Hence, it is a closed surjective map. For such a map, a closed set is empty if and only if is empty. Applying this to for , we get if and only if , i.e. if and only if .
Lemma 2. Let be an injective ring homomorphism such that the associated map (induced by ) is closed. Then is integral.
Proof. Let . Consider the ideal , and let . Note that , and is the image of the composite map (first isomorphism theorem). Write for the inclusion . Note that the map
induced by is just the restriction of to the closed subsets , hence it is a closed map. Since is injective, Lemma 1 asserts that
But is invertible in (its inverse is ), and it lies in since it is the image of under . Hence, it is invertible in , i.e. . That is, we can write
for certain . Hence, in . Hence, some multiple is in , proving that is integral over .
Corollary. Let be a ring homomorphism such that the associated map is proper. Then is finite.
Proof. Let be the kernel of ; then lands in . Then is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map
is injective, hence integral by Lemma 2. Since is proper, it is of finite type. Hence, is integral and of finite type, hence finite. Hence so is .
A more geometric version is the following:
Theorem. Let be a morphism of schemes that is both affine and proper. Then is finite.
Proof. Let be an affine open in . Then is affine (since is an affine morphism); say . Then the restriction is proper (properness is local on the target), hence is finite over by the theorem above. This proves that is finite.
Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into for some , i.e. it embeds into a trivial projective bundle).
I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let be a field of infinite degree of imperfection, i.e. and is an infinite extension of . Then Theorem 2 in [B-McL] says that for each there exists a finite field extension of which cannot be generated by fewer than elements¹. Correspondingly, cannot be embedded into for . Then consider an infinite disjoint union of (labelled by ), and the disjoint union of for all . Then is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any .
[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.
¹It is probably not very hard to actually come up with an example of such a field with extensions . I think that and should do the trick.