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К . J, Nowak

It is easy to check that a commutative ring A is a Jacobson ring (i.e. each prime ideal in A is the intersection of a family of maximal ideals) iff Spec(A) with the Zariski topology is a Jacobson space. We say that a scheme X is a Jacobson scheme if X with the Zariski topology is a Jacobson space. When a morphism / : У -^ X is of finite type and when X is a Jacobson scheme, Y is also a Jacobson scheme and /(У°) g X°; therefore we can define the restriction fo.yo _^ Xo Qf tt^e morphism /. In the proof of our main theorem we shall need an auxiliary proposition on injective and surjective moфhisms between Jacobson schemes.

Proposition . Let f:Y -^ X be a morphism of finite type between noetherian Jacobson schemes Y and X, and f\'Y° -^ X^ the restriction of f. Then the mapping f is injective (surjective) iff so is the mapping f°.

Proof Obviously, if / is injective, so is /°. Conversely, assume that /° is an injective mapping and fix two different points у^,У2 G Y, Ух ^ Уг- Supposing f{y\) = /(2/2) =• ^» we want to get a contradiction.

Since schemes are To-spaces, the closures Fj and F^ of the sets {^J and {1/2} are different subsets of Y whence {F^XF^) U {F^XF^) 4 0; say F^XF^ 4 0. By virtue of Chevalley's constructibility theorem (cf. [5, Theorem 1.7.1.4]), /(Fi\F2) and /(F2) are constructible subsets of X. The closures of both these sets coincide with the closure of {x} (because they both contain the generic point x), and thus f(Fx\F2) П /(F2) 4 0. Since X is a Jacobson scheme, this implies that

/ ( FAF2 ) n / ( F2 ) nXM 0.

Consequently , there exist points z^ G Fx\F2 and Z2 G F2 such that fiz^) = /(^2) G X°. If the set {zj (г = 1,2) were not closed, its closure E would contain two different points z[,z[' G F°. Otherwise, F П F° = {z[} whence (£'\{г^})ПУ° = 0, E\{z[} = 0 (as У is a Jacobson scheme) and E = {z[} Ç Y° - against the hypothesis. But then f%z[) = f%z'^) = /(2;J, which contradicts the injectivity of /°. Therefore 2:1,2:2 G F° and (as /° is injective) 2:^ = 2:2 - contrary to the choice of these points.

Now , observe that if / is surjective, so is /°. Conversely, supposing that /° is a surjective mapping, f(Y) is a constructible subset of X such that /(У) n X° = X°. Hence f(Y) = X (because X is a Jacobson scheme), and thus / is surjective, concluding the proof. D

Corollary . Under the assumptions of the above proposition, f is an isomorphism (open immersion) iff so is f°.

Every Z-scheme X of finite type is a noetherian Jacobson scheme (cf. [5, Chap. I, Sect. 6.4]). Besides, x £ X"" iff the residue field к(х) of the local ring of ж is finite (this is an immediate consequence of Hilbert's Nullstellensatz). The following lemma will play a key role in the proof of our main theorem. It gives some decomposition of the set Y° onto a countable family of finite sets.