Proper + Affine = Finite

Another argument I really like. We first present two auxiliary lemmata.

Lemma 1. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \Spec B \ra \Spec A is closed. Then f^{-1}(B\x) = A\x.

Proof. Since f is injective, the map \phi is dominant. Hence, it is a closed surjective map. For such a map, a closed set Z \sbq \Spec A is empty if and only if \phi^{-1}(Z) is empty. Applying this to Z = V(a) for a \in A, we get V(a) = \varnothing if and only if V(f(a)) = \varnothing, i.e. a \in A\x if and only if f(a) \in B\x. \qedsymbol

Lemma 2. Let f \colon A \ra B be an injective ring homomorphism such that the associated map \phi \colon \A^1_B \ra \A^1_A (induced by A[x] \ra B[x]) is closed. Then f is integral.

Proof. Let b \in B. Consider the ideal I = (xb - 1) \sbq B[x], and let J = I \cap A[x]. Note that B[x]/I = B[\tfrac{1}{b}], and A[x]/J is the image C of the composite map A[x] \ra B[x] \ra B[\tfrac{1}{b}] (first isomorphism theorem). Write g for the inclusion C \ra B[\tfrac{1}{b}]. Note that the map

    \[ \psi \colon \Spec B\left[\tfrac{1}{b}\right] \ra \Spec C \]

induced by g is just the restriction of \phi to the closed subsets V(I) \ra V(J), hence it is a closed map. Since g is injective, Lemma 1 asserts that

    \[ g^{-1}\left(B\left[\tfrac{1}{b}\right]\x\right) = C\x. \]

But \tfrac{1}{b} is invertible in B[\tfrac{1}{b}] (its inverse is b), and it lies in C since it is the image of x under A[x] \ra B[\tfrac{1}{b}]. Hence, it is invertible in C, i.e. b \in C. That is, we can write

    \[ b = a_0 + a_1 \left(\frac{1}{b}\right) + \ldots + a_n \left(\frac{1}{b}\right)^n \]

for certain a_0, \ldots, a_n \in A. Hence, b^{n+1} - a_0 b^n - \ldots - a_n = 0 in B\left[\tfrac{1}{b}\right]. Hence, some multiple b^m \left(b^{n+1} - a_0 b^n - \ldots - a_n\right) is 0 in B, proving that b is integral over A. \qedsymbol

Corollary. Let f \colon A \ra B be a ring homomorphism such that the associated map \Spec B \ra \Spec A is proper. Then f is finite.

Proof. Let I be the kernel of f; then \Spec B \ra \Spec A lands in \Spec A/I. Then \Spec B \ra \Spec A/I is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map

    \[ \bar{f} \colon A/I \ra B \]

is injective, hence integral by Lemma 2. Since \Spec B \ra \Spec A/I is proper, it is of finite type. Hence, \bar{f} is integral and of finite type, hence finite. Hence so is f. \qedsymbol

A more geometric version is the following:

Theorem. Let \phi \colon X \ra Y be a morphism of schemes that is both affine and proper. Then \phi is finite.

Proof. Let U = \Spec A be an affine open in Y. Then V = \phi^{-1}(U) is affine (since \phi is an affine morphism); say V = \Spec B. Then the restriction \phi|_V \colon V \ra U is proper (properness is local on the target), hence B is finite over A by the theorem above. This proves that \phi is finite. \qedsymbol

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 \P^n_Y for some n, 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 K be a field of infinite degree of imperfection, i.e. \char K = p and K is an infinite extension of K^p. Then Theorem 2 in [B-McL] says that for each n \in \N there exists a finite field extension L_n of K which cannot be generated by fewer than n elements┬╣. Correspondingly, \Spec L_n \ra \Spec K cannot be embedded into \P^m_K for m < n. Then consider Y an infinite disjoint union of \Spec K (labelled by \N), and X the disjoint union of \Spec L_n for all n. Then Y 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 \P^n_X.

[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.

┬╣It is probably not very hard to actually come up with an example of such a field K with extensions L_n. I think that K = \F_p(x_1, x_2, \ldots) and L_n = K(\sqrt[\uproot{3}p]{x_1}, \ldots, \sqrt[\uproot{3}p]{x_n}) should do the trick.

3 thoughts on “Proper + Affine = Finite

    • I think the following is a counterexample. Let Y be an infinite chain of \mathbf P^1s over a finite field k, with irreducible components Y_0, Y_1, \ldots. Say that the point \infty in Y_i is attached to the point 0 in Y_{i+1}. For notational reasons, it is easier to remove 0 from Y_0, so let’s do so. I will define X as another tree of \mathbf P^1s based on some local model.

      For a subset J \subseteq \mathbf N, write Y_J for the open subset Y \setminus \bigcup_{i \not\in J} Y_i. Note that Y_{\{i\}} is not the irreducible component Y_i, but rather the open subset \mathbf P^1 \setminus \{0,\infty\} of Y_i = \mathbf P^1. Moreover, Y is the union of the opens Y_{i,i+1} := Y_{\{i,i+1\}}.

      For the local model, note that Y_{i,i+1} is isomorphic to S = \operatorname{Spec} k[x,y]/(xy), where the x-axis V(y) corresponds to Y_i and the y-axis V(x) to Y_{i+1}. Write T = \operatorname{Spec} k[x,y,z]/(xy,xz,yz), the union of the three axes in \mathbf A^3. The map \mathbf A^3 \to \mathbf A^2 given by (x,y,z) \mapsto (x,y+z) induces a finite map T \to S. Note that over Y_{\{i\}}, this is an isomorphism, whereas over Y_{\{i+1\}} it is two copies of Y_{\{i+1\}}.

      Now to construct X, let X_{i,i+1} be the disjoint union of 2^i copies of T, with its natural map to S = Y_{i,i+1}. Above Y_{\{i\}}, this is 2^i copies of Y_{\{i\}}, and above Y_{\{i+1\}} it is 2^{i+1} copies of Y_{\{i+1\}}. This gives the glueing maps between X_{i,i+1} and X_{i+1,i+2}, and we let X be the union.

      Then X \to Y is finite since this is a local condition. But it cannot be embedded into \mathbf P^n_Y for any n, since \mathbf P^n(k) is finite and the fibres of X_{\{i\}} \to Y_{\{i\}} get arbitrarily many k-points. \qedsymbol

    • That said, in cases of interest this is of course true. For instance, if Y is quasi-projective over an affine scheme, then the two notions of projectivity agree; see [Tag 087S].

      If Y has the resolution property [Tag 0F85], then every quasi-coherent sheaf of finite type is a quotient of a locally free sheaf, so then at least you can always embed X into \mathbf P(\mathscr E) for some locally free sheaf \mathscr E. If Y is connected, then \mathscr E has constant rank.

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