Math 256B. Some Lemmas on Curves, and the End of the Proof of (II, 6.8)

This handout gives a few more lemmas on (nonsingular) curves, leading up to a more rigorous proof that the map of (II, 6.8) is finite if it is dominant.

Throughout this note, k is an algebraically closed field. Also, t(C_{K}) is as in (II, 6.7) (i.e., a nonsingular projective curve over k whose function field is isomorphic to a given field K, finitely generated and of transcendence degree 1 over k).

Lemma 1. Let A be an entire k-algebra of finite type such that Spec A is a nonsingular curve over k. Let K = Frac A, and let X = t(C_{K}) . Then there is an open embedding i: Spec A ↪ X that induces the isomorphism Frac A → K(X) .

Proof. Let X_{0} be a projective closure of Spec A (I, Ex. 2.9). (In the language of schemes, this is done as follows. Choose a finite generating set a_{1}, …, a_{n} ∈ A for A over k; then the associated ring surjection k[x_{1}, …, x_{n}] → A over k defined by x_{i} ↦ a_{i} for all i determines a closed embedding Spec A ↪ 𝔸_{k}^{n}. Then X_{0} is the closure of the image of this map into ℙ_{k}^{n}, with reduced induced subscheme structure (where we identify 𝔸_{k}^{n} with the complement of the “hyperplane at infinity” H_{0} as on p. 10 of Hartshorne).)

Moreover, letting U_{0} = ℙ_{k}^{n} \ H_{0} , we have that Spec A is a closed subscheme of U_{0} (via the usual identification of U_{0} with 𝔸_{k}^{n}), and the open subscheme X_{0} ∩ U_{0} of X_{0} is isomorphic to Spec A, since both schemes are reduced and have the same (closed) image in U_{0} .

Now let π: X̃ → X be the normalization of X_{0} (II, Ex. 3.8). Since Spec A is normal (it is nonsingular), π is an isomorphism over X_{0} ∩ U_{0} , so Spec A is also an open subscheme of X̃. Also, since X̃ is birational to Spec A, we have

K(X) ≅ K(Spec A) = Frac A .

Finally, since X̃ is smooth and complete (it is finite, hence proper, over X_{0} ), we have X̃ ≅ t(C_{K}) , so one obtains the desired open embedding Spec A ↪ t(C_{K}) = X. □

Lemma 2. Let X be a variety over k, let U = Spec A be a nonempty open affine subset of X, and let P ∈ X be a point not in U . Then there is a function f ∈ A that does not extend to a regular function at P (i.e., f ∉ 𝒪_{X,P} ).

Proof. This is Question 3a on Homework 10. □

This next lemma is central to the main result of this handout.

Lemma 3. Let X be a complete nonsingular curve over k, let Y be any curve over k, and let f: X → Y be a morphism with f(X) = Y . Let V = Spec B be any nonempty open affine subset of Y , and let A be the integral closure of B in K(X) . Then Spec A is isomorphic to an open subset U of X, compatible with the inclusion A ⊆ K(X) , and U = f^{−1}(V) .

Proof. Clearly A is integrally closed with fraction field K(X) . Then dim Spec A = 1 and Spec A is normal, hence nonsingular. Also A is finite over B, hence of finite type over k. Thus Spec A is a nonsingular curve over k. The existence of U is then immediate from Lemma 1.

Next, since X ≅ t(C_{K(X)}) , the inclusion B ⊆ A is compatible with the inclusion K(Y) ⊆ K(X) induced by f . Therefore the natural map Spec A → Spec B coincides with f|_{U} . In particular, f(U) ⊆ V . In other words, U ⊆ f^{−1}(V) .

Therefore, if U ≠ f^{−1}(V) , then there is a point x ∈ X \ U such that f(x) ∈ V . Assume that x is such a point. Then, by Lemma 2, there is an a ∈ A such that a ∉ 𝒪_{X,x}. Since a is integral over B, there are elements b_{0}, …, b_{n−1} ∈ B such that

a^{n} + b_{n−1}a^{n−1} + ⋯ + b_{0} = 0 .

But since f(x) ∈ Spec B, the local ring 𝒪_{Y,f(x)} contains B, so b_{i} ∈ 𝒪_{Y,f(x)} ⊆ 𝒪_{X,x} for all i. Thus a is integral over 𝒪_{X,x}.

However, since X is nonsingular, it is normal, so 𝒪_{X,x} is integrally closed. In particular, a ∈ 𝒪_{X,x}, a contradiction. Hence U = f^{−1}(V) . □

Finally, we get to the main result of this handout.

Proposition 4. Let X be a complete nonsingular curve over k, let Y be any curve over k, and let f: X → Y be a morphism with f(X) = Y . Then f is a finite morphism.

Proof. Let V = Spec B be any nonempty open affine subset of Y . By Lemma 3, f^{−1}(V) = Spec A, where A is the integral closure of B in K(X) . By (I, 3.9A), A is finite as a module over B, and we are done. □