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Generically finite morphisms

I’m going to take a quick break from representation theory and discuss my solution of Exercise II.3.7 from Hartshorne’s book, which is pretty technical but I think has an interesting proof. For one thing, it applies Noether normalization to the generic fiber of a morphism, which is a very non-classical sort of move.

Hartshorne defines a morphism f : X \to Y of irreducible schemes to be generically finite provided that the set-theoretic preimage f^{-1}(\eta) of the generic point \eta \in Y is finite. The name for this term might suggest another definition, namely that there exists a nonempty open subscheme U \subset Y over which f is finite. The exercise is that under some assumptions, namely that X and Y are integral and f is dominant and of finite type, Hartshorne’s definition implies this one

First let us assume X \cong \text{Spec } A and Y \cong \text{Spec } B are affine and prove that the function field K of X is a finite extension of the function field L of Y. The key here is to consider the scheme-theoretic fiber \text{Spec } K \times_Y X, which is the spectrum of K \otimes_B A. Note that if we write S = B \setminus \{ 0 \}, we can identify the localization S^{-1}A with K \otimes_B A, whence K \otimes_B A is a domain. Now by the finite type assumption K \otimes_B A is a finitely generated K-algebra, so Noether normalization provides an injection K[x_1,\cdots,x_n] \to K \otimes_B A which makes K \otimes_B A into a finitely generated module for K[x_1,\cdots,x_n]. Geometrically this corresponds to a surjection \text{Spec } K \times_Y X \to \mathbb{A}^n_K, and since \text{Spec } K \times_Y X is homeomorphic to the finite space f^{-1}(\eta) this forces n = 0. But this means K \otimes_B A is a finite-dimensional K-algebra, and a finite-dimensional domain over a field is a field. Thus K \otimes_B A = L is actually the fraction field of A and finite over K.

To finish the affine case, choose a finite set of generators for A over B, which by the previous paragraph satisfy some polynomials with coefficients in K. By clearing denominators, we see that these polynomials can actually be taken to have coefficients in B. Write b \in B for the product of the leading coefficients of these polynomials, so the localization A_b is integral over B_b. But A_b is a finitely generated B_b-algebra, hence a finitely generated B_b-module. Thus we can take \text{Spec } B_b \to \text{Spec } A_b to be the desired finite morphism.

Now we reduce from the general situation to the case that X and Y are affine: choose an open affine V \subset Y and use the finite type hypothesis to cover f^{-1}(V) with finitely many open affines U_i. From the proof of the affine case, we can find principal open sets V_i \subset V such that each f^{-1}(V_i) \cap U_i \to V_i is finite. In particular, W = \cap V_i is affine, so let us write W \cong \text{Spec } B. Also each W_i = f^{-1}(W) \cap U_i is principal in U_i, hence affine, say W_i \cong \text{Spec } A_i. By the usual trick we can find a_i \in A such that \text{Spec } (A_i)_{a_i} \subset \cap W_i for each i and \text{Spec } (A_i)_{a_i} = \text{Spec } (A_j)_{a_j} for all i,j. Now since the A_i are finitely generated modules for B, each a_i satisfies a monic polynomial with coefficients in B, which may be taken to have nonzero constant term b_i. Put b = \prod b_i, so any prime ideal of A_i which contains a_i also contains b_i, hence b. Thus \text{Spec } (A_i)_b \subset \text{Spec } (A_i)_{a_i}, from which it follows that f^{-1}(\text{Spec } B_b) = \text{Spec } (A_i)_b, and \text{Spec } (A_i)_b \to \text{Spec } B_b is the desired finite morphism.

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