Home > algebraic geometry > Generically finite morphisms

## 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.