I’ve recently attempted exercise II.3.7 of Hartshorne’s Algebraic Geometry. The statement is as follows:
Let \(X, Y\) be integral schemes. Let \(f : X \to Y\) be a dominant, generically finite morphism of finite type. Show that there is a dense open \(U \subseteq Y\) such that \(f^{-1}(U) \to U\) is finite.
Here’s my solution and thought process:
We first observe that there is no harm in assuming \(\DeclareMathOperator{Spec}{Spec}X = \Spec B\) and \(Y = \Spec A\) are affine. Let \(\phi : A \to B\) denote the corresponding map on global sections. Let’s now translate our assumptions on \(f\):
\(f\) is dominant is equivalent to \(f\) sending generic point to generic point, which translates to \(\phi\) being injective. Thus we shall assume
- \(A\) is a subring of \(B\).
\(f\) is of finite type translates to
- \(B\) is a finitely generated \(A\)-algebra.
\(f\) is generically finite translates to
- There are only finitely many primes \(P\) of \(B\) with \(P \cap A = (0)\).
We want to show that there is a \(g \in A\) such that \(B_g\) is a finite \(A_g\)-algebra. The hint in Hartshorne suggests we show that \(\DeclareMathOperator{Frac}{Frac}\Frac(B)\) is a finite extension of \(\Frac(A)\), which we will follow. By Zariski’s lemma, one way to show this is to just prove that \(\Frac(B)\) is a finitely generated \(\Frac(A)\)-algebra. Here’s where I got stuck for a while. I thought of using finite generation in towers:
Easy Fact. If \(B\) is a finitely generated \(A\)-algebra and \(C\) is a finitely generated \(B\)-algebra, then \(C\) is a finitely generated \(A\)-algebra.
The appropriate tower soon came to mind: \((A \setminus 0)^{-1} A \subseteq (A \setminus 0)^{-1} B \subseteq (B \setminus 0)^{-1} B\). The primes in \((A \setminus 0)^{-1} B\) correspond to primes \(P\) of \(B\) such that \(P \cap A \setminus 0 = \emptyset\), i.e. \(P \cap A = (0)\), but there are finitely many of these by assumption! Thus I am lead to the following guess:
Guess. If \(A\) is an integral domain with finitely many primes, then \(\Frac(A)\) is a finitely generated \(A\)-algebra.
This is easy to prove:
Proof. For each non-zero prime \(P_i\) of \(A\), pick \(x_i \in P_i \setminus 0\) and set \(x = \prod_i x_i\). Then \(A_x\) is an integral domain with only one prime \((0)\) hence is a field. Thus \(\Frac(A) = A_x\).
Thus we have shown that \(\Frac(B)\) is a finite extension of \(\Frac(A)\). Here’s another crucial observation: our tower above then tells us that \((A \setminus 0)^{-1} B\) is an integral domain and a finite dimensional \(\Frac(A)\)-vector space, hence a field, so \(\Frac(B) = (A \setminus 0)^{-1} B\). Hence:
- \(\Frac(B)\) is spanned by \(B\) as a \(\Frac(A)\)-vector space.
Combining this with 2, there are \(b_1, \ldots, b_n \in B\) which generate \(B\) as an \(A\)-algebra, and span \(\Frac(B)\) as a \(\Frac(A)\)-vector space. How do we get our element \(g \in A\)? Playing around with the problem reminded me of Proposition 7.8 in Atiyah-Macdonald. We’ll use a similar idea here. For \(1 \le i, j, k \le n\), there are \(a_{ijk} \in A\) and \(g \in A \setminus 0\) such that
\[b_i b_j = \sum_{k=1}^n \frac{a_{ijk}}{g} b_k\]
where we put every term in \(\Frac(A)\) occuring above over a common denominator \(g\). Now we claim that the \(b_i\) generate \(B_g\) as an \(A_g\)-module. They certainly generate \(B_g\) as an \(A_g\)-algebra, and the above equation expresses a product of \(b_i\) as an \(A_g\)-linear combination of \(b_i\). This concludes the proof!