A morphism \(f: X \to Y\), with \(Y\) irreducible, is generically finite if \(\inverseof{f}(\eta)\) is a finite set, where \(\eta\) is the generic point of \(Y\). A morphism \(f: X \to Y\) is dominant if \(f(X)\) is dense in \(Y\).
Now let \(f: X \to Y\) be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset \(U \subseteq Y\) such that the induced morphism \(\inverseof{f}(U) \to U\) is finite.
First show that the function field of \(X\) is a finite field extension of the function field of \(Y\).
Solution
Let \[ K=K(Y), \qquad L=K(X). \]
1. After replacing \(Y\) by a nonempty affine open and \(X\) by its inverse image, we may assume\[ Y=\Spec B \]with \(B\) a domain, while all the hypotheses remain true.
Every nonempty open subset of the integral scheme \(Y\) is dense. If \[ V\subseteq Y \] is nonempty affine, then \(f^{-1}(V)\) is a nonempty open subset of the integral scheme \(X\), hence is integral. Dominance, finite type, and the generic fibre are unchanged after this restriction. Any dense open found inside \(V\) is also dense in the original \(Y\).
2. There is a finite affine cover\[ X=U_1\cup\cdots\cup U_r, \qquad U_i=\Spec A_i, \]such that every \(A_i\) is a finitely generated \(B\)-algebra.Moreover\[ B\hookrightarrow A_i, \qquad \operatorname{Frac}(B)=K, \qquad \operatorname{Frac}(A_i)=L. \]
By Hartshorne II.3.3(b), finite type over the affine scheme \(Y=\Spec B\) gives a finite affine cover with each \(A_i\) finitely generated over \(B\).
Discard any empty member. Since \(X\) is irreducible, every nonempty open \(U_i\) contains the generic point \(\xi\) of \(X\). Dominance gives \[ f(\xi)=\eta, \] the generic point of \(Y\). Hence the ring map \(B\to A_i\) has zero kernel, and II.3.6 identifies the fraction fields with \(K(Y)\) and \(K(X)\) respectively.
3. The field extension\[ \boxed{L/K\text{ is finite}.} \]
Fix one affine chart \(U_i=\Spec A_i\) and let \[ S=B\setminus\{0\}. \] Its generic fibre is \[ (U_i)_\eta = \Spec(S^{-1}A_i). \] This is an open subscheme of the finite set \(X_\eta\), hence has finitely many points. The ring \[ R=S^{-1}A_i \] is a finitely generated \(K\)-algebra and a domain.
By Noether normalization, there exist algebraically independent elements \[ t_1,\ldots,t_d\in R \] such that \(R\) is finite over \[ K[t_1,\ldots,t_d]. \] If \(d>0\), lying over would give a surjection \[ \Spec R\longrightarrow\Spec K[t_1,\ldots,t_d], \] but the target has infinitely many prime ideals, contradicting finiteness of \(\Spec R\). Hence \(d=0\).
Therefore \(R\) is finite-dimensional over \(K\). Since \(R\) is a domain, it is a field. It contains \(A_i\), so it contains the fraction field of \(A_i\); conversely it lies inside that fraction field. Thus \[ R=\operatorname{Frac}(A_i)=L. \] Hence \([L:K]<\infty\).
4. After replacing \(Y\) by a smaller dense distinguished open, we may assume that every \(A_i\) is finite as a \(B\)-module.
For each \(i\), choose algebra generators \[ A_i=B[\alpha_{i1},\ldots,\alpha_{in_i}]. \] By 3, every \(\alpha_{ij}\in L\) is algebraic over \(K=\operatorname{Frac}(B)\). For each generator choose a monic polynomial over \(K\) which it satisfies. Clearing the finitely many denominators occurring in all these polynomials, there is one nonzero element \[ d\in B \] such that every \(\alpha_{ij}\) is integral over \(B_d\).
Consequently \[ (A_i)_d = B_d[\alpha_{i1},\ldots,\alpha_{in_i}] \] is a finite \(B_d\)-module, because an algebra generated by finitely many integral elements is module-finite.
Replace \(Y\) by the dense open \(D(d)\) and \(X\) by its inverse image, and replace each \(A_i\) by \((A_i)_d\). Renaming the rings, we may assume from now on that every \(A_i\) is finite over \(B\).
5. Let\[ W=U_1\cap\cdots\cap U_r. \]Then \(W\) is a nonempty open subset of \(X\).
Each \(U_i\) is nonempty and \(X\) is irreducible, so every \(U_i\) contains the generic point \(\xi\). Hence their finite intersection also contains \(\xi\).
6. For every \(i\), write\[ U_i\setminus W=V(J_i) \]for an ideal \(J_i\subseteq A_i\). Then\[ J_i\cap B \]contains a nonzero element.
The complement \(U_i\setminus W\) is closed in the affine scheme \(U_i\), so it is \(V(J_i)\) for some ideal. Since \(W\) is nonempty, \(V(J_i)\ne U_i\), and because \(A_i\) is a domain this implies \[ J_i\ne(0). \] Choose \[ 0\ne a\in J_i. \] The element \(a\in L\) is algebraic over \(K\). Its minimal polynomial over \(K\) has nonzero constant coefficient, since \(a\ne0\). Clearing denominators gives a relation \[ c_na^n+c_{n-1}a^{n-1}+\cdots+c_1a+c_0=0 \] with \(c_j\in B\) and \[ c_0\ne0. \] Therefore \[ c_0=-a(c_na^{n-1}+\cdots+c_1)\in J_i\cap B. \] Thus the intersection contains a nonzero element.
7. Choose\[ 0\ne b_i\in J_i\cap B \]for each \(i\), and put\[ b=b_1\cdots b_r\in B. \]Then\[ f^{-1}(D(b))\subseteq W. \]
The product \(b\) is nonzero because \(B\) is a domain.
Let \(x\in f^{-1}(D(b))\). Choose \(i\) with \(x\in U_i\). Since \(b\) is a unit at \(f(x)\), so is every factor \(b_i\). Thus \[ x\in D_{U_i}(b_i). \] But \(b_i\in J_i\), hence \[ V(J_i)\subseteq V(b_i), \] so \[ D_{U_i}(b_i)\subseteq U_i\setminus V(J_i)=W. \] Therefore \(x\in W\).
8. The morphism\[ f^{-1}(D(b))\longrightarrow D(b) \]is finite.
By 7, the entire inverse image lies in every \(U_i\). Fix \(U_1=\Spec A_1\). Inside this affine chart, \[ f^{-1}(D(b))=D_{U_1}(b) \cong \Spec (A_1)_b. \] The target is \[ D(b)\cong\Spec B_b. \] By 4, \(A_1\) is a finite \(B\)-module, so localization gives \((A_1)_b\) finite over \(B_b\). Hartshorne II.3.4 therefore shows that the restricted morphism is finite.
9. The open subset\[ U=D(b)\subseteq Y \]is dense and has the required property.
Since \(B\) is a domain and \(b\ne0\), the distinguished open \(D(b)\) contains the generic point of \(Y\) and is therefore nonempty and dense. Step 8 proves that \[ f^{-1}(U)\to U \] is finite.
10. Q.E.D.
Step 9 is the required dense-open finiteness statement.