Show that a morphism \(f: X \to Y\) is finite if and only if for every open affine subset \(V = \Spec B\) of \(Y\), the preimage \(\inverseof{f}(V)\) is affine, equal to \(\Spec A\), where \(A\) is a finite \(B\)-module.
Solution
1. If the stated condition holds for every affine open \(V\subseteq Y\), then \(f\) is finite.
Choose any affine open cover of \(Y\). On every member \(V=\Spec B\) of that cover, the hypothesis says \[ f^{-1}(V)=\Spec A \] with \(A\) finite as a \(B\)-module. This is exactly the definition of a finite morphism.
2. Conversely, suppose \(f\) is finite and fix an arbitrary affine open\[ V=\Spec B\subseteq Y. \]Choose an affine cover\[ Y=\bigcup_iV_i, \qquad V_i=\Spec B_i, \]witnessing finiteness, so\[ f^{-1}(V_i)=\Spec A_i \]with \(A_i\) a finite \(B_i\)-module.
This is the defining affine-cover condition for a finite morphism.
3. Every point \(y\in V\) has an open neighborhood\[ W_y\subseteq V\cap V_i \]for some \(i\) which is distinguished in both \(V\) and \(V_i\).
Choose \(i\) with \(y\in V_i\). Inside the affine \(V\), choose \[ y\in D_V(b)\subseteq V\cap V_i. \] Inside the affine \(V_i\), choose \[ y\in D_{V_i}(c)\subseteq D_V(b). \] The regular function \(c\) on \(D_V(b)\) is an element of \(B_b\), so write \[ c=d/b^m, \qquad d\in B. \] Then \[ D_{V_i}(c)=D_V(b)\cap D_V(d)=D_V(bd). \] Thus this neighborhood is distinguished in both affines. Denote it by \(W_y\).
4. For every neighborhood \(W_y\) from 3, the inverse image \(f^{-1}(W_y)\) is affine and finite over \(W_y\).
Write \(W_y=D_{V_i}(c)\) inside the witnessing affine \(V_i=\Spec B_i\). Then \[ f^{-1}(W_y)=D_{\Spec A_i}(\phi_i(c)) \cong \Spec (A_i)_c, \] where \(\phi_i:B_i\to A_i\) is the structural ring map.
Since \(A_i\) is a finite \(B_i\)-module, localization gives \((A_i)_c\) as a finite \((B_i)_c\)-module. But \[ (B_i)_c\cong\Gamma(W_y,\mathcal O_Y). \] Thus \(f^{-1}(W_y)\) is affine and finite over \(W_y\).
5. There are finitely many such common distinguished opens\[ W_j=D_V(b_j) \]which cover \(V\).
The neighborhoods from 3 cover \(V\). Since \(V\) is affine, it is quasi-compact, so a finite subcover suffices. Express each chosen member in its distinguished form inside \(V\).
6. The elements \(b_1,\ldots,b_r\in B\) generate the unit ideal.
The distinguished opens \(D(b_j)\) cover \(\Spec B\), so \[ V(b_1,\ldots,b_r)=\varnothing. \] Hence \[ (b_1,\ldots,b_r)=B. \]
7. Put\[ X_V=f^{-1}(V), \qquad A=\Gamma(X_V,\mathcal O_{X_V}), \]and let \(a_j\in A\) be the pullback of \(b_j\).Then\[ (a_1,\ldots,a_r)=A \]and\[ (X_V)_{a_j}=f^{-1}(W_j) \]is affine for every \(j\).
Choose \(d_j\in B\) with \[ \sum_jd_jb_j=1. \] Pulling this identity back to \(A\) gives \[ \sum_jf^*(d_j)a_j=1, \] so the \(a_j\) generate the unit ideal.
A point of \(X_V\) belongs to the nonvanishing locus of \(a_j=f^*b_j\) exactly when its image in \(V\) belongs to \(D(b_j)=W_j\). Hence \[ (X_V)_{a_j}=f^{-1}(W_j), \] which is affine by 4.
8. The scheme \(X_V=f^{-1}(V)\) is affine.
7. Hartshorne II.2.17(b) therefore gives\[ X_V\cong\Spec A. \]
9. For every \(j\), the localization \(A_{a_j}\) is a finite \(B_{b_j}\)-module.
Since \(X_V=\Spec A\), the distinguished open \((X_V)_{a_j}\) has coordinate ring \(A_{a_j}\). By 7 it equals \(f^{-1}(W_j)\), and by 4 the coordinate ring of that inverse image is finite over \[ \Gamma(W_j,\mathcal O_Y)=B_{b_j}. \] Thus \(A_{a_j}\) is finite as a \(B_{b_j}\)-module.
10. We use the following module lemma.Let \(B\to A\) be a ring map and let \(b_1,\ldots,b_r\in B\) generate the unit ideal.If every\[ A_{b_j} \]is a finite \(B_{b_j}\)-module, then \(A\) is a finite \(B\)-module.
For each \(j\), choose finitely many generators of the \(B_{b_j}\)-module \(A_{b_j}\) and write them as fractions whose numerators lie in \(A\). Let \[ M\subseteq A \] be the \(B\)-submodule generated by all these finitely many numerators. Then \[ M_{b_j}=A_{b_j} \] for every \(j\).
Set \(Q=A/M\). Then \(Q_{b_j}=0\) for every \(j\). For \(q\in Q\), there is therefore an exponent \(N_j\) with \[ b_j^{N_j}q=0. \] Choose a common \(N\) at least all \(N_j\). Then \(b_j^Nq=0\) for all \(j\). Since the \(b_j\) generate the unit ideal, the powers \(b_j^N\) do as well. Hence some \(c_j\in B\) satisfy \[ \sum_jc_jb_j^N=1. \] It follows that \[ q=\sum_jc_jb_j^Nq=0. \] Thus \(Q=0\), so \(A=M\) is generated by finitely many elements as a \(B\)-module.
11. The ring \(A=\Gamma(X_V,\mathcal O_{X_V})\) is a finite \(B\)-module.
6. Localizing \(A\) at \(b_j\) is the same as localizing at its image \(a_j\), and 9 shows that this localization is finite over \(B_{b_j}\). Apply the module lemma
10.
12. Hence for every affine open \(V=\Spec B\subseteq Y\),\[ f^{-1}(V)=\Spec A \]with \(A\) finite as a \(B\)-module.
Step 8 proves affineness and 11 proves module finiteness. The affine open \(V\) was arbitrary.
13. Q.E.D.
Step 1 proves one implication and steps <1>2–<1>12 prove the converse.