Let \(f: X \to Y\) be a proper morphism of affine varieties over \(k\). Then \(f\) is a finite morphism. Hint: Use (4.11A).
Solution
Write \[ X=\Spec A, \qquad Y=\Spec B, \] where \(A\) and \(B\) are finitely generated integral \(k\)-algebras, and let \[ \phi:B\longrightarrow A \] be the ring homomorphism corresponding to \(f\). Put \[ K=\operatorname{Frac}(A). \]
1. Let \(R\subseteq K\) be any valuation ring containing the subring \(\phi(B)\). Then there is a commutative diagram\[ \begin{array}{ccc} \Spec K&\longrightarrow&X\\ \downarrow&&\downarrow{\scriptstyle f}\\ \Spec R&\longrightarrow&Y. \end{array} \]
The top morphism is induced by the inclusion \[ A\hookrightarrow K. \] The bottom morphism is induced by the ring homomorphism \[ B\xrightarrow{\phi}\phi(B)\hookrightarrow R. \] After composing with the inclusion \(R\hookrightarrow K\), both routes on coordinate rings are the same map \[ B\xrightarrow{\phi}A\hookrightarrow K. \] Hence the square commutes.
2. Properness of \(f\) gives a lift\[ \Spec R\longrightarrow X \]in the diagram of
1.
The valuative criterion for properness applies to every valuation ring and every such commutative square. Since \(f\) is proper, a unique lift exists.
3. The lift in 2 forces\[ \boxed{A\subseteq R.} \]
The lift corresponds to a ring homomorphism \[ A\longrightarrow R \] whose composite with the inclusion \[ R\hookrightarrow K \] equals the original inclusion \[ A\hookrightarrow K, \] because the lift restricts on the generic point to the top arrow in 1.
Therefore every \(a\in A\) maps to the same element \(a\in K\), now lying in \(R\). Hence \(A\subseteq R\).
4. Every element of \(A\) is integral over \(\phi(B)\).
Hartshorne II.4.11A states that if \(C\) is a subring of a field \(K\), then the integral closure of \(C\) in \(K\) is the intersection of all valuation rings of \(K\) which contain \(C\).
Take \[ C=\phi(B)\subseteq K. \] By 3, every valuation ring \(R\) of \(K\) containing \(C\) also contains \(A\). Therefore \[ A \subseteq \bigcap_{R\supseteq C}R = \overline C^{\,K}, \] the integral closure of \(C\) in \(K\). Thus every \(a\in A\) is integral over \(C=\phi(B)\).
5. The \(B\)-algebra \(A\) is finitely generated.
The morphism \(f\) is proper, hence of finite type by definition. Since both source and target are affine, the affine criterion for finite type gives \[ A=B[a_1,\ldots,a_n] \] for finitely many elements \(a_i\in A\).
6. A finitely generated integral algebra is finite as a module. Hence \(A\) is a finite \(B\)-module.
By 4, each generator \(a_i\) from 5 is integral over the image of \(B\). Thus for each \(i\) there is a monic equation \[ a_i^{d_i}+b_{i,d_i-1}a_i^{d_i-1}+\cdots+b_{i,0}=0 \] with coefficients from the image of \(B\).
Consequently every sufficiently high power of \(a_i\) is a \(B\)-linear combination of lower powers. Therefore \(A\) is generated as a \(B\)-module by the finitely many monomials \[ a_1^{e_1}\cdots a_n^{e_n}, \qquad 0\le e_i<d_i. \] Hence \(A\) is finite as a \(B\)-module.
7. Therefore\[ \boxed{f:X\longrightarrow Y\text{ is finite}.} \]
For an affine morphism \[ \Spec A\longrightarrow\Spec B, \] finiteness is exactly the condition that \(A\) be finite as a \(B\)-module. This is 6.
8. Q.E.D.
Step 7 is the required conclusion.