Show that a finite morphism is proper.
Solution
Let \[ f:X\longrightarrow Y \] be finite. We verify the three defining properties of a proper morphism.
1. The morphism \(f\) is of finite type.
Let \[ V=\Spec B\subseteq Y \] be affine. Since \(f\) is finite, \[ f^{-1}(V)=\Spec A \] with \(A\) finite as a \(B\)-module.
If \(a_1,\ldots,a_n\) generate \(A\) as a \(B\)-module, then they also generate \(A\) as a \(B\)-algebra: \[ A=B[a_1,\ldots,a_n], \] because the right side is a \(B\)-subalgebra containing the module generators and hence all of \(A\).
Thus the affine criterion of Hartshorne II.3.3 shows that \(f\) is of finite type.
2. The morphism \(f\) is separated.
Separatedness is local on the target, so work over \[ V=\Spec B\subseteq Y, \qquad f^{-1}(V)=\Spec A. \] Then \[ X\times_YX \] restricted over \(V\) is \[ \Spec(A\otimes_BA). \] The diagonal \[ \Delta_f:X\longrightarrow X\times_YX \] is induced by the multiplication homomorphism \[ \mu:A\otimes_BA\longrightarrow A, \qquad a\otimes a'\longmapsto aa'. \] This homomorphism is surjective because \[ a=\mu(a\otimes1). \] A surjective ring homomorphism induces a closed immersion of spectra. Hence the diagonal is a closed immersion affine-locally on \(Y\), and therefore globally. Thus \(f\) is separated.
3. Every base change of \(f\) is finite.
Let \[ Y'\longrightarrow Y \] be any morphism and put \[ X'=X\times_YY'. \] This can be checked affine-locally. If \[ Y=\Spec B, \qquad X=\Spec A, \qquad Y'=\Spec B', \] with \(A\) finite over \(B\), then \[ X'=\Spec(A\otimes_BB'). \] If \(a_1,\ldots,a_n\) generate \(A\) as a \(B\)-module, then \[ a_1\otimes1,\ldots,a_n\otimes1 \] generate \(A\otimes_BB'\) as a \(B'\)-module. Hence the base-changed morphism is finite.
4. Every base change of \(f\) is a closed map.
By 3 every base change is finite. Hartshorne II.3.5(b) proves that finite morphisms are closed. Hence every base change of \(f\) is closed.
5. Thus \(f\) is universally closed.
Universal closedness means exactly that every base change of \(f\) is a closed map. This is 4.
6. Therefore\[ \boxed{f\text{ is proper}.} \]
By definition, a morphism is proper if it is separated, of finite type, and universally closed. These three properties are 2, 1, and 5, respectively.
7. Q.E.D.
Step 6 is the required conclusion.