(Finite)
\(f : X \to Y\) is finite if \(Y\) has an affine cover by \(\Spec B_i\) with \(f^{-1}(\Spec B_i) = \Spec A_i\) affine and \(A_i\) a \(B_i\)-module of finite type (D-DEFFGMOD).
(Maps of proper curves)
Let \(C\) and \(C'\) be irreducible curves proper over a field \(k\). Every \(k\)-morphism \(f: C \to C'\) is proper, so \(f(C)\) is an irreducible closed subset of \(C'\): \(f\) is either constant with value a closed point or surjective. If \(C\) is nonsingular, \(C'\) is integral and \(f\) is surjective, then \(f\) is finite and \(k(C')\subseteq k(C)\) is a finite extension. [@Har10a, Corollary II.4.8, Proposition II.6.8]
A module generated by \(a_1,\ldots,a_n\) is generated as an algebra by the same elements, so finite morphisms are of finite type. The converse fails: \(k[x]\) has one algebra generator over \(k\) and infinite dimension as a \(k\)-vector space, so \(\AA^1_k \to \Spec k\) is of finite type and not finite.
Finite morphisms are affine and proper, and their fibres are finite sets. Closedness of a finite morphism follows from lying over and going up for the integral ring map \(B_i\to A_i\). If \(f\) is finite and flat and \(Y\) is Noetherian, then \(f_*\OO_X\) is locally free, and its rank at \(y\) is \(\dim_{\kappa(y)}\Gamma(X_y,\OO_{X_y})\). A nonconstant morphism of nonsingular projective curves over a field is finite, and its degree is the degree of the extension of function fields.
Dependencies
D-MORAFFAffine morphismsuses
Backlinks
PR-MORFINCHARFinite equals proper plus quasi-finiteusesPR-MIRACLEFLATFinite morphisms of smooth varieties of equal dimension are flatusesD-MORQFQuasi-finite morphismsuses