Finite morphism

A morphism f: X → Y of schemes is a finite morphism if Y has an open cover by affine schemes

${\displaystyle V_{i}={\mbox{Spec}}\;B_{i}}$

such that for each i,

${\displaystyle f^{-1}(V_{i})=U_{i}}$

is an open affine subscheme Spec A_i, and the restriction of f to U_i, which induces a ring homomorphism

${\displaystyle B_{i}\rightarrow A_{i},}$

makes A_i a finitely generated module over B_i.[3] One also says that X is finite over Y.

In fact, f is finite if and only if for every open affine subscheme V = Spec B in Y, the inverse image of V in X is affine, of the form Spec A, with A a finitely generated B-module.[4]

For example, for any field k, ${\displaystyle {\text{Spec}}(k[t,x]/(x^{n}-t))\to {\text{Spec}}(k[t])}$ is a finite morphism since ${\displaystyle k[t,x]/(x^{n}-t)\cong k[t]\oplus k[t]\cdot x\oplus \cdots \oplus k[t]\cdot x^{n-1}}$ as ${\displaystyle k[t]}$-modules. Geometrically, this is obviously finite since this is a ramified n-sheeted cover of the affine line which degenerates at the origin. By contrast, the inclusion of A¹ − 0 into A¹ is not finite. (Indeed, the Laurent polynomial ring k[y, y⁻¹] is not finitely generated as a module over k[y].) This restricts our geometric intuition to surjective families with finite fibers.

• The composition of two finite morphisms is finite.
• Any base change of a finite morphism f: X → Y is finite. That is, if g: Z → Y is any morphism of schemes, then the resulting morphism X ×_Y Z → Z is finite. This corresponds to the following algebraic statement: if A and C are (commutative) B-algebras, and A is finitely generated as a B-module, then the tensor product A ⊗_B C is finitely generated as a C-module. Indeed, the generators can be taken to be the elements a_i ⊗ 1, where a_i are the given generators of A as a B-module.
• Closed immersions are finite, as they are locally given by A → A/I, where I is the ideal corresponding to the closed subscheme.
• Finite morphisms are closed, hence (because of their stability under base change) proper.[5] This follows from the going up theorem of Cohen-Seidenberg in commutative algebra.
• Finite morphisms have finite fibers (that is, they are quasi-finite).[6] This follows from the fact that for a field k, every finite k-algebra is an Artinian ring. A related statement is that for a finite surjective morphism f: X → Y, X and Y have the same dimension.
• By Deligne, a morphism of schemes is finite if and only if it is proper and quasi-finite.[7] This had been shown by Grothendieck if the morphism f: X → Y is locally of finite presentation, which follows from the other assumptions if Y is Noetherian.[8]
• Finite morphisms are both projective and affine.[9]

• Glossary of algebraic geometry
• Finite algebra

• The Stacks Project Authors, The Stacks Project