Finite-Type Morphisms

by iarnab_kundu, Dec 9, 2018, 7:04 PM

Definition- Let $f:Y\to X$ be a morphism of schemes. Then for every open affine sub-scheme $U=\text{Spec}(A)$ of $X$ we have that the pre-image $f^{-1}(U)$ can be covered by affine opens $V_i=\text{Spec}(B_i)$ of $Y$ such that for each $i$ we have that $B_i$ is a finitely generated $A$ algebra via the induced morphism.

Example- Let $A$ be a ring. Then any basic open affine $\text{Spec}(A_f)\to\text{Spec}(A)$ is locally of finite type. In fact it is of finite-type.

Proposition- Let $f:Y\to X$ be a morphism of schemes. Suppose there is an open affine cover $U_i=\text{Spec}(A_i)$ such that for each $i$ we have that $f^{-1}(U_i)=\cup (V_{ij}=\text{Spec}(B_{ij}))$ where $B_{ij}$ is a finitely-generated $A_i$ algebra for each $j$ . Then $f$ is locally of finite type.
Lemma- Let $f:Y=\text{Spec}(B)\to X=\text{Spec}(A)$ be a morphism of affine schemes, such that there are $f_i\in A$ for which $B_{f(t_i)}$ is a finite-type algebra over $A_{t_i}$ . Then $B$ is of finite type over $A$ .
Proof- Since $X$ is quasi-compact, we extract a finite sub-cover $\{D(t_i)\}_{i=1\ldots n}$ of $X$ . Then $D(t_1,\ldots,t_n)=X$ implies that the radical of the ideal $(t_1,\ldots,t_n)$ equals $1$ .
Let $\{\alpha_{ij}\}_{j=1\ldots N_i}\in B$ be a finite generating set of the algebra $B_{f(t_i)}$ over $A_{t_i}$ . We consider the set $\{\alpha_{ij}\}_{i=1\ldots n, j=1\ldots N_i}$ , which we claim to be the generating set of $B$ over $A$ . Let $b\in B$ . By the definition of the generators, for each $i$ there exists $m_i$ such that $f^{m_i}(b-b_i)=0$ where $b_i=\sum_j\beta_i\alpha_{ij}$ . Since radical of the ideal $(t_1^{m_1},\ldots,t_n^{m_n})$ equals the radical of the ideal $(t_1,\ldots,t_n)$ . Thus there are elements $a_i\in A$ such that $\sum a_if_i^{m_i}=1$ , and therefore $b=\sum a_if_i^{m_i}b_i$ . Thus we are done.
Proof- Let $U=\text{Spec}(A)$ be an affine open of $X$ . Then $g$

Proposition- Let $f:Y\to X$ be a finite type morphism of schemes. Then for any open affine subset $U=\text{Spec}(A)$ of $X$ and $V=\text{Spec}(B)$ of $Y$ such that $f(V)\subset U$ we have that $B$ is an A-algebra of finite type.

Definition: Let $f:Y\to X$ be a morphism of schemes. Then $f$ is said to be of finite type if it is locally of finite type and quasi compact.

Proposition: Let $f:Y\to X$ be a morphism of schemes. Suppose there is an open affine cover $U_i=\text{Spec}(A_i)$ such that for each $i$ we have that $f^{-1}(U_i)=\cup_{j=1}^{N_i} (V_{ij}=\text{Spec}(B_{ij}))$ (pre-image covered by finitely many such affine opens) where $B_{ij}$ is a finitely-generated $A_i$ algebra for each $j$ . Then $f$ is of finite type.
Proof- Follows from the above proposition and the proposition about quasi-compact maps.

Proposition: Let $f:Y\to X$ be a finite type morphism of schemes. Then for any open affine subset $U=\text{Spec}(A)$ of $X$ and $V=\text{Spec}(B)$ of $Y$ such that $f(V)\subset U$ we have that $B$ is an A-algebra of finite type.
Proof- By the definition

This post has been edited 2 times. Last edited by iarnab_kundu, Dec 9, 2018, 9:35 PM

algebraic geometry

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Cracking the Nut Open

Finite-Type Morphisms

by iarnab_kundu, Dec 9, 2018, 7:04 PM

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