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Chapter 2: Problem 5

A morphism \(f: X \rightarrow Y\) is quasi-finite if for every point \(y \in Y, f^{-1}(y)\) is a finite set. (a) Show that a finite morphism is quasi-finite. (b) Show that a finite morphism is closed, i.e., the image of any closed subset is closed. (c) Show by example that a surjective, finite-type, quasi-finite morphism need not be finite.

Short Answer

Expert verified
A finite morphism is quasi-finite since the pre-image of every point under a finite morphism is a finite set. It's closed because it's universally closed meaning the image of any closed subset is closed. However, a counterexample which is surjective, finite-type, quasi-finite but not finite is given by the morphism from the affine line with double origin to the usual affine line.

Step by step solution

01

Understanding Finite and Quasi-finite morphisms

In algebraic geometry, a morhphism between schemes \(f: X \rightarrow Y\) is said to be finite if it satisfies two conditions: it is schematically proper (that means it is universally closed) and quasi-compact. A morphism is said to be quasi-finite if for every point \(y \in Y\), the pre-image \(f^{-1}(y)\) is a finite set.
02

Proving finite morphisms are quasi-finite

For a finite morphism \(f: X \rightarrow Y\), consider any point \(y \in Y\). Since the morphism is a finite morphism, the pre-image \(f^{-1}(y)\) is a finite set, by definition, thus satisfying the condition to be a quasi-finite morphism.
03

Demonstrating finite morphisms are closed

For a finite morphism \(f: X \rightarrow Y\), consider any closed subset \(C\) of \(X\). Since the morphism is finite, it is also universally closed. This means that the image of any closed set under \(f\) is closed in \(Y\). Therefore, \(f(C)\) is a closed set in \(Y\), demonstrating that the finite morphism is closed.
04

Showing a counterexample of finite-type, quasi-finite, surjective morphism

Consider a morphism \(f: X \rightarrow Y\) from the affine line with double origin to the usual affine line. This morphism is surjective and of finite type. Each pre-image is a finite set (of one point) so the morphism is quasi-finite as well. However, this morphism is not finite because it fails to fulfill the condition of being universally closed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quasi-finite Morphism
Let's delve into one of the fundamental concepts in algebraic geometry, the quasi-finite morphism. A morphism is seen as a mathematical function between two spaces, and it replicates the idea of a function in basic algebra, connecting elements from one set to another. However, in the context of schemes, which are the primary objects of study in algebraic geometry, it becomes a more nuanced relationship.

Imagine a morphism as a bridge connecting every point in one geometric space to a point in another space. A quasi-finite morphism, specifically, is a bridge that ensures you only have a limited number of destinations in the target space for each starting point. In mathematical terms, a quasi-finite morphism is one for which the pre-image of every point in the target space consists of a finite number of points.

Now, building from the textbook solution, we understand that a finite morphism is inherently quasi-finite, because it maps points to a restricted or finite set of points in the target space. This finite interaction is central to its definition, and is crucial for comprehending the basic structure of algebraic varieties and the ways they can interact.
Algebraic Geometry
Algebraic geometry is a branch of mathematics that serves as a meeting ground for algebra, geometry, and number theory. It examines the shapes and properties of spaces that are defined by polynomial equations鈥攖hese spaces are known as algebraic varieties. Think of it as the study of geometric figures you can draw using algebraic equations. This can range from simple lines and curves to more intricate shapes like hyperbolas, ellipses, and even higher-dimensional analogues.

When dealing with algebraic geometry, morphisms are akin to translators that help us understand how one shape can transform or relate to another. Through the lens of this subject, a morphism can take an algebraic variety and stretch it, shrink it, twist it, or fold it into a new shape. Understanding how these morphisms work aids in solving various problems in mathematics and even in theoretical physics.
Universally Closed Morphism
In the grand tapestry of shapes and structures in algebraic geometry, certain morphisms act like the perfect weavers, able to tie things together without losing any threads. A universally closed morphism is one such morphism鈥攊t's like a magic net that catches all the points in a closed set, and when it throws them to the target space, they still remain within a closed set.

Consider the concept of 'closed' in mathematics鈥攊t typically refers to something that is complete, contains all its limit points, and has certain desirable properties that are preserved under various operations. Now picture a morphism that, when you apply it to a closed set within one geometric space, the resulting image is also closed in the target space. As we gleaned from the problem's solution, this feature is significant when you're evaluating the nature of a finite morphism. It offers a reassurance of sorts that the structure and integrity of the closed set are preserved upon mapping. This characteristic is critically important in topology, a key part of algebraic geometry, as it affects the continuity of shapes and ensures that our mathematical 'weaving' is flawless.

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Most popular questions from this chapter

Espace Etale of a Presheuf. (This exercise is included only to establish the connection between our definition of a sheaf and another definition often found in the literature. See for example Godement [1. Ch. II, \$1.2].) Given a presheaf \(\mathscr{F}\) un \(X\), we define a topological space Sp茅( \(\bar{y}\) ), called the espuce 茅ral茅 of \(\mathscr{F},\) as follows. As a set. Sp茅(. \(\overline{\mathscr{F}})=\cup_{P e}, x^{-} \overline{\mathscr{H}}_{P} .\) We define a projection map \(\pi: \operatorname{Sp茅}(\mathscr{F}) \rightarrow X\) by sending \(s \in \overline{\mathscr{H}}_{p}\) to \(P\). For each open set \(U \subseteq X\) and each section \(s \in \overline{\mathscr{F}}(\mathcal{L}),\) we obtain a \(\operatorname{map} \bar{\Im}: L \rightarrow \operatorname{Spei}(\mathscr{F})\) by sending \(P \mapsto s_{P},\) its germ at \(P .\) This map has the property that \(\pi \quad \bar{s}=\) id \(_{l},\) in other words, it is a "section" of \(\pi\) over \(U\). We now make Sp茅(. \(\overline{\mathscr{H}}\) ) into a topological space by giving it the strongest topology such that show that the sheaf \(\bar{y}^{+}\) associated to \(\bar{y}\) can be described as follows: for any open set \(l \subseteq X, \overline{\mathscr{H}}^{+}(U)\) is the set of continuous sections of \(\operatorname{Spei}(\mathscr{F})\) over \(U\). In particular, the original presheaf \(\mathscr{I}\) was a sheaf if and only if for each \(U, \mathscr{F}(U)\) is equal to the set of all continuous sections of Sp茅(. \(\overline{\mathcal{F}}\) ) over \(U\)

Constructible Sets. Let \(X\) be a Zariski topological space. A constructible subset of \(X\) is a subset which belongs to the smallest family \(\mathfrak{Y}\) of subsets such that (1) every open subset is in \(\mathfrak{F},(2)\) a finite intersection of elements of \(\mathfrak{F}\) is in \(\mathfrak{F},\) and (3) the complement of an element of \(\mathfrak{F}\) is in \(\mathfrak{F}\) (a) A subset of \(X\) is locally closed if it is the intersection of an open subset with a closed subset. Show that a subset of \(X\) is constructible if and only if it can be written as a finite disjoint union of locally closed subsets. (b) Show that a constructible subset of an irreducible Zariski space \(X\) is dense if and only if it contains the generic point. Furthermore, in that case it contains a nonempty open subset. (c) A subset \(S\) of \(X\) is closed if and only if it is constructible and stable under specialization. Similarly, a subset \(T\) of \(X\) is open if and only if it is constructible and stable under generization. (d) If \(f: X \rightarrow Y\) is a continuous map of Zariski spaces, then the inverse image of any constructible subset of \(Y\) is a constructible subset of \(X\)

Let \(A\) be a ring, let \(S=A\left[x_{0}, \ldots, x_{r}\right]\) and let \(X=\) Proj \(S\). We have seen that a homogeneous ideal \(I\) in \(S\) defines a closed subscheme of \(X\) (Ex. 3.12 ), and that conversely every closed subscheme of \(X\) arises in this way (5.16) (a) For any homogeneous ideal \(I \subseteq S\), we define the saturation \(I\) of \(I\) to be \(\left\\{s \in S | \text { for each } i=0, \ldots, r, \text { there is an } n \text { such that } x_{i}^{n} s \in I\right\\} .\) We say that \(I\) is saturated if \(I=I .\) Show that \(T\) is a homogeneous ideal of \(S\). (b) Two homogeneous ideals \(I_{1}\) and \(I_{2}\) of \(S\) define the same closed subscheme of \(X\) if and only if they have the same saturation. (c) If \(Y\) is any closed subscheme of \(X\), then the ideal \(\Gamma_{*}\left(\mathscr{I}_{Y}\right)\) is saturated. Hence it is the largest homogeneous ideal defining the subscheme \(Y\) (d) There is a \(1-1\) correspondence between saturated ideals of \(S\) and closed subschemes of \(X\).

If \(X\) is a scheme of finite type over a field, show that the closed points of \(X\) are dense. Give an example to show that this is not true for arbitrary schemes.

Let \(f: X \rightarrow Y\) be a morphism of separated schemes of finite type over a noetherian scheme \(S\). Let \(Z\) be a closed subscheme of \(X\) which is proper over \(S\). Show that \(f(Z)\) is closed in \(Y,\) and that \(f(Z)\) with its image subscheme structure (Ex. \(3.11 d\) ) is proper over \(S .\) We refer to this result by saying that "the image of a proper scheme is proper." [Hint: Factor \(f\) into the graph morphism \(\Gamma_{f}: X \rightarrow X \times_{s} Y\) followed by the second projection \(\left.p_{2}, \text { and show that } \Gamma_{f} \text { is a closed immersion. }\right]\)
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