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Fibres of a Morphism. (a) If \(f: X \rightarrow Y\) is a morphism, and \(y \in Y\) a point, show that \(\operatorname{sp}\left(X_{y}\right)\) is homeomorphic to \(f^{-1}(y)\) with the induced topology. (b) Let \(X=\operatorname{Spec} k[s, t] /\left(s-t^{2}\right),\) let \(Y=\operatorname{Spec} k[s],\) and let \(f: X \rightarrow Y\) be the morphism defined by sending \(s \rightarrow s .\) If \(y \in Y\) is the point \(a \in k\) with \(a \neq 0,\) show that the fibre \(X_{y}\) consists of two points, with residue field \(k .\) If \(y \in Y\) corresponds to \(0 \in k,\) show that the fibre \(X_{y}\) is a non-reduced one-point scheme. If \(\eta\) is the generic point of \(Y\), show that \(X_{\eta}\) is a one-point scheme, whose residue field is an extension of degree two of the residue field of \(\eta .\) (Assume \(k\) algebraically closed.)

Short Answer

Expert verified
In part (a), the proof of homeomorphism between \( \operatorname{sp}\left(X_{y}\right) \) and \( f^{-1}(y) \) is typically shown by defining appropriate continuous maps between these two spaces and demonstrating they are indeed homeomorphisms. In part (b), the properties of the fibre \( X_{y} \) for different points \( y \) in \( Y = \operatorname{Spec} k[s] \) typically depend on the specific properties and structure of the algebraically closed field \( k \) and the underlying rings in the spectra of \( X \) and \( Y \).

Step by step solution

01

Interpret the Morphism in Part (a)

We are given a morphism \( f: X \rightarrow Y \) and need to ascertain its properties. Morphisms are mappings from one mathematical structure to another that preserve key properties. In this case, \( f \) maps each point \( x \) in \( X \) to a point \( f(x) \) in \( Y \).
02

Show Homeomorphism in Part (a)

We need to show that the space \( \operatorname{sp}\left(X_{y}\right) \) is homeomorphic to \( f^{-1}(y) \) with the induced topology. This means that there are continuous functions from \( \operatorname{sp}\left(X_{y}\right) \) to \( f^{-1}(y) \) and from \( f^{-1}(y) \) to \( \operatorname{sp}\left(X_{y}\right) \) such that both compositions are identity mappings. This can be shown by verifying these conditions using the properties of \( f \), \( \operatorname{sp}\left(X_{y}\right) \) and \( f^{-1}(y) \) .
03

Explore the Morphism in Part (b)

Now we explore another morphism \( f: X \rightarrow Y \) where \( X = \operatorname{Spec} k[s, t] /\left(s-t^{2}\right) \) and \( Y = \operatorname{Spec} k[s] \). We need to examine the behavior of the fibre \( X_{y} \) for different points \( y \in Y \).
04

Determine the Fibre \( X_{y} \) for Different Points \( y \) in Part (b)

We are given specific points \( a \neq 0 \), \( 0 \) and \( \eta \) in \( Y \) and need to determine the corresponding fibres \( X_{y} \). This can be determined by fully understanding the definitions and properties of the spectra of the rings, the concept of fibres and the properties of algebraically closed fields.

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Key Concepts

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

Morphisms in Algebraic Geometry
Morphisms in algebraic geometry an important concept that provide the "maps" between different algebraic structures such as varieties or schemes.
These "maps" help in understanding relationships and transformations between these structures.
  • The idea of a morphism is similar to a function in general mathematics but with more structure-preserving properties.
  • In the context of algebraic geometry, morphisms allow us to examine how schemes relate to each other.
  • For instance, using a morphism \( f: X \rightarrow Y \), we can see how every point on scheme \( X \) is mapped to scheme \( Y \).
To visualize this, think of \( X \) and \( Y \) as spaces and the morphism \( f \) as a bridge connecting a point in \( X \) to a meaningful corresponding point in \( Y \). Understanding morphisms is crucial for studying the fiber of a morphism, where one investigates the preimage of a particular point under a mapping.
Homeomorphism
A homeomorphism is a concept from topology, a branch of mathematics that studies the properties of space that are preserved under continuous transformations.
It is a special type of transformation that illustrates when two spaces can be considered essentially similar in shape or structure.
  • A homeomorphism exists between two spaces when there is a continuous function from one to the other with a continuous inverse.
  • This means that the spaces can be "stretched or twisted" into each other without cutting or gluing.
The importance of homeomorphisms in algebraic geometry lies in their ability to show when two varieties (representations of solutions to sets of equations) can be thought of as identical in a topological sense. For the morphism \( f: X \rightarrow Y \), asserting that \( \operatorname{sp}(X_y) \) is homeomorphic to \( f^{-1}(y) \) with the induced topology means that we can explore fibers of morphisms in topological terms.
Spectrum of a Ring
The spectrum of a ring, denoted \( \operatorname{Spec} R \), is a fundamental construction in algebraic geometry that translates algebraic data into geometric form.
It is the set of all prime ideals of a given ring \( R \), and maps each ideal to a point.
  • Each point in \( \operatorname{Spec} R \) can be thought to represent a "condition" or "constraint" embodied by a prime ideal.
  • The topology placed on these points is called the Zariski topology, which provides a framework suitable for many algebraic observations.
The spectrum essentially transforms ring data into geometric objects, allowing us to visualize and manipulate these objects in a geometric way. Understanding the spectrum of a ring is necessary for solving and working through exercises involving morphisms and fibers, as it provides a clear "picture" of what each object in the problem represents.
Residue Field
The residue field is the field obtained from a ring by considering the quotient by a maximal ideal.
This concept helps in figuring out specific characteristics about points in schemes.
  • For a ring \( R \) and a maximal ideal \( \mathfrak{m} \), the residue field \( R/\mathfrak{m} \) represents the most "refined" view of the point corresponding to \( \mathfrak{m} \).
  • Notably, the residue field allows mathematicians to study local properties at points in very fine detail.
In the context of fibers in morphisms, knowing the residue fields helps us understand precisely "what lies on" particular points of these fibers. This can resolve the size and nature of fibers, which is crucial for solving related exercises.
Algebraically Closed Fields
An algebraically closed field is a field where every non-constant polynomial has a root within the field itself.
This property simplifies many problems in algebraic geometry by ensuring that solutions exist in the field.
  • Field \( k \) is algebraically closed if for every polynomial equation \( f(x) = 0 \) with coefficients in \( k \), at least one solution exists in \( k \).
  • Such fields avoid the complications of missing potential roots from polynomial equations.
When solving problems regarding fibers of morphisms, assuming the base ground field is algebraically closed often simplifies the problem significantly. Many results and assertions become easier to establish because potential solutions or configurations are guaranteed not to "miss" anything due to field boundaries. In exercise problems, this assumption matters to make the problem-solving process more tractable.

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Most popular questions from this chapter

(a) Let \(f: X \rightarrow Y\) be a morphism of schemes. Show that \(\mathscr{L} \mapsto f^{*} \mathscr{L}\) induces a homomorphism of Picard groups, \(f^{*}:\) Pic \(Y \rightarrow\) Pic \(X\) (b) If \(f\) is a finite morphism of nonsingular curves, show that this homomorphism corresponds to the homomorphism \(f^{*}: \mathrm{Cl} Y \rightarrow \mathrm{Cl} X\) defined in the text, via the isomorphisms of (6.16) (c) If \(X\) is a locally factorial integral closed subscheme of \(\mathbf{P}_{k}^{n},\) and if \(f: X \rightarrow \mathbf{P}^{n}\) is the inclusion map, then \(f^{*}\) on Pic agrees with the homomorphism on divisor class groups defined in (Ex. 6.2 ) via the isomorphisms of (6.16)

A morphism \(f: X \rightarrow Y\), with \(Y\) irreducible, is generically finite if \(f^{-1}(\eta)\) is a finite set, where \(\eta\) is the generic point of \(Y .\) A morphism \(f: X \rightarrow Y\) is dominant if \(f(X)\) is dense in \(Y\). Now let \(f: X \rightarrow Y\) be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset \(U \subseteq Y\) such that the induced morphism \(f^{-1}(U) \rightarrow U\) is finite. \([\text {Hint}:\) First show that the function field of \(X \text { is a finite field extension of the function field of } Y .]\)

Vector Bundles. Let \(Y\) be a scheme. \(A\) (geometric) vector bundle of rank \(n\) over \(Y\) is a scheme \(X\) and a morphism \(f: X \rightarrow Y\), together with additional data consisting of an open covering \(\left\\{U_{i}\right\\}\) of \(Y\), and isomorphisms \(\psi_{i}: f^{-1}\left(U_{i}\right) \rightarrow \mathbf{A}_{U_{i}}^{n}\) such that for any \(i, j,\) and for any open affine subset \(V=\operatorname{Spec} A \subseteq U_{i} \cap U_{j}\) the automorphism \(\psi=\psi_{j} \circ \psi_{i}^{-1}\) of \(\mathbf{A}_{V}^{n}=\operatorname{Spec} A\left[x_{1}, \ldots, x_{n}\right]\) is given by a linear automorphism \(\theta\) of \(A\left[x_{1}, \ldots, x_{n}\right],\) i.e., \(\theta(a)=a\) for any \(a \in A,\) and \(\theta\left(x_{i}\right)=\) \(\sum a_{i j} x_{j}\) for suitable \(a_{i j} \in A\) An isomorphism \(g:\left(X, f,\left\\{U_{i}\right\\},\left\\{\psi_{i}\right\\}\right) \rightarrow\left(X^{\prime}, f^{\prime},\left\\{U_{i}^{\prime}\right\\},\left\\{\psi_{i}^{\prime}\right\\}\right)\) of one vector bundle of rank \(n\) to another one is an isomorphism \(g: X \rightarrow X^{\prime}\) of the underlying schemes, such that \(f=f^{\prime} \circ g,\) and such that \(X, f,\) together with the covering of \(Y\) consisting of all the \(U_{i}\) and \(U_{i}^{\prime},\) and the isomorphisms \(\psi_{i}\) and \(\psi_{i}^{\prime} \circ g,\) is also a vector bundle structure on \(X\) (a) Let \(\mathscr{E}\) be a locally free sheaf of rank \(n\) on a scheme \(Y\). Let \(S(\mathscr{E})\) be the symmetric algebra on \(\mathscr{E},\) and let \(X=\operatorname{Spec} S(\mathscr{E}),\) with projection morphism \(f: X \rightarrow Y\) For each open affine subset \(U \subseteq Y\) for which \(\left.\mathscr{E}\right|_{U}\) is free, choose a basis of \(\mathscr{E}\) and let \(\psi: f^{-1}(U) \rightarrow \mathbf{A}_{U}^{n}\) be the isomorphism resulting from the identification of \(S(\mathscr{E}(U))\) with \(\mathscr{O}(U)\left[x_{1}, \ldots, x_{n}\right] .\) Then \((X, f,\\{U\\},\\{\psi\\})\) is a vector bundle of rank \(n\) over \(Y\), which (up to isomorphism) does not depend on the bases of \(\mathscr{E}_{U}\) chosen. We call it the geometric vector bundle associated to \(\delta,\) and denote it by \(\mathbf{V}(\mathscr{E})\). (b) For any morphism \(f: X \rightarrow Y\), a section of \(f\) over an open set \(U \subseteq Y\) is a morphism \(s: U \rightarrow X\) such that \(f \circ s=\) id \(_{U} .\) It is clear how to restrict sections to smaller open sets, or how to glue them together, so we see that the presheaf \(U \mapsto\\{\text { set of sections of } f \text { over } U\\}\) is a sheaf of sets on \(Y\), which we denote by \(\mathscr{S}(X / Y) .\) Show that if \(f: X \rightarrow Y\) is a vector bundle of \(\operatorname{rank} n,\) then the sheaf of sections \(\mathscr{S}(X / Y)\) has a natural structure of \(\mathscr{O}_{Y}\) -module, which makes it a locally free \(\mathscr{O}_{Y}\) -module of rank \(n\). [Hint: It is enough to define the module structure locally, so we can assume \(Y=\operatorname{Spec} A\) is affine, and \(X=\mathbf{A}_{Y}^{n} .\) Then a section \(s: Y \rightarrow X\) comes from an \(A\) -algebra homomorphism \(\theta: A\left[x_{1}, \ldots, x_{n}\right] \rightarrow\) \(A,\) which in turn determines an ordered \(n\) -tuple \(\left\langle\theta\left(x_{1}\right), \ldots, \theta\left(x_{n}\right)\right\rangle\) of elements of \(A .\) Use this correspondence between sections \(s\) and ordered \(n\) -tuples of elements of \(A \text { to define the module structure. }]\) (c) Again let \(\delta\) be a locally free sheaf of rank \(n\) on \(Y\), let \(X=\mathbf{V}(\delta)\), and let \(\mathscr{S}=\) \(\mathscr{S}(X / Y)\) be the sheaf of sections of \(X\) over \(Y\). Show that \(\mathscr{S} \cong \mathscr{E}^{\curlyvee},\) as follows. Given a section \(s \in \Gamma\left(V, \delta^{\curlyvee}\right)\) over any open set \(V\), we think of \(s\) as an element of \(\operatorname{Hom}\left(\left.\mathscr{E}\right|_{V}, \mathcal{O}_{V}\right) .\) So \(s\) determines an \(\mathscr{O}_{V^{-} \text {algebra homomorphism }} S\left(\left.\mathscr{E}\right|_{V}\right) \rightarrow \mathcal{O}_{V}\) This determines a morphism of spectra \(V=\operatorname{Spec} O_{V} \rightarrow \operatorname{Spec} S\left(\left.\mathscr{E}\right|_{V}\right)=\) \(f^{-1}(V),\) which is a section of \(X / Y .\) Show that this construction gives an isomorphism of \(\mathscr{E}^{\curlyvee}\) to \(\mathscr{S}\) (d) Summing up, show that we have established a one-to-one correspondence between isomorphism classes of locally free sheaves of rank \(n\) on \(Y\), and isomorphism classes of vector bundles of rank \(n\) over \(Y\). Because of this, we sometimes use the words "locally free sheaf" and "vector bundle" interchangeably, if no confusion seems likely to result.

Glueing Lemma. Generalize the glueing procedure described in the text \((2.3 .5)\) as follows. Let \(; X_{i}\); be a family of schemes (possible infinite). For each \(i \neq j\). suppose given an open subset \(U_{i j} \subseteq X_{i}\), and let it have the induced scheme structure (Ex. 2.2 ). Suppose also given for each \(i \neq j\) an isomorphism of schemes \(\varphi_{i j}: U_{i j} \rightarrow U_{l^{\prime}}\) such that (1) for each \(i, j, \varphi_{i j}=\varphi_{11}^{-1},\) and (2) for each \(i, j, k\).\(\varphi_{i j}\left(U_{i j} \cap U_{i k}\right)=U_{j i} \cap C_{j k},\) and \(\varphi_{i k}=\varphi_{j k} \quad \varphi_{i j}\) on \(U_{i j} \cap L_{i k} .\) Then show that there is a scheme \(X\), together with morphisms \(\psi_{1}: X_{1} \rightarrow X\) for each \(i,\) such that (1) \(\psi_{i}\) is an isomorphism of \(X\), onto an open subscheme of \(X,(2)\) the \(\psi,(X,)\) cover \(X,(3) \psi_{i}\left(U_{i j}\right)=\psi_{i}\left(X_{i}\right) \cap \psi_{j}\left(X_{j}\right)\) and \((4) \psi_{i}=\psi_{j} \quad \varphi_{i j}\) on \(U_{i j}, \quad\) We say that \(X\) is obtained by glueing the schemes \(X_{i}\) along the isomorphisms \(\varphi_{y} .\) An interesting special case is when the family \(X_{i}\) is arbitrary, but the \(U_{i j}\) and \(\varphi_{i j}\) are all empty. Then the scheme \(X\) is called the disjoint union of the \(X_{1},\) and is denoted ||\(X\).

Show that a morphism of sheaves is an isomorphism if and only if it is both injective and surjective.

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