/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Describe Spec \(\mathbf{Z}\), an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 5

Describe Spec \(\mathbf{Z}\), and show that it is a final object for the category of schemes. i.e., each scheme \(X\) admits a unique morphism to Spec \(\mathbf{Z}\).

Short Answer

Expert verified
Spec \( \mathbf{Z} \) can be understood as the set \(\{ (0), (2), (3), (5), (7), \ldots \}\), the set of all prime numbers and 0, considering that \(\mathbf{Z}\) only has two prime ideals. It is a final object in the category of schemes because for any scheme X there is a unique morphism \( X \rightarrow \) Spec \( \mathbf{Z} \).

Step by step solution

01

Describe Spec \( \mathbf{Z} \)

Since \(\mathbf{Z}\) only has two prime ideals, the ideal (0) and (p) for any prime p. The Spec \( \mathbf{Z} \) is the set of these ideals, that can be seen as \(\{ (0), (2), (3), (5), (7), \ldots \}\), i.e., the set of all prime numbers and 0.
02

Proving Spec \( \mathbf{Z} \) as a Final Object

To show Spec \( \mathbf{Z} \) is a final object in the category of schemes, we need to prove that for every scheme \( X \), there is a unique morphism \( X \rightarrow \) Spec \( \mathbf{Z} \). By definition, over any scheme \( X \), the structure sheaf of Spec \( \mathbf{Z} \) is constant, it's the sheaf associated to the constant presheaf with value \( \mathbf{Z} \). Hence, there is a unique morphism of schemes \( X \rightarrow \) Spec \( \mathbf{Z} \), given on an open set \( U \subseteq X \) by the unique ring homomorphism \( \mathbf{Z} \rightarrow \mathcal{O}(U) \). This shows that Spec \( \mathbf{Z} \) is a final object in the category of schemes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Scheme Theory
Scheme theory forms the backbone of modern algebraic geometry, building upon the concept of a variety by allowing 'singularities' and incorporating prime ideals as points. Think of schemes as geometric spaces that are defined not only by their points but also by the 'functions' that live on them. In essence, a scheme is a topological space teamed up with a structure sheaf, which assigns rings of functions to every open set.
The specific example of Spec \(\mathbb{Z}\) illustrates these ideas vividly. It consists of points corresponding to prime ideals, giving us an elegant geometric picture of arithmetic properties of integers.
Prime Ideals
Prime ideals are a cornerstone of scheme theory and algebraic geometry. A prime ideal in a ring like \(\mathbb{Z}\) is a subset of elements that is not the whole ring, and it observes a key property: if the product of two elements is in the ideal, then at least one of these elements is also in the ideal. The prime ideals of \(\mathbb{Z}\) correspond to the zero ideal and the ideals generated by prime numbers, representing the 'points' of Spec \(\mathbb{Z}\).
Understanding the role of prime ideals in schemes helps in visualizing algebraic structures as geometric entities and forms the foundation for discussing morphisms and mappings between different schemes.
Category of Schemes
In the category of schemes, each object is a scheme and morphisms between these objects follow specific algebraic rules. It's an abstract framework that helps mathematicians to deal with schemes in a structured way.
Within this setting, Spec \(\mathbb{Z}\) holds a special place as a final object, meaning that every scheme maps to it in a unique way鈥攋ust like all roads lead to Rome. This uniqueness is crucial for understanding how schemes relate to each other and how they can be transformed through morphisms.
Structure Sheaf
A structure sheaf is an essential part of a scheme that links the geometrical with the algebraic. For each open subset of a scheme, the structure sheaf assigns a ring of 'functions,' allowing us to 'do algebra' locally on the scheme.
The structure sheaf of Spec \(\mathbb{Z}\) is particularly simple: it assigns the integer ring \(\mathbb{Z}\) to any open set. This simplicity is what enables the unique morphisms we find in Spec \(\mathbb{Z}\) as a final object in the category of schemes, neatly tying together the algebraic structure with the geometrical intuition.
Morphism of Schemes
A morphism of schemes is a way of transforming one scheme into another, involving both a continuous function between the underlying topological spaces and a compatible family of ring homomorphisms for the structure sheaves.
Every scheme 'wants' to map to Spec \(\mathbb{Z}\) in its own unique way through such a morphism. It is akin to saying there is a specific and unique perspective from which to view all other schemes from the vantage point of Spec \(\mathbb{Z}\). This unique mapping perspective is not only foundational for understanding how different algebraic structures interact with one another but also emphasizes the universality of \(\mathbb{Z}\) within the world of schemes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Support. Let \(\mathscr{F}\) be a sheaf on \(X\), and let \(s \in \mathscr{F}(U)\) be a section over an open set \(U\) The support of \(s\), denoted Supp s, is defined to be \(\left\\{P \in U | s_{P} \neq 0\right\\},\) where \(s_{P}\) denotes the germ of s in the stalk \(\overline{\mathscr{F}}_{p}\). Show that Supp s is a closed subset of \(U\). We define the support of \(\overline{\mathscr{F}}, \operatorname{Supp}, \overline{\mathscr{F}},\) to be \(\left\\{P \in X | \mathscr{F}_{P} \neq 0\right\\},\) It need not be a closed subset.

Let \(X\) be a scheme of finite type over a field \(k\) (not necessarily algebraically closed). (a) Show that the following three conditions are equivalent (in which case we say that \(X\) is geometrically irreducible). (i) \(X \times_{k} \bar{k}\) is irreducible, where \(\bar{k}\) denotes the algebraic closure of \(k .\) abuse of notation, we write \(X \times_{k} \bar{k}\) to denote \(X \times_{\text {spec } k}\) Spec \(\bar{k} .\) (ii) \(X \times_{k} k_{s}\) is irreducible, where \(k_{s}\) denotes the separable closure of \(k\) (iii) \(X \times_{k} K\) is irreducible for every extension field \(K\) of \(k\) (b) Show that the following three conditions are equivalent (in which case we say \(X\) is geometrically reduced) (i) \(X \times_{k} \bar{k}\) is reduced. (ii) \(X \times_{k} k_{p}\) is reduced, where \(k_{p}\) denotes the perfect closure of \(k\) (iii) \(X \times_{k} K\) is reduced for all extension fields \(K\) of \(k\) (c) We say that \(X\) is geometrically integral if \(X \times_{k} \bar{k}\) is integral. Give examples of integral schemes which are neither geometrically irreducible nor geometrically reduced.

Let \(A\) be a ring. Show that the following conditions are equivalent: (i) Spec \(A\) is disconnected : (ii) there exist nonzero elements \(e_{1}, e_{2} \in A\) such that \(e_{1} e_{2}=0, e_{1}^{2}=e_{1}, e_{2}^{2}=e_{2}\) \(e_{1}+e_{2}=1\) (these elements are called orthogonal idempotents): (iii) \(A\) is isomorphic to a direct product \(A_{1} \times A_{2}\) of two nonzero rings.

Closed Subschemes. (a) Closed immersions are stable under base extension: if \(f: Y \rightarrow X\) is a closed immersion, and if \(X^{\prime} \rightarrow X\) is any morphism, then \(f^{\prime}: Y \times_{X} X^{\prime} \rightarrow X^{\prime}\) is also a closed immersion. (b) If \(Y\) is a closed subscheme of an affine scheme \(X=\operatorname{Spec} A\), then \(Y\) is also affine, and in fact \(Y\) is the closed subscheme determined by a suitable ideal \(\mathfrak{a} \subseteq A\) as the image of the closed immersion \(\operatorname{Spec} A / \mathfrak{a} \rightarrow \operatorname{Spec} A\). [Hints: First show that \(Y\) can be covered by a finite number of open affine subsets of the form \(D\left(f_{i}\right) \cap Y,\) with \(f_{i} \in A .\) By adding some more \(f_{i}\) with \(D\left(f_{i}\right) \cap Y=\varnothing\) if necessary, show that we may assume that the \(D\left(f_{i}\right)\) cover \(X .\) Next show that \(f_{1}, \ldots, f_{r}\) generate the unit ideal of \(A .\) Then use (Ex. 2.17 b) to show that \(Y\) is affine, and (Ex. \(2.18 \mathrm{d}\) ) to show that \(Y\) comes from an ideal \(\mathfrak{a} \subseteq\) A. .] Note: We will give another proof of this result using sheaves of ideals later (5.10). (c) Let \(Y\) be a closed subset of a scheme \(X\), and give \(Y\) the reduced induced subscheme structure. If \(Y^{\prime}\) is any other closed subscheme of \(X\) with the same underlying topological space, show that the closed immersion \(Y \rightarrow X\) factors through \(Y^{\prime} .\) We express this property by saying that the reduced induced structure is the smallest subscheme structure on a closed subset. (d) Let \(f: Z \rightarrow X\) be a morphism. Then there is a unique closed subscheme \(Y\) of \(X\) with the following property: the morphism \(f\) factors through \(Y\), and if \(Y^{\prime}\) is any other closed subscheme of \(X\) through which \(f\) factors, then \(Y \rightarrow X\) factors through \(Y^{\prime}\) also. We call \(Y\) the scheme-theoretic image of \(f\). If \(Z\) is a reduced scheme, then \(Y\) is just the reduced induced structure on the closure of the image \(f(Z)\)

In this exercise, we compare some properties of a ring homomorphism to the induced morphism of the spectra of the rings. (a) Let \(A\) be a ring, \(X=\operatorname{Spec} A,\) and \(f \in A .\) Show that \(f\) is nilpotent if and only if \(D(f)\) is empty. (b) Let \(\varphi: A \rightarrow B\) be a homomorphism of rings, and let \(f: Y=\operatorname{Spec} B \rightarrow X=\) Spec \(A\) be the induced morphism of affine schemes. Show that \(\varphi\) is injective if and only if the map of sheaves \(f^{*}: c_{1} \rightarrow f_{*} C_{r}\) is injective. Show furthermore in that case \(f\) is dominant, i.e., \(f(Y)\) is dense in \(X\). (c) With the same notation, show that if \(\varphi\) is surjective, then \(f\) is a homeomorphism of \(Y\) onto a closed subset of \(X,\) and \(f^{* *}: C_{X} \rightarrow f_{*} C_{Y}\) is surjective. (d) Prove the converse to (c). namely. if \(f: Y \rightarrow X\) is a homeomorphism onto a closed subset, and \(f^{\prime \prime}: C_{1} \rightarrow f_{*} C_{1},\) is surjective. then \(\varphi\) is surjective. [Hint: Consider \(\left.X^{\prime}=\operatorname{Spec}(.4 \mathrm{ker} \varphi) \text { and use }(\mathrm{b}) \text { and }(\mathrm{c}) .\right]\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.