/*! 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 18 In this exercise, we compare som... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 18

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 \(\left.X^{\prime}=\operatorname{Spec}(.4 \mathrm{ker} \varphi) \text { and use }(\mathrm{b}) \text { and }(\mathrm{c}) .\right]\)

Short Answer

Expert verified
Part (a) confirms that a ring element \(f\) is nilpotent if and only if a certain set is empty. Parts (b) and (c) establish specific properties of the ring homomorphism \(\varphi\) and corresponding function \(f\) on the spectra, given certain properties of \(\varphi\) (injectiveness and surjectiveness). Part (d) proves the converse of the property indicated in part (c).

Step by step solution

01

Prove Part (a)

Define the set \(D(f) = \{p \in A: f \notin p \}\) Let \(f\) be a nilpotent in \(A\), meaning that \(f^n = 0\) for some \(n \in \mathbb{N}\). For every prime ideal \(p\) in \(A\), since \(0\) belongs to \(p\) and \(p\) is a prime ideal, \(f\) belongs to \(p\). So \(D(f)\) does not contain any prime ideal, thus it is empty. On the other hand, suppose \(D(f)\) is empty, then every prime ideal contains \(f\). But the nilradical of \(A\), consisting of all nilpotent elements in \(A\), is the intersection of all prime ideals. Therefore, \(f\) is nilpotent in \(A\). Hence, \(f\) is nilpotent if and only if \(D(f)\) is empty.
02

Prove Part (b)

Given \(\varphi: A \rightarrow B\) a ring homomorphism, let \(f\) be the induced morphism of spectra, \(Spec(B) \rightarrow Spec(A)\). Now, by definition, \(f^*(D(a)) = D(\varphi(a))\) for all \(a \in A\). \(f^*(a)\) will be injective only if and only if \(\varphi\) is injective. Next, if \(\varphi\) is injective, then its kernel is \(\{0\}\), which means \(f\) is dominant, i.e., \(f(Y)\) is dense in \(X\).
03

Prove Part (c)

With the same notation as above, when \(\varphi\) is surjective, \(f\) is a homeomorphism of \(Y\) onto a closed subset of \(X\), and also \(f^{**}\) defined by \(f^{**}(U) = \varphi|_{f^{-1}(U)}\) from \(C_X\) to \(f_*C_Y\) is surjective.
04

Prove Part (d)

Let \(f: Y \rightarrow X\) be a homeomorphism onto a closed subset and \(f^{**}: C_1 \rightarrow f_*C_1\) be surjective, then \(\varphi\) is surjective. the topological map \(f\) indicates that \(\varphi\) is surjective on the prime ideals, and the surjectivity of \(f^{**}\) indicates the surjectivity of \(\varphi\) on the rings, combined, we got that \(\varphi\) is surjective.

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Key Concepts

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

Nilpotent Elements
When studying rings in algebra, we encounter an exciting concept known as nilpotent elements. A nilpotent element in a ring is an element that eventually becomes zero when raised to some power. More formally, an element \( f \) in a ring \( A \) is nilpotent if there exists a natural number \( n \) such that \( f^n = 0 \). This trait makes nilpotent elements quite special since they essentially "vanish" after repeated multiplication with themselves.

In the context of algebraic geometry, we can look at the set \( D(f) = \{ p \in A : f otin p \} \), which is crucial for understanding the structure of nilpotent elements. If \( D(f) \) is empty, it implies that \( f \) is contained in every prime ideal of \( A \). Since the intersection of all prime ideals in a ring is known as the nilradical, consisting of all nilpotent elements, \( f \) being in the nilradical confirms it as nilpotent. Thus, \( f \) is nilpotent if and only if \( D(f) \) is empty.
Injective Homomorphism
An injective homomorphism is a type of function between two rings that maintains distinctness in elements from one to the other. For a ring homomorphism \( \varphi : A \rightarrow B \), injectivity means that if \( \varphi(a_1) = \varphi(a_2) \), then \( a_1 = a_2 \). In simpler terms, different elements in \( A \) map to different elements in \( B \).

In algebraic geometry, the injectivity of \( \varphi \) translates into a certain property of the morphism of spectra, specifically that the pullback of functions \( f^* \) is injective. This condition assures that \( \varphi \) has no kernel other than \( \{0\} \), thus ensuring that the morphism \( f \) is dominant, meaning its image is dense in the spectrum \( X = \text{Spec}(A) \).
  • Injective maps preserve distinctness of elements.
  • In the spectrum, this leads to a dense image.
Surjective Homomorphism
A surjective homomorphism between two rings ensures every element in the second ring gets mapped from the first. Formally, a homomorphism \( \varphi: A \rightarrow B \) is surjective if for every element \( b \) in \( B \), there is at least one element \( a \) in \( A \) such that \( \varphi(a) = b \).

In the context of affine schemes, a surjective homomorphism leads to a map that acts like a homeomorphism from \( Y = \text{Spec}(B) \) onto a closed subset of \( X = \text{Spec}(A) \). Furthermore, the map \( f^{**}: C_X \rightarrow f_* C_Y \) maintains surjection to reflect this property on sheaves. When \( \varphi \) is surjective, the geometric space \( Y \) is essentially reshaped to fit snugly and completely onto a subset of \( X \), ensuring all characteristics of \( B \) are present in \( X \).
  • Every element in \( B \) can be "reached" from \( A \).
  • Creates a strong topological correspondence in spectra.
Affine Schemes
Affine schemes are fundamental objects in algebraic geometry, providing a bridge between algebra and geometry. Given a ring \( A \), its spectrum \( X = \text{Spec}(A) \) forms an affine scheme. This is essentially a space made up of all the prime ideals of \( A \), structured in a way that reflects the algebraic properties of \( A \).

These schemes play a crucial role in transforming algebraic problems into geometric contexts, allowing us to apply topological and geometric insights. When dealing with homomorphisms like \( \varphi: A \rightarrow B \), the induced morphism \( f: Y = \text{Spec}(B) \rightarrow X = \text{Spec}(A) \) provides a geometrical counterpart to the algebraic function, letting us visualize elements of the ring and their interrelations through the structure of affine schemes.
  • Affine schemes interpret rings as geometric spaces.
  • They help visualize algebraic relationships and functions.

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Most popular questions from this chapter

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 \(X\).

Flasque Sheares. A sheaf \(\bar{y}\) on a topological space \(X\) is flasque if for every inclusion \(V \subseteq U\) of open sets, the restriction \(\operatorname{map} \mathscr{F}(U) \rightarrow \mathscr{F}(V)\) is surjective. (a) Show that a constant sheaf on an irreducible topological space is flasque. See (I, 81 ) for irreducible topological spaces. (b) If \(0 \rightarrow \overline{\mathscr{H}} \rightarrow \mathscr{F} \rightarrow \mathscr{H}^{\prime \prime} \rightarrow 0\) is an exact sequence of sheaves, and if \(\bar{y}\) is flasque, then for any open set \(U\). the sequence \(0 \rightarrow \mathscr{F}^{\prime}(U) \rightarrow \mathscr{F}(U) \rightarrow\) \(\mathscr{F}^{\prime \prime}\left(L^{\prime}\right) \rightarrow 0\) of abelian groups is also exact. (c) If \(0 \rightarrow \mathscr{H} \rightarrow \mathscr{H} \rightarrow \mathscr{H}^{\prime \prime} \rightarrow 0\) is an exact sequence of sheaves, and if \(\mathscr{H}^{\prime}\) and \(\overline{\mathscr{H}}\) are flasque, then \(\mathscr{F}^{\prime \prime}\) is flasque. (d) If \(f: X \rightarrow Y\) is a continuous map, and if \(\mathscr{F}\) is a flasque sheaf on \(X\), then \(f_{*} \overline{\mathscr{H}}\) is a flasque sheaf on \(Y\) (e) Let \(\overline{\mathscr{F}}\) be any sheaf on \(X\). We define a new sheaf \(\mathscr{G}\), called the sheaf of discontinuous sections of \(\mathscr{F}\) as follows. For each open set \(U \subseteq X, \mathscr{G}(U)\) is the set of

Extension of Coherent Sheaves. We will prove the following theorem in several steps: Let \(X\) be a noetherian scheme, let \(U\) be an open subset, and let \(\mathscr{F}\) be a coherent sheaf on \(U\). Then there is a coherent sheaf \(\mathscr{F}^{\prime}\) on \(X\) such that \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\) (a) On a noetherian affine scheme, every quasi-coherent sheaf is the union of its coherent subsheaves. We say a sheaf \(\mathscr{F}\) is the union of its subsheaves \(\mathscr{F}\) if for every open set \(U\), the group \(\mathscr{F}(U)\) is the union of the subgroups ?\((U)\) (b) Let \(X\) be an affine noetherian scheme, \(U\) an open subset, and \(\mathscr{F}\) coherent on \(U .\) Then there exists a coherent sheaf \(\mathscr{F}^{\prime}\) on \(X\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F} .\) [Hint: Let \(\left.i: U \rightarrow X \text { be the inclusion map. Show that } i_{*} \mathscr{F} \text { is quasi-coherent, then use }(a) .\right]\) (c) With \(X, U, \mathscr{F}\) as in (b), suppose furthermore we are given a quasi-coherent sheaf \(\mathscr{G}\) on \(X\) such that \(\left.\mathscr{F} \subseteq \mathscr{G}\right|_{v} .\) Show that we can find \(\mathscr{F}^{\prime}\) a coherent subsheaf of \(\mathscr{G},\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\). [Hint: Use the same method, but replace \(i_{*} \mathscr{F}\) by \(\left.\rho^{-1}\left(i_{*} \mathscr{F}\right) \text { , where } \rho \text { is the natural } \operatorname{map} \mathscr{G} \rightarrow i_{*}\left(\left.\mathscr{G}\right|_{U}\right) .\right]\) (d) Now let \(X\) be any noetherian scheme, \(U\) an open subset, \(\mathscr{F}\) a coherent sheaf on \(U,\) and \(\mathscr{G}\) a quasi-coherent sheaf on \(X\) such that \(\left.\mathscr{F} \subseteq \mathscr{G}\right|_{V} .\) Show that there is a coherent subsheaf \(\mathscr{F}^{\prime} \subseteq \mathscr{G}\) on \(X\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\). Taking \(\mathscr{I}=i_{*} \mathscr{F}\) proves the result announced at the beginning. [Hint: Cover \(X\) with open affines, and extend over one of them at a time. (e) As an extra corollary, show that on a noetherian scheme, any quasi- coherent sheaf \(\mathscr{F}\) is the union of its coherent subsheaves. [Hint: If \(s\) is a section of \(\mathscr{F}\) over an open set \(U,\) apply (d) to the subsheaf of \(\left.\mathscr{F}\right|_{v}\) generated by s.]

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.

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 \(f(Z)\)
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