/*! 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 10 Let \(A\) be a ring, let \(S=A\l... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 10

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 \(X\).

Short Answer

Expert verified
The saturation of a homogeneous ideal is a homogeneous ideal. Two homogeneous ideals define the same closed subscheme if they have the same saturation. The ideal of any closed subscheme is saturated, implying that it is the largest homogeneous ideal defining the subscheme. Therefore, there exists a bijection between saturated ideals of \( S \) and closed subschemes of \( X \).

Step by step solution

01

Show that Saturation of an Ideal is a Homogeneous Ideal

To show this, one needs to primarily show that if \( s \) lies in the saturation \( I \) of a homogeneous ideal \( I \subseteq S \), and \( s \) is homogeneous, then for any \( i \) there exists \( n \) such that \( x_i^n s \) is in \( I \). This establishes \( I \) to be a homogeneous ideal of \( S \).
02

Prove Two Homogeneous Ideals Define the Same Closed Subscheme If They Have Same Saturation

In order to show that \( I_1 \) and \( I_2 \) define the same closed subscheme of \( X \) if and only if they have the same saturation, we need to show that if two ideals have the same saturation, i.e., \(I_1 = I_2\) then \( V(I_1) = V(I_2) \). Thus, they define the same subscheme.
03

Demonstrate that Ideal of Any Closed Subscheme is Saturated

Here, we show that if \(Y\) is any closed subscheme of \(X\), then the ideal \(\Gamma_{*}\left(\mathscr{I}_{Y}\right)\) is saturated. This is evidence that it is the largest homogeneous ideal defining the subscheme \(Y\). This involves showing that the homogeneous coordinate ring of \( Y \) is \( S/\Gamma_{*}\left(\mathscr{I}_{Y}\right) \). Thus, this ideal is saturated since it equals its own saturation.
04

Establish a 1-1 Correspondence Between Saturated Ideals and Closed Subscheme

Lastly, we state that there exists a one-to-one (1-1) correspondence between saturated ideals of \( S \) and closed subschemes of \( X \). This indicates that for every saturated ideal, there exists a unique closed subscheme, and vice versa. This can be established by showing that the mapping from saturated ideals to closed subschemes is both injective and surjective.

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Key Concepts

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

Homogeneous Ideal
Understanding the concept of a homogeneous ideal requires familiarity with polynomial rings and the notion of homogeneity. In algebraic geometry and commutative algebra, a polynomial is called homogeneous if all its terms are of the same degree. For example, the polynomial x^2 + xy is homogeneous because each term is of degree 2.

Similarly, an ideal I within a polynomial ring S is called a homogeneous ideal if it can be generated by homogeneous elements. This implies that for any element in the ideal, if that element can be expressed as a sum of terms, each term on its own must also be in the ideal. This special structure ensures that the geometric object it defines, such as a variety or subscheme, has certain symmetries.

For instance, consider an ideal I generated by x^2 and xy in the polynomial ring k[x, y], where k is our field. Since both x^2 and xy are homogeneous of degree 2, I is a homogeneous ideal. Any polynomial in I will be a k-linear combination of these generators, and thus also homogeneous of degree 2 or higher.

A saturated homogeneous ideal has a direct geometric interpretation. By capturing certain 'boundary' elements of the ideal, it helps define more nuanced properties of the geometric objects. In the exercise, the saturation process involves ensuring that if any term of a polynomial can be made to lie in the ideal after multiplying by some power of each variable, then this polynomial should also be included in the ideal itself.
Closed Subscheme
In algebraic geometry, schemes are the fundamental objects we study, generalizing varieties to include 'extra data' for more refined geometrical interpretation. A closed subscheme is a subset of a scheme that corresponds to an intuitively 'closed' shape in geometric contexts, resembling the idea of a closed set in topology, but with algebraic structure.

A classic example is the closed subscheme defined by a single polynomial equation within the spectrum of a polynomial ring, such as all points (x, y) in the affine space A^2 satisfying x^2 + y^2 - 1 = 0, which gives us the unit circle.

When we speak about the closed subscheme defined by a homogeneous ideal, we are essentially talking about shapes defined by certain 'equations' in projective space. For example, a homogeneous ideal generated by the polynomial x^2 + y^2 - z^2 can define a conic section in projective space. The saturation condition discussed in the exercise ensures that the closed subscheme being described is complete in a sense, not missing any points that should belong to it due to allowed multiplicities by variables in the polynomial ring defining the scheme structure.
Proj of a Ring
The notion of Proj is a construction in algebraic geometry that allows us to create a geometric object from a graded ring. Specifically, for a ring A and a graded algebra S over A, Proj S represents a space of 'lines,' or more formally, equivalence classes of nonzero points, when thought of in terms of the homogeneous coordinates given by S. It's akin to creating a 'projective' version of Spec, the prime spectrum of a ring, which captures the 'points' or prime ideals of a ring.

In the context of the exercise, Proj S refers to the projective space formed from the graded ring S = A[x_0, ..., x_r], representing the set of all prime ideals of S that do not contain the irrelevant ideal formed by the homogeneous elements of positive degree.

The relationship between Proj and closed subschemes is encapsulated by the correspondence between saturated ideals and geometric objects. Each saturated homogeneous ideal of S can be thought to 'carve out' a unique subscheme in Proj S, encapsulating the interplay between algebra and geometry that is central to algebraic geometry. In practice, this means that by studying the algebraic properties of specific ideals within our ring S, we can make conclusions about the geometric shapes, or closed subschemes, they define in projective space.

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Most popular questions from this chapter

(a) Let \(\mathscr{F}^{\prime}\) be a subsheaf of a sheaf \(\mathscr{F}\). Show that the natural map of \(\mathscr{F}\) to the quotient sheaf \(\mathscr{F}^{\prime} \mathscr{F}^{\prime}\) is surjective, and has kernel \(\mathscr{F}\) : Thus there is an exact sequence $$0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{H} / \mathscr{F}^{\prime} \rightarrow 0$$ (b) Conversely, if \(0 \rightarrow \overline{\mathscr{F}}^{\prime} \rightarrow \widetilde{\mathscr{F}} \rightarrow \widetilde{\mathscr{F}}^{\prime \prime} \rightarrow 0\) is an exact sequence, show that \(\mathscr{F}^{\prime}\) is isomorphic to a subsheaf of \(\overline{\mathscr{F}},\) and that \(\mathscr{F}^{\prime \prime}\) is isomorphic to the quotient of \(\mathscr{F}\) by this subsheaf.

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.

Affine Morphisms. A morphism \(f: X \rightarrow Y\) of schemes is affine if there is an open affine cover \(\left\\{V_{i}\right\\}\) of \(Y\) such that \(f^{-1}\left(V_{i}\right)\) is affine for each \(i\) (a) Show that \(f: X \rightarrow Y\) is an affine morphism if and only if for every open affine \(V \subseteq Y, f^{-1}(V)\) is affine. [Hint: Reduce to the case \(Y\) affine, and use (Ex. 2.17).] (b) An affine morphism is quasi-compact and separated. Any finite morphism is affine (c) Let \(Y\) be a scheme, and let \(\mathscr{A}\) be a quasi-coherent sheaf of \(\emptyset_{Y}\) -algebras (i.e., a sheaf of rings which is at the same time a quasi-coherent sheaf of \(\mathscr{O}_{Y}\) -modules. Show that there is a unique scheme \(X\), and a morphism \(f: X \rightarrow Y\), such that for every open affine \(V \subseteq Y, f^{-1}(V) \cong \operatorname{Spec} \mathscr{A}(V),\) and for every inclusion \(U\) s \(V\) of open affines of \(Y\), the morphism \(f^{-1}(U)\) s \(f^{-1}(V)\) corresponds to the restriction homomorphism \(\mathscr{A}(V) \rightarrow \mathscr{A}(U)\). The scheme \(X\) is called Spec \(\mathscr{A}\). [Hint: Construct \(X\) by glueing together the schemes Spec \(\mathscr{A}(V)\) for \(V \text { open affine in } Y .]\) (d) If \(\mathscr{A}\) is a quasi-coherent \(\mathscr{V}_{\boldsymbol{r}}\) -algebra, then \(f: X=\operatorname{Spec} \mathscr{A} \rightarrow Y\) is an affine morphism, and \(\mathscr{A} \cong f_{*} \mathscr{O}_{X} .\) Conversely, if \(f: X \rightarrow Y\) is an affine morphism, then \(\mathscr{A}=f_{*} \mathscr{O}_{X}\) is a quasi-coherent sheaf of \(\mathscr{O}_{Y^{-}}\) algebras, and \(X \cong\) Spec \(\mathscr{A}\) (e) Let \(f: X \rightarrow Y\) be an affine morphism, and let \(\mathscr{A}=f_{*} O_{X} .\) Show that \(f_{*}\) induces an equivalence of categories from the category of quasi-coherent \(\mathscr{O}_{X}\) -modules to the category of quasi-coherent \(\mathscr{A}\) -modules (i.e., quasi-coherent \(\mathcal{O}_{Y}\) -modules having a structure of \(\mathscr{A}\) -module). [Hint: For any quasi- coherent \(\mathscr{A}\) -module \(\mathscr{M},\) construct a quasi-coherent \(\mathscr{O}_{X}\) -module \(\mathscr{M},\) and show that the functors \(f_{*}\) and \(^{\sim}\) are inverse to each other.

Tensor Operations on Sheaves. First we recall the definitions of various tensor operations on a module. Let \(A\) be a ring, and let \(M\) be an \(A\) -module. Let \(T^{\prime \prime}(M)\) be the tensor product \(M \otimes \ldots \otimes M\) of \(M\) with itself \(n\) times, for \(n \geqslant 1\). For \(n=0\) we put \(T^{0}(M)=A .\) Then \(T(M)=\bigoplus_{n \geqslant 0} T^{\prime \prime}(M)\) is a (noncommutative) \(A\) -algebra, which we call the tensor algebra of \(M .\) We define the symmetric algebra \(S(M)=\bigoplus_{n \geqslant 0} S^{\prime \prime}(M)\) of \(M\) to be the quotient of \(T(M)\) by the two-sided ideal generated by all expressions \(x \otimes y-y \otimes x,\) for all \(x, y \in M .\) Then \(S(M)\) is a commutative \(A\) -algebra. Its component \(S^{n}(M)\) in degree \(n\) is called the \(n\) th symmetric product of \(M .\) We denote the image of \(x \otimes y\) in \(S(M)\) by \(x y,\) for any \(x, y \in M .\) As an example, note that if \(M\) is a free \(A\) -module of rank \(r,\) then \(S(M) \cong\) \(A\left[x_{1}, \ldots, x_{r}\right]\). We define the exterior algebra \(\wedge(M)=\bigoplus_{n \geqslant 0} \wedge^{\prime \prime}(M)\) of \(M\) to be the quotient of \(T(M)\) by the two- sided ideal generated by all expressions \(x \otimes x\) for \(x \in M .\) Note that this ideal contains all expressions of the form \(x \otimes y+y \otimes x\) so that \(\wedge(M)\) is a skew commutative graded \(A\) -algebra. This means that if \(u \in\) \(\wedge^{r}(M)\) and \(v \in \Lambda^{s}(M),\) then \(u \wedge v=(-1)^{r s} v \wedge u\) (here we denote by \(\wedge\) the multiplication in this algebra; so the image of \(x \otimes y\) in \(\wedge^{2}(M)\) is denoted by \(x \wedge y\) ). The \(n\) th component \(\wedge^{\prime \prime}(M)\) is called the \(n\) th exterior power of \(M\). Now let \(\left(X, O_{X}\right)\) be a ringed space, and let \(\mathscr{F}\) be a sheaf of \(\mathcal{O}_{X}\) -modules. We define the tensor algebra, symmetric algebra, and exterior algebra of \(\mathscr{F}\) by taking the sheaves associated to the presheaf, which to each open 'set \(U\) assigns the corresponding tensor operation applied to \(\mathscr{F}(U)\) as an \(\mathscr{O}_{X}(U)\) -module. The results are \(\mathcal{O}_{X^{-}}\) algebras, and their components in each degree are \(\mathscr{C}_{X}\) -modules. (a) Suppose that \(\mathscr{F}\) is locally free of rank \(n\). Then \(T^{\prime}(\mathscr{F}), S^{\prime}(\mathscr{F})\), and \(\wedge^{\prime}(\mathscr{F})\) are also locally free, of ranks \(n^{\prime},\left(\begin{array}{c}m+r-1 \\ n-1\end{array}\right),\) and \(\left(\begin{array}{c}m \\ 2\end{array}\right)\) respectively. (b) Again let \(\mathscr{F}\) be locally free of rank \(n\). Then the multiplication \(\operatorname{map} \wedge \mathscr{F} \otimes\) \(\wedge^{n-r} \mathscr{F} \rightarrow \wedge^{n} \cdot \mathscr{F}\) is a perfect pairing for any \(r,\) i.c., it induces an isomorphism of \(\wedge^{\prime \prime} \mathscr{F}\) with \(\left(\wedge^{n-r} \mathscr{F}\right)^{\sim} \otimes \wedge^{\prime \prime} \mathscr{F}\). As a special case, note if \(\mathscr{F}\) has rank 2 then \(\mathscr{F} \cong \mathscr{F}^{\sim} \otimes \wedge^{2} \mathscr{F}\) (c) Let \(0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{F}^{\prime \prime} \rightarrow 0\) be an exact sequence of locally free sheaves. Then for any \(r\) there is a finite filtration of \(S^{\prime}(\mathscr{F})\) \\[ S^{\prime}(\mathscr{F})=F^{0} \supseteq F^{1} \supseteq \ldots \supseteq F^{\prime} \supseteq F^{r+1}=0 \\] with quotients \\[ F^{p} / F^{p+1} \cong S^{p}\left(\mathscr{F}^{\prime}\right) \otimes S^{r-p}\left(\mathscr{F}^{\prime \prime}\right) \\] for each \(p\). (d) Same statement as (c), with exterior powers instead of symmetric powers. In particular, if \(\mathscr{F}^{\prime}, \mathscr{F}, \mathscr{F}^{\prime \prime}\) have ranks \(n^{\prime}, n, n^{\prime \prime}\) respectively, there is an isomorphism \(\wedge^{n} \mathscr{F} \cong \wedge^{n^{\prime} \mathscr{F}^{\prime}} \otimes \wedge^{n^{\prime \prime}} \mathscr{F}^{\prime \prime}\) (e) Let \(f: X \rightarrow Y\) be a morphism of ringed spaces, and let \(\mathscr{F}\) be an \(\mathscr{U}_{Y}\) -module. Then \(f^{*}\) commutes with all the tensor operations on \(\mathscr{F},\) i.e., \(f^{*}\left(S^{n}(\mathscr{F})\right)=\) \(S^{\prime \prime}\left(f^{*} \mathscr{F}\right)\) etc.

Let \(\varphi: \overline{\mathscr{H}} \rightarrow \mathscr{G}\) be a morphism of sheaves. (a) Show that im \(\varphi \cong \mathscr{F} /\) ker \(\varphi\) (b) Show that coker \(\varphi \cong \mathscr{G} / \mathrm{im} \varphi\)
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