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Chapter 2: Problem 14

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.

Short Answer

Expert verified
The support of a section \(s\) is a closed subset of \(U\), while the support of the sheaf itself need not be closed.

Step by step solution

01

Definition of Support

The support of a section \(s\) of a sheaf \(\mathscr{F}\) on \(X\) over an open set \(U\), denoted by Supp \(s\), is defined as the set \(\left\{P \in U | s_{P} \neq 0\right\}\), where \(s_{P}\) denotes the germ of \(s\) at \(P\) in the stalk \(\overline{\mathscr{F}}_{p}\). This is the set of points in \(U\) where \(s\) is 'felt' or non-zero. Similarly, the support of a sheaf, denoted by Supp \(\overline{\mathscr{F}}\), is defined as the set \(\left\{P \in X | \mathscr{F}_{P} \neq 0\right\}\).
02

Show that Supp s is closed in \(U\)

To show that Supp \(s\) is closed, it's enough to show that its complement in \(U\) is open. The complement of Supp \(s\) in \(U\) is \(\left\{P \in U | s_{P} = 0\right\}\). This is the set of all points \(P\) in \(U\) where the germ of \(s\) is zero. Because the section is zero, this set is open by the definition of a sheaf. Therefore, Supp \(s\) is closed in \(U\).
03

Contrast with Support of a Sheaf

For contrast, note that the support of a sheaf \(\mathscr{F}\) itself, Supp \(\overline{\mathscr{F}}\), need not be a closed subset of \(U\). This is because we are considering all sections of the sheaf, not just a single section. Its support is the set of points where some section of the sheaf is nonzero, so it could be that at some points, some sections vanish while others do not.

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Key Concepts

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

Sheaf
In the language of topology and abstract algebra, a sheaf is a tool for systematically tracking locally defined data attached to the open sets of a topological space. Think of it as a way to understand how local pieces of information fit together to form global data on a space.

For instance, a sheaf might assign to each open set of a space a collection of functions defined on that set. These functions should behave nicely - if you have a function on a larger open set and you restrict it to a smaller open set, it should be part of the collection assigned to the smaller set. This property is known as 'locality'. Likewise, if you have a bunch of functions on overlapping sets that agree where they overlap, you should be able to stitch them together into a single function on the union of those sets. This is called 'gluing'.

These properties make sheaves exceptionally useful in many areas of mathematics, including differential geometry, algebraic geometry, complex analysis, and more. They allow mathematicians to patch local solutions to differential equations or piece together local bits of shapes to understand the larger structure of geometrical objects.
Stalk of a Sheaf
The 'stalk' of a sheaf might conjure up images of a plant stem, but in mathematics, it's a bit more abstract. A stalk at a point 'P' within the topological space X provides a bird's-eye view of all the information the sheaf has about 'P'.

You can think of it this way: for every open set containing 'P', look at the data the sheaf assigns to that open set and then zoom in on what that data says specifically about 'P'. The stalk at 'P' collects all these little bits of information into a single algebraic structure, typically a group or a ring.

Now, why is this useful? The stalk allows us to understand the 'behavior' of sheaf data right at the point 'P' without getting distracted by what's happening away from 'P'. It's all about local data. To calculate the stalk, we actually look at 'germs' of sections 鈥� essentially, the relevant elements of the sheaf that are attached to neighborhoods of 'P'.

Within the context of our exercise, the germ of a section is what tells us whether the support of a section will include the point 'P' or not. The support, therefore, is essentially the 'footprint' of the section where it has a non-zero presence.
Closed Subset
When we step back into the broader field of topological spaces, a subset is 'closed' if it contains all its limit points; it's complete and 'sealed off' in a sense. Intuitively, if you have a sequence of points in the closed subset that gets closer and closer to some boundary point, that boundary point must also be in the set.

In our scenario, proving that the support of a section is a closed subset involves showing that 'outside' the support, the section is consistently zero. In topological terms, the complement of the support (the part of the open set 'U' not in the support) has to be an open set itself. Since sections of a sheaf obey the locality and gluing properties, if a section is zero at some point it must be zero in some vicinity of that point - which allows us to see how the whole complement is open.

This is an attractive property because it falls in line with how we naturally think of 'vanishing': if a function is zero at a point and is continuous, we expect it to be zero in a surrounding region. That is precisely what's reflected in the topology when we proclaim that the support is closed. It affirms that within 'U', the behavior of the section has clear boundaries - it's present or 'felt' up to a particular frontier and beyond that, it simply doesn't exist.

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Most popular questions from this chapter

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.

A topological space is quasi-compact if every open cover has a finite subcover. (a) Show that a topological space is noetherian (I, \(\$ 1)\) if and only if every open subset is quasi-compact. (b) If \(X\) is an affine scheme. show that \(\operatorname{sp}(X)\) is quasi- compact. but not in general noetherian. We say a scheme \(X\) is quati-ciompact if \(\operatorname{sp}(X)\) is. (c) If \(A\) is a noetherian ring. show that spiSpec 1 ) is a nocthcrian topological space. (d) Give an example to show that sp(Spec \(A\) ) can be noetherian even when \(A\) is not.

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 \(f(Z)\)

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 \(X\).

Zariski Spaces. A topological space \(X\) is a Zariski space if it is noetherian and every (nonempty) closed irreducible subset has a unique generic point (Ex. 2.9 ). For example, let \(R\) be a discrete valuation ring, and let \(T=\operatorname{sp}(\operatorname{Spec} R)\). Then \(T\) consists of two points \(t_{0}=\) the maximal ideal, \(t_{1}=\) the zero ideal. The open subsets are \(\varnothing,\left\\{t_{1}\right\\},\) and \(T .\) This is an irreducible Zariski space with generic point \(t_{1}\). (a) Show that if \(X\) is a noetherian scheme, then \(\operatorname{sp}(X)\) is a Zariski space. (b) Show that any minimal nonempty closed subset of a Zariski space consists of one point. We call these closed points. (c) Show that a Zariski space \(X\) satisfies the axiom \(T_{0}\) : given any two distinct points of \(X\), there is an open set containing one but not the other (d) If \(X\) is an irreducible Zariski space, then its generic point is contained in every nonempty open subset of \(X\) (e) If \(x_{0}, x_{1}\) are points of a topological space \(X,\) and if \(x_{0} \in\left\\{x_{1}\right\\}^{-},\) then we say that \(x_{1}\) specializes to \(x_{0},\) written \(x_{1} \leadsto \rightarrow x_{0} .\) We also say \(x_{0}\) is a specialization of \(x_{1},\) or that \(x_{1}\) is a generization of \(x_{0} .\) Now let \(X\) be a Zariski space. Show that the minimal points, for the partial ordering determined by \(x_{1}>x_{0}\) if \(x_{1} \leadsto x\) \(x_{0},\) are the closed points, and the maximal points are the generic points of the irreducible components of \(X .\) Show also that a closed subset contains every specialization of any of its points. (We say closed subsets are stable under specialization. . Similarly, open subsets are stable under generization. (f) Let \(t\) be the functor on topological spaces introduced in the proof of (2.6) If \(X\) is a noetherian topological space, show that \(t(X)\) is a Zariski space. Furthermore \(X\) itself is a Zariski space if and only if the \(\operatorname{map} \alpha: X \rightarrow t(X)\) is a homeomorphism.
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