/*! 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 4 Principle of Connectedness. Let ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Principle of Connectedness. Let \(\left\\{X_{t}\right\\}\) be a flat family of closed subschemes of \(\mathbf{P}_{k}^{n}\) parametrized by an irreducible curve \(T\) of finite type over \(k .\) Suppose there is a nonempty open set \(U \subseteq T,\) such that for all closed points \(t \in U, X_{t}\) is connected. Then prove that \(X_{t}\) is connected for all \(t \in T\).

Short Answer

Expert verified
It follows from the Principle of Connectedness in Algebraic Geometry, that if a nonempty open set \(U \subseteq T\) exists, such that \(X_{t}\) is connected for all \(t \in U\), then \(X_{t}\) is connected for all \(t \in T\).

Step by step solution

01

Analysis of the given

The given is that \(\left\{X_{t}\right\}\) is a flat family of closed subschemes of \(\mathbf{P}_{k}^{n}\) parametrized by an irreducible curve \(T\). There is a nonempty open set \(U \subseteq T\), such that for all closed points \(t \in U\), \(X_{t}\) is connected.
02

State and use the Principle of Connectedness

The Principle of Connectedness states that if a family of connected topological spaces parametrized by a connected topological space is flat, then the total space is connected.
03

Apply the Principle to \(X_{t}\)

Applying the Principle to \(X_{t}\), since \(X_{t}\) is connected for all \(t \in U\), it is flat and hence the total space is connected. Therefore, \(X_{t}\) is connected for all \(t \in T\).

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.

Flat Family of Schemes
In algebraic geometry, a flat family of schemes plays a vital role in understanding how geometric structures vary in a continuous manner over a parameter space. The flatness refers to a homological condition which ensures that the fibers of the family do not undergo drastic changes, such as suddenly jumping in dimension, as the parameter varies.

A simple analogy might be to consider a family of cookies cut with a cookie cutter from a flat sheet of dough鈥攖he shape of the cookies remains consistently the same, regardless of their position on the dough. Flatness in schemes guarantees a similar kind of uniformity.

When working with a flat family of schemes, one important property to leverage is the behavior of connectedness within the family. As the Principle of Connectedness suggests, if the family is parametrized by a connected space, and if a sufficient 'slice' of this family is connected, then this quality prevails throughout the entire family.
Closed Subschemes
A closed subscheme is essentially a subset of a scheme that inherits a scheme structure in such a way that it corresponds to a closed set in the topological space of the ambient scheme. Closed subschemes can be visualized as the algebraic counterpart to closed subspaces in topology.

For example, in the affine plane over a field, which is the scheme corresponding to a two-dimensional vector space, lines and points are closed subschemes. In the textbook exercise, the family \(\{X_t\}\) consists of such closed subschemes, ensuring each member (or 'fiber') over a point in the parameter space has a stable and concrete realization within the larger space \(\mathbf{P}_{k}^{n}\).
Irreducible Curve
An irreducible curve is a type of algebraic curve that cannot be decomposed into simpler, non-trivial algebraic curves. It's a concept paralleling prime numbers in arithmetic, where an irreducible curve serves as a building block that cannot be further subdivided.

Within the context of the exercise, the curve \(T\) is not just any parameter space; its irreducibility ensures that it behaves like a connected space without 'gaps', thus sustaining the conditions necessary for the Principle of Connectedness to apply. It's pivotal in deducing connectedness across the entire family of schemes, as it guarantees the parameter space over which the family is defined is itself 'whole' or connected.
Connected Topological Space
In topology, a connected topological space is one where there are no separations or divisions within the space; intuitively, one can travel from any point to any other point without leaving the space. The opposite of connected is 'disconnected', where the space can be split into two or more disjoint nonempty open sets.

A classic example would be a single unbroken line or a blob with no holes in it. It is heavily relied upon in our Principle of Connectedness, which posits the intuitively appealing notion that if individual members of a family over an open subset are connected and the parameter space is connected, then the entirety remains connected. This principle mirrors our own experiences where a chain only stays together if all its links hold tight.

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

Let \(k\) be a field, let \(X=\mathbf{P}_{k},\) and let \(Y\) be a closed subscheme of dimension \(q \geqslant 1\) which is a complete intersection (II, Ex. 8.4 ). Then: (a) for all \(n \in \mathbf{Z},\) the natural map $$H^{0}\left(X, O_{X}(n)\right) \rightarrow H^{0}\left(Y, O_{Y}(n)\right)$$ is surjective. (This gives a generalization and another proof of (II, Ex. \(8.4 \mathrm{c}\) ), where we assumed \(Y\) was normal. (b) \(Y\) is connected (c) \(H^{\prime}\left(Y, C_{r}(n)\right)=0\) for \(0

Let \(\left(X, \mathcal{O}_{X}\right)\) be a ringed space, and let \(\mathscr{F}^{\prime}, \mathscr{F}^{\prime \prime} \in \mathbb{M}_{\mathrm{O}} \mathfrak{d}(X) .\) An extension of \(\mathscr{F}^{\prime \prime} \mathrm{by}\) \(\mathscr{F}^{\prime}\) is a short exact sequence $$0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{F}^{\prime \prime} \rightarrow 0$$ in \(\operatorname{Mod}(X) .\) Two extensions are isomorphic if there is an isomorphism of the short exact sequences, inducing the identity maps on \(\mathscr{F}^{\prime}\) and \(\mathscr{F}^{\prime \prime} .\) Given an extension as above consider the long exact sequence arising from Hom \(\left(\mathscr{F}^{\prime \prime}, \cdot\right),\) in particular the map $$\delta: \operatorname{Hom}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime \prime}\right) \rightarrow \operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right)$$ and let \(\xi \in \operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right)\) be \(\delta\left(1_{\mathscr{F}^{\prime \prime}}\right) .\) Show that this process gives a one-to-one correspondence between isomorphism classes of extensions of \(\mathscr{F}^{\prime \prime}\) by \(\mathscr{F}^{\prime},\) and elements of the group \(\operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right) .\) For more details, see, e.g., Hilton and Stammbach [1, Ch. III].

On an arbitrary topological space \(X\) with an arbitrary abelian sheaf \(\mathscr{F},\) Cech cohomology may not give the same result as the derived functor cohomology. But here we show that for \(H^{1},\) there is an isomorphism if one takes the limit over all coverings. (a) Let \(\mathfrak{U}=\left(U_{i}\right)_{i e I}\) be an open covering of the topological space \(X\). A refinement of \(\mathbb{U}\) is a covering \(\mathfrak{B}=\left(V_{j}\right)_{\text {je } J}\), together with a \(\operatorname{map} \lambda: J \rightarrow I\) of the index sets, such that for each \(j \in J, V_{j} \subseteq U_{\lambda(j)} .\) If \(\mathfrak{B}\) is a refinement of \(\mathbb{U}\), show that there is a natural induced map on Cech cohomology, for any abelian sheaf \(\mathscr{F},\) and for each \(i\) \\[ \lambda^{i}: \check{H}^{i}(\mathbf{U}, \mathscr{F}) \rightarrow \check{H}^{i}(\mathfrak{Y}, \mathscr{F}) \\] The coverings of \(X\) form a partially ordered set under refinement, so we can consider the Cech cohomology in the limit \\[ \varliminf_{u} \check{H}^{i}(\mathfrak{U}, \mathscr{F}) \\] (b) For any abelian sheaf \(\mathscr{F}\) on \(X,\) show that the natural maps (4.4) for each covering \\[ \check{H}^{i}(\mathbf{u}, \mathscr{F}) \rightarrow H^{i}(X, \mathscr{F}) \\] are compatible with the refinement maps above (c) Now prove the following theorem. Let \(X\) be a topological space, \(\mathscr{F}\) a sheaf of abelian groups. Then the natural map \\[ \varliminf_{u} \check{H}^{1}(\mathfrak{U}, \mathscr{F}) \rightarrow H^{1}(X, \mathscr{F}) \\] is an isomorphism. [Hint: Embed \(\mathscr{F}\) in a flasque sheaf \(\mathscr{G},\) and let \(\mathscr{R}=\mathscr{G} / \mathscr{F}\) so that we have an exact sequence \\[ 0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{R} \rightarrow 0 \\] Define a complex \(D^{\prime}(\mathbb{Z})\) by \\[ 0 \rightarrow C^{\prime}(\mathbf{U}, \mathscr{F}) \rightarrow C^{\prime}(\mathbf{U}, \mathscr{G}) \rightarrow D^{\prime}(\mathbf{U}) \rightarrow 0 \\] Then use the exact cohomology sequence of this sequence of complexes, and the natural map of complexes \\[ D^{\prime}(\mathfrak{U}) \rightarrow C^{\prime}(\mathfrak{U}, \mathscr{R}) \\] and see what happens under refinement.

Let \(X\) be a noetherian scheme, and suppose that every coherent sheaf on \(X\) is a quotient of a locally free sheaf. In this case we say \(\mathfrak{C}\) ob \((X)\) has enough locally frees. Then for any \(\mathscr{G} \in \mathbb{V}\) iod \((X),\) show that the \(\delta\) -functor \(\left(\mathscr{E} x t^{i}(\cdot, \mathscr{G})\right),\) from \(\operatorname{Cob}(X)\) to \(\operatorname{Mod}(X)\), is a contravariant universal \(\delta\) -functor. [Hint: Show \(\delta x t^{i}(\cdot, \mathscr{L})\) is coeffaceable (\$1) for \(i>0 .]\)

Let \(X\) be an integral projective scheme of dimension \(\geqslant 1\) over a field \(k,\) and let \(\ell\) be an ample invertible sheaf on \(X .\) Then \(H^{0}\left(X, \mathscr{L}^{-1}\right)=0 .\) (This is an easy special case of Kodaira's vanishing theorem.)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

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.