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Let \(P, Q \in X, X\) a variety. Show that there is an affine open set \(V\) on \(X\) that contains \(P\) and \(Q\). (Hint: See the proof of the Corollary to Proposition 5 , and use Problem 1.17(c).)

Short Answer

Expert verified
Question: Prove that for any two points P and Q in a variety X, there exists an affine open set V that contains both points. Answer: By constructing an affine open set V = U_P ⋃ U_Q using Problem 1.17(c), we have shown that V contains both P and Q, proving that there exists an affine open set that contains any two points in a variety X.

Step by step solution

01

SetUp/Notation

Let X be a variety, and P, Q be two distinct points in X. We aim to find an affine open set V such that P, Q ∈ V.
02

Recall Relevant Properties

Recall that X is a variety, meaning it is an irreducible and reduced closed subset of some affine space A^n. So, X is algebraic and connected.
03

Use Problem 1.17(c) to Find Affine Open Sets

By Problem 1.17(c), we can find affine open sets U_P and U_Q in X, such that P ∈ U_P and Q ∈ U_Q.
04

Intersection of Affine Open Sets

Since X is connected (given it is a variety), we can claim the intersection of U_P and U_Q is non-empty: U_P ⋂ U_Q ≠ ∅
05

Construct Affine Open Set V

Let S = U_P ⋂ U_Q. Now, construct an affine open set V = U_P ⋃ U_Q. We have P ∈ U_P ⊆ V and Q ∈ U_Q ⊆ V. This shows that V is an affine open subset of X containing both P and Q.
06

Conclusion

In conclusion, we have shown that there exists an affine open set V on X that contains P and Q, as requested.

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Key Concepts

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

Affine Open Sets
Affine open sets are crucial in algebraic geometry as they form the basic building blocks for understanding more complex varieties. An affine open set within a variety is essentially a subset that allows for local analysis using algebraic equations. These sets resemble open sets in standard topology but are grounded in algebraic equations defining the space.

In the exercise, we are tasked with finding an affine open set that contains two specific points, \( P \) and \( Q \). This highlights the concept that such open sets can be constructed in a way that accommodates any pair of points on a variety.

Key features of affine open sets include:
  • They contain a dense subset where the coordinate functions are well-defined.
  • They play a significant role in patching together to form entire varieties, much like pieces of a puzzle.
  • Affine open sets can often intersect in useful ways, allowing for constructive overlaps that aid in proving various properties.
Understanding affine open sets strengthens insight into the structure and behavior of varieties in algebraic geometry.
Algebraic Geometry
Algebraic geometry is the branch of mathematics combining algebra, especially commutary algebra, with geometry. Its primary aim is understanding the solution sets of systems of polynomial equations as geometric objects.

Varieties, central objects in algebraic geometry, are defined as the solution sets to these polynomial equations. They offer a rich structure where algebra and geometry interplay.

In the context of the exercise, we consider the variety \( X \) as a closed space that can be analyzed using these algebraic principles. Some traits of algebraic geometry include:
  • Utilizes rings and fields to derive geometric concepts.
  • Supports the translation of geometric problems into algebraic questions that are often easier to solve.
  • Deals with affine and projective varieties, offering a diverse perspective on geometric objects.
Overall, algebraic geometry provides a comprehensive framework for understanding varieties like \( X \) and the structure of open sets within them.
Irreducible Spaces
In algebraic geometry, an irreducible space is one that cannot be decomposed into two smaller non-empty closed subsets. This is a crucial concept when defining a variety, as it underpins much of its structure and significance.

An irreducible variety, such as \( X \) in our exercise, serves as a "whole" entity, making it fundamental to support operations like locating intersecting affine open sets.

Characteristics of irreducible spaces include:
  • These spaces ensure that every non-empty open subset is dense, reinforcing their unified nature.
  • They provide a basis for ensuring connectedness, essential for confirming intersections of open sets.
  • Their behavior often guarantees non-trivial intersections that facilitate demonstrating various algebraic properties.
Understanding irreducible spaces enhances comprehension of how varieties maintain their integrity and support algebraic processes, such as finding affine open sets encompassing given points.

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Most popular questions from this chapter

If there is a dominating rational map from \(X\) to \(Y\), then \(\operatorname{dim}(Y) \leq \operatorname{dim}(X)\).

Let \(U_{i} \subset \mathbb{P}^{n}, \varphi_{i}: \mathbb{A}^{n} \rightarrow U_{i}\) as in Chapter 4. Give \(U_{i}\) the topology induced from \(\mathbb{P}^{n}\). (a) Show that \(\varphi_{i}\) is a homeomorphism. (b) Show that a set \(W \subset \mathbb{P}^{n}\) is closed if and only if each \(\varphi_{i}^{-1}(W)\) is closed in \(\mathbb{A}^{n}, i=1, \ldots, n+1\). (c) Show that if \(V \subset \mathbb{A}^{n}\) is an affine variety, then the projective closure \(V^{*}\) of \(V\) is the closure of \(\varphi_{n+1}(V)\) in \(\mathbb{P}^{n}\).

(a) Let \(X\) be a topological space, \(X=\bigcup_{\alpha \in \mathscr{A}} U_{\alpha}, U_{\alpha}\) open in \(X\). Show that a subset \(W\) of \(X\) is closed if and only if each \(W \cap U_{\alpha}\) is closed (in the induced topology) in \(U_{\alpha}\). (b) Suppose similarly \(Y=\bigcup_{\alpha \in \mathscr{A}} V_{\alpha}, V_{\alpha}\) open in \(Y\), and suppose \(f: X \rightarrow Y\) is a mapping such that \(f\left(U_{\alpha}\right) \subset V_{\alpha}\). Show that \(f\) is continuous if and only if the restriction of \(f\) to each \(U_{\alpha}\) is a continuous mapping from \(U_{\alpha}\) to \(V_{\alpha}\).

Show that \(\operatorname{dim} \mathbb{A}^{n}=\operatorname{dim} \mathbb{P}^{n}=n\).

Let \(K\) be a field of a characteristic zero, \(L\) a finite (algebraic) extension of \(K\). Then there is a \(z \in L\) such that \(L=K(z)\). Outline of Proof. Step (i): Suppose \(L=K(x, y)\). Let \(F\) and \(G\) be monic irreducible polynomials in \(K[T]\) such that \(F(x)=0, G(y)=0\). Let \(L^{\prime}\) be a field in which \(F=\) \(\prod_{i=1}^{n}\left(T-x_{i}\right), G=\prod_{j=1}^{m}\left(T-y_{i}\right), x=x_{1}, y=y_{1}, L^{\prime} \supset L\) (see Problems 1.52, 1.53). Choose \(\lambda \neq 0\) in \(K\) so that \(\lambda x+y \neq \lambda x_{i}+y_{j}\) for all \(i \neq 1, j \neq 1\). Let \(z=\lambda x+y, K^{\prime}=K(z)\). Set \(H(T)=G(z-\lambda T) \in K^{\prime}[T] .\) Then \(H(x)=0, H\left(x_{i}\right) \neq 0\) if \(i>0 .\) Therefore \((H, F)=\) \((T-x) \in K^{\prime}[T]\). Then \(x \in K^{\prime}\), so \(y \in K^{\prime}\), so \(L=K^{\prime}\) Step (ii): If \(L=K\left(x_{1}, \ldots, x_{n}\right)\), use induction on \(n\) to find \(\lambda_{1}, \ldots, \lambda_{n} \in k\) such that \(L=K\left(\sum \lambda_{i} x_{i}\right)\)

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