/*! 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 1 If \(f\) and \(g\) are regular f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 1

If \(f\) and \(g\) are regular functions on open subsets \(U\) and \(V\) of a variety \(X,\) and if \(f=g\) on \(U \cap V\). show that the function which is \(f\) on \(U\) and \(g\) on \(V\) is a regular function on \(U \cup V\). Conclude that if \(f\) is a rational function on \(X\). then there is a largest open subset \(U\) of \(X\) on which \(f\) is represented by a regular function. We say that \(f\) is defined at the points of \(U\).

Short Answer

Expert verified
The function which equals \(f\) on \(U\) and \(g\) on \(V\) is regular on \(U \cup V\). If \(f\) is a rational function on \(X\), the largest open subset \(U\) of \(X\) where \(f\) can be represented by a regular function is the open set where the denominator doesn't vanish.

Step by step solution

01

Analysis and Assumptions

Assume \(f\) and \(g\) to be regular functions on open subsets \(U\) and \(V\) of \(X\) respectively, that are equal on the intersection of \(U\) and \(V\). A regular function is a function that is locally the quotient of two polynomials.
02

Proving the Regularity of the Defined Function

Consider a point \(x\) in \(U \cup V\). If \(x\) is in \(U\), but not in \(V\) then by our function definition, at \(x\) the function is \(f\), which is regular, because \(f\) is regular on \(U\). Similar argument goes if \(x\) is in \(V\), but not in \(U\). If \(x\) is in both \(U\) and \(V\), which means \(x\) is in \(U \cap V\), at \(x\) the function is either \(f\) or \(g\), but since \(f\) equals \(g\) on \(U \cap V\), it doesn't matter, and is also regular as both \(f\) and \(g\) are regular.
03

Rational Function Analysis

Now, if \(f\) is a rational function on \(X\), by the definition of rational functions, we can express \(f\) as the quotient of two polynomials and thus, \(f\) is a regular function on the domain \(D\) where the denominator doesn't vanish. As a regular function, \(U\) is the complementary of the zero set of the denominator polynomial, which is open and hence, is the largest open subset of \(X\) where \(f\) can be represented as a regular function.

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.

Regular Functions
Regular functions are a fundamental concept in algebraic geometry. They are essentially the building blocks that allow us to understand more complex structures and varieties. A regular function is locally the quotient of two polynomials. This means, around every point in its domain, we can express it as \( \frac{p(x)}{q(x)} \) where both \( p(x) \) and \( q(x) \) are polynomials.
What makes a function regular is that the denominator \( q(x) \) does not equal zero on the domain. Thus, regular functions emphasize the local polynomial-like nature that helps in studying geometric objects in algebraic geometry.
  • Defined locally as polynomial quotients.
  • Denominator never zero on the domain.
Considering this, regular functions play a crucial role in constructing and composing functions within algebraic varieties, opening up avenues for deeper analysis and geometric interpretations.
Open Subsets
In algebraic geometry, the concept of open subsets is key to understanding the topological properties of varieties. An open subset of a variety can be thought of as a portion of the space where certain properties or conditions hold true. Even though varieties themselves might be complex structures, we often simplify problems by looking at them on open subsets.
Open subsets are used to determine where functions (like regular functions) are defined and behave nicely, ensuring continuity and differentiability.
  • Open aspects allow layered analysis of spaces.
  • Facilitates localization of function properties.
For instance, in the problem given, understanding the interaction of regular functions on different open subsets \( U \) and \( V \) helps us knit together their properties over the union \( U \cup V \). This method provides important insights into how functions can "patch" together over complex varieties.
Rational Functions
Rational functions are closely related to regular functions but have their own unique characteristics. They can be expressed as the quotient of two polynomials, similar to regular functions. The difference, however, is that rational functions are defined more broadly and might have points where the denominator does vanish, leading to potential undefined spots.
While regular functions are defined in open sets where the denominator never zeroes, rational functions can still exist outside these sets as long as the expression is defined.
  • Expressed as a quotient: numerator and denominator polynomials.
  • Can have points of undefinedness where denominator zeroes.
In our exercise, if \( f \) is a rational function, it depicts the general form of representation, but we find the largest open subset where it acts like a regular function, providing a zero-free denominator region. This approach enables determining regions where the algebraic structure maintains smooth and predictable behavior, critical for geometrical analysis.

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 \(Y\) be a variety of dimension \(r\) in \(\mathbf{P}^{n}\), with Hilbert polynomial \(P_{Y}\). We define the arithmetic genus of \(Y\) to be \(p_{a}(Y)=(-1)^{r}\left(P_{Y}(0)-1\right) .\) This is an important invariant which (as we will see later in (III, Ex. 5.3)) is independent of the projective embedding of \(Y\). (a) Show that \(p_{a}\left(\mathbf{P}^{n}\right)=0\). (b) If \(Y\) is a plane curve of degree \(d\), show that \(p_{a}(Y)=\frac{1}{2}(d-1)(d-2)\). (c) More generally, if \(H\) is a hypersurface of degree \(d\) in \(\mathbf{P}^{n}\), then \(p_{a}(H)=\left(\begin{array}{c}d-1 \\\ n\end{array}\right)\). (d) If \(Y\) is a complete intersection (Ex. 2.17 ) of surfaces of degrees \(a, b\) in \(\mathbf{P}^{3}\), then \(p_{a}(Y)=\frac{1}{2} a b(a+b-4)+1\). (e) Let \(Y^{\prime} \subseteq \mathbf{P}^{n}, Z^{s} \subseteq \mathbf{P}^{m}\) be projective varieties, and embed \(Y \times Z \subseteq \mathbf{P}^{n} \times\). \(\mathbf{P}^{m} \rightarrow \mathbf{P}^{N}\) by the Segre embedding. Show that \\[p_{a}(Y \times Z)=p_{a}(Y) p_{a}(Z)+(-1)^{s} p_{a}(Y)+(-1)^{r} p_{a}(Z)\\].

Intersection Multiplicity. If \(Y, Z \subseteq \mathbf{A}^{2}\) are two distinct curves, given by equations \(f=0, g=0,\) and if \(P \in Y \cap Z,\) we define the intersection multiplicity \((Y \cdot Z)_{P}\) of \(Y\) and \(\mathcal{L}\) at \(P\) to be the length of the \(\mathscr{O}_{p}\) -module \(\mathscr{C}_{p} f(f, g)\) (a) Show that \((Y \cdot Z)_{P}\) is finite, and \((Y \cdot Z)_{P} \geqslant \mu_{P}(Y) \cdot \mu_{P}(Z)\) (b) If \(P \in Y\), show that for almost all lines \(L\) through \(P\) (i.e., all but a finite number) \((L \cdot Y)_{Y}=\mu_{P}(Y)\) (c) If \(Y\) is a curve of degree \(d\) in \(\mathbf{P}^{2},\) and if \(L\) is a line in \(\mathbf{P}^{2}, L \neq Y,\) show that \((L \cdot Y)=d .\) Here we define \((L \cdot Y)=\sum(L \cdot Y)_{P}\) taken over all points \(P \in\) \(L \cap Y,\) where \((L \cdot Y)_{p}\) is defined using a suitable affine cover of \(\mathbf{P}^{2}\)

Let \(Y \subseteq \mathbf{A}^{3}\) be the curve given parametrically by \(x=t^{3}, y=t^{4}, z=t^{5} .\) Show that \(I(Y)\) is a prime ideal of height 2 in \(k[x, y, z]\) which cannot be generated by 2 elements. We say \(Y\) is not a local complete intersection - cf. (Ex. 2.17 ).

Vormal Varielies. A variety \(Y\) is normal all a point \(P \in Y\) if \((p\) is an integrally closed ring. \(Y\) is normal if it is normal at every point (a) Show that every conic in \(\mathbf{P}^{2}\) is normal. (b) Show that the quadric surfaces \(Q_{1}, Q_{2}\) in \(\mathbf{P}^{3}\) given by equations \(Q_{1}: x y=z w\) \(Q_{2}: x y=z^{2}\) are normal (cf. (II. Ex. 6.4) for the latter.) (c) Show that the cuspidal cubic \(y^{2}=x^{3}\) in \(A^{2}\) is not normal (d) If \(Y\) is affine. then \(Y\) is normal \(\Leftrightarrow A(Y)\) is integrally closed. (e) Let \(Y\) be an affine variety. Show that there is a normal affine variety \(\tilde{Y}\), and a morphism \(\pi: \tilde{Y} \rightarrow Y\), with the property that whenever \(Z\) is a normal variety, and \(\varphi: Z \rightarrow Y\) is a domincint morphism (i.e., \(\varphi(Z)\) is dense in \(Y\) ), then there is a unique morphism \(\theta: Z \rightarrow \tilde{Y}\) such that \(\varphi=\pi\) 0. \(\tilde{Y}\) is called the normalization of \(Y\). You will need \((3.9 \mathrm{A})\) above.

A projective variety \(Y \subseteq \mathbf{P}^{n}\) has dimension \(n-1\) if and only if it is the zero set of a single irreducible homogeneous polynomial \(f\) of positive degree. \(Y\) is called a hypersurfuce in \(\mathbf{P}^{\prime \prime}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

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.