/*! 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 25 Let \(E\) be a field, and \(F \s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 25

Let \(E\) be a field, and \(F \subseteq E\) a subfield of \(E\). Show that \(E\) is a finite extension of \(F\) if and only if there exist elements \(\alpha_{i} \in E\) with each \(\alpha_{i}\) algebraic over \(F\) and \(E=F\left(\alpha_{1}, \ldots, \alpha_{r}\right)\)

Short Answer

Expert verified
鉁� Finite extension iff \(E = F(\alpha_1, \ldots, \alpha_r)\) with each \(\alpha_i\) algebraic over \(F\).

Step by step solution

01

Understanding Finite Extensions

In a finite extension, the field extension \(E/F\) has finite degree, meaning the dimension of \(E\) as a vector space over \(F\) is finite. So, assume \( [E:F] = n < \infty \), which indicates that there exists a finite basis of \(E\) over \(F\).
02

Expressing E using Algebraic Elements

Assume \(E = F(\beta_1, \beta_2, \ldots, \beta_r)\) where each \(\beta_i\) is an element in \(E\). Since \(E/F\) is finite, each element \(\beta_i\) is algebraic over \(F\) 鈥� there exists a non-zero polynomial in \(F[x]\) with \(\beta_i\) as a root. Thus, the \(\beta_i\) provide the necessary elements \(\alpha_i\).
03

Proving Necessity

Conversely, assume there exist elements \(\alpha_i\), each algebraic over \(F\), such that \(E = F(\alpha_1, \ldots, \alpha_r)\). Since each \(\alpha_i\) is algebraic over \(F\), \(F(\alpha_1, \ldots, \alpha_r)/F\) is a finite extension. This verifies that \(E/F\) is a finite extension because a finite extension over a field is itself finite.
04

Conclusion

In both directions (if and only if), we've shown that \(E\) being a finite extension is equivalent to \(E\) being generated by a finite number of algebraic elements over \(F\). This completes the proof.

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.

Field Theory
Field theory is a branch of mathematics that studies fields, which are sets equipped with two operations: addition and multiplication. These operations are associative, commutative, and distributive, and the set contains both additive and multiplicative identities. An essential concept in field theory is the field extension.

A field extension, written as \(E/F\), is when field \(E\) contains field \(F\) as a subset. Field extensions are crucial for understanding how different fields relate and merge. A finite field extension means the dimension of \(E\) as a vector space over \(F\), noted as \([E:F]\), is finite.
  • If \(E/F\) is a finite extension, then there is a finite set of elements in \(E\) that, when added to the elements of \(F\), allow you to express every element in \(E\).
  • This implies the extension is manageable, as it consists of a finite basis like a vector space.
Algebraic Elements
An algebraic element over a field \(F\) is an element in a larger field \(E\) that is a root of some non-zero polynomial with coefficients in \(F\). Put simply, these are elements whose behavior is predictable by polynomials coming from \(F\).

When considering an extension \(E/F\), it's crucial to determine which elements in \(E\) are algebraic over \(F\). Why are these elements important?
  • Every algebraic element is part of a finite extension because each satisfies a polynomial with coefficients from the base field \(F\).
  • The algebra involved in handling algebraic elements helps build and control extensions in a systematic way, ensuring that the entire extension is finite if constructed using a finite number of such elements.
Vector Space Dimension
In the context of field extensions, the concept of vector space dimension is pivotal. When you have a field extension \(E/F\), you can also look at \(E\) as a vector space over \(F\).

The degree \([E:F]\) is equal to the number of elements in a basis for this vector space.
  • A basis consists of vectors in \(E\) such that every element of \(E\) can be expressed as a linear combination of these basis vectors with coefficients from \(F\).
  • The finite dimension \([E:F] = n < \infty\) implies that \(E\) is a finite-dimensional vector space over \(F\), meaning only a finite number of elements are sufficient to span the entire field \(E\).
Understanding the vector space dimension aids in visualizing abstract field extensions by simplifying them into understandable, finite components.
Polynomial Root
A polynomial root in field theory is a solution to a polynomial equation. When considering field extensions, polynomial roots play a fundamental role.

In our context with field \(F\), an element \(\alpha\) in an extended field \(E\) is said to be algebraic over \(F\) if there exists a non-zero polynomial \(p(x) \in F[x]\) such that \(p(\alpha) = 0\). This is where the magic happens:
  • Being able to express elements of \(E\) as roots of polynomials in \(F[x]\) means each algebraic element satisfies some polynomial relationship with \(F\).
  • Algebraic elements being the roots ensures that these extensions can be controlled and described completely with their associated polynomials, turning potentially infinite issues into finite, solvable problems.
Thus, polynomial roots are at the heart of understanding how different fields can be systematically enlarged or extended while keeping the extensions manageable.

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

Find the splitting field \(K\) in \(C\) of the indicated polynomial \(f(x)\) over \(Q\), and determine \([K: \mathbb{Q}]\). $$ f(x)=x^{5}+x^{4}+x^{3}-x^{2}-x-1 $$

Find the minimal polynomial of \((\sqrt{2}+\sqrt{2} i) / 2\) over the indicated field \(F\) : (a) \(F=Q\) (b) \(\mathbb{R}\) (c) \(Q(i)\)

In Exercises 8 through 10 show that the indicated \(\alpha \in C\) is algebraic over \(Q\), and determine \(\operatorname{deg}_{\mathrm{O}}(\alpha) .\) $$ \sqrt{3}-i $$

Let \(p\) and \(q\) be two distinct prime integers. (a) Show that \(Q(\sqrt{p}+\sqrt{q})=Q(\sqrt{p}, \sqrt{q})\). (b) Find the minimal polynomial of \(\sqrt{p}+\sqrt{q}\) over \(Q\).

In Exercises 13 through 17 find a basis for the indicated extension field of \(Q\) over \(Q\). $$ Q\left(2^{1 / 3}, 7^{1 / 2}\right) $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.