Page 251 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra Richa Nandra, Lovely Professional University

Notes Unit 26: Splitting Fields, Existence and Uniqueness

CONTENTS
Objectives

Introduction
26.1 Extension Field
26.2 Summary
26.3 Keywords
26.4 Review Questions

26.5 Further Readings

Objectives

After studying this unit, you will be able to:
Discuss splitting field

Describe extension field and theorem related to extension

Introduction

Beginning with a field K, and a polynomial f(x)  K, we need to construct the smallest possible
extension field F of K that contains all of the roots of f(x). This will be called a splitting field for
f(x) over K. The word “the” is justified by proving that any two splitting fields are isomorphic.

Let F be an extension field of K and let u  F. If there exists a non-zero polynomial f(x)  K[x]
such that f(u) = 0, then u is said to be algebraic over K. If not, then u is said to be transcendental
over K.

26.1 Extension Field

If F is an extension field of K, and u  F is algebraic over K, then there exists a unique monic
irreducible polynomial p(x)  K[x] such that p(u) = 0. It is the monic polynomial of minimal
degree that has u as a root, and if f(x) is any polynomial in K[x] with f(u) = 0, then
p(x) | f(x).

Alternate proof: The proof in the text uses some elementary ring theory. Then decided to include
a proof that depends only on basic facts about polynomials.
Assume that u  F is algebraic over K, and let I be the set of all polynomials f(x)  K[x] such that
f(u) = 0. The division algorithm for polynomials can be used to show that if p(x) is a non-zero
monic polynomial in I of minimal degree, then p(x) is a generator for I, and thus p(x) | f(x)
whenever f(u) = 0.
Furthermore, p(x) must be an irreducible polynomial, since if p(x) = g(x)h(x) for g(x); h(x)  K[x],
then g(u)h(u) = p(u) = 0, and so either g(u) = 0 or h(u) = 0 since F is a field. From the choice of p(x)
as a polynomial of minimal degree that has u as a root, we see that either g(x) or h(x) has the
same degree as p(x), and so p(x) must be irreducible.

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