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Q10E

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Abstract Algebra: An Introduction

Found in: Page 422

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Two intermediate field E and L are said to be conjugate if there exists $σ \in G a l_{F} K$ auch that $σ (E) = L$ . Prove that E and L are conjugate if and only if $G a l_{L} K$ and $G a l_{L} k$ are conjugate subgroup of role="math" localid="1657965474587" $G a l_{F} K$ .

$G a l_{E} K$ and $G a l_{L} K$ are conjugate subgroup of $G a l_{F} K$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of the abelian group.

An abelian group is a commutative group and satisfy the condition for all $a, b$ in the field extension is $a . b = b . a$ .

Step-2: Showing that GalEK and GalLK are conjugate subgroup of GalFK .

Consider the fixed subgroups $K$ of the extension field $F$ and this normal fixed subgroup consist two subgroup $E$ and $L$ here the conditions is given both subgroup over the fixed are conjugate.

Now the property of conjugate between two subgroup if it is exist so the Galois group of the fixed field over the extension field is conjugate. Consider the two element $a, β$ of the subgroup $E$ over the $K$ if the subgroup are conjugate so all the element to the subgroup are conjugate so both element are conjugate and the automorphism on $E$ is define as $σ (a) = β$ and minimal polynomial of $a$ over $k$ is role="math" localid="1657965967250" $p (x)$ assume both element are zeros of the same irreducible polynomial $p (x)$ over the fixed field $k$ .

Then $K$ is isomorphism of the element is

$ϕ (a) = β$

And $k$ automorphism of the element is

$σ (a) = ϕ (a) = β$

Same for another subgroup $L$ consist the two element $a, b \in L$ and automorphism is

$σ (a) = ϕ (a) = b$

Since the element of the both subgroup are conjugate and subgroup are also conjugate so the automorphism over the fixed field is

$σ (E) = ϕ (E) = L$

Therefore $σ (E) = L$

If both subgroup over the fixed field are conjugate so Galois field of the both subgroup over the extension field are also conjugate by using the galois theory.

Since $E$ and $L$ are conjugate if and only if $G a l_{E} K$ and $G a l_{L} K$ are conjugate of the field over the extension field $F$ is $G a l_{F} K$ .

Most popular questions for Math Textbooks

Find $G a l_{Q} Q (\sqrt{2}, i)$ .

Let $ζ$ be a primitive fifth root of unity. Prove that ${Gal}_{Q} Q (ζ)$ the Galois group of $x^{5} - 1$ is cyclic of order 4.

Determine whether the given equation over $ℚ$ is solvable by radicals:

$(a) x^{6} + 2 x^{3} + 1 = 0 (b) 3 x^{5} - 15 x + 5 = 0 (c) 2 x^{5} - 5 x^{4} + 5 = 0 (d) x^{5} = x^{4} - 16 x + 16 = 0$

If $[K : F]$ is finite, $σ \in G a l_{F} K$ , and $u \in K$ is such that $σ (u) = u$ , show that $σ \in G a l_{F} K$ .

List all the nth roots of unity in $ℂ$ when $n =$

(a) 2

(b) 3

(c) 4

(d) 5

(e) 6

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