Let Q be the field of rational numbers and let F Q(√2, √7). You may assume that √2, √7, √14 are irrational. (a) Show that √7 Q(√2). Hence obtain that [F : Q] = 4.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 36E
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I don't understand 2 points in the solution for this problem. Can you please explain why? Thanks.

1. Why : {1, √2} is a basis for Q(√2) over Q then we conclude 2rs = 0 and r2 + 2s2 = 0 ???

2. Why: [F : Q(√2) ] = 2

Let be the field of rational numbers and let F = Q(√2, √7). You may assume that
√2, √7, √14 are irrational.
(a) Show that √7 ¢ Q(√2). Hence obtain that [F : Q] = 4.
Answer
=
(a) Assume that √7 € Q(√2). Then there are r, s € Q such that √7 = r + s√√2. Thus
7 = r² +2s² +2rs√2. Since {1, √2} is a basis for Q(√2) over Q we obtain that 2rs = 0
and r² + 2s² 7. Therefore r = 0, 2s²
7, or s = 0, p²
7. The former implies
(2s)² = 14. In either case we have a contradiction because √14 and √7 are irrational.
By the degree formula [F : Q] = [F : Q(√2)] · [Q(√2) : Q] = 2 ⋅ 2 = 4.
=
=
Transcribed Image Text:Let be the field of rational numbers and let F = Q(√2, √7). You may assume that √2, √7, √14 are irrational. (a) Show that √7 ¢ Q(√2). Hence obtain that [F : Q] = 4. Answer = (a) Assume that √7 € Q(√2). Then there are r, s € Q such that √7 = r + s√√2. Thus 7 = r² +2s² +2rs√2. Since {1, √2} is a basis for Q(√2) over Q we obtain that 2rs = 0 and r² + 2s² 7. Therefore r = 0, 2s² 7, or s = 0, p² 7. The former implies (2s)² = 14. In either case we have a contradiction because √14 and √7 are irrational. By the degree formula [F : Q] = [F : Q(√2)] · [Q(√2) : Q] = 2 ⋅ 2 = 4. = =
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