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EXTENSION FIELD The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 } Can someone please explain the highlight parts in his works for me? I don't get it: - Why if F is a field then F is the smallest ? - Why  √2 not in Q

EXTENSION FIELD The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 } Can someone please explain the highlight parts in his works for me? I don't get it: - Why if F is a field then F is the smallest ? - Why  √2 not in Q

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 24RE
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EXTENSION FIELD


The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 }

Can someone please explain the highlight parts in his works for me? I don't get it:


- Why if F is a field then F is the smallest ?

- Why  √2 not in Q, then we have those stuff = 0 ?

- Similarly, why  √3 not in Q, then we have those stuff = 0. ?

Thank you.


 

QQ (√₂₁ √3)= { a+b√₂ +c√3 + d√6 | a, b, c, d = Q } F •F is cubring of R, and FCQ (√2, √3) • Is F a field? If yes, then F is smallest and F = Q(√2₁ √3) Check: 1 a+b√₂+c√3+d√6 1 (a+b√₂) + √3 (c+d√₂) (a+b√z) - (c+d)√3 (a+b√² ) ² = 3 [c+d√₂) ² (a+b√₂ = √3 (c+d√₂)).(a²-3c² + 26²-6d²-√₂ (2ab-6cd)) EF 2 (a²-3c² +26²-66²) ² EQ 2 2 (2ab-6cd) ² 1 Hence +b√√₂ +c√3+d√√6 the denominator is 0. (*) € F as long as
Transcribed Image Text:QQ (√₂₁ √3)= { a+b√₂ +c√3 + d√6 | a, b, c, d = Q } F •F is cubring of R, and FCQ (√2, √3) • Is F a field? If yes, then F is smallest and F = Q(√2₁ √3) Check: 1 a+b√₂+c√3+d√6 1 (a+b√₂) + √3 (c+d√₂) (a+b√z) - (c+d)√3 (a+b√² ) ² = 3 [c+d√₂) ² (a+b√₂ = √3 (c+d√₂)).(a²-3c² + 26²-6d²-√₂ (2ab-6cd)) EF 2 (a²-3c² +26²-66²) ² EQ 2 2 (2ab-6cd) ² 1 Hence +b√√₂ +c√3+d√√6 the denominator is 0. (*) € F as long as
Assume √₂ & Q denominator Hence = 0 Qab-6cd = 0 ) ab-3cd=0 a² +2b²-3c² - 6d² = 0 √3 & Q = ab = cd = 0 ¢ => a²+ 26² = 0 = c² + 2d² a = b = c = d <contradiction,
Transcribed Image Text:Assume √₂ & Q denominator Hence = 0 Qab-6cd = 0 ) ab-3cd=0 a² +2b²-3c² - 6d² = 0 √3 & Q = ab = cd = 0 ¢ => a²+ 26² = 0 = c² + 2d² a = b = c = d <contradiction,
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