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
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 58RE
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Question
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.

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

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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