6.2 Algebraic Extensions
1.
a. 
whence (u2 − 8)2 = 60 . Hence u4 − 16u + 4 = 0.
whence (u2 − 8)2 = 60 . Hence u4 − 16u + 4 = 0.
c. 
so 
that is 
Thus (u4 + 7)2 = 12u4, so u8 + 2u4 + 49 = 0.
so that is Thus (u4 + 7)2 = 12u4, so u8 + 2u4 + 49 = 0.
2.
a. 
so (u2 − 5)2 = 24, u4 − 10u2 + 1 = 0. We claim m = x4 − 10x2 + 1 is the minimal polynomial; it suffices to prove it is irreducible over 
. It has no root in 
so suppose
so (u2 − 5)2 = 24, u4 − 10u2 + 1 = 0. We claim m = x4 − 10x2 + 1 is the minimal polynomial; it suffices to prove it is irreducible over . It has no root in so suppose
Then
Then c = − a and b = d = ± 1 . Hence 2b − a2 = − 10, so a2 = 10 + 2b = 12 or 8, a contradiction.
c. 
so (u2 − 1)2 = 3; u4 − 2u2 − 2 = 0. The minimal polynomial is x4 − 2
so (u2 − 1)2 = 3; u4 − 2u2 − 2 = 0. The minimal polynomial is x4 − 2Get Introduction to Abstract Algebra, Solutions Manual, 4th Edition now with the O’Reilly learning platform.
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