SolutionsChap6-10

# SolutionsChap6-10 - Solutions Chapters 610 Section 6.1 1....

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Solutions Chapters 6–10 Section 6.1 1. We have r 1 =2 , r 2 =1 , r 3 =1so t 1 , t 2 =0 , t 3 = 1. The algorithm terminates in one step after after subtraction of ( X 1 + X 2 + X 3 )( X 1 X 2 X 3 ). The given polynomial can be expressed as e 1 e 3 . 2. We have r 1 ,r 2 3 =0s o t 1 ,t 2 3 = 0. At step 1, subtract ( X 1 + X 2 + X 3 )( X 1 X 2 + X 1 X 3 + X 2 X 3 ). The result is 3 X 1 X 2 X 3 +4 X 1 X 2 X 3 = X 1 X 2 X 3 . By inspection (or by a second step of the algorithm), the given polynomial can be expressed as e 1 e 2 + e 3 . 3. Equation (1) follows upon taking σ 1 ( h ) outside the summation and using the linear dependence. Equation (2) is also a consequence of the linear dependence, because σ i ( h ) σ i ( g )= σ i ( hg ). 4. By hypothesis, the characters are distinct, so for some h G we have σ 1 ( h ) 6 = σ 2 ( h ). Thus in (3), each a i is nonzero and σ 1 ( h ) σ i ( h ) ( =0 if i =1; 6 i . This contradicts the minimality of r . (Note that the i = 2 case is important, since there is no contradiction if σ 1 ( h ) σ i ( h ) = 0 for all i .) 5. By (3.5.10), the Galois group consists of the identity alone. Since the identity Fxes all elements, the Fxed Feld of G is Q ( 3 2). 6. Since C = R [ i ], an R -automorphism σ of C is determined by its action on i . Since σ must permute the roots of X 2 + 1 by (3.5.1), we have σ ( i i or i . Thus the Galois group has two elements, the identity automorphism and complex conjugation. 7. The complex number z is Fxed by complex conjugation if and only if z is real, so the Fxed Feld is R . Section 6.2 1. The right side is a subset of the left since both E i and E p i +1 are contained in E i +1 . Since E i is contained in the set on the right, it is enough to show that α i +1 E i ( E p i +1 ). By 1

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2 hypothesis, α i +1 is separable over F , hence over E i ( α p i +1 ). By Section 3.4, Problem 3, α i +1 E i ( α p i +1 ) E i ( E p i +1 ). 2. Apply Section 3.4, Problem 7, with E = F ( E p ) replaced by E i +1 = E i ( E p i +1 ), to conclude that E i +1 is separable over E i . By the induction hypothesis, E i is separable over F . By transitivity of separable extensions (Section 3.4, Problem 8), E i +1 is separable over F . By induction, E/F is separable. 3. Let f i be the minimal polynomial of α i over F . Then E is a splitting Feld for f = f 1 ··· f n over F , and the result follows. 4. This is a corollary of part 2 of the fundamental theorem, with F replaced by K i 1 and G replaced by Gal( E/K i 1 )= H i 1 . 5. E ( A ) is a Feld containing E F and A , hence E ( A ) contains E and K , so that by deFnition of composite, EK E ( A ). But any Feld (in particular ) that contains E and K contains E and A , hence contains E ( A ). Thus E ( A ) . 6. If σ G , deFne Ψ( σ )( τ ( x )) = τσ ( x ), x E . Then ψ ( σ ) G 0 . [If y = τ ( x ) F 0 with x F , then Ψ( σ ) y =Ψ( σ ) τx = ( x τ ( x y .] Now Ψ( σ 1 σ 2 ) τ ( x 1 σ 2 ( x ) and Ψ( σ 1 )Ψ( σ 2 ) τ ( x )=Ψ ( σ 1 ) 2 ( x 1 σ 2 ( x ), so Ψ is a group homomorphism. The inverse of Ψ is given by Ψ 0 ( σ 0 ) τ 1 y = τ 1 σ 0 ( y ), σ 0 G 0 , y E 0 . To see this, we compute Ψ 0 (Ψ( σ )) τ 1 y = τ 1 Ψ( σ ) y = τ 1 Ψ( σ ) = τ 1 ( x σ ( x σ ( τ 1 y ) .
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## This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.

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SolutionsChap6-10 - Solutions Chapters 610 Section 6.1 1....

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