Explanation Given information: 9 3 − 3 3 + 2 in ℚ ( 3 3 ) , where ℚ is the field of rational numbers. Explanation: Let α = 3 3 Taking cube from both sides, we get, α 3 = 3 Therefore, p ( α ) = α 3 − 3 . Hence, p ( x ) = x 3 − 3 is a polynomial which is irreducible over a field of rational number ℚ by the Eisenstein criteria. Let 9 3 − 3 3 + 2 be written as ( 3 3 ) 2 − 3 3 + 2 Here, α = 3 3 implies that f ( α ) = α 2 − α + 2 Hence, f ( x ) = x 2 − x + 2 When p ( x ) is divided by f ( x ) , we get p ( x ) = f ( x ) ( x + 1 ) + ( − x − 5 ) This implies that − x − 5 = p ( x ) − f ( x ) ( x + 1 ) Divide f ( x ) by − x − 5 , we get f ( x ) = ( − x − 5 ) ( − x + 6 ) + 32 Divide − x − 5 by 32 , we get − x − 5 = 32 ( − 1 32 x − 5 32 ) Hence, 1 is the greatest common divisor of p ( x ) and f ( x ) 1 = 1 32 ⋅ 32 As f ( x ) = ( − x − 5 ) ( − x + 6 ) + 32 implies that 32 = f ( x ) − ( − x − 5 ) ( − x + 6 ) , 1 = 1 32 ⋅ [ f ( x ) − ( − x − 5 ) ( − x + 6 ) ] 1 = 1 32 ⋅ [ f ( x ) + ( x + 5 ) ( − x + 6 ) ] As p ( x ) = f ( x ) ( x + 1 ) + ( − x − 5 ) implies that x

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Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 8.6, Problem 11E

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The multiplicative inverse of 93 − 33 + 2 in ℚ ( 33 ), where ℚ is the field of rational numbers.

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Chapter 8.6, Problem 10E

Chapter 8.6, Problem 12E

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Chapter 8 Solutions

Elements Of Modern Algebra

Ch. 8.1 - True or False Label each of the following...Ch. 8.1 - Prob. 2TFE Ch. 8.1 - Prob. 3TFE Ch. 8.1 - Prob. 4TFE Ch. 8.1 - Prob. 5TFE Ch. 8.1 - Prob. 6TFE Ch. 8.1 - Prob. 7TFE Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E

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The multiplicative inverse of 9 3 − 3 3 + 2 in ℚ ( 3 3 ) , where ℚ is the field of rational numbers.

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The multiplicative inverse of 93 − 33 + 2 in ℚ ( 33 ), where ℚ is the field of rational numbers.

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Chapter 8 Solutions