Abstract Algebra: Theory and Applications · Problem 21.5.22

Finite Fields — Thomas Judson

Mathematics — Abstract Algebra · field-equality-adjoining-sum

Problem

Show that Q(3,7)=Q(3+7)\mathbb{Q}(\sqrt{3}, \sqrt{7}) = \mathbb{Q}(\sqrt{3} + \sqrt{7}). Extend your proof to show that Q(a,b)=Q(a+b)\mathbb{Q}(\sqrt{a}, \sqrt{b}) = \mathbb{Q}(\sqrt{a} + \sqrt{b}), where a≠ba \neq b and neither aa nor bb is a perfect square. Hint. Since {1,3,7,21}\{1, \sqrt{3}, \sqrt{7}, \sqrt{21}\} is a basis for Q(3,7)\mathbb{Q}(\sqrt{3}, \sqrt{7}) over Q\mathbb{Q}, Q(3,7)⊃Q(3+7)\mathbb{Q}(\sqrt{3}, \sqrt{7}) \supset \mathbb{Q}(\sqrt{3} + \sqrt{7}). Since [Q(3,7):Q]=4[\mathbb{Q}(\sqrt{3}, \sqrt{7}) : \mathbb{Q}] = 4, [Q(3+7):Q]=2[\mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q}] = 2 or 44. Since the degree of the minimal polynomial of 3+7\sqrt{3} + \sqrt{7} is 44, Q(3,7)=Q(3+7)\mathbb{Q}(\sqrt{3}, \sqrt{7}) = \mathbb{Q}(\sqrt{3} + \sqrt{7}).

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