Exercise 6.18

Show that there exist algebraic numbers of arbitrarily high degree.

Answers

Proof. As 1 + x + + x p 1 is irreducible on [ x ] (Prop. 6.4.1), the numbers ζ p = e 2 p , with p prime number, are algebraic numbers of arbitrary large degree. □

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2022-07-19 00:00
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