Exercise 6.2

Let α be an algebraic number. Show that there’s an integer n such that is an algebraic integer.

Answers

( 0 is a valid answer to this sentence ! More seriously, we search a positive integer n .)

Proof.

Let α an algebraic number. By definition, there exist a 0 , a 1 , , a n , a n 0 , such that

a n α n + a n 1 α n 1 + + a k α k + + a 0 = 0 .

(Up to multiply this equation by 1 , we can suppose that a n > 0 ).

Multiplying by a n n 1 , we obtain

a n n α n + a n n 1 a n 1 α n 1 + + a n n 1 a k α k + + a n n 1 a 0 = 0 .

So

( a n α ) n + a n 1 ( a n α ) n 1 + + a n n k 1 a k ( a n α ) k + + a n n 1 a 0 = 0 .

Soit p ( x ) = x n + k = 0 n 1 a n n k 1 a k x k . Then p ( x ) [ x ] , p ( x ) is monic, and p ( a n α ) = 0 . So a n α is an algebraic integer, with m = a n .

Conclusion : if α is an algebraic number, there exists an integer m > 0 such that is an algebraic integer. □

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2022-07-19 00:00
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