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Exercise 6.2
Let be an algebraic number. Show that there’s an integer such that is an algebraic integer.
Answers
( is a valid answer to this sentence ! More seriously, we search a positive integer .)
Proof.
Let an algebraic number. By definition, there exist such that
(Up to multiply this equation by , we can suppose that ).
Multiplying by , we obtain
So
Soit . Then , is monic, and . So is an algebraic integer, with .
Conclusion : if is an algebraic number, there exists an integer such that is an algebraic integer. □