Are all algebraic integers with absolute value 1 roots of unity? If we have an algebraic number α with (complex) absolute value 1, it does not follow that α is a root of unity (i.e., that for some n). For example, is not a root of unity.But if we assume that is an algebraic integer with absolute value 1, does it follow that is a root of unity?
Answer & Explanation
blindhaedzgs
Beginner2022-03-19Added 2 answers
Step 1 Let me first mention an example in Character Theory. Let G be a finite group of order n and assume is a representation with character which is defined by . Since G is a finite group then, by invoking facts from linear algebra, one can show . For abelian groups, it is easy to see is a root of unity, when is irreducible, but what about non-abelian groups? In other words let , what can we say about ? This relates to your question. Let assume be an abelian Galois extension inside , and take an algebraic integer such that , then for any we have Since is abelian then so Then norm of all its conjugate is one so it must be a root of unity. This answer to the question was posed, therefore if then is root of unity.
Quentin Olsen
Beginner2022-03-20Added 5 answers
Step 1 Let x be an algebraic number with absolute value 1. Then x and its complex conjugate have the same minimal polynomial. Writing f(T) for the minimal polynomial of x over , with degree n, the polynomials and are irreducible over with root , so the polynomials are equal up to a scaling factor: . Setting Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so f(1) is nonzero and thus . Therefore , so f(T) has symmetric coefficients. In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even. Partial conclusion: an algebraic number other than 1 or -1 which has absolute value 1 has even degree over and its minimal polynomial has constant term 1. In particular, if x is an algebraic integer then it must be a unit. There are no examples of algebraic integers with degree 2 and absolute value 1 that are not roots of unity, since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for and ). Thus the smallest degree x could have over 𝕢 is 4 and there are examples with degree 4: the polynomial has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1).