# Are all algebraic integers with absolute value 1 roots of unity?

If we have an algebraic number $$\alpha$$ with (complex) absolute value $$1$$, it does not follow that $$\alpha$$ is a root of unity (i.e., that $$\alpha^n = 1$$ for some $$n$$). For example, $$(3/5 + 4/5 i)$$ is not a root of unity.

But if we assume that $$\alpha$$ is an algebraic integer with absolute value $$1$$, does it follow that $$\alpha$$ is a root of unity?

I know that if all conjugates of $$\alpha$$ have absolute value $$1$$, then $$\alpha$$ is a root of unity by the argument below:

The minimal polynomial of $$\alpha$$ over $$\mathbb{Z}$$ is $$\prod_{i=1}^d (x-\alpha_i)$$, where the $$\alpha_i$$ are just the conjugates of $$\alpha$$. Then $$\prod_{i=1}^d (x-\alpha_i^n)$$ is a polynomial over $$\mathbb{Z}$$ with $$\alpha^n$$ as a root. It also has degree $$d$$, and all roots have absolute value $$1$$. But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that $$\alpha^n=\sigma(\alpha)$$ for some Galois conjugation $$\sigma$$. If $$\sigma^m(\alpha) = \alpha$$, then $$\alpha^{n^m} = \alpha$$.

Thus $$\alpha^{n^m – 1} = 1$$.

Let $x$ be an algebraic number with absolute value $1$. Then $x$ and its complex conjugate $\overline{x} = 1/x$ have the same minimal polynomial. Writing $f(T)$ for the minimal polynomial of $x$ over $\mathbb{Q}$, with degree $n$, the polynomials $T^nf(1/T)$ and $f(T)$ are irreducible over $\mathbb{Q}$ with root $\overline{x}$, so the polynomials are equal up to a scaling factor: $$T^nf(1/T) = cf(T).$$ Setting $T = 1$, $f(1) = cf(1)$.
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 $c = 1$. Therefore $$T^nf(1/T) = f(T),$$ 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 $\mathbb{Q}$ 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 $\mathbb{Q}(i)$ and $\mathbb{Q}(\omega)$). Thus the smallest degree $x$ could have over $\mathbb{Q}$ is $4$ and there are examples with degree $4$: the polynomial $$x^4 – 2x^3 – 2x + 1$$ has two roots on the unit circle and two real roots (one between $0$ and $1$ and the other greater than $1$).