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

Question

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?

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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 $\alpha^{n^m - 1} = 1$.

Answer 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 $1$).

Answer 2

No. There are algebraic integers on the unit circle which aren't roots of unity.

This paper by Ryan Daileda provides some useful information and references.

Also see Salem numbers.

Answer 3

The answer to your question is no - not all algebraic integers with absolute value 1 are roots of unity - as you may have learned by now from the 2005 paper of Daileda that was referenced in the answer supplied by Robin Chapman. However, you should be aware that Daileda has in fact rediscovered these simple folklore results on unimodular units. These results are probably at least a half-century old - if not much older. Indeed, I recall reading similar results in papers published by Iwasawa in the fifties or sixties. I don't have the time now to locate said Iwasawa reference, but here is another reference that suffices to prove my point. This 1979 paper of Nakahata includes the same results and the same simple proofs found in Daileda's paper. Moreover, Nakahata places these results naturally into a more general context. Thus I highly recommend that you consult Nakahata's paper in addition to Daileda's. I've appended its Zbl review below.

Answer 4

Let me first mention an example in Character Theory. Let $G$ be a finite group of order $n$ and assume $\rho$ is a representation with character $\chi:=\chi_\rho$ which is defined by $\chi(g)=Tr(\rho(g))$. Since $G$ is a finite group then, by invoking facts from linear algebra, one can show $\chi(g)\in\mathbb{Z}[\zeta_n]$. For abelian groups, it is easy to see $\chi(g)$ is a root of unity, when $\chi$ is irreducible, but what about non-abelian groups? In other words let $|\chi(g)|=1$, what can we say about $\chi(g)$?

This relates to your question. Let assume $K/\mathbb{Q}$ be an abelian Galois extension inside $\mathbb{C}$, and take an algebraic integer $\alpha\in\mathcal{O}_K$ such that $|\alpha|=1$, then for any $\sigma\in Gal(K/\mathbb{Q})$ we have $$ |\sigma(\alpha)|^2=\sigma(\alpha)\overline{\sigma(\alpha)} $$ Since $K/\mathbb{Q}$ is abelian then $\overline{\sigma(\alpha)}=\sigma(\overline{\alpha})$ so $$ |\sigma(\alpha)|^2=\sigma(|\alpha|)=1 $$ 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 $|\chi(g)|=1$ then $\chi(g)$ is root of unity.

Answer 5

Notice however that if all roots of a polynomial with integer coefficients lie on the unit complex circle (i.e., all roots z have |z|=1), then, by a theorem of Kronecker, these roots are indeed roots of unit.

Answer 6

As pointed out above, an algebraic integer with absolute value equal to 1 does not have to be a root of unity. But if all the conjugates of the algebraic integer have absolute value 1, then it is indeed the case.

Let $\alpha$ be an algebraic integer with minimum polynomial $f\in \mathbb{Z}[X]$, which is monic and say of degree $n$. We can assume $n \geq 2$ otherwise $\alpha \in \{-1,1\}$. Assume all the roots of $f$ have absolute value 1. Let us concentrate on the coefficient of $X^i$: the sum of the roots taking $i$ of them each time is in absolute value bounded by ${n}\choose{i}$ by the triangle inequality. Thus the coefficient of $X^i$ is in absolute value bounded by ${n}\choose{i}$, hence for any $n$ there are only finitely many algebraic integers of degree $n$ such that all conjugates have absolute value $1$, since there are only finitely many polynomials in $\mathbb{Z}[X]$ with given bounded coefficients.
Next, consider the powers of $\alpha$. They are all algebraic integers of degree at most $n$, and furthermore all their conjugates also have absolute value $1$ since the Galois actions map powers of $\alpha$ to powers of its conjugates. Thus, the powers of $\alpha$ are elements of a finite set. This implies $\alpha$ must be a root of unity.

In the case of a finite group $G$, $g \in G$ and complex (not necessarily irreducible) character $\chi$ with $|\chi(g)|=1$ all the Galois conjugates of the algebraic number $\chi(g)$ also have absolute value $1$. Let $n=o(g)$ and $K=\Omega_{\mathbb{Q}}^{X^n-1} \subseteq \mathbb{C}$ the splitting field. Let $\mathfrak{G}=Gal(K/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^*$. If $\sigma \in \mathfrak{G}$ and $\varepsilon$ is an $n$th-root of unity, then $\sigma(\varepsilon)=\varepsilon^m$, for some $m \in \mathbb{Z}$, with gcd$(m,n)=1$. Now, $\chi(g)=\varepsilon_1 + \cdots + \varepsilon_{\chi(1)}$, a sum of $n$th-roots of unity. Hence, $\sigma(\chi(g))=\varepsilon_1^m + \cdots + \varepsilon_{\chi(1)}^m=\chi(g^m)$. Note that $\mathfrak{G}$ is abelian and that the restriction of complex conjugation to $K$ induces an element of $\mathfrak{G}$ of order $2$. Hence, $|\sigma(\chi(g))|^2=\sigma(\chi(g)) \cdot \overline{\sigma(\chi(g))}=\sigma(\chi(g)) \cdot \sigma(\overline{\chi(g)})=\sigma(|\chi(g)|^2)=\sigma(1)=1$, which yields $|\chi(g^m)|=1$.

Answer 7

Following the paper provided by Robin Chapman, I give an explicit algebraic integer that is not a root of unity:

Let $\alpha = \left(1-\dfrac{1}{2}\sqrt[3]{3}-\dfrac{1}{2}\sqrt[3]{3^2}\right) + \left(\dfrac{1}{2}\sqrt[6]{3}-\dfrac{1}{2}\sqrt[6]{3^5}\right)i$, then $|\alpha| = 1$, and $\alpha$ is a root of $x^6 - 6x^5 + 6x^4 + 10x^3 + 6x^2 - 6x + 1$ (thanks to Mathematica), but $\alpha$ is not a root of unity because that polynomial is not a cyclotomic polynomial.