Character values with absolute value $1$ are roots of unity

Question

An algebraic integer with absolute value $1$ may not be root of unity.

It is an exercise in Isaac's Charcter theory of finite groups that

if a character value $\chi(x)$ has absolute value $1$ (for some $x$ in a finite group and its irreducible character $\chi$) then $\chi(x)$ is a root of unity.

I did not get any direction although I completed hint in the book.

The hint is that: there are finitely many monic polynomials of a given degree over $\mathbb{Z}$ whose roots have absolute value $1$. Since $\chi(x)$ is $\lambda_1+\cdots+\lambda_k$ where $\lambda_i$ are roots of unity, we can see that $\chi(x)$ is root of a monic polynomial of degree $k$ say $f(x)=x^k+a_1x^{k-1}+\cdots + a_k$ over $\mathbb{Z}$ where $|a_i|\le \binom{k}{2}$.

How to proceed to show that $\chi(x)$ is a root of unity if $|\chi(x)|=1$?

Answer 1

Here is a proof and in fact more information.

Theorem Let $G$ be a finite group and $\chi$ a complex character of $G$, and $g \in G$. Then the following hold.

(a) $|\chi(g)|=1$ if and only if $\chi(g)$ is a root of unity.
(b) If $|\chi(g)|=1$, then $\chi(g) \in \{-1,1\}$ or $\chi(g)$ has even degree.
(c) If $|\chi(g)|=1$, then $\chi(1) \mid |Cl_G(g)|$.

Proof In general, an algebraic integer with absolute value equal to 1 does not have to be a root of unity, see: http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf. But if all the Galois conjugates of the algebraic integer have absolute value 1, then it is the case indeed.

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.

The only thing left to show is that if $|\chi(g)|=1$ 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 $|\chi(g^m)|=1$.

(b) By (a) $\chi(g)$ is a root of unity, hence the minimum polynomial of $\chi(g)$ is a cyclotomic polynomial, which has even degree, unless $\chi(g)=\pm1$.

(c) Let $K$ be the sum of the class $Cl_G(g)$ in $\mathbb{C}[G]$. Then $\omega_{\chi}(K)=\frac{\chi(g) \cdot |Cl_G(g)|}{\chi(1)}$ is an algebraic integer (see CTFG, Theorem (3.7)). Hence, the absolute value is an algebraic integer as well, and $|\omega_{\chi}(K)|=\frac{|Cl_G(g)|}{\chi(1)}$ is rational algebraic integer and hence an element of $\mathbb{Z}$.

Answer 2

Reduce to: let $\omega$ be a root of $1$ and $P(x)\in \mathbb{Z}[x]$ such that $|P(\omega)|=1$. Then $P(\omega)$ is again a root of $1$.

For $\omega^n=1$ for some $n$, so $\omega$ is a root of an (monic) irreducible polynomial $ \mathbb{Q}[x]\ni C(x)\mid x^n-1$, which will have integral coefficients (Gauss lemma).

Now $$1=|P(\omega)|^2 = P(\omega) \cdot \overline{P(\omega)}= P(\omega) \cdot P(\bar \omega)= P(\omega) \cdot P(\frac{1}{\omega})$$

We conclude that for all of the roots $\omega'$ of $C(x)$ we have

$$1=|P(\omega')|^2 = P(\omega) \cdot \overline{P(\omega')}= P(\omega') \cdot P(\bar \omega')= P(\omega') \cdot P(\frac{1}{\omega'})$$

Consider the monic polynomial with roots $P(\omega')$, where $\omega'$ are the roots of $C(x)$. We got ourselves a monic polynomial in $\mathbb{Z}[x]$ and all roots of absolute value $1$. We conclude ( Kronecker lemma, on this site) that all of its roots $P(\omega')$ are roots of $1$.