TITLE: The group of roots of unity in an algebraic number field
QUESTION [10 upvotes]: Is the following proposition true? If yes, how would you prove this?
Proposition.
Let $K$ be an algebraic number field.
The group of roots of unity in $K$ is finite.
In other words, the torsion subgroup of $K^*$ is finite.
Motivation.
Let $A$ be the ring of algebraic integers in $K$.
A root of unity in $K$ is a unit (i.e. an invertible element of $A$). It is important to determine the structure of the group of units in $K$ to investigate the arithmetic properties of $K$.
Remark.
Perhaps, the following fact can be used in the proof.
Every conjugate of a root of unity in $K$ has absolute value 1.
Related question:
The group of roots of unity in the cyclotomic number field of an odd prime order
Is an algebraic integer all of whose conjugates have absolute value 1 a root of unity?

REPLY [4 votes]: Lemma. Let $f(x)$ be a degree $n$ polynomial with rational coefficients all of whose roots have absolute value 1. Then the coefficient of $x^k$ in $f$ has absolute value $\leq \binom{n}{k}$.
Proof: Apply Vieta's formulas and the triangle inequality to conclude. $\square$
Proposition. There are only finitely many $n$-th roots of unity in an algebraic number field $K$.
Proof. It will be enough to prove that there are only finitely many algebraic integers $\alpha$ of fixed degree $n$, all of whose Galois conjugates (including $\alpha$) have absolute value $1$. Let $f_\alpha(x)$ be the minimal polynomial of $\alpha$ with coefficients in $\Bbb{Z}$.  Now for each $k$ with $1 \leq k\leq n$ there are only finite many integers of absolute value $\leq \binom{n}{k}$. Since given any $\alpha$ every Galois conjugate of $\alpha$ has absolute value $1$ we apply the Lemma above to conclude that only finitely many such $f_\alpha$ can exist. Thus only finitely many such $\alpha$ can exist which completes the proof of the proposition. $\square$