The group of roots of unity in an algebraic number field 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?

This would pop out of the unit theorem, since part of that theorem is that the unit group of $\mathcal O_K$ is finitely generated. You probably don't need the whole proof, but I'd have to set aside time to check that.
Alternatively, suppose that $K$ contains a primitive $n$-th root of unity, and let $p^r$ be a term in the prime factorization of $n$. Then
\[
\varphi(p^r) = p^{r - 1}(p - 1) \leq \varphi(n) \leq [K : \mathbb Q]
\]
gives bounds for both $p$ and $r$ which depend only on $[K : \mathbb Q]$. So $n$ came from a finite list of numbers, and we are done.
