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?
Best Answer
The degree of $e^{2\pi i/n}$ goes to infinity with $n$. If $K$ had an infinity of roots of unity, it would have elements of arbitrarily high degree, and thus would not be of finite degree over the rationals, and thus would not, in fact, be an algebraic number field.