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?

Edit(Jan. 18, 2013) To the downvoters, why don't you reset your votes? The question is clearly important in algebraic number theory. I'm saying this not because I care my reps, but because the negative votes are sending wrong signals to the users.

• Please explain the reason for the downvotes. Unless you make it clear, it is hard to improve my question. – Makoto Kato Jul 28 '12 at 0:52
• What does the importance or lack thereof of a question's topic have to do with how it is voted?!?!?!?! – Hurkyl Jan 18 '13 at 16:35
• @Hurkyl [What does the importance or lack thereof of a question's topic have to do with how it is voted?!?!?!?!] The answer is that the negative votes are sending wrong signals to the users. – Makoto Kato May 25 '14 at 3:38

The degree of 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.

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.

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.

$\hspace{5in} \square$

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.

$\hspace{5in} \square$