## Finite extensions of the rationals contain only finitely many roots of unity

Let $K$ be a finite extension of $\mathbb{Q}$. Prove that $K$ contains only finitely many roots of unity.

Suppose to the contrary that $K$ contains infinitely many roots of unity. Now for each $n$, there are only finitely many primitive roots of unity (in fact $\varphi(n)$ of them). So for each $m$, the number of primitive roots of unity of order at most $m$ is finite. In particular, for any $m$, there exists a primitive $n$th root of unity for some $n > m$.

Let $m$ be the degree of $K$ over $\mathbb{Q}$. If $\zeta \in K$ is a primitive $k$th root of unity, then $[\mathbb{Q}(\zeta):\mathbb{Q}] = \varphi(k)$ by Corollary 42 on page 555, where $\varphi$ denotes the Euler totient. By this previous exercise, since $k$ is arbitrarily large, $\varphi(k)$ is arbitrarily large. So there exists a primitive $k$th root $\zeta$ such that $\varphi(k) > m$, a contradiction since $\mathbb{Q}(\zeta) \subseteq K$.

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