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 nth root of unity for some n > m.

Let m be the degree of K over \mathbb{Q}. If \zeta \in K is a primitive kth 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 kth root \zeta such that \varphi(k) > m, a contradiction since \mathbb{Q}(\zeta) \subseteq K.

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