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Two questions; the hint I've been provided is that they are, in fact, related.

  1. Prove that a finite extension of $\mathbb{Q}$ contains finitely many roots of unity.
  2. What is the largest (finite) order an element of $GL_{10}(\mathbb{Q})$ can have?

I am not sure how to approach the first problem. Does the second have something to do with the relationship between field extensions and vector spaces over a base field?

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Suppose $K$ is a number field of degree $d$ which contains infinitely many roots of unity. It then contains roots of unity of arbitrary high order (beecause there are finitely many of each order), and therefore it contains roots of cyclotomic polynomials $\Phi_n$ for infinitely many $n$. Let $S$ be the set of such $n$'s

Now the degree of $\Phi_n$ is $\phi(n)$ and $\phi(n)\geq\sqrt n$ for $n>6$. Since $S$ is infinite, there is an element $m\in S$ such that $\phi(m)>d$.

But then $K$ contains the field generated by a root of the polynomial $\Phi_m$, which is of degree $\phi(m)$. This is absurd.

For the second problem, notice that if a matrix in $GL_{10}(\mathbb Q)$ has finite order, it generates a subring of the ring of matrices $M_{10}(\mathbb Q)$ isomorphic to an extension of $\mathbb Q$ by a root of unity. This extension has degree at most 100, because it is contained in $M_{10}(\mathbb Q)$ (of course you can do much better in bounding this...)

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  • $\begingroup$ I know this is a bit of a delay to add a comment, but I just came across this post recently. In your first paragraph, you make the observation that “…and therefore it contains roots of cyclotomic polynomials $\Phi_n$ for infinitely many $n$”. Why must this be? Why can’t I decide to avoid primitive roots of unity? $\endgroup$ – Sam Y. May 16 '17 at 23:50
  • $\begingroup$ [continued] Then I could still have a set $S$, but the definition of $S$ would be \begin{equation*} S = \{n \in \mathbb{N}: K \text{ contains an $n$th root of unity that is not primitive} \} \end{equation*} $\endgroup$ – Sam Y. May 16 '17 at 23:51
  • $\begingroup$ @Sam, for every $N$ there is a finite number of roots of unity of order bouned by $N$. $\endgroup$ – Mariano Suárez-Álvarez May 17 '17 at 2:49
  • $\begingroup$ Thanks for taking the time to reply. Unfortunately, I'm still a bit lost. Could you elaborate a bit more on the sentence: "for every $N$, there is a finite number of roots of unity of order bounded by $N$"? I do not understand this statement. Should there be an additional word between "order" and "bounded"? $\endgroup$ – Sam Y. May 17 '17 at 17:14
  • $\begingroup$ No, there is no word missing. $\endgroup$ – Mariano Suárez-Álvarez May 18 '17 at 1:22

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