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This is exercise 13.6.5 from Dummit and Foote.

Let me first note that I realize there is already a post on this question here. However, I am a bit confused about the explanation given in that post, and I want to discuss my confusion about it here.

The following is the beginning of an argument made by the author (@Mariano Suarez-Alvarez) of the answer to the question I linked to.

  • Suppose $K$ is a number field of degree $d$ which contains infinitely many roots of unity. It then contains roots of unity of arbitrarily high order (because 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

I am fine with everything until the line: and therefore if contains roots of cyclotomic polynomials $\Phi_n$ for infinitely many $n$. It seems like the author is assuming that the following implication holds: \begin{equation*} \Big[ K \text{ contains roots of $x^n - 1$ for arbitrarily large $n$} \Big] \implies \Big[ K \text{ contains roots of $\Phi_n(x)$ for arbitrarily large $n$} \Big] \end{equation*}

While the implication certainly might hold, I don’t understand why it does. Why couldn’t I make an infinite set of $n$th roots of unity in such a way that I explicitly avoid primitive $n$th roots of unity, i.e. something like: \begin{equation*} A = \{ \zeta \in \mathbb{C}: \zeta \text{ is an $n$th root of unity that is not a primitive $n$th root of unity} \} \end{equation*}

It is not obvious to me why $A$ could not be contained in a finite field extension $K/\mathbb{Q}$.

Any thoughts on this would be much appreciated.


EDIT:

Perhaps some more context might be helpful: The paragraph I quoted above is only the beginning part of the proof. The remainder of the proof that $K$ must be a finite extension of $\mathbb{Q}$ relies on there being a primitive $m$th root of unity in $K$, where $\phi(m) > [K: \mathbb{Q}]$. Once it is proved that $K$ contains this primitive $m$th root of unity, call it $\alpha$, then we get that $[\mathbb{Q}(\alpha): \mathbb{Q}] = \phi(m) > n = [K : \mathbb{Q}]$, which is an obvious contradiction.

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  • $\begingroup$ have you tried math.stackexchange.com/questions/512112/…? $\endgroup$ – Bogdan Simeonov May 18 '17 at 16:36
  • $\begingroup$ @Bogdan Yes, I have looked at that question. I think that one uses a different approach than the question I have linked to. I am more interested in understanding why the particular approach I have outlined above works than in finding any solution that works. But thanks for taking the time to mention this link! $\endgroup$ – Sam Y. May 18 '17 at 16:42
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Every $n$-th root of unity is a primitive $m$-th root of unity for some $m$ (which divides $n$). There are only finitely many primitive $m$-th roots of unity for any given $m$. So if you have infinitely many roots of unity, there cannot be a bound on their orders.

But there's another standard argument that a number field has finitely many roots of unity. Let $d$ be its degree. The characteristic polynomial of a root of unity in $k$ has bounded $r$-th coefficient (bounded by $\binom{d}{r}$). So there are finitely many such polynomials, and so only finitely many roots of unity in $K$.

ADDED IN EDIT

To be more explicit, let $X_m$ be the set of primitive $m$-th roots of unity. Each $X_m$ is a finite set. Now $\bigcup_{m=1}^\infty X_m$ is the set of all roots of unity. If $K$ has infinitely many roots of unity, they cannot all lie in $\bigcup_{m=1}^M X_m$, for that is a finite set, so $K$ would have elements in $X_m$ for infinitely many $m$.

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  • $\begingroup$ I think your first sentence is really the heart of my confusion. Suppose that $\zeta$ is a root of $x^n - 1$. Could you elaborate a little more on why we can find an $m | n$ such that $\zeta$ is a root of $\Phi_m(x)$? $\endgroup$ – Sam Y. May 18 '17 at 16:40
  • $\begingroup$ @SamY. $m$ is the order of $\zeta$: the least number with $\zeta^m=1$. $\endgroup$ – Lord Shark the Unknown May 18 '17 at 16:41
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    $\begingroup$ I feel a bit embarrassed now. That is very simple. I'm sorry to bother you with such a triviality. $\endgroup$ – Sam Y. May 18 '17 at 16:44
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    $\begingroup$ Sorry to bother you again...I thought I understood perhaps a bit prematurely. I agree that if $K$ contains infinitely many roots of unity, then $K$ contains $n$th roots of unity for arbitrarily large $n$. And I also agree that for each $n$th root of unity - call this $n$th root of unity $\alpha_n$ - we have that $\alpha_n$ is a primitive $m$th root of unity for some $m | n$, i.e. $\alpha_n = \zeta_m$. We want to argue that the set $\{m \in \mathbb{N}: \zeta_m \text{ is a primitive $m$th root of unity}\}$ is unbounded. This is the part where I am lost... $\endgroup$ – Sam Y. May 18 '17 at 17:37
  • $\begingroup$ [continued] I am confused because there could be two distinct $n$th roots of unity - call them $\alpha_{n_1}$ and $\alpha_{n_2}$ that both "correspond" to the same $\zeta_m$, i.e. $\alpha_{n_1}$ and $\alpha_{n_2}$ are both primitive $m$th roots of unity for the same $m$, even though $n_1 \neq n_2$. So I am worried that although the set $\{n \in \mathbb{N}: \alpha \text{ is an $n$th root of unity} \}$ is unbounded, there might be so much "collapsing" that the set $\{m \in \mathbb{N}: \zeta_m \text{ is a primitive $m$th root of unity} \}$ is bounded. $\endgroup$ – Sam Y. May 18 '17 at 17:41

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