I must show that any finite field extension $K\supset \mathbb Q$ can only contain finitely many roots of unity.

I reasoned in the following manner: Let $n<\infty$ be the degree of $K$ over $\mathbb Q$. Now, for any $\alpha\in K$ we must clearly have $\mathbb Q(\alpha)\subseteq K$. So the degree of $\alpha$ over $\mathbb Q$ is at most $n$. Hence, if we can show that there are only finitely many roots of unity of degree less than or equal to $n$, we are done.

Now, since any root of unity of degree less than or equal to $n$ is a primitive $k$'th root of unity for some $k\leq n$ it follows that $K$ can at most contain all roots of the finite list of cyclotomic polynomials $\Phi_1,\Phi_2,...,\Phi_n$ which each has finitely many roots. All together this will be finitely many, which proves the claim.

So my questions are:

  • Is this correct/enough?
  • When this exercise was on the blackboard in class a rather lengthy argument came up. So am I missing some important point here?

My final version

Having understood the point I missed before, I now see that it will suffice to argue that $\operatorname{deg}(\Phi_k)=\varphi(k)$ eventually exceeds $n$. We know that $$ \varphi(k)=\prod_ j\varphi(p_j^{a_j}) $$ where $p_j$ are the prime factors of $k$ (similarly to what Ewan Delanoy gave in his answer). Now first note that $\varphi(p^k)$ is stricly increasing both with the prime $p$ and the multiplicity $k$. So we may find a multiplicity $M$ such that $\varphi(2^M)>n$ thus implying $\varphi(p^M)>n$ for all primes. Also any prime $p>n+1$ must have $\varphi(p^k)\geq p-1>n$ for all $k$.

This shows that $\varphi(k)>n$ if one of the prime factors of $k$ is larger than $n+1$ or one of the multiplicities is at least $M$. This will certainly happen for $k$ large enough.

  • $\begingroup$ It looks just fine to me...Perhaps the exercise in class was lengthy since you had to prove stuff about the cyclotomic polynomials and etc. $\endgroup$ – DonAntonio Oct 2 '13 at 11:59
  • $\begingroup$ @DonAntonio: The cyclotomic polynomials were covered in the previous chapter already. It was a (usually quite bright) student elaborating on divisors of $n$ and such things. The lecturer also suggested we could find a much less conservative bound than that, still he mentioned induction. $\endgroup$ – String Oct 2 '13 at 12:08
  • $\begingroup$ But if the above holds then why use induction? My argument immediately shows that there will be less than $n!$ roots of unity in $K$. $\endgroup$ – String Oct 2 '13 at 12:09
  • $\begingroup$ 'Any root of degree $\leq$ n is a primitive $k$ root for some $k \leq n$' : this is wrong (consider $j=\exp(2i\pi/3)$). $\endgroup$ – user10676 Oct 2 '13 at 12:23
  • $\begingroup$ I don't get your point, @user10676: the root $\;j\;$ of degree three indeed is a primitive root of degree $\;3\le 3\;$... $\endgroup$ – DonAntonio Oct 2 '13 at 12:33

Your proof is correct except on one point.

It is not true that “a root of unity of degree $\leq n$ is a primitive $k$-th root of unity for $k\leq n$.”, Indeed, $z=exp(\frac{2\pi i}{3})$ has degree $\leq 2$ but is a primitive $3$-rd of unity (as noted in user10676’s comment).

In general, a $k$-th primitive root of unity has degree $\phi(k)$ where $\phi$ is Euler’s totient function.

What we need to show is that for any integer $M\gt 0$ the set $X=\lbrace k | \phi(k) \leq M\rbrace$ is finite.

We will use the well-known formula :

$$ \phi(\prod_{j}p_j^{a_j})=\prod_{j}(p_j-1)p_j^{a_j-1} \tag{1} $$

Let $p$ be the smallest prime $ > M$. If $k$ has a prime divisor $q$ that is $> p$, then by (1) $q-1$ divides $\phi(x)$, so $\phi(x) \geq q-1 >M$. So if $x\in X$, $x$ can only be divisible by the primes $p_1,p_2, \ldots ,p_r$ that are $\leq p$. So we can write

$$ x=p_1^{a_1}p_2^{a_2} \ldots p_r^{a_r} $$

Now for each $j$, $p_j^{a_j-1}$ divides $\phi(x)$ by (1), so $\phi(k) \geq p_j^{a_j-1}$. We deduce the bound $p_j^{a_j-1} \leq M$, and hence $a_j \leq 1+\frac{\log(M)}{\log(p_j)}$. So there are at most $2+\frac{\log(M)}{\log(p_j)}$ values for the exponent $a_j$.

This shows that $X$ is finite : there are at most

$$ N=\prod_{j=1}^{r} ( 2+\frac{\log(M)}{\log(p_j)}) $$

elements in $X$.

  • 1
    $\begingroup$ Yeah, that's the point, Ewan:+1. $\endgroup$ – Georges Elencwajg Oct 2 '13 at 12:50
  • $\begingroup$ That makes sense! It was a stupid mistake of me to 'forget' that the degree of $\Phi_k$ is actually $\varphi(k)$. $\endgroup$ – String Oct 2 '13 at 20:50
  • $\begingroup$ Finiteness of $\{k:\phi(k)<M\}$ follows from $\phi(k)\gg k/\log\log k$. $\endgroup$ – Sungjin Kim Mar 21 '14 at 17:10

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