1
$\begingroup$

Which roots of unity are contained in the fields: $\mathbb{Q}[i]$, $\mathbb{Q}[\sqrt2]$, $\mathbb{Q}[\sqrt3]$, $\mathbb{Q}[\sqrt5]$, $\mathbb{Q}[\sqrt{-2}]$ and $\mathbb{Q}[\sqrt{-3}]$?

I know that the roots of unity in $\mathbb{Q}[i]$ are $1$, $-1$, $i$, and $-i$. I'm having a hard time finding the roots of unity in the other fields. If anyone could offer any advice, it would be greatly appreciated.

4 Answers 4

6

The first three are easy enough so I will tackle the last two. Recall that the degree of the extension $\Bbb{Q}(\zeta_n)/\Bbb{Q}$ is $\varphi(n)$ where $\varphi$ is the Euler Totient Function. Now it is not hard to see that the only values of $n$ for which $\varphi(n) = 2$ is when $n = 2,3,4$ and $6$.

Now when you look at $\Bbb{Q}(\sqrt{-2})$ and $\Bbb{Q}(\sqrt{-3})$, if you have an $n^{th}$ root of unity in there it can only be for those stipulated values of $n$ above, because otherwise you have a $\Bbb{Q}$ - subspace of dimension greater than 2 sitting inside of a $\Bbb{Q}$ - vector space of dimension 2 which is impossible. Now let us write out $\zeta_n$ for these values of $n$, we have: $\zeta_2 = - 1$, $\zeta_3 = \frac{-1 + \sqrt{3}i}{2}$, $\zeta_4 = i$ and $\zeta_6 = \frac{1 + \sqrt{3}i}{2}.$

Can you now complete your problem? I leave the rest for you since this is a homework problem. By applying degree arguments, etc. you should be able to eliminate cases. For example, you should be able to work out for yourself why $\pm i$ is not in $\Bbb{Q}(\sqrt{-3})$ say.

  • 0
    $n=2$ cannot make $\phi(n)=2$, I think you mean, we have to have $\phi(n)\leq 2$, so that we have those four integers, (as we do not really care about $n=1$)2018-04-21
4

Hint: What is a big field that $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{5})$ are all subfields of? You should know what this big field's roots of unity are, and then the roots of unity in each of these three fields will have to be among the roots of unity of this big field.

Hint: For $\mathbb{Q}(\sqrt{-3})$, you should write out what the cube root of unity $e^{2\pi i/3}$ is using Euler's identity.

3

The only roots of unity in ${\bf Q}(\sqrt d)$ are $1$ and $-1$, with the following exceptions: if $d=-1$, you get 4th roots, and if $d=-3$, you get 6th roots. I'm assuming $d$ is squarefree.

  • 0
    How to prove it depends on what you know. If you know that every root of unity is an algebraic integer, and that the algebraic integers in ${\bf Q}(\sqrt{-2})$ are the numbers $a+bi\sqrt2$ where $a$ and $b$ are integers, then the square of the modulus of $a+bi\sqrt2$ is $a^2+2b^2$, and the only way to make that 1 is to take $a=\pm1$, $b=0$. But maybe this is using things you haven't done yet.2011-04-13
2

Hint: $\mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{5})$ are all subfields of $\mathbb{R}$, so any root of unity in them is a root of unity in $\mathbb{R}$. What are the roots of unity in $\mathbb{R}$?

For the other two, they are both subfields of $\mathbb{C}$ and so any root of unity in them is a root of unity in $\mathbb{C}$. Let $U_n$ be the set of $n^{th}$ roots of unity of $\mathbb{C}$. What are the elements of $U_n$ (for arbitrary $n$)? What are the elements of $U_n\cap \mathbb{Q}(\sqrt{-2})$ and $U_n\cap \mathbb{Q}(\sqrt{-3})$?

  • 0
    @user8771: the point is that you're not getting them in the fields $Q(\sqrt{-2})$ and $Q(\sqrt{-3}); you're getting them in the larger field. For all the roots of unity in the larger field, you should be able to write them out in terms of 'simplest terms' - that will help you see what is in each of your subfields.2011-04-13