# Roots of Unity in fields

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.

-

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.

-

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.

-
well, yes, this is the answer. But probably the OP is seeking a method of proof? –  Pete L. Clark Apr 13 '11 at 1:32
@ Gerry thanks. I understand that. Is that enough to answer the question? For the fields Q(sqrt -2) and Q(sqrt -3), again your correct..I just have to have all of them, that is where I am stuck. I don't know how to get those roots of unity and show why. –  user8771 Apr 13 '11 at 1:41
@ pete. yes I will need some sort of method of proof. Once I can find those roots of unity in the last two fields. –  user8771 Apr 13 '11 at 1:45
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. –  Gerry Myerson Apr 13 '11 at 4:54
@GerryMyerson Please see my answer above. –  user38268 Jul 21 '12 at 7:15

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 $U_n\cap \mathbb{Q}(\sqrt{-3})$?

-
@ alex, thanks for your hint. I just don't know how to get those roots of unity in the fields Q(sqrt -2) and Q(sqrt -3). Once I understand how to get them, then I see where your hint helps out. –  user8771 Apr 13 '11 at 1:43
@user8771: the point is that you're not getting them in the fields $Q(\sqrt{-2})$ and $Q(\sqrt{-3}); you're getting them in the larger field. For all the roots of unity in the larger field, you should be able to write them out in terms of 'simplest terms' - that will help you see what is in each of your subfields. – Steven Stadnicki Apr 13 '11 at 3:24 The first three are easy enough so I will tackle the last two. Recall that the degree of the extension$\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 isimpossible. Now let us write out$\zeta_n$for these values of$n$, we have:$\zeta_2 = \pm 1$,$\zeta_3 = \frac{-1 + \sqrt{3}i}{2}$,$\zeta_4 = \pm 1, \pm i$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.

-