Determine the group of Automorphisms of $\mathbb{Q}(\zeta_8)$, where $\zeta_8=\exp(2\pi i/8)$

Question

I've been tasked with the following:

Let $\zeta_8=\exp(2\pi i/8)$. Compute the group of field automorphisms of $\mathbb{Q}(\zeta_8)$.

This seems easy enough, I assume that the problem reduces to showing that automorphisms of $\mathbb{Q}(\zeta_8)$ simply permute the roots of the polynomial $x^7+x^6+\cdots +x+1$, and hence it'll probably be isomorphic to $S_7$ or something similar.

However I've hit a snag: from my knowledge, we only know that a field automorphism $\sigma$ of $\mathbb{Q}(\zeta_8)$ permutes the roots of the cyclotomic polynomial if $\sigma$ fixes $\mathbb{Q}$. I don't quite know how to show this though. We've shown that the only automorphism of $\mathbb{Q}$ is the trivial automorphism, so I assume that ties in somewhere, but I can't see where

Any help is truly appreciated. If you can't tell, this is a homework exercises. So just hints if possible. Thanks in advance.

Edit: is it because we know that $\sigma(\mathbb{Q})\cong\mathbb{Q}/\ker\sigma=\mathbb{Q}$ for each $\sigma\in\text{Aut}\left(\mathbb{Q}(\zeta_8)\right)$? In this case how do we know that $\mathbb{Q}$ doesn't map onto an isomorphic copy of $\mathbb{Q}$ distinct from $\mathbb{Q}$?

Answer 1

If $K$ is any field containing $\mathbb{Q}$ and $f:K\to K$ is any automorphism, then $f(q)=q$ for all $q\in\mathbb{Q}$. First, $f(0)=0$ and $f(1)=1$ by definition of a homomorphism. It follows that $f(2)=f(1)+f(1)=1+1=2$, $f(3)=f(2)+f(1)=2+1=3$, and so on, so $f(n)=n$ for all $n\in\mathbb{N}$. Since $f$ preserves additive inverses it follows that $f(n)=n$ for all $n\in\mathbb{Z}$, and since $f$ preserves multiplicative inverses $f(1/n)=1/n$ for all nonzero $n\in\mathbb{Z}$. Thus if $n/m\in\mathbb{Q}$, $f(n/m)=f(n)f(1/m)=n/m$.

Answer 2

Let us start with the basics. Let $f: E\rightarrow E$ be a field isomorphism with $\mathbb{Q}\subset E$ being a subfield. Then we have $$ f(x+y)=f(x)+f(y); f(xy)=f(x)f(y) $$ It is now easy to see that $$ f(nx)=nf(x), f(1)=1 $$ Therefore we have $$ f(p*\frac{1}{p})=p*f(\frac{1}{p})=1\rightarrow f(\frac{1}{p})=\frac{1}{p} $$ and thus $f$ must leave $\mathbb{Q}$ unchanged. If you got stuck with why $f(1)$ must be $1$, think about $f(1^{n})=f(1)^{n}$ for any $n$.

Answer 3

$Q(2i\pi/8)$ is the splitting field of the irreducible polynomial $x^4+1=0$.

https://en.wikipedia.org/wiki/Cyclotomic_polynomial#Examples

So it is a Galois extension of degree 4. The elements of the Galois group are defined by $f_n(2il\pi/8)= 2i ln\pi/8$ where $n=1,3,5,7$. which is isomorphic to $Z/2\times Z/2$.

Answer 4

I just took this class last term, so I'm familiar but not necessarily fluent yet. It looks like you may already be past where we got so if I'm just ignorant of the distinction of Cyclotomic Polynomials feel free to say so.

However, here is where I think this is going.

1) Remember that $\mathbb{Q}(\zeta_8)$ is the extension field over $\Bbb Q$, which we are constructing using Kronecker's Theorem.

And recall that by earlier Proposition that for an extension field F over field K our automorphism has been constructed so that $\phi: F \to F\;\; s.t.\; \phi (a)=a$ for all $a \in K$

Then for $\Theta \in Gal(F/K)$, a further proposition shows

$\theta (f(u))=\theta (a_0 + a_1 u+a_2u^2+...a_nu^n)$

$=\theta (a_0) + \theta (a_1 u)+\theta (a_2u^2)+...\theta (a_nu^n)$

$=a_0 + a_1\theta (u)+a_2\theta (u^2)+...a_n\theta (u^n)$

This is the proof that our automorphism leaves elements of the base field fixed. We are only permutting roots in the extension field.

2) $\zeta_8=\exp(2\pi i/8)$ are the eighth roots of unity, which means the extension field is

{$a+bi+c$ ${\sqrt 2}\over 2$$i$ $| a,b,c$ $\in \Bbb Q$}

So $\pm {\sqrt 2}\over 2$$i$ and $\pm i$ are the roots which lie outside of $\Bbb Q$. This makes me think the correct answer will involve a group of 4 elements.

3) Consider the much simpler example of $\Bbb Q(i) \cong \Bbb Z_2$