<s>[INST] All automorphisms of splitting fields Let $M \le \mathbb{C} $ be the splitting field of polynomial $ f(x)
 \in \mathbb{Q}[x] $. Find all automorphisms of field $ M $ in cases: 
1) $ f(x) = x^6 - 1 $ 
2) $ f(x) = x^{2011} - 1 $
1) In the first case I found all complex roots of the 6th degree of 1. It is $$ 1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, -\frac{1}{2} + i \frac{\sqrt{3}}{2}, -1, -\frac{1}{2} - i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2} $$
Hence splitting field is $ M = \mathbb{Q}(i\sqrt{3}) $ and $ \dim(\mathbb{Q}(i\sqrt{3}):\mathbb{Q}) = 2 $
All elements in $ M $ have the form $ a + b \cdot i\sqrt{3} $, where $ a, b \in \mathbb{Q} $
Let $f$ is our automorphism and $ f(i\sqrt{3}) = \alpha $.
$$ -3 = f(-3) = f((i\sqrt{3})^2) = \alpha^2 $$
And $ \alpha $ can be $ i\sqrt{3} $ or $ -i\sqrt{3} $.
If $ \alpha = i\sqrt{3} $ then it is identity automorphism and $f(a + b \cdot i\sqrt{3}) = a + b \cdot i\sqrt{3}$
If $ \alpha = -i\sqrt{3} $ then $f(a + b \cdot i\sqrt{3}) = a - b \cdot i\sqrt{3}$
And there are exactly 2 automorphisms. Is that correct solution?
2) But I have no solution in the second case. I have an idea to take one of the roots of 2011th degree of 1: $ \phi = \cos{\frac{2\pi}{2011}} + i\sin{\frac{2\pi}{2011}}$ and find degree of minimal irreducible polynomial for $ \phi $ for find $ n $ in form of element $ a = a_0 + a_1\phi + ...+a_n\phi^n $
And then write that $1 = f(1) = f(\phi^{2011}) = \alpha^{2011}$
But I have no finished solution...
Thanks for the help!
 [/INST] For $(2)$, we have that $2011$ is a prime number taking the thousand and eleventh root of unity, that is $\xi \neq 1$. Then $L = \mathbb{Q}[\xi]$ and as 
$$\xi^{2010}+\xi^{2009}+\ldots +\xi^2 + \xi + 1 = 0$$
and $p(x) = x^{2010}+x^{2009}+\ldots +x^2 + x + 1$ is irreducible*  over $\mathbb{Q}$ then $[L:\mathbb{Q}] = 2010$. 
To find the automorphism notice that $\sigma \in Aut_{\mathbb{Q}} L $ is complete determined by $\sigma(\xi)$. That is the possibilies are $\{\xi,\xi^2\,\ldots,\xi^{2010}\}$. 
(*)It's possible to prove for any $p$ prime that 
$$x^{p-1} + x^{p-2} + \ldots + x + 1$$
is irreducible over $\mathbb{Q}$. Show that $$q(x+1) = (x+1)^{p-1}+ (x+1)^{p-2}+\ldots+ (x+1)+ 1$$
is irreducible by using Eiseinstein Criterion. 
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