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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!

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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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  • $\begingroup$ But $ x^5 + x^4 + x^3 + x^2 + x + 1 = (x+1)(x^2 + x + 1)(x^2 - x + 1) $ and not irreducible over $\mathbb{Q}$ $\endgroup$ – Stanislav Morozov Jan 25 '15 at 15:53
  • $\begingroup$ Yes, the number must be prime to do what I did. The first I you got it right. $\endgroup$ – Aaron Maroja Jan 25 '15 at 16:18

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