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!

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.

• 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 $\mathbb{Q}$ – Stanislav Morozov Jan 25 '15 at 15:53
• Yes, the number must be prime to do what I did. The first I you got it right. – Aaron Maroja Jan 25 '15 at 16:18