Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$ Please, help me to understand this problem:
Let $\alpha=\sqrt[3]{2}$ be a root of the polynomial $x^3-2$. 
a) Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$;
b) If $L$ is the Galois Field of $x^3-2$ over $\mathbb Q$ ($Gal(x^3-2,\mathbb Q))$. Prove that the number of automorphisms in $L$ equals $6$ $(|Aut L|=6)$.
My solution:
a) I claim that there exists only one authomorphism in $\mathbb {Q}[\alpha]$: the identity function. In fact, an arbitrary element of $\mathbb{Q}[\alpha]$ is of the form $a_0+a_1\alpha+a_2\alpha^2$. Then , if $f$ is an automorphism, then $f(a_0+a_1\alpha+a_2\alpha^2)=a_0+a_1f(\alpha)+a_2f(\alpha^2)=a_0+a_1\alpha+a_2\alpha^2$.
Now, if I try to apply this method in "b", it fails. A generic element of $L$ is of the form $a_0+a_1\alpha+a_2\alpha^2+a_3u+a_4\alpha u+a_5\alpha^2u$, where "u" is the primitive 3th root of unity. Then $f(a_0+a_1\alpha+a_2 \alpha^2 +a_3u+a_4\alpha u+a_5\alpha^2 u)=a_0+a_1\alpha+a_2\alpha^2+a_3f(u)+a_4\alpha f(u)+a_5\alpha^2 f(u)$. There is no way to get 6 automorphisms from here. By the way: what is $f(u)$?
 A: For (a), I'm not quite sure whether your proof is correct, as you seem to have assume that $f$ is the identity in the proof. Basically, if you want $f$ to fix the rationals, then $f(2)=2$. But $f(\alpha)^3=f(\alpha^3)=2,$ so $f(\alpha)=\alpha$, which forces $f$ to be the identity on everything else by the properties of automorphisms.
For b, note that the roots are $\alpha, \omega \alpha, \omega^2 \alpha$, and check that any permutation of the roots by $f$ makes $f$ into an automorphism of $\Bbb{Q}[\alpha, \omega]$ (do you want me to write down explicitly the check?). Thus $\text{Aut}L=S_3$.
A: In a) how do you make the assumption that $f(\alpha)=\alpha$? In b) you are asked to compute the Galois group of $L$ over $\mathbb{Q}$.
Hint for both parts of the exercise: First show that any automorphism of the fields in question fixes $\mathbb{Q}$. Then try to find out where it could possibly map $\alpha$ and $u$; since $\alpha$ and $u$ satisfy a polynomial equation over $\mathbb{Q}$ the same is true for $\phi(\alpha)$ and $\phi(u)$.
