# 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 $\alpha=\sqrt{2}$ 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 $f(u)$?

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 $\text{Aut}L=S_3$.
• Hello, Alyosha! I understand your explanation for item (a). For item (b), do you mean I must prove that, for example, $f(\alpha)=\omega \alpha$? – Walter r Oct 24 '14 at 17:09
• @Walterr I mean that one of $f(\alpha)=\alpha, f(\alpha)=\omega \alpha, f(\alpha)=\omega^2 \alpha$ will occur, and for each of these options you have $2$ choices as to what $f(\alpha \omega)$ is (which, along with $f(\alpha)$, determines $f(\omega^2 \alpha)$). – Meow Oct 24 '14 at 17:14
• The problem is that $L$ contais not only the roots of the polynomial, but also products of powers of these roots. A basis of $L$ has 6 elements. – Walter r Oct 24 '14 at 17:25
• @Walterr if $\alpha, \beta, \gamma$ are the roots, then $f(a_1 \alpha+a_2 \alpha^2 +b_1 \beta +b_2 \beta^2 +c_1 \gamma +c_2 \gamma^2+d)=a_1 f(\alpha)+a_2 f(\alpha)^2+b_1 f(\beta)+b_2 f(\beta)^2 +c_1 f(\gamma)+c_2 f(\gamma)^2+d.$ Thus $f(l)$ for any $l \in L$ is determined by the action of $f$ on the three roots. – Meow Oct 24 '14 at 18:31
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 $\phi(u)$.