What are the automorphisme that permute roots of $X^3-1$? What is $\text{Gal}(K/\mathbb Q)$?

Question

1) What are the automorphisms that permute the roots of $X^3-1$ ?

Let denote $\alpha =e^{2i\pi/3}$. To me, there is $$id, (1\ \alpha ), \ (\alpha \ \alpha ^2),\ (1\ \alpha ^2), \ (1 \ \alpha\ \alpha ^2)\quad \text{and}\quad (1\ \alpha^2 \ \alpha),$$

and thus, if $K$ is the splitting field of $X^3-1$, then $\text{Gal}(K/\mathbb Q)\cong S_3$.

Is it correct ?

2) Why an automorphism of $\mathbb Q(\sqrt 2,i)\longrightarrow \mathbb Q(i,\sqrt 2)$ can't send $i\longmapsto \sqrt 2$ ?

Answer 1

1. Gal$(K/\mathbb{Q}) = \mathbb{Z}/2\mathbb{Z}$. Any element of $\mathbb{Q}$ must be fixed by any automorphism in the Galois group, so the "automorphisms" $$ (1 \>\alpha) (1 \>\alpha^2) (1 \>\alpha \>\alpha^2) (1 \>\alpha^2 \>\alpha)$$ cannot actually be automorphisms since they do not fix $1$. That leaves only two possible automorphisms left, so the Galois group cannot be larger than $\mathbb{Z}/2\mathbb{Z}$. It cannot be trivial since $K \neq \mathbb{Q}$, so the Galois group must be exactly $\mathbb{Z}/2\mathbb{Z}$.

2. If $\phi$ is an automorphism and $x^2 + 1 = 0$, $\phi(x)^2 + 1 = 0$, so $\phi(x)$ can only be $i$ or $-i$. In general, automorphisms permute root sets of polynomials.

Answer 2

1) The set of roots of any polynomial is fixed under an automorphism. In this case, since the roots are $\alpha$, $\alpha^2$, and $1$ each automorphism, $\varphi$, is determined by the image of $\alpha$.

That is $\varphi(\alpha^n)=[\varphi(\alpha)]^n$.

Since there are two possible images of $\alpha$ we have only two automorphisms and $\operatorname{Gal}(K/\mathbb{Q})\cong\mathbb{Z}/2\mathbb{Z}$.

2) Since an automorphism must fix the set of roots of a polynomial we have that $\sqrt{2}$ must be sent to another root of $x^2-2$ and $i$ must be sent to another root of $x^2+1$.