# Galois extension and Galois group

Let $x$ be a real root of the polynomial $X^3-X+1$, and $y,\overline{y}$ two other roots in $\mathbb{C}$, and $K$ be the cubic field $\mathbb{Q}[x]$.

Show that $y+\overline{y}=-x$ , $y\overline{y}=-1/x$, and $[(y-x)(\overline{y}-x)(y-\overline{y})]^2=-23$.

Show that $L=K[\sqrt{-23}]=\mathbb{Q}[x,y,\overline{y}]$ and that field is a Galois extension of degree $6$ of $\mathbb{Q}$. Determine its Galois group $G$ and subfields of $L$ which are Galois over $\mathbb{Q}$.

I don't now how to solve this. I need help.

• Are you familiar with what some people call Vieta relations? If $$p(X)=(X-x_1)(X-x_2)(X-x_3)=X^3-a_1X^2+a_2X-a_3,$$ then $a_1=x_1+x_2+x_3$, $a_2=x_1x_2+\cdots$, $a_3=x_1x_2x_3$. Jun 16, 2015 at 18:28
• Ok. I solved the first part. I also use that discriminant $D$ for polynomial is $[\sqcap _{i<j}(x_i -x_j)]^2$ (by Vandermonde). On the other side discriminant is $-27\cdot 1 - 4\cdot (-1)^3=-23$ so I can conclude that $x,y, \overline{y}$ is base.
– JJMM
Jun 17, 2015 at 9:01
• Because polynomial is irreducible and the discriminat is not square in $Q$, Galois group is $S_3$.
– JJMM
Jun 17, 2015 at 9:49
• You're getting there. Next: what do you know about intermediate fields of a Galois extension? When are they Galois over the small field themselves? Jun 17, 2015 at 10:12

Hint: Every extension that contains a complex number has even order, since if it contains $i$ it can be written as $\mathbb Q (\alpha, i) / \mathbb Q (i) / \mathbb Q$. By the tower law, the order of this field over $\mathbb Q$ is $[\mathbb Q (\alpha, i) : \mathbb Q(i)] [\mathbb Q (i) : \mathbb Q]$. Since $[\mathbb Q(i) : \mathbb Q] = 2$, the order of the extension is even.