Galois Group of a cubic polynomial

Question

Let $P=X^3+a_1X^2+a_2X+a_3 \in \mathbb{Q}[X]$ be irreducible, $x_1,x_2,x_3$ the roots of $P$ and $L:=\mathbb{Q}[x_1,x_2,x_3]$. The galois group of $P$ is isomorph to $S_3$.

Now we define $z:=(x_1-x_2)(x_1-x_3)(x_2-x_3)$. How can I show that $$\text{Gal}(L|\mathbb{Q[z]})=\text{Aut}(L|\mathbb{Q[z]}) \cong A_3, \ \text{with} \ [\mathbb{Q}[z]:\mathbb{Q}]=2.$$

Answer 1

It is easy to check that $\mathbb Q(z)$ is fixed by $A_3,$ and hence $A_3\subset \text{Gal}(L/\mathbb Q(z)).$
Finally, you can show that the Galois group is not the whole $S _3,$ by finding an automorphism not fixing $z.$ This finishes the proof.
Hope this helps.