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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 $$\text{Gal}(L|\mathbb{Q[z]})=\text{Aut}(L|\mathbb{Q[z]}) \cong A_3, \ \text{with} \ [\mathbb{Q}[z]:\mathbb{Q}]=2.$$

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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.

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    $\begingroup$ @user152623 It has to be $2$, since $[L: \Bbb Q(z)] = 3$, $[L:\Bbb Q] = 6$ and $[L:\Bbb Q] = [L:\Bbb Q(z)]\cdot [\Bbb Q(z):\Bbb Q]$ $\endgroup$ – Arthur May 23 '14 at 13:25
  • $\begingroup$ @user152623 Exactly as Arthur said, this follows immediately from the field theory; if needed, I could add an explanation. Thanks for your attention. $\endgroup$ – awllower May 23 '14 at 15:00

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