 Solution

The rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant iswhich is not a square in $\mathbb{Q}$. The Galois group of an irreducible cubic therefore gives

Quadratic subfields correspond to order-three subgroups. The unique such subgroup is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is

Let the roots be $\alpha ,\beta ,\gamma$ and choose
$\delta =(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$, so ${\delta }^2=-23$. Put $\theta =\alpha +\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta -\alpha =2\delta$. Cancelling the factor $\beta -\alpha$ in the expression for $\delta$ would giveBut $f^{\prime }(\gamma )=3{\gamma }^2-1$, so this would make ${\gamma }^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal polynomial. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so