Primitive element for splitting field of irreducible polynomial $x^3+18x+1$ I am dealing with an irreducible polynomial $p(x)=x^3+18x+1$ over $\Bbb Q[x]$. Using Cardano's formula I have found the roots as
$$r_1=p+q$$
$$r_2=\omega p+(\omega^2)q$$
$$r_3=(\omega^2)p+\omega q$$
where $\omega$ is a primitive cube root of unity, 
$p=\sqrt[3]{-\frac12+\sqrt{216+\frac14}}$ and $q=\sqrt[3]{-\frac12-\sqrt{216+\frac14}}$.
$\Bbb Q[r_1,r_2,r_3]$ is the splitting field of $p(x)$ and its extension over $\Bbb Q$ is normal because it is the splitting field of $p(x)$ over $\Bbb Q$.
I know that the Galois group $\text{Gal}(\Bbb Q[r_1,r_2,r_3]/\Bbb Q)$ has 6 elements, or in other words, the degree of the splitting field over $\Bbb Q$ is 6. But I am not able to find a primitive element t such that $[\Bbb Q[t]:\Bbb Q]=6$. Please guide me.   
 A: The discriminant of the polynomial is $ -23355 $, which is not a perfect square, therefore the splitting field is of degree $ 6 $ over $ \mathbf Q $. In particular, if $ \lambda $ is a root of $ x^3 + 18x + 1 $ and $ \omega = \sqrt{-23355} $, then the splitting field is $ L = \mathbf Q(\lambda, \omega) $.
Assume that an automorphism $ \sigma $ in $ G(L/\mathbf Q) \cong S_3 $ fixed $ \lambda + \omega $. Then, we would have
$ \lambda - \lambda' = \omega' - \omega = q $ 
for some conjugates $ \lambda', \omega' $ of resp. $ \lambda, \omega $. Assume that $ q \neq 0 $ so that $ \lambda \neq \lambda' $, and let $ \lambda_1, \lambda_2, \lambda_3 $ be the roots of $ x^3 + 18x + 1 $. Without loss of generality, assume $ \lambda = \lambda_1 $ and $\lambda' = \lambda_2 $. We check that the following are all distinct conjugates of $ \lambda_1 - \lambda_2 $:
$$ \lambda_1 - \lambda_2, \lambda_2 - \lambda_1, \lambda_3 - \lambda_2 $$
The only non-obvious inequality is the inequality of the second and the third conjugates. Assume that they were equal, then $ 2 \lambda_2 = \lambda_1 + \lambda_2 $ and $ 3\lambda_2 = \lambda_1 + \lambda_2 + \lambda_3 = 0 $, so $ \lambda_2 = 0 $, which is absurd. We have shown that $ \lambda - \lambda' $ has at least 3 distinct $ \mathbf Q $-conjugates, whereas $ \omega' - \omega $ can only have 2, which is the desired contradiction.
Therefore, the only automorphism fixing $ \lambda + \omega $ is the identity, and it is a primitive element, i.e $ L = \mathbf Q(\lambda + \omega) $.
