# 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.

• I find "cardon's formula" and "galios group" quite funny ;-) Commented Aug 21, 2016 at 13:49
• The name is Cardano, and he published results by Tartaglia and del Ferro. The name of the course from which you're taking this problem is Galois Theory. Besides this tiny picking, please do try to follow the easy instructions to properly write mathematics in this site. Commented Aug 21, 2016 at 13:54
• @watson: `Cardoon's formula’ in English – a stinging formula, for sure :o) Commented Aug 21, 2016 at 13:55

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 $L = \mathbf Q(\lambda + \omega)$.