Did I find the splitting field of $x^3-3x+1$ over $\Bbb Q$?

Question

I want to find the splitting field of $x^3-3x+1$ over $\Bbb Q$. I think I've got it but I'm not sure …

Here's what I did :

Using the formula for cubic roots I said that

$$x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}=\dfrac{3\pm\sqrt6}{3}=\dfrac {1\pm\sqrt{2}}{\sqrt3}.$$

The field $\Bbb Q(\sqrt2/\sqrt{3})$ contains both roots so the splitting field is a subfield of $\Bbb Q(\sqrt2/\sqrt{3})$. The polynomial $x^2-\frac23$ is irreducible over $\Bbb Q$. So $|\Bbb Q(\sqrt{2/3}):\Bbb Q|=2.$

So $\Bbb Q (\sqrt2/\sqrt{3})$ is the splitting field of $x^3-3x+1$ with degree $2$.

Is this correct ?

Answer 1

As the polynomial is of degree 3 and having no roots in field of rationals so the polynomial is irreducible over rationals, so degree of splitting field can not be less than 3 over rationals.To find the roots of the cubic , we can use cardan method and so the splitting field.