Let $K$ be a field with $char(K)=0$, $f(x)\in K[x]$ an irreducible polynomial of degree $3$ and $L$ its splitting field. Let $\alpha\in L$ be a root of $f$. We want to show $L=K(\alpha, \sqrt{\Delta (f)})$, where $\Delta$ denotes a polynomial's discriminant.
As $char(K)=0$ and $f$ is irreducible, it must separable and hence all its roots are different. In other words, $\Delta (f)$ is not zero. It's also clear that $K(\alpha, \sqrt{\Delta (f)})\subset L$, as both $\alpha$ and $\sqrt{\Delta (f)}$ are in $L$, but how can I prove that these two elements are enough to generate the whole extension? I have tried to directly write the two other roots in terms of the discriminant and $\alpha$, but with no success.
Best Answer
So your polynomial is $f(x)=x^3+px^2+qx+r$ with $p,q,r\in K$, and its roots are $\alpha$, $\alpha_2$ and $\alpha_3$, and you've extended the field with $\alpha$ and with $\sqrt{\Delta(f)}=(\alpha-\alpha_2)(\alpha_3-\alpha)(\alpha_2-\alpha_3)$. How do you get $\alpha_2$ and $\alpha_3$?
Note that
$$\sqrt{\Delta(f)}=\alpha^2(\alpha_3-\alpha_2)-\alpha(\alpha_2+\alpha_3)(\alpha_3-\alpha_2)+\alpha_2\alpha_3(\alpha_3-\alpha_2)=(\alpha_3-\alpha_2)\left(\alpha^2+\alpha(p+\alpha)+\frac{r}{\alpha}\right)$$
(Vieta formulae) so $\alpha_3-\alpha_2\in K\left(\alpha,\sqrt{\Delta(f)}\right)$ and knowing that $\alpha_2+\alpha_3=-(p+\alpha)\in K\left(\alpha,\sqrt{\Delta(f)}\right)$ you find that both $\alpha_2,\alpha_3\in K\left(\alpha,\sqrt{\Delta(f)}\right)$.