# Splitting field of a polynomial over $\mathbb{C}(t)$

Consider the polynomial $f=X^{3}-2tX+t$ over $K=\mathbb{C}(t)$, where $t$ is transcendental over $\mathbb{C}$. Let $\alpha$ be a root of $f$ and $L$ be the splitting field of $f$. I proved that $K(\alpha) \subsetneq L=K(\alpha, \beta)$, where $\beta$ is a root of the polynomial $X^{2}+3\alpha^{2}-8t \in K(\alpha)[X]$. So the degree $[L:K]$ must be 6.

I am asked to show further that $Gal(L/K) \simeq S_{3}$ and to determine a a polynomial $g \in K[X]$ such that $M$ is the splitting field of $g$, where $M$ is the unique intermediate field of the extension $L/K$ of degree 2. Additionally, is $M/K$ Galois? From what I proved so far, I know that the Galois group of $f$ is either $\mathbb{Z}/6\mathbb{Z}$ or $S_{3}$, but how do I show that it can't be abelian? Also, I dont know how to answer the other questions.

I would appreciate any help. Thank you!

The galois group permutes the roots of the polynomial, and an element is completely determined the images of the roots, so it must be a subgroup of $S_3$, as it has order $6$, it is the whole of $S_3$. Alternatively, if the extension was abelian, every subgroup the Galois group would be normal, hence by Galois correspondence, every subextension would be Galois, but from what you computed, $K(\alpha)$ is not Galois as it does not contain all roots of $f$.