Galois group and degree of splitting field over complex rational functions. Suppose $F=\mathbb C (t)$ the field of rational functions over $\mathbb C$.  let \begin{equation*}f(x)=x^6-t^2\in F[x]\end{equation*}Denote $K$ as the spliting field of $f$ over $F$. I'm trying to calculate $[K:F]$ and $\textrm{Gal}(K/F)$. I noticed that \begin{equation*}f(x)= (x^3-t)(x^3+t)\end{equation*}
but I'm not sure how to determine $[K:F]$. also, why is it obvious that $t^\frac13\not\in F$?
 A: For your last question, let $u = v/w \in F$ in irreducible form. Then $u^3 = v^3/w^3$ is also in irreducible form, and can't equal $t$ (or $-t$) because the degree of its numerator is a multiple of $3$.
Now if $\alpha$ is a root of $f$, then so is $-\alpha$ because $(-\alpha)^3 = -\alpha^3$ so if $\alpha$ is a root of one factor, then $-\alpha$ is a root of the other. Hence $K$ is the splitting field of $x^3-t$ (or $x^3+t$). Now, if $\alpha$ is a root of $x^3-t$, then so are $\alpha e^{2i\pi/3}$ and $\alpha e^{4i\pi/3}$, so $K = F(\alpha)$ and $[K:F] = 3$. And thus $\mathrm{Gal}(K/F)$ has order $3$, and is isomorphic to $\mathbf{Z}/3\mathbf{Z}$.
A: First notice that $g(x) = x^3-t$ is irreducible over $\mathbb{C}$, since it's Eisenstein with respect to the prime ideal $\langle t \rangle$.  But this means it's also irreducible over $F$ by Gauss's lemma.  All the roots of $g(x)$ are third roots of unity times $t^{1/3}$, so the splitting field of $g(x)$ is $F(t^{1/3})$, since the roots of unity are already in $\mathbb{C}$.  Since this is a degree three extension of $F$ which contains all the roots of $g(x)$.  But the roots of $h(x) = x^3+t$ are just the opposites of the roots of $g(x)$, so they are already in the field we have just created.  Therefore the splitting field of $F(x)$ is just $K = F(t^{1/3})$.
Now it's easy to say that the Galois group is just cyclic of order three, since $[K:F] = 3$, and there is only one group of order 3.
