Galois group of splitting field of $X^{14}-tX^7+1 \in \mathbb{C}(t)[X]$ over $\mathbb{C}(t)$ I need to find the Galois group of the polynomial:
$$X^{14}-tX^7+1 \in \mathbb{C}(t)[X]$$
So far I know how to find the polynomial's roots. I put $ Y = X^7 $ and solve the equation, then take both roots and take all $7$ roots of each.
I'm not sure what to do next, and how to see where the roots are mapped to.
 A: First, we find the degree of the splitting field $ L $ over $ K = \mathbf C(t) $. The discriminant of the polynomial, viewed as quadratic in $ Y = X^7 $, is $ t^2 - 4 $, which is not a perfect square in $ K $. Writing $ \omega = \sqrt{t^2 - 4} $, we see that $ [K(\omega) : K] = 2 $. The roots of the polynomial are then the seventh roots of 
$$ \frac{t \pm \omega}{2} $$
However, we have
$$ \frac{t + \omega}{2} \cdot \frac{t - \omega}{2} = 1 $$
and since $ K $ contains the seventh roots of unity, the extension obtained by adjoining just one seventh root is normal: it is a Kummer extension. (Note that the seventh roots are not actually in $ K(\omega) $ - if you do not see why, prove this.) Denote one such root by $ \alpha $, then we have a tower of fields
$$ K \subset K(\omega) \subset K(\alpha) = L $$
where all extensions are Galois. This gives rise to a short exact sequence of Galois groups
$$ 0 \to C_7 \to \textrm{Gal}(L/K) \to C_2 \to 0 $$
We find that $ G = \textrm{Gal}(L/K) $ is a group of order $ 14 $. From general group theory, we know that there are only two such groups: $ C_{14} $ and $ D_7 $. Therefore, it suffices to ascertain whether $ G $ is abelian or not. Let $ \sigma \in G $ be the element fixing $ \omega $ and mapping $ \alpha \to \zeta \alpha $ for some primitive seventh root of unity $ \zeta $, and let $ \tau \in G $ be the element sending $ \alpha \to 1/\alpha $ and $ \omega \to -\omega $. (How do we know that these automorphisms exist?) Then, we have
$$ \sigma^{-1} \tau \sigma(\alpha) = \sigma^{-1} \tau(\zeta \alpha) = \sigma^{-1}(\zeta/\alpha) = \zeta^2/\alpha \neq \tau(\alpha) = 1/\alpha $$
This implies that $ \sigma $ and $ \tau $ do not commute, and thus $ G $ is not abelian: we have that $ G \cong D_7 $.
