# 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.

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 $G \cong D_7$.

• Thanks!!! What do you mean extension obtained by adjoining just one seventh root is normal ? – user378396 Oct 14 '16 at 0:15
• I am referring to a seventh root of $(t + \omega)/2$. Since $K$ contains the seventh roots of unity, this is a Kummer extension of $K(\omega)$. From the relation I've posted, this also implies that such an extension contains all seventh roots of $(t - \omega)/2$ (they are just the reciprocals), so it contains all of the roots of $X^{14} - t X^7 + 1$, and is the smallest extension to do so: it is the splitting field, thus normal. – Starfall Oct 14 '16 at 0:17
• Thanks! I'm sorry but I have bad english. What do you mean by "Short exact sequence of Galois groups" ? And what that graph you put in means ? – user378396 Oct 14 '16 at 0:20
• And what is the group $C14$ ? Cyclic of 14 ? – user378396 Oct 14 '16 at 0:23
• @user378396 Yes, that is correct. – Starfall Oct 14 '16 at 0:25