Splitting field of $X^5-t$ in $\mathbb{Q}(t)[X]$ 
Given $K = \mathbb{Q}(t)$, $f(X) = X^5-t \in K[X]$ with $t$ trancedental over $\mathbb{Q}$.
a) Is $f$ irreducible?
b) Determine the degree of the splitting field of $f$ over $K$.
c) Determine the Galois group of $f$ over $K$.

My solutions so far:
a) I was thinking, $f$ is irreducible, since it is an eistenstein polynomial at $t$.
b) $\zeta^k\sqrt[5]{t}$ are the roots of $f$ for $k \in \{0,1,2,3,4\}$.
Let $L$ the splitting field of $f$ over $K$. So $\sqrt[5]{t}$ is a element of the splitting field, and has minimal polynomial $X^5-t$. So $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}] = 5$. Now $\zeta$ has minimal polynomial $X^4+X^3+X^2+X+1$ which is degree 4. So $[\mathbb{Q}(\zeta):\mathbb{Q}] = 4$. But since $gcd(4,5) =1$ and $L = \mathbb{Q}(\zeta,\sqrt[5]{t})$. Then $[L:\mathbb{Q}] = 4\cdot 5 =20$. Is this correct reasoning?
c) Since $f$ has distinct roots, the given extension is a Galois extension and therefore $|Gal(L/K)| =20$.
I don't know how to continue, but found something in my book: Consider $\sigma(\sqrt[5]{t}) = \zeta\sqrt[5]{t}, \sigma(\zeta) = \zeta$, and consider $\tau(\sqrt[5]{t})=\sqrt[5]{t}, \tau(\zeta) = \zeta^2$. These have properties $ord(\sigma)=5, ord(\tau) =4$. But how to continue? or is it not possible to apply this method on this problem?
 A: First just a short remark concerning b): You accidentally wrote $\mathbb{Q}$ in some places where it should have been $\mathbb{Q}(t)$ instead, for example, $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}(t)]=5$, and, accordingly, you have to show that $\phi_5(X)=X^4+X^3+X^2+X+1$ is indeed the minimal polynomial for $\zeta$ over $K=\mathbb{Q}(t)$. 
Now for c). I'm not sure if this is what the the hint in the book aimed for, but here is a suggestion: 
From the main theorem of Galois theory, we know that as $\mathbb{Q}(t)\subset \mathbb{Q}(t)(\zeta)$ is a normal extension (of degree $4$), the subgroup $\mathrm{Gal}(L/\mathbb{Q}(t)(\zeta))=:U\subset G:=\mathrm{Gal}(L/K)$ is normal (of index $4$, in particular, $\lvert U\rvert = 5$). Then, as $\lvert G\rvert = 20$ by b), $G$ must also have some subgroup $V$ of order $4$ (a $2$-Sylowsubgroup). As $\mathrm{gcd}(4,5)=1$, we have $U\cap V=\{\mathrm{Id}\}$, so we know that $G$ is a semidirect product,
$$G\simeq U\rtimes_{\varphi}V,\quad\text{with }\begin{aligned}[t]
\varphi\colon V&\rightarrow\mathrm{Aut}(U)\\v&\mapsto (u\mapsto vuv^{-1}).\end{aligned}$$
Regarding the hint of the book, $U=\langle \sigma\rangle $ and $V$ might be chosen as $V=\langle\tau\rangle$. The semidirect product is then completely determined by the relation established by $\varphi(\tau)$, i.e. by knowing the value of $\varphi(\tau)(\sigma)=\tau\sigma\tau^{-1}$:
$$G=\langle\sigma,\tau\vert\mathrm{ord}(\sigma)=5,~\mathrm{ord}(\tau)=4,~\tau\sigma\tau^{-1}=\sigma^2\rangle$$
