Let $f(x)=x^4 + px +p \in \mathbb{Q}[x]$ for prime $p$. Then, we can see $f(x)$ is irreducible over $\mathbb{Q}[x]$. Then, show that
(1) for $p \neq 3, 5$ , Galois group is $S_4$.
(2) for $p = 3 $ , Galois group is $D_8$ (Dihedral group of order $8$)
(3) for $p = 5 $ , Galois group is $\mathbb{Z}_4$ (cyclic of oder $4$)
[My attempt]
First, the resolvent cubic of $f(x)$ is $h(y) = y^3 - 4py + p^2$. Also, the discriminant $D$ of $h(y) = -27(p^2)^2 - 4(-4p)^3 = 256p^3 - 27p^4$ and $\sqrt{D}\notin \mathbb{Q}$.
(1) When $p \neq 3, 5$ , we can know $h(y)$ is irreducible, so Galois group is $S_4$ by order argument.
(2),(3) My question is here. Since $D_8 \cap A_4$ is transitive subgroup of $S_4$ and $\mathbb{Z}_4 \cap A_4$ is not , if $f(x)$ is irreducible over $\mathbb{Q}(\sqrt{D})$ then galois is $D_8$ and if $f(x)$ is reducible over $\mathbb{Q}(\sqrt{D})$ then galois is $\mathbb{Z}_4$. In the case (2) (that is , $p = 3$) , $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\sqrt{21})$ , i have to know that $f(x)$ is irreducible or not over $\mathbb{Q}(\sqrt{21})$. However, i don't know whether $\mathbb{Q}(\sqrt{q})$ is U.F.D or not ,where $q$ is prime (to use gauss lemma and eisenstein's criterion).
Similary, in the case (3) ( $p = 5$ ) , $f(x)$ is reducible over $\mathbb{Q}(\sqrt{5})$ but i don't know how can i find this fact.
[My question]
(i) How can I test irreducibility of $f(x)$ over $\mathbb{Q}(\sqrt{D})$.
(ii) If exsist, is there another method to solve this problem?
Thank you for your attention!
