Prove for $a,b \in \mathbb{F}_{p^n}$, if $p(x) = x^3 + ax +b$ is irreducible, then $-4a^3 - 27b^2$ is a square in $\mathbb{F}_{p^n}$.

Question

The problem is as the title states. We know that in this case determinant $D = -4a^3 -27b^2$, and also I know that if $G$ is the Galois group of $x^3 + ax + b$, then

$$G \subset A_n \, \iff \sqrt{D} \in \mathbb{F}_{p^n}.$$

So I am trying to show $G \subset A_n$, and I suppose it is probably something easy having to do with finite fields, but I am not seeing it. Any help is appreciated :)

CK

Answer 1

Actually, as $p(x)=x^3+ax+b$ is irreducible, $G$ can be viewed as a subgroup of $S_3,$ not $S_n,$ and $G\subset A_3\iff \sqrt D\in \mathbb F_{p^n}.$
By the theory of finite fields, if $p$ is irreducible, then the splitting field of $p$ over $\mathbb F_{p^n}$ is isomorphic to $\mathbb F_{p^{3n}},$ thus $|G|=[\mathbb F_{p^{3n}}:\mathbb F_{p^n}]=3.$ Notice that $S_3$ has only one subgroup of order $3,$ by Sylow's theorem(s), therefore $G=A_3.$
Hope this helps.