Exercise 7.4.7

Proof. By hypothesis, $f\in F[x]$ is a monic irreducible separable polynomial of degree 3, the characteristic of $F$ is not 2, and $L$ is the splitting field of $f$ over $F$.

We suppose here that $\mathrm{\Delta }=\mathrm{\Delta }(f)$ is not a square in $F$. Theorem 7.4.2 give then the result

Therefore $[L:F]=|\mathrm{Gal}(L∕F)|=6$. Since $\sqrt{\mathrm{\Delta }}\notin F$, $[F(\sqrt{\mathrm{\Delta }}):F]=2$, and so $[L:F(\sqrt{\mathrm{\Delta }})]=3$.

By the Galois correspondence, the extension $F(\sqrt{\mathrm{\Delta }})$ of degree 2 over $F$ corresponds to the subgroup $H=\mathrm{Gal}(L∕F(\sqrt{\mathrm{\Delta }}))$ of $G=\mathrm{Gal}(L∕F)$, of index 2 in $G\simeq S_3$. As $S_3$ has a unique subgroup of index 2, which is $A_3\simeq \mathbb{Z}∕3\mathbb{Z}$, we can conclude

Let $\alpha \in L$ be a root of $f$. Since $f$ is irreducible over $F$, $f$ is the minimal polynomial of $\alpha$ over $F$. Let $p\in F(\sqrt{\mathrm{\Delta }})[x]$ the minimal polynomial of $\alpha$ over $F(\sqrt{\mathrm{\Delta }})$. As $\alpha$ is a root of $f\in F[x]\subset F(\sqrt{\mathrm{\Delta }})[x]$, $p$ divides $f$ in $F(\sqrt{\mathrm{\Delta }})[x]$. Moreover $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=[L:F(\sqrt{\mathrm{\Delta }})]=3=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$.

$p\mid f,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$, and $f,p$ are monic, thus $f=p$. Therefore $f$ is irreducible over $F(\sqrt{\mathrm{\Delta }})$. □