Degree of a splitting field of a polynomial

Question

If $F \subseteq E$ is a field extension, and $E$ is a splitting field of a polynomial $f(x)\in F[x]$ of degree n over $F$. Does that mean $|E:F|\le n$ ? Could we conclude something more? If someone please could explain it so I understand.

Answer 1

The splitting field $E$ of $f\in F[x]$ is the smallest field containing $F$ and all the roots of $f$. You can see that $|E:F|\leq n!$ Indeed, take the quotient field $F_1=F[x]/(f)$, then $f$ factors as $f=(x-\xi)f_1$ over $F_1$. By induction, we have $|E:F_1|\leq (n-1)!$, so we get the desired inequality.

You mentioned a stronger inequality $|E:F|\leq n$, but it is not generally true. You can see that the polynomial $g=x^3-2$ does not split over $\mathbb{Q}(\sqrt{2})$, so the splitting field of $g$ has degree six over $\mathbb{Q}$.