Let $f ∈ F[x]$ have degree $n > 0$, and let $L$ be the splitting field of $f$ over $F$. I wish to show if $[L : F] = n!$ then $f$ is irreducible over $F$.
Also, the converse of this is false with some counterexample?
I know $[L:F]$ divides $n!$ from this result but i can't make the same method work to prove this~ Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.
Let $f = gh$ with $\deg(g) = m$ and $\deg(h) = k$ be a reducible polynomial. Let $F(a) = K$ and $K(b) = L$ where $a$ is a root of $g$ and $b$ is a root of $h$. We know that $[L :F] = [L : K][K : F] \mid m! k!$, but also that $m!k! < (m + k)!$ (if $m, k > 0$). Hence, $[L : K] < (m + k)! = n!$; which is a contradiction, therefore our original assumption must have been false, meaning that $f$ is irreducible.