Let $\mathbb K$ be a field, $\,f \in \mathbb K[X]$, $n :=$ deg(f). Let $L$ be a splitting field of $\,f$ over $\mathbb K$. Then:
- $[ L : \mathbb K] \,\mid\, n!$
- $[ L : \mathbb K] = n! \implies f$ is irreducible
This is my attempt at proving the two statements in a single induction over $n$. Since $L$ is a splitting field $n \geq 1$.
Base step ($n = 1$):
- In this case $f$ has a single root $\alpha \in \mathbb K$ and $\frac{f}{l(f)}$ is the minimal polynomial of $\alpha$ over $\mathbb K$. Now $L = \mathbb K(\alpha) = \mathbb K \implies [L : \mathbb K] = 1$.
Inductive step($n \Rightarrow n+1$):
Let $g \in \mathbb{K}[X]$ be an irreducible factor of $f$ (therefore $f = g \cdot h$ for some $h \in \mathbb{K}[X]$), with degree $m = $ deg($g$) $\in \{1, ..., n\}$. Let $\tilde{L}$ be the splitting field of $g$ over $\tilde{L}$ over $\mathbb{K}$.
Now $L$ is also a splitting field of $\frac{f}{g}$ over $\tilde{L}$, because:
- $L$ is generated by the roots of $\frac{f}{g}$ over $\tilde{L}$, since $L$ is generated by the roots of $\,f$ over $\mathbb K$ and every root $\alpha$ of $f$ is either (i) a root $\frac{f}{g}$ or (ii) a root of $g$ and therefore already included in $\tilde{L}$.
- $\frac{f}{g}$ factors over $L$, since $f$ can be factored into linear terms over $L$.
$$ \implies \hspace{25px}[L : \mathbb K] = \underbrace{[L : \tilde{L}]}_{\,\mid\, m! \text{ (by induction)}}\cdot\hspace{15px}\underbrace{[\tilde{L} : \mathbb{K}]}_{\,\mid\, (n + 1 - m)! \text{ (by induction)}} $$
Since for $a, b \in \mathbb N: a!\cdot b! \,\mid\, (a+b)!$ (and $a!\cdot b! < (a+b)!$ if $a \neq 0$ or $b \neq 0$), it follows: $$ \implies \hspace{25px}[L : \mathbb K] \,\mid\, (n + 1)! \\ \text{(and}\hspace{15px} [L : \mathbb K] < (n + 1)\hspace{15px}\text{if $f$ wasn't irreducible already)} $$
I'm just wondering whether this proof is correct or am I missing something?