I have just proven the following statement:  Let $K$ be a field and $L$ the splitting field of a separable polynomial $f\in K[X]$ of degree $n$. Denote the zeros of $f$ in $L$ by $\alpha_1,\alpha_2,...,\alpha_n$. Then for $i=1,2,...,n$: $$ [K(\alpha_1,...,\alpha_i):K]\leq n(n-1)...(n-i+1). $$  I now need to show that $[L:K]$ divides $n!$.  What I figured out and I think I have to use:   $|\text{Gal}(L/K)|=[L:K]$  $L/K$ a normal and separable extension $|\text{Gal}(L'/K)|$ equals the number of distinct zeros of $f$ which lie in $L$ if $L'/K$ is a simple extension And as a hint: The Galoisgroup permutes zeros of polynomials
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
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