Given a field $\mathbb{F}$ with an irreducible, separable polynomial $f(x),$ let $E$ denote the splitting field of $f$ over $\mathbb{F},$ and assume that $\alpha$ and $\alpha + 1$ are roots of $f(x).$ Prove that the characteristic of $\mathbb{F}$ is not zero. Further, prove that there exists a field $\mathbb{F} \leq L \leq E$ such that $\operatorname{char} (\mathbb{F}) = [E : L].$
Could anyone provide any direction on this problem? Unfortunately, I have been thinking about it for a few hours, and I have not come up with anything close to a solution; however, I have several ingredients with which I have worked. Considering that $f(x)$ is separable, $f(x)$ and $f'(x)$ are relatively prime so that by Bezout's Theorem, there exist polynomials $a(x), b(x) \in \mathbb{F}[x]$ such that $a(x) f(x) + b(x) f'(x) = 1.$ Using the fact that $\alpha$ and $\alpha + 1$ are roots of $f(x)$ but not roots of $f'(x),$ we have that $b(\alpha) f'(\alpha) = 1$ and $b(\alpha + 1) f'(\alpha + 1) = 1.$ We have not really used any facts about the roots yet, so I wonder if that might be a direction worth exploring, but I am stuck at the moment.
The Galois group $G$ acts transitively on the roots of $f$ since $\Pi_{s\in G}(X-s(\alpha))\in \mathbb{F}[X]$, there exists $s$ such that $s(\alpha) =\alpha+1$, we deduce that $s(\alpha+1)=s(\alpha)+1=\alpha+2$, and recursively $\alpha+p$ is a root for every integer. Since the number of root is finite, there exists $n, m$ distinct such that $\alpha+n=\alpha+m$, we deduce that $m-n=0$ and the characteristic is finite.