$|E:F|=1$ if $\operatorname{char}(k)=0$ and $p^m$ for some $m \in \mathbb N \cup \{0\}$ if $\operatorname{char}(k)=p$. 
Let $k$ be a field and let $f(x)\in k[x]$ be a separable polynomial. Let $E$ be the splitting field of $f(x)$ over $k$ and let $F$ be the subfield of $E$ generated over $k$ by all elements $(\alpha - \beta)$, where $\alpha, \beta$ are roots of $f(x)$ in $E$. Then I have to show that:$|E:F|=1$ if $\operatorname{char}(k)=0$ and $p^m$ for some $m \in \mathbb N \cup \{0\}$ if $\operatorname{char}(k)=p$.

I solved it for $\operatorname{char}(k)=0$ as following:
Let $\deg(f)=n$ and $Z(f)=\{\alpha_1,\ldots,\alpha_n\} \subset E$. Then by definition $E=k(\alpha_1,\ldots,\alpha_n)$ and $F=k(\alpha_i-\alpha_j:1\leq i<j\leq n)$. For $\sigma \in \text{Gal}(E/F)$ we have $\sigma(\alpha_1-\alpha_i)=\alpha_1-\alpha_i$ for $i=2,\ldots,n$. Adding these $n-1$ equations and using the fact $\sum_{i=1}^{n}\alpha_i\in k$ we obtain  $n\sigma(\alpha_1)=n\alpha_1$. Using $\operatorname{char}(k)=0$ we see that $\sigma(\alpha_1)=\alpha_1.$ Now each of the above $n-1$ equations give that $\sigma(\alpha_i)=\alpha_i$ for all $i>1$ as well. Thus $\sigma=Id$ and $|E:F|=1$
I need some help for the $\operatorname{char}(k)=p$ situation. Thank you.
 A: Let $\text{char}(k)=p>0$ and take any arbitrary $\sigma \in Gal(E/F)$. Note that if $\sigma$ fixes any $\alpha_i$ it will fix the remaining $\alpha_i's$. Since $\sigma(\alpha_1)$ is also a root of $f(x)$, let $\sigma(\alpha_1)=\alpha_{\ell}$ for some $\ell \in \{1,\ldots,n\}$. If $\ell=1$, we are done with $\sigma=Id$. Now suppose $\ell >1$. Since $\sigma$ fixes $F$, $\sigma(\alpha_1-\alpha_{\ell})=\alpha_1-\alpha_{\ell}$ implying that $\sigma(\alpha_1)-\sigma(\alpha_{\ell})=\alpha_1-\alpha_{\ell}$, i.e., $\sigma(\alpha_{\ell})=2\alpha_{\ell}-\alpha_1$, i.e., $\sigma^2(\alpha_1)=2\alpha_{\ell}-\alpha_1$.
One can rewrite $\sigma^2(\alpha_1)=2\alpha_{\ell}-\alpha_1$ as $\sigma^2(\alpha_1)=2\alpha_{\ell}-2\alpha_1+\alpha_1=2(\alpha_{\ell}-\alpha_1)+\alpha_1$. Again after applying $\sigma$ on both sides we have $\sigma^3(\alpha_1)=2(\alpha_{\ell}-\alpha_1)+\alpha_{\ell}=3\alpha_{\ell}-2\alpha_1$. Proceeding in this way we finally have $\sigma^p(\alpha_1)=p\alpha_{\ell}-(p-1)\alpha_1=\alpha_1$. Thus $\sigma^p$ fixes all the $\alpha_i's$ and hence $Order(\sigma)=p$ in $Gal(E/F)$. Thus every nontrivial element in $Gal(E/F)$ has order $p$. Thus $Gal(E/F)$ is a $p$-group, and hence $|E:F|=p^m$ for some $m\in \mathbb N\cup \{0\}$.
