Show that the polynomial $t^q+x_1t^{q-1}+\dots+x_{q-1}t+x_q$ is separable?

Question

I am reading "Déformations formelles des revêtements sauvagement ramifiés de courbes algébriques" by Bertin and Me'zard https://link.springer.com/article/10.1007/s002220000071 and trying to study the deformation on page 219. In order to understand the deformation, I want to show that the polynomial $a(t)=t^q+x_1t^{q-1}+\dots+x_{q-1}t+x_q$ over an algebraic closure of $k[x_1,\ldots,x_q][t]$ is separable where $k$ is an algebraically closed field of characteristic $p>0$. Could anyone help?

Answer 1

In order to show that $a(t) = t^{q} + x_{1}t^{q-1} + \dotsb + x_{q-1}t + x_{q}$ is separable over $K := k(x_{1},\dotsc,x_{q})$, it suffices to show that it remains separable over $L$ for some field extension $L/K$. We will take $L := k(s_{1},\dotsc,s_{q})$ where $K$ is viewed as a subfield of $L$ by the $k$-algebra homomorphism sending $x_{i} \mapsto \sigma_{i}(s_{1},\dotsc,s_{q})$, the $i$th elementary symmetric polynomial in $s_{1},\dotsc,s_{q}$. Then over $L$ the polynomial $a(t) \in L[t]$ factors as $(t-s_{1}) \dotsb (t-s_{q})$, which is separable.

(Remark: The extension $L/K$ is finite of degree $[L:K] = q!$.)