Powers of linearly independent separable elements are linearly independent Let $F$ be a field of characteristic $p$, and let $v_1, ... , v_n \in \overline{F}$ be separable over $F$ and linearly independent.  Is it true that $v_1^{p^e}, ... , v_n^{p^e}$ are also linearly independent for all $e \geq 1$?  I know that $F(v_1, ... , v_n) = F(v_1^{p^e}, ... , v_n^{p^e})$, but this doesn't seem like it's enough.
 A: I found an explanation in Hungerford.  Let $M$ be the field $F[v_1, ... ,v_n]$.  Since $M/F$ is finite dimensional, and $v_1, ... , v_n$ are linearly independent, we can extend this collection to a basis $v_1, ... , v_n, v_{n+1}, ... , v_m$ for $M/F$.  We claim that $v_1^{p^e}, ... , v_m^{p^e}$ spans $M$ as an $F$-vector space.  Since $M/F$ is $m$-dimensional, this will implies that $v_1^{p^e}, ... , v_m^{p^e}$ is a basis, and in particular these things are linearly independent over $F$.
Let $\alpha \in M$.  We need to write $\alpha$ as an $F$-linear combination of the $v_i^{p^e}$.  First, for every $k \geq 0$, we can write $\alpha^k$ as an $F$-linear combination of the $v_i$: $$\alpha^k = \sum\limits_{i=1}^m c_{ik} v_i$$ and hence $$\alpha^{p^e k } = \sum\limits_{i=1}^m c_{ik}^{p^e} v_i^{p^e}$$ Also, $\alpha$ is separable over $F$, so a standard result gives us that $F[\alpha] = F[\alpha^{p^e}]$.  Therefore, $\alpha$ is the evaluation of a polynomial $d_0 + d_1X + \cdots + d_sX^s \in F[X]$ at $\alpha^{p^e}$: $$\alpha = \sum\limits_{k=0}^s d_k \alpha^{p^e k} = \sum\limits_{k=0}^s \sum\limits_{i=1}^m d_k c_{ik}^{p^e} v_i^{p^e} = \sum\limits_{i=1}^m (\sum\limits_{k=0}^s d_k c_{ik}^{p^e})v_i^{p^e} $$
