Suppose that $F $ is a field with characteristic $p >0$, that $K/F$ is a finite extension, and that $K=F (K^p) $. If $\{x_1,...,x_n \} \subset K $ is linearly independent over $F$, then so is $\{x_1^p,...,x_n^p\} $.
Here, $K^p $ is the set of all $x^p $ such that $x \in K $.
I am not sure how to prove this problem. Here is what I know so far, though, which I tried to use for proving but failed: $K/F $ is finite iff $K=F (a_1,...,a_n) $ with each $a_i \in K $ algebraic over $F$.
I believe that I may need the following lemma, which I have not been able to solve (but was a question right before this problem):
Suppose $F $ is a field with positive characteristic $p$, and that $K/F $ is an extension such that evey irreducible polynomial is separable. Prove that $K=F (K^p) $.
Any help would be great!
Use the fact that $f(x)=x^p$ is an automorphism. ($f$ is linear, and is injective, since $K/F$ is finite its is bijective). $f(a_1x_1+..a_nx_n)=a_1x_1^p+..+a_nx_n^p=0=0$ implies that $f^-1f(a_1x_1+..+a_nx_n)=a_1x_1+..+a_px_n=0$.