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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!

darij grinberg
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Squirrel-Power
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  • I have now answered this at https://math.stackexchange.com/questions/4184233/if-the-elements-y-1-y-2-dots-y-r-in-e-are-linearly-independent-over-f-sho/4188099#4188099 . – darij grinberg Jul 01 '21 at 20:19

1 Answers1

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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 $f^-1f(a_1x_1+..+a_nx_n)=a_1x_1+..+a_px_n=0$.

Tsemo Aristide
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  • A linear injective map $f:E\rightarrow E$ is bijective if $E$ is finite dimensional, because if $(e_1,...,e_n)$ is a base and $a_1f(e_1)+..+a_nf(e_n)=f(a_1e_1+..+a_ne_n)=0$, then $a_1e_1+...+a_ne_n=0$. So $f(e_1),...,f(e_n)$ is also a base. The characteristic $p$ is used to show that $f$ is liinear. – Tsemo Aristide May 13 '18 at 00:04
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    I just came across this but for posterity I want it known that this proof is wrong. The issue is that $f$ is not actually $F$-linear. Indeed, if it were, then $f(a) = a f(1) = a$ for all $a \in F$. However, the equation $x^p = x$ has exactly $p$ solutions in characteristic $p$. In fact, this defines the finite field $\mathbb F_p$. Hence, this is only $\mathbb F_p$ linear - not $F$ linear. Indeed, the result is clearly problematic. $f$ being an automorphism means that every element of $K$ has a $p^{th}$ root. This is false for $K = \mathbb F_p(t)$ wherein $t$ has no $p^{th}$ root. – paul blart math cop Jun 10 '20 at 01:31