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Let $E/k$ be a finite field extension, $\operatorname{char}(k)=p>0$. Suppose that $E^p k = E$. Is it then true that $E^{p^n}k = E$ for any positive integer $n$? If yes, why?

Thanks.

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What do you mean by $\,E^pk\,$? Is this the field of all elements in $\,E\,$ raised to the p-th power? Then why did you write there that $\,k\,$? –  DonAntonio Aug 6 '12 at 15:52
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$E^p$ is the field of all elements of $E$ raised to $p$ power. $E^pk$ is the compositum of $E^p$ and $k$, i.e. the smallest field containing both $E^p$ and $k$. –  Manos Aug 6 '12 at 15:57
    
Dear @DonAntonio, suppose $k=\mathbb F_p(x,y)\subset E=\mathbb F_p(x,y^{1/p})$ where $x,y$ are independent indeterminates. Then $E^p=\mathbb F_p(x^p,y)\subsetneq E^pk=\mathbb F_p(x,y)=k$ –  Georges Elencwajg Aug 6 '12 at 18:57
    
Thank you both Manos and Georges. I know what the compositum of two fields is, I'm just used to see it as $\,F\vee K\,$ . –  DonAntonio Aug 6 '12 at 19:29

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up vote 6 down vote accepted

Yes, it is true. I will show that $E=E^{p^2}k$ and leave to you the proof of the general case $E=E^{p^n}k$.

Since $E=E^{p}k$, any element $e\in E$ can be written as $e=\sum q_ie_i^p\;$ (for some $e_i\in k, q_i\in k$) .
[This is due to the fact that the ring formed by the sums on the right is already a field, because that ring is a $k$-subalgebra of the algebraic extension $E/k$]
In the same way, each $e_i$ can be written as $e_i=\sum q_{ij}e_{ij}^p$.
Substituting yields $$e=\sum q_i(\sum q_{ij}e_{ij}^p)^p=\sum q_iq_{ij}^pe_{ij}^{p^2}$$
which shows that indeed $E=E^{p^2}k$

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