Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
$E^p$ 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
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 $E=E^{p^2}k$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.