# The $p^{th}$ power of the elements of a basis of a finite separable field extension.

I came across the following claim.

Let $L/K$ be a finite, separable extension of characteristic $p$ fields. Suppose $a_1,\dots,a_d$ is a basis. Then, so is $a_1^p,\dots,a_d^p$.

To prove this, one just needs to consider $\sum_{i=1}^d c_ia_i^p = 0$, $c_i\in K$ and show that each $c_i$ is a $p^{th}$ power. If so, then $\sum_{i=1}^d c_ia_i^p = \sum_{i=1}^d b_i^pa_i^p = (\sum_{i=1}^d b_ia_i)^p = 0$. Then each $b_i = 0$ by linear independence of the set $\{a_1,\dots,a_d\}$.

So, my question is, why does separability imply that $c_i = b_i^p$ for some $c_i,b_i\in K$?

Thanks.

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Separability of the extension does not imply that the $c_i=b_i^p$. Else by taking $L=K$ you could prove that every field is perfect! –  Georges Elencwajg Jun 26 '12 at 22:47
This actually makes me feel better. I thought of the reason Georges gave but figured I must be wrong. –  student Jun 26 '12 at 22:55
@Dylan: They were in some unpublished notes I was looking at. Perhaps the author assumed perfect way back and I just couldn't find it. –  student Jun 26 '12 at 22:56
How about: Instead of assuming that all $p$-th roots of the $c_i$ lie in $K$, simply extend $K$ and $L$ by adjoining roots of the $c_i$, since the resulting extension $L'/K'$ will still be separable and your argument applies. –  Alastair Litterick Jun 26 '12 at 23:12
@AlastairLitterick I think that works, maybe after arguing that because $K'/K$ is purely inseparable we'll have $[L' : K'] = [L : K]$? –  Dylan Moreland Jun 27 '12 at 1:06

The claim in the question is true . Here is a proof:

Let $[L:K]=d$. Since $L/K$ is separable, the number of $K$-algebra embeddings $u_i :L\to K^{alg}$ of $L$ into an algebraic closure of of $K$ is $d$.
Now if $a_1,...,a_d$ is a $K$-basis of $L$ we have an intermediate field $K\subset L'=K[a_1^p,...,a^p_d]\subset L=K[a_1,...,a_d]$
Since the extension $L'\subset L$ is purely inseparable the restrictions $u_i\mid L':L'\to K^{alg}$ are still distinct which proves that $[L':K]\geq [L':K]_{sep}\geq d=[L:K]\geq [L':K]$ so that $L'=L$.
This in particular implies that any element $\lambda \in L$ can be written as a sum of terms $q\cdot A^p$ with $A$ a monomial in the $a_1,...,a_d$ and $q\in K$.

Lemma The vectors $a_1^p,...,a^p_d$ generate $L$ as a $K$ vector space.
Proof of lemma Write as above any element $\lambda \in L$ as a sum of terms of the form $qA^p$.
Since in particular $A\in L$, we can write $A=\sum_{i=1}^{d} q_ia_i\;(q_i\in K)$ and thus $\lambda$ is a sum of terms of the form $q\cdot (\sum_{i=1}^{d} q_i a_i)^p=q\cdot (\sum_{i=1}^{d} q_i^pa_i^p)=\sum_{i=1}^{d}(q\cdot q_i^p) a_i^p$, which proves that the $a_i$'s generate $L$

End of proof of claim Since the $d$ elements $a_1^p,...,a^p_d$ generate the $d$-dimensional $K$-vector space $L$, thay are a $K$-basis of $L$.

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Thank you for your help. –  student Jun 27 '12 at 2:28