Let $E$ be a finite field extension of a field $F$ of a prime characteristic $p$, and assume that $E=F(E^p)$. If the elements $y_1,y_2,\dots,y_r \in E$ are linearly independent over $F$, show that $y_1^p,y_2^p,\dots,y_r^p$ are linearly independent over $F$.
Take note, (I) there is one related exercise showing the following result, which I already know.
Let $E$ be a finite field extension of a field $F$ of a prime characteristic $p$, and assume that $K=F(E^p)$ be the subfield of $E$ obtained by adjoining the $p$th powers of all elements in $E$. Show that $F(E^p)$ consists of all finite linear combinations of elements in $E^p$ with coefficients in $F$.
(II) There are two similar posts found on MSE, but i am not satisfied with them.
I want a more detailed answer by showing that if $\sum_{i=0}^ra_iy_i^p=0$, then $a_i=0$ for all $i\in\{1,2,\dots,r\}$.
Best Answer
The following is a detailed proof following the hint posted as an answer by @user26857. We will use the following fact:
Lemma 1 is the particular case of the Theorem 1 from https://math.stackexchange.com/a/3161397/ when $g$ is an inclusion map. (Of course, this particular case is easily seen to be equivalent to the general case.)
Let us now solve your question.
Assume that $y_{1},y_{2},\ldots,y_{r}$ are $r$ elements of $E$ that are linearly independent over $F$. We must show that the $r$ elements $y_{1} ^{p},y_{2}^{p},\ldots,y_{r}^{p}$ of $E$ are linearly independent over $F$ as well.
The list $\left( y_{1},y_{2},\ldots,y_{r}\right) $ of vectors in $E$ is linearly independent over $F$; thus, we can extend this list to a basis $\left( y_{1},y_{2},\ldots,y_{n}\right) $ of the $F$-vector space $E$ (since $E$ is a finite-dimensional $F$-vector space). Consider such a basis $\left( y_{1},y_{2},\ldots,y_{n}\right) $. Since $\left( y_{1},y_{2},\ldots ,y_{n}\right) $ is a basis of $E$, there is a unique endomorphism $\phi$ of the $F$-vector space $E$ that sends $y_{1},y_{2},\ldots,y_{n}$ to $y_{1} ^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Consider this $\phi$.
The image $\phi\left( E\right) $ of $E$ is an $F$-vector subspace of $E$ (since $\phi$ is $F$-linear), but it has further properties. To wit, let us first show that $E^{p}\subseteq\phi\left( E\right) $.
Indeed, let $w\in E^{p}$. Then, $w=e^{p}$ for some $e\in E$. Consider this $e$. Then, $e=\sum_{i=1}^{n}a_{i}y_{i}$ for some $a_{1},a_{2},\ldots,a_{n}\in F$ (since $\left( y_{1},y_{2},\ldots,y_{n}\right) $ is a basis of the $F$-vector space $E$). Consider these $a_{1},a_{2},\ldots,a_{n}$. Now, \begin{align*} w & =e^{p}=\left( \sum_{i=1}^{n}a_{i}y_{i}\right) ^{p}\qquad\left( \text{since }e=\sum_{i=1}^{n}a_{i}y_{i}\right) \\ & =\sum_{i=1}^{n}a_{i}^{p}y_{i}^{p} \end{align*} (since we are in characteristic $p>0$, so that the map $E\rightarrow E,\ z\mapsto z^{p}$ is a ring endomorphism of $E$). On the other hand, the definition of $\phi$ yields \begin{align*} \phi\left( \sum_{i=1}^{n}a_{i}^{p}y_{i}\right) =\sum_{i=1}^{n}a_{i}^{p} y_{i}^{p} \end{align*} (since $a_{1},a_{2},\ldots,a_{n}$ are scalars in $F$). Comparing these two equalities, we find $w=\phi\left( \sum_{i=1}^{n}a_{i}^{p}y_{i}\right) \in\phi\left( E\right) $.
Forget that we fixed $w$. We thus have shown that $w\in\phi\left( E\right) $ for each $w\in E^{p}$. In other words, $E^{p}\subseteq\phi\left( E\right) $. Hence, $1\in E^{p}\subseteq\phi\left( E\right) $, so that $F=F\cdot 1\subseteq\phi\left( E\right) $ (since $\phi\left( E\right) $ is an $F$-vector subspace of $E$ and since $1\in\phi\left( E\right) $).
Next, we shall show that the set $\phi\left( E\right) $ is closed under multiplication. Indeed, $\phi\left( E\right) $ is the image of the $F$-linear map $\phi$, which sends the basis elements $y_{1},y_{2} ,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, respectively. Thus, $\phi\left( E\right) $ is the $F$-linear span of the elements $y_{1} ^{p},y_{2}^{p},\ldots,y_{n}^{p}$. Hence, in order to show that $\phi\left( E\right) $ is closed under multiplication, it will suffice to prove that the pairwise products $y_{i}^{p}y_{j}^{p}$ of the latter elements $y_{1}^{p} ,y_{2}^{p},\ldots,y_{n}^{p}$ belong to $\phi\left( E\right) $. But this is easy: For any $i,j\in\left\{ 1,2,\ldots,n\right\} $, we have $y_{i}^{p} y_{j}^{p}=\left( y_{i}y_{j}\right) ^{p}\in E^{p}\subseteq\phi\left( E\right) $. Thus, we have shown that $\phi\left( E\right) $ is closed under multiplication. Since $\phi\left( E\right) $ is furthermore an $F$-vector subspace of $E$ and contains $1$ (since $1\in\phi\left( E\right) $), we thus conclude that $\phi\left( E\right) $ is an $F$-subalgebra of $E$. In particular, $\phi\left( E\right) $ is a subring of the field $E$, and thus is an integral domain (since any subring of a field is an integral domain). Hence, Lemma 1 (applied to $K=F$ and $L=E$ and $R=\phi\left( E\right) $) yields that $\phi\left( E\right) $ is a field. Therefore, $\phi\left( E\right) $ is a subfield of $E$ that contains both $F$ and $E^{p}$ as subfields (since $F\subseteq\phi\left( E\right) $ and $E^{p}\subseteq \phi\left( E\right) $).
However, $F\left( E^{p}\right) $ is the smallest such subfield (by the definition of $F\left( E^{p}\right) $). Hence, $F\left( E^{p}\right) \subseteq\phi\left( E\right) $. Now, recall that $E=F\left( E^{p}\right) $. Hence, $E=F\left( E^{p}\right) \subseteq\phi\left( E\right) $. Combining this with $\phi\left( E\right) \subseteq E$, we obtain $E=\phi\left( E\right) $. In other words, the map $\phi$ is surjective.
However, a surjective endomorphism of a finite-dimensional $F$-vector space must necessarily be injective. Thus, $\phi$ is injective (since $\phi$ is a surjective endomorphism of the finite-dimensional $F$-vector space $E$). Since $\phi$ is the $F$-linear map that sends the basis vectors $y_{1},y_{2} ,\ldots,y_{n}$ to $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$, this entails that the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{n}^{p}$ are $F$-linearly independent. Hence, the vectors $y_{1}^{p},y_{2}^{p},\ldots,y_{r}^{p}$ are $F$-linearly independent (since $n\geq r$). Qed.