Show that $F(E^p)$ consists of all linear combinations of elements in $E^p$ with coefficients in $F$. Let $E$ be a finite extension of a field $F$ of prime characteristic $p$, and let $K=F(E^p)$ be the subfield of $E$ obtained from $F$ by adjoining the $p$th power of all elements of $E$. Show that $F(E^p)$ consists of all linear combinations of elements in $E^p$ with coefficients in $F$.
Since $[E:F]< \infty$, let $\alpha_1, \alpha_2 ,...,\alpha_n$be a basis for $E$ over $F$. Then, I claim that $F(E^p)=F(\alpha_1 ^p, \alpha_2 ^p,...,\alpha_n ^p)$. Note that since char$(F)=p$, let $x \in F $ ,by Frobenius Automorphism, we have $x^p \in F(\alpha_1 ^p, \alpha_2 ^p,...,\alpha_n ^p) \Rightarrow F(E^p) \subset F(\alpha_1 ^p, \alpha_2 ^p,...,\alpha_n ^p)$. The direction $F(\alpha_1 ^p, \alpha_2 ^p,...,\alpha_n ^p) \subset F(E^p)$ is trivial since all $\alpha_i^p \in E$ for $1 \leq i \leq n$.
Then I don know how to proceed. Can anyone guide me?
 A: Notice that it is enough to show that $F[E^p] = F(E^p)$.
To see this, let $$f_1 e_1^p + \cdots + f_n e_n^p$$ a linear combination of elements of $E^p$ with coefficients in $F$. Then it is clear that
$$f_1 e_1^p + \cdots + f_n e_n^p = f\left(e_1,\ldots,e_n^p\right),$$
where $$f\left( x_1,\ldots,x_n \right) = f_1 x_1 + \cdots + f_n x_n \in F\left[x_1,\ldots,x_n\right].$$
Thus $$f_1 e_1^p + \cdots + f_n e_n^p\in F\left[ E^p\right].$$
Let's denote $$[n] := \{0,\ldots,n\},$$ and for $i = (i_1,\ldots,i_n)\in\Bbb Z_{\geq 0}^n$ and $e_1,\ldots,e_n\in E$ write 
$$\left( e_1,\ldots,e_n \right)^i = e_1^{i_1}\cdots e_n^{i_n}.$$
Let $\alpha\in F[E^p]$. Then
$$\alpha = \sum_{i\in \left[r_1\right]\times\cdots\times \left[r_n\right]} c_i \left( e_1^p,\ldots,e_n^p \right)^i,$$
for some $r_1,\ldots,r_n\in \Bbb Z_{\geq 0}$, $c_i\in F$ and $e_1,\ldots,e_n\in E$. But then
\begin{align*}
\alpha &= \sum_{i\in \left[r_1\right]\times\cdots\times \left[r_n\right]} c_i \left( e_1^p,\ldots,e_n^p \right)^i \\
       &= \sum_{i\in \left[r_1\right]\times\cdots\times \left[r_n\right]} c_i \tilde{e}_i^p,\tag{$\ast$}
\end{align*}
where
$$\tilde{e}_i^p = \left( e_1^p,\ldots,e_n^p \right)^i = e_1^{p i_1}\cdots e_n^{p i_n}
= \left( e_1^{i_1}\cdots e_n^{i_n} \right)^p,\in E^p.$$
In view of $(\ast)$, $\alpha$ is a linear combination of elements of $E^p$ with coefficients in $F$.
To show $F[E^p] = F(E^p)$ it's enough to show that $F[E^p]$ is a field. If $F[E^p]$ is a field, since it contains $F$ and $E^p$ we must have $F(E^p)\subseteq F[E^p]$, and since the other inclusion: $F[E^p]\subseteq F(E^p)$ always holds, we have $F[E^p] = F(E^p)$.
To show that $F[E^p]$ is a field, I'll use the following
Lemma. Let $E$ an algebraic extension of $F$. If $R$ is a ring satisfying $$F\subseteq R\subseteq E,$$ then $R$ is a field.
Proof. $R$ is already a commutative ring with unity, so it's enough to see that the elements of $R\setminus F$ are units of $R$.
Let $\alpha\in R\setminus F$. Let $${p(x) = a_0 + a_1x + \cdots + x^n}\in F[x]$$ be the minimal polynomial of $\alpha$. Notice that it must be that $n\gt 1$.
If $a_0 = 0$, then $$p(x) = x\left(a_1 + \cdots + x^{n-1}\right),$$ which will contradict the irreducibility of $p(x)$ in $F[x]$.
Now, let
$$g(x) = -\frac1{a_0}\left(a_1+\cdots+x^{n-1}\right).$$
Then $g(\alpha)\neq 0$ because otherwise the minimality of $p(x)$ will be contradicted. Thus, $g(\alpha)\in R$, $g(\alpha)\neq 0$ and
\begin{align*}
g(\alpha)\alpha &= -\frac1{a_0}\left(a_1+\cdots+\alpha^{n-1}\right)\cdot \alpha \\
&= -\frac1{a_0}\left(p(\alpha) - a_0\right) \\
&= -\frac1{a_0}\left(0 - a_0\right) = 1,
\end{align*}
that is $\alpha^{-1} = g(\alpha)\quad\square$.
Now, back to the question, if $B\subseteq E$ then $F[B]$ is ring with $$F\subseteq F[B]\subseteq E$$ and since $E$ is an algebraic extension, the Lemma implies $F[B]$ is a field.
In particular $F[E^p]$ is a field, as we wanted.
A: Let more generally $E/F$ be a finite extension (actually we only need that it is algebraic) and $B \subseteq E$ be a subset. Then $F(B) = F[B]$ is the set of all polynomial expressions in $B$ with coefficients in $F$. If $B$ is multiplicatively closed, all monomials reduce to elements of $B$, and we see that $F(B)$ consists of all linear combinations in $B$ with coefficients in $F$.
