In fields $K/F$ of characteristic not 2, an $F$-subspace of $K$ closed under taking positive powers is actually a subfield of $K$ 
Let $F\subset K$ be an algebraic extension of fields. Let $F\subset R\subset K$, where $R$ is an $F$-subspace of $K$ with the property that for any $a \in R$, $a^k\in R$ for any $k\geq 2$.

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*Assume that $\operatorname{char}(F)$ is not equal to 2. Show that $R$ is a subfield of $K$.


*Give an example such that $R$ may not be a field if $\operatorname{char}(F)=2.$

Firstly, I can prove that $R$ is a subring of $K$. Because $(x+y)^2=x^2+2xy+y^2$, based on $\operatorname{char}(F)$ is not equal to $2$, we have $xy\in R$. If so, the multiplication and addition are closed in $R$. But how can I prove $R$ is a field?
Also, how can I give an example, if $\operatorname{char}(F)$ is not equal to $2$, then the result of part 1 does not necessarily hold?
 A: For $r\in R\setminus 0$ consider the minimal polynomial $f:=X^n+a_{n-1}X^{n-1}+\ldots +a_0\in F[X]$ of $r$ over $F$. Note that $a_0\neq 0$, hence $-a_0$ is invertible in $R$. The equation
$r(r^{n-1}+a_{n-1}r^{n-2}+\ldots +a_1)=-a_0$
then shows that $r$ is invertible in $R$.
A: I will give a counter example in the case $\operatorname{char} F = 2$: 
Consider $F = \mathbb F_2(X^2,Y^2)$ and $K = \mathbb F_2(X,Y)$, where $X$ and $Y$ are free variables. Then we have $[K:F] = 4$. Let $R$ be the $3$-dimensional $F$-space spanned by $1, X, Y$. Clearly, $R$ is not an intermediate field of $K/F$ (otherwise $4 = [K:F] = [K:R]\cdot [R:F]$, i. e. $\dim_FR = 3$ would be a divisor of $4$).
Hence, it suffices to prove that $R$ is closed under taking positive powers. Let $a+bX+cY\in R$, where $a,b,c\in F$, and $n\in \mathbb N$. Then $(a+bX+cY)^n$ can be uniquely expressed in the $F$-basis $\{1, X, Y, XY\}$ of $K$ and we need only show that the coefficient of $XY$ is zero. Using the Multinomial theorem, the coefficient of $XY$ in $(a+bX+cY)^n$ is the sum of expressions of the form
$$
\binom{n}{k_a,k_b,k_c}a^{k_a}b^{k_b}c^{k_c}, \tag{1}
$$
where $k_b$ and $k_c$ are both odd integers and $k_a+k_b+k_c = n$. Now, we can write
$$
\binom{n}{k_a,k_b,k_c} = \binom{n}{k_a}\cdot \binom{k_b+k_c}{k_b}
$$
and this answer shows that $\binom{k_b+k_c}{k_b}$ is even, since there is at least one carry when adding $k_b$ and $k_c$ in base $2$. Therefore, all the expressions in $(1)$ are zero in $F$.
Hence, $(a+bX+cY)^n \in R$ and $R$ is closed under taking positive powers.
