When a finite extension $E$ of a finite prime field $F$ can be linear spaned by a subset of $E$? Let $F$ be a finite field of order $p$, and $E$ be an extension of $F$ with $[E:F]=2n$, then
$|E|=p^{2n}:=q$.
Now denote $\Omega=\{ x\in E:x^{\sqrt{q}+1}=-1\}$. My question is:
(1) Is it true that $\sum_{x\in\Omega}Fx=E$?
(2) Generally, for a subset $\Delta$ of $E^{\times}$, when $\sum_{x\in\Delta}Fx=E$?
 A: Concentrating on (1), and answering it in the affirmative.
Let's denote $r=p^n=\sqrt{q}, K=\Bbb{F}_r\subset E$, and
$$
S=\{x\in E\mid x^{r+1}=1\}.
$$
The mapping $N=N_{E/K}:E\to K, x\mapsto x^{r+1}$, is the relative norm. It is known to be a surjective homomorphism of multiplicative groups $E^*\to K^*$. Its kernel $S$ is a cyclic group of order $r+1$ consists of the roots of unity of order $d\mid r+1$.
The claim follows for example from the following argument:

*

*Because $S$ is closed under multiplication, so is its $F$-span
$$L=\sum_{s\in S}Fs\subseteq E.$$

*As $1\in S$, this implies that $L$ is a subring of $F$. As a finite integral domain $L$ is then necessarily also a subfield.

*Clearly $|L|\ge |S|=r+1>p^n$. This is so large a number that the only remaining alternative is $L=E$.

*By surjectivity of the relative norm there exists an element $\epsilon\in E$ such that $N(\epsilon)=-1$. Clearly $\Omega$ is the coset $\Omega=\epsilon S$.

*Consequently
$$\sum_{x\in\Omega}Fx=\epsilon L=\epsilon E= E.$$

The question (2) is a bit too general to say anything very specific about it. Of course, linear algebra tells us that
$$\sum_{x\in\Delta}Fx=E$$
if and only if $\Delta$ contains an $F$-basis of $E$.
