Why if $E$ has characteristic $p$ then $F=\{a\in E:a^{p^n}=a\}$ is a subfield? We want to prove there exists a finite field of $p^n$ elements ($p$ is prime and $n>0$). Take $q=p^n$ and $g(x)=x^q-x\in\mathbb{Z}_p[x]$, and let $E$ be a field that contains $\mathbb{Z}_p$ and all roots of $g(x)$. Let $F=\{a\in E:a^q=a\}$.
I understand that $F$ is closed under addition (since $E$ has characteristic $p$), multiplication and multiplicative inverse.
However, I don't understand why if $a^q=a$ then $(-a)^q=-a$. I would guess this is not true if $q=2^n$. Am I wrong? Can anyone explain to me?
 A: When $q=2^n$, the characteristic is $2$. So, $2=0\Rightarrow 1=-1$. Hence $(-a)^q=a^q=a=-a$.
In general, if $a,b\in F$, then $(ab)^{p^n}=a^{p^n}b^{p^n}=a\cdot b\Rightarrow ab\in F$. Also, $(a+b)^{p^n}=\sum_{k=0}^{p^n}\binom{p^n}{k}a^{p^n-k}b^k=a^{p^n}+b^{p^n}=a+b$, since $p|\binom{p^n}{k}$ for $1\leqslant k\leqslant p^n-1$. Thus, $a+b\in F$. Moreover, $(a^{-1})^{p^n}=(a^{p^n})^{-1}=a^{-1}\Rightarrow a^{-1}\in F$.
A: If a subset $F$ of a field $E$ of characteristic $p$ is closed under addition, then it is automatically also closed under taking negatives.  For if $x\in F$ then $px=0$ so $(p-1)x=-x$, and so $-x$ can be found by just taking a sum of $p-1$ copies of $x$.
A: Let $g(x)=x^{p^n}-x$. You want to prove that if $g(\beta)=0$, then $g(-\beta)=0$.  So suppose $g(\beta)=\beta^{p^n}-\beta=0$. Then $g(-\beta) = (-\beta)^{p^n}-(-\beta)=(-\beta)^{p^n}+\beta$.  
If $p=2$, then $g(-\beta)=(-\beta)^{p^n}+\beta = (\beta)^{p^n}+\beta$ (because we are raising to an even power).  This equals $\beta+\beta=2\beta=0$ where the last equality holds because $p=2$.
If $p$ is odd, then $g(-\beta) = (-\beta)^{p^n}+\beta = -(\beta^{p^n})+\beta=-\beta+\beta=0$. 
