$F_q$ is a finite field of $q$ (which equals $p^n$，p is a prime number). $F_q^*$ denotes the elements in $F_q$ which has inverse. Prove that:
- The multiply of all the elements in $F_q^*$ equals -1.
- $p＞2$,The sum of all elements in $F_q$ equal 0.
Here is my proof,and I want to konw wehther it is true?
Proof: $F_q$ is a finite field,then $F_q^*$ is a cyclic group generated by $u$ (which $u^{q-1}=1$) with respect to mulitplication in $F_q$ . $F_q=\{0,1,u,…,u^{p^n-2}\}$.
1 .Let $S=1×u×u^2×…×u^{p^n-2}$,then $S=u^{1+2+…+(p^n-2)}= u^{\frac{{(p^n-2)(p^n-1)}}{2}}$
Case (1). $p=2$,then $p^n-2$ could divided by 2,then $S=1$ and $1=-1$ (Does this hold in $F_{2^n}$?) in $F_q$ . So $S=-1$
Case (2). $p＞2$ , then $p$ is odd,$S=u^{\frac{(p^n-2)(p^n-1)}{2}}=u^{\frac{-(p^n-1)}{2}}$,we know $S^2=1$ .Thus $(S+1)(S-1)=0$. $S≠1$ because the order of $u$ is $p^n-1$ (Is this right or wrong?).Then $S+1=0$(in the the field )
2.This is directly by the order of $u$: Let $T=0+1+u+…+u^{p^n-2}$.$u≠1$ in this case. Thus $T=\frac{1-u^{p^n-1}}{1-u}=0$
(1) $p=2$ case: yes, $v=u^{(p^n-2)/2}$ makes sense and $v^{p^n-1}=1$ because $\mathbb{F}_{2^n}^\times$ has order $p^n-1$, and indeed $1=-1$ in any field of characteritic two.
(1) $p>2$ case: you have a typo, should be $S=u^{-(p^n-1)/2}$. And yes, we know $S\ne1$ since $u$ has order $p^n-1$, so you can conclude $S=-1$ from $S^2=1$.
(2) The proof you edited in is correct, and even works for $p=2$ as long as $n>1$.
You can also say that if $p>2$ then $x\ne -x$ for all $x\in\mathbb{F}_{p^n}$, which means each element will cancel with its additive inverse in the sum.