Given a field $F_{p^n}$ with $char(F)=p$ and $p$ prime. And thus our main misunderstanding: Why is the arithmetic in $F_{p^n}$ modular p? Why is it that it's prime field $F_{p}$ forces upon $F_{p^n}$ this particular arithmetic? I have seen that $Z/pZ \cong$ primefield of $F_{p^n}$. But i doesn't convince that the binary operation + on $F_{p^n}$ is mod p.
EDIT: the example earlier included is deleted as it weren't that important for the problem itself
If $\mathbb F$ is a field (or more general a ring) with $\text{char}({\mathbb F})=p>0$ then for each $x\in\mathbb F$ we have: $$\stackrel{\overbrace{p\text{ times}}}{x+\cdots+x}=\stackrel{\overbrace{p\text{ times}}}{\left(1_{\mathbb F}+\cdots+1_{\mathbb F}\right)}.x=0_{F}.x=0_{F}$$
This works in field $\mathbb F_{p^n}$ since $\text{char}({\mathbb F_{p^n}})=p>0$.
The information $\text{char}(\mathbb F)=p$ is actually saying us that the unique homomorphims $f:\mathbb Z\to\mathbb F$ has ideal $\langle p\rangle$ as its kernel. Based on that we find: $$\stackrel{\overbrace{p\text{ times}}}{1_{\mathbb{F}}+\cdots+1_{\mathbb{F}}}=f\left(\stackrel{\overbrace{p\text{ times}}}{1+\cdots+1}\right)=f\left(p\right)=0_{F}$$