When we work with polynomials modulo $m(x)$ where m(x) is a primitive polynomial over $GP(p^{m})$, i know that i can take every coefficient of any polynomial and replace it with it's congruent modulo $p$. (i don't know what's the point in doing this,i think it's a definition).
Moreover, is this also valid for exponents? And why ?
For example, we have $p(X) = x^{3}+x+1$ over $GP(2^{3})$ and a is the primive element. Then why $a^{8} \equiv a$ ?
When evaluating polynomials over a finite field of size $q$ at a point $b\ne0$, you can replace $b^{q-1}$ by $1$, by Lagrange's theorem applied to the multiplicative group of the field. In other words, you can reduce exponents mod $q-1$.
For instance, $a^{8} = a$ because $a^{7} = 1$ in $GP(2^{3})$.
The point is: $x^8-x$ is the zero function but not the zero polynomial.