Prove that polynomial x^2+x+2 is a primitive over Z_5 (The polynomial p(x)=x is the generator of multiplicative group of the field
Z_5[x]/<x^2+x+2>)
Don't have a clue?
it's very simple: an element of $\displaystyle \mathbb{F}=\frac{\mathbb{Z}_5[x]}{<x^2+x+2>}$ is in the form $\displaystyle ax + b \ + <x^2+x+2>, \ a,b \in \mathbb{Z}_5,$ which for simplicity i'll write it as $\displaystyle ax+b.$ so in $\displaystyle \mathbb{F}$ we have: $\displaystyle x^2=-x-2.$ we want to show
that $\displaystyle x^n \neq 1, \ \text{for} \ n=1,2,3,4,6,12.$ it's clear for $\displaystyle n=1.$ now we have: $\displaystyle x^2=-x-2 \neq 1, \ \ x^3=2-x \neq 1, \ \ x^4 = 3x+2 \neq 1, \ \ x^6 = 2 \neq 1, \ \ x^{12}=4 \neq 1.$ hence the order of $\displaystyle x$ in $\displaystyle \mathbb{F}^{\times}$ is 24. Q.E.D.