# Prove Primitive over Z_5

• Dec 6th 2008, 11:44 AM
mandy123
Prove Primitive over Z_5
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?
• Dec 6th 2008, 12:44 PM
NonCommAlg
Quote:

Originally Posted by mandy123
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 $\mathbb{F}=\frac{\mathbb{Z}_5[x]}{}$ is in the form $ax + b \ + , \ a,b \in \mathbb{Z}_5,$ which for simplicity i'll write it as $ax+b.$ so in $\mathbb{F}$ we have: $x^2=-x-2.$ we want to show

that $x^n \neq 1, \ \text{for} \ n=1,2,3,4,6,12.$ it's clear for $n=1.$ now we have: $x^2=-x-2 \neq 1, \ \ x^3=2-x \neq 1, \ \ x^4 = 3x+2 \neq 1, \ \ x^6 = 2 \neq 1, \ \ x^{12}=4 \neq 1.$ hence the order of $x$ in $\mathbb{F}^{\times}$ is 24. Q.E.D.