# Proving a primitive element

• Apr 1st 2011, 06:43 AM
dollydaggerxo
Proving a primitive element
Hi, I need to show that $\alpha+1$=[x] is a primitive element of GF(9)= $\mathbb{Z}_3[x]/$
I have already worked out that the function in the < > is irreducible but I do not know where to go from this.

there are 8 elements in the multiplicative group, what would they be? I guess it would be: 0, 2, x+2, what else? Im very unsure how to do this.

• Apr 1st 2011, 07:16 AM
Swlabr
The elements of your multiplicative group are,

1, 2, x, x+1, x+2, 2x, 2x+1 and 2x+2. Not zero (as then your multiplicative group would not be a group...). Do you see where I got these elements from? (Strictly speaking, I should have brackets around these, like [1], [2], etc, but that doesn't really matter.)

So, what is the order of x+1 (I presume that is the element you are trying to show is primitive)? That is, is $(x+1)^2=1$? $(x+1)^4=1$? What can you conclude?
• Apr 1st 2011, 02:09 PM
Deveno
right, in Z3[x]/(x^2 + x + 2) = Z3[x]/I, we really ought to write x + I, or α instead of x, but it is a common abuse of notation.

let's start with (α+1)^2 = α^2 + 2x + 1 = (-α - 2) + 2α + 1 = 2α + 1 + 2α + 1 = α + 2.

make life easier on yourself. GF(9) is a field with 9 elements. thus the multiplicative group of GF(9) is cyclic of order 8.

so what are the possible orders of α+1? if there are n possibilities, can you see it suffices to just check n-1 of them?

basically we identify GF(9) with Z3[α] where α is a root of x^2 + x + 2. since Z3[α] is an extension field of Z3, it is a vector space

of dimension deg(x^2 + x + 1) = 2 over Z3, with basis {1, α}. we can explictly list the elements of Z3 x Z3:

(0,0) <--> 0 + 0α = 0
(1,0) <--> 1 + 0α = 1
(2,0) <--> 2 + 0α = 2
(0,1) <--> 0 + 1α = α
(1,1) <--> 1 + 1α= 1 + α
(2,1) <--> 2 + 1α = 2 + α
(0,2) <--> 0 +2α = 2α
(1,2) <--> 1 + 2α
(2,2) <--> 2 + 2α (since vector addition is abelian, we can re-write c+dα as dα + c, if we wish).

the multiplicative group elements are all the non-zero ones.