What are the primitive elements in GF(9)?

[Dollydaggerxo]

Homework Statement

Hi, I need to show that $$\alpha+1$$=[x] is a primitive element of GF(9)= $$\mathbb{Z}_3[x]/<x^{2}+x+2>$$
I have already worked out that the function in the < > is irreducible but I do not know where to go from this.

Homework Equations

there are 8 elements in the multiplicative group, what would they be?

The Attempt at a Solution

I guess it would be: 0, 2, x+2, what else? I am very unsure how to do this.

please help me would be appreciated thanks

 

[micromass]

You'll need to show that [x] generates the elements of the multiplicative group. The way of doing this is just compute all the powers of [x] and see if they are indeed the elements that you want... So, you'll need to compute [x]⁰, [x]¹, [x]², [x]³, [x]⁴, [x]⁵, [x]⁶ and [x]⁷.

 

[Dollydaggerxo]

brilliant thank you.
how do i show that these elements that i have generated are the actual elements though? or is it enough just to say they are?

and the fact that I haven't got the same result twice from 0 to 7 means that it is primitive?
thanks so much

 

[micromass]

brilliant thank you.
how do i show that these elements that i have generated are the actual elements though? or is it enough just to say they are?

I don't know what you mean with this? The elements that you generated are the actual elements because [x]ⁿ is just n-times multiplication in the group GF(9). So the elements remain in GF(9) and are the actual elements.

I probably misunderstand something...

and the fact that I haven't got the same result twice from 0 to 7 means that it is primitive?

Exactly!

 

[Dollydaggerxo]

Haha ok just ignore me, i think i get it now anyway. many thanks for your help its much appreciated!