What are the primitive elements in GF(9)?

Click For Summary
SUMMARY

The discussion centers on demonstrating that the element \(\alpha + 1 = [x]\) is a primitive element of the Galois Field GF(9), defined as \(\mathbb{Z}_3[x]/\langle x^{2}+x+2 \rangle\). Participants confirm that the irreducibility of the polynomial \(x^2 + x + 2\) allows for the generation of all elements in the multiplicative group of GF(9). The key conclusion is that computing the powers of \([x]\) from \([x]^0\) to \([x]^7\) yields distinct elements, confirming \([x]\) as a primitive element.

PREREQUISITES
  • Understanding of Galois Fields, specifically GF(9)
  • Knowledge of polynomial irreducibility in \(\mathbb{Z}_3[x]\)
  • Familiarity with group theory concepts, particularly multiplicative groups
  • Ability to compute powers of elements in finite fields
NEXT STEPS
  • Study the properties of Galois Fields, focusing on GF(p^n) structures
  • Learn about polynomial factorization in \(\mathbb{Z}_p[x]\)
  • Explore the concept of primitive elements in finite fields
  • Investigate the application of Galois Fields in coding theory and cryptography
USEFUL FOR

This discussion is beneficial for students and researchers in abstract algebra, particularly those studying finite fields and their applications in mathematics and computer science.

Dollydaggerxo
Messages
60
Reaction score
0

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
 
Physics news on Phys.org
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.
 
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
 
Dollydaggerxo said:
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!
 
Haha ok just ignore me, i think i get it now anyway. many thanks for your help its much appreciated!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K