What is the Basis of a Quotient Ring?

Click For Summary
SUMMARY

The basis of the quotient ring E = \frac{\mathbb{Z}_{3}[X]}{(X^2 + X + 2)} is established as [1, \bar{X}]. This conclusion arises from the ability to express higher powers of \bar{X}, such as \bar{X}^2, in terms of the basis elements. Specifically, \bar{X}^2 can be rewritten as 2\bar{X} + \bar{1} in \mathbb{Z}_3, demonstrating that the basis does not require additional elements beyond 1 and \bar{X}. The misconception regarding negativity in \mathbb{Z}_3 was clarified, emphasizing that addition of multiples of 3 can be utilized to simplify expressions.

PREREQUISITES
  • Understanding of Abstract Algebra concepts
  • Familiarity with quotient rings
  • Knowledge of polynomial rings over finite fields
  • Basic operations in modular arithmetic, specifically in \mathbb{Z}_3
NEXT STEPS
  • Study the structure and properties of quotient rings in Abstract Algebra
  • Explore polynomial ring operations in \mathbb{Z}_p for various primes p
  • Learn about linear combinations and bases in vector spaces over finite fields
  • Investigate the implications of modular arithmetic on polynomial identities
USEFUL FOR

Students of Abstract Algebra, mathematicians exploring ring theory, and anyone interested in the properties of polynomial rings and finite fields.

BVM
Messages
9
Reaction score
0
In my Abstract Algebra course, it was said that if
E := \frac{\mathbb{Z}_{3}[X]}{(X^2 + X + 2)}.<br />
The basis of E over \mathbb{Z}_{3} is equal to [1,\bar{X}].
But this, honestly, doesn't really make sense to me. Why should \bar{X} be in the basis without it containing any other \bar{X}^n? How did they arrive at that exact basis?
 
Physics news on Phys.org
Can you write \overline{X}^2 in terms of \overline{X} and 1??
 
Thank you for replying.

I've solved the problem. Whereas I previously thought that I couldn't write any \bar{X}^n in terms of 1 and \bar{X} I've since realized that for instance: \bar{X}^2 = -\bar{X}-\bar{2} = 2\bar{X}+\bar{1}.

My initial mistake as to think that in \mathbb{Z}_3 we can't define the negativity resulting in subtracting that polynomal, but obviously you can just add any 3n to it.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
988
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K