What is the Basis of a Quotient Ring?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
BVM
Messages
9
Reaction score
0
In my Abstract Algebra course, it was said that if
[tex]E := \frac{\mathbb{Z}_{3}[X]}{(X^2 + X + 2)}.[/tex]
The basis of E over [itex]\mathbb{Z}_{3}[/itex] is equal to [itex][1,\bar{X}][/itex].
But this, honestly, doesn't really make sense to me. Why should [itex]\bar{X}[/itex] be in the basis without it containing any other [itex]\bar{X}^n[/itex]? How did they arrive at that exact basis?
 
Physics news on Phys.org
Thank you for replying.

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

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