- Thread starter
- #1

- Jun 22, 2012

- 2,918

I tried to first deal with the rings [TEX] \mathbb{R}[x]/<x^2> [/TEX] and [TEX] \mathbb{R}[x]/<x^2 +1> [/TEX] as they seemed easier to deal with ... my thinking ... and my problems are as follows: (Would really appreciate clarification)

Following an example I found in Gallian (page 257), first consider [TEX] \mathbb{R}[x]/<x^2> [/TEX] where [TEX] \mathbb{R}[x] [/TEX] is the ring of polynomials with real co-efficients.

Then [TEX] \mathbb{R}[x]/<x^2> = \{ g(x) + <x^2> | g(x) \in \mathbb{R}[x] \} [/TEX]

But [TEX] \mathbb{R}[x] [/TEX] is a Euclidean Domain and hence possesses a Division Algorithm, so we may write:

[TEX] g(x) = q(x)(x^2) + r(x) [/TEX] where r(x) = 0 or r(x) has degree less than 2.

so we can write r(x) = ax + b where a, b [TEX] \in \mathbb{R} [/TEX]

Thus [TEX] g(x) + <x^2> = q(x)(x^2) + r(x) + <x^2> [/TEX]

= [TEX] r(x) + <x^2> [/TEX] since the ideal [TEX] <x^2> [/TEX] absorbs the term [TEX] q(x)(x^2) + r(x) [/TEX]

= [TEX] ax + b + <x^2> [/TEX]

Thus [TEX] \mathbb{R}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{R} \}[/TEX]

Now, by a similar argument we can demonstrate that

[TEX] \mathbb{R}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{R} \} [/TEX]

which makes the two rings [TEX] \mathbb{R}[x]/<x^2> [/TEX] and [TEX] \mathbb{R}[x]/<x^2 +1> [/TEX] look to have the same structure?

One of my questions is how exactly are these two ring structures different?

A second worry is that the above demonstration works because [TEX] \mathbb{R}[x] [/TEX] is a Euclidean Domain ... so the same argument as above does not apply to

[TEX] \mathbb{Z}[x] [/TEX] because [TEX] \mathbb{Z}[x] [/TEX] is not a Euclidean Domain and hence we cannot use the Division algorithm.

How do we rigorously demonstrate that

[TEX] \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} [/TEX] and

[TEX] \mathbb{Z}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z} \} [/TEX]

Can someone please help clarify the above problems and issues?

Peter