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Finding elements in a quotient ring

hmmm16

Member
Feb 25, 2012
31
I have to describle the elements of the quotient ring $$\frac{\mathbb{Z}[x]}{2\mathbb{Z}[x]+x^2\mathbb{Z}[x]}$$ do I use the division algorithm to write any element as $$f(x)=(x^2+2)q(x)+r(x)$$ where $\operatorname{deg}(r(x))<2$ so the elements of the ring are the linear polynomials over $\mathbb{Z}$?

Thanks for any help
 
Last edited:

Swlabr

New member
Feb 21, 2012
27
I have to describle the elements of the quotient ring $$\frac{\mathbb{Z}[x]}{2\mathbb{Z}[x]+x^2\mathbb{Z}[x]}$$ do I use the division algorithm to write any element as $$f(x)=(x^2+2)q(x)+r(x)$$ where $deg(r(x))<2$ so the elements of the ring are the linear polynomials over $\mathbb{Z}$?

Thanks for any help
What you suggest doing will find the coset representative of your given polynomial. Remember, you are no longer dealing with polynomials, but with cosets!

So, the elements of your ring are the cosets $[r(x)]=\{f(x): (x^2+2)q(r)+r(x)\}$.

(As a LaTeX side-point, instead of writing deg, write \operatorname{deg}. This is, strictly speaking, correct, because it looks nicer and is distinguishes the writing from the variables. It will look like $\operatorname{deg}(r(x))$ as opposed to $deg(r(x))$.)
 

hmmm16

Member
Feb 25, 2012
31
Ok, thanks so my ring would be the infinite ring with elements $[r(x)]$ with $r(x)$ a linear polynomial, which is the coset representative for all $f(x)\in\mathbb{Z}[x]$ such that $f(x)=(x^2+1)q(x)+r(x)$ for some $q(x)$

(Thanks for the LaTex tip, I've edited it but the LaTex doesn't seem to be rendering properly for me at the moment)
 

Swlabr

New member
Feb 21, 2012
27
Ok, thanks so my ring would be the infinite ring with elements $[r(x)]$ with $r(x)$ a linear polynomial, which is the coset representative for all $f(x)\in\mathbb{Z}[x]$ such that $f(x)=(x^2+1)q(x)+r(x)$ for some $q(x)$

(Thanks for the LaTex tip, I've edited it but the LaTex doesn't seem to be rendering properly for me at the moment)
Basically, yeah. The coset is "what is happening", but for all practical purposes your ring is the set of all polynomials of degree $1$ where multplication is modulo $x^2+1$. I mean, when you are working in the integers mod $n$ you are really working with cosets of numbers, but you don't ever think about it that way...

The LaTeX here takes a while to render properly, I've found.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
I have to describle the elements of the quotient ring $$\frac{\mathbb{Z}[x]}{2\mathbb{Z}[x]+x^2\mathbb{Z}[x]}$$ do I use the division algorithm to write any element as $$f(x)=(x^2+2)q(x)+r(x)$$ where $\operatorname{deg}(r(x))<2$ so the elements of the ring are the linear polynomials over $\mathbb{Z}$?

Thanks for any help
If I understand this question correctly, you need to be careful in interpreting the ideal $2\mathbb{Z}[x]+x^2\mathbb{Z}[x]$, which is not the same as the ideal $(2+x^2)\mathbb{Z}[x].$ An element of $(2+x^2)\mathbb{Z}[x]$ looks like $(x^2+2)q(x)$, but an element of $2\mathbb{Z}[x]+x^2\mathbb{Z}[x]$ looks like $x^2q_1(x) + 2q_2(x).$

When you form the quotient ring $\dfrac{\mathbb{Z}[x]}{2\mathbb{Z}[x]+x^2\mathbb{Z}[x]}$, the intuitive way to think about it is that you start with $\mathbb{Z}[x]$, and you identify everything in the ideal to 0. The fact that $x^2$ is in the ideal means that the quotient ring will only contain (cosets of) linear polynomials. And the fact that $2$ is in the ideal means that the scalar ring gets reduced to the two-element field $\mathbb{Z}/2\mathbb{Z}$. So the quotient ring will just consist of the four elements $0,\,1,\,x,\,x+1.$
 

Swlabr

New member
Feb 21, 2012
27
If I understand this question correctly, you need to be careful in interpreting the ideal $2\mathbb{Z}[x]+x^2\mathbb{Z}[x]$, which is not the same as the ideal $(2+x^2)\mathbb{Z}[x].$ An element of $(2+x^2)\mathbb{Z}[x]$ looks like $(x^2+2)q(x)$, but an element of $2\mathbb{Z}[x]+x^2\mathbb{Z}[x]$ looks like $x^2q_1(x) + 2q_2(x).$

When you form the quotient ring $\dfrac{\mathbb{Z}[x]}{2\mathbb{Z}[x]+x^2\mathbb{Z}[x]}$, the intuitive way to think about it is that you start with $\mathbb{Z}[x]$, and you identify everything in the ideal to 0. The fact that $x^2$ is in the ideal means that the quotient ring will only contain (cosets of) linear polynomials. And the fact that $2$ is in the ideal means that the scalar ring gets reduced to the two-element field $\mathbb{Z}/2\mathbb{Z}$. So the quotient ring will just consist of the four elements $0,\,1,\,x,\,x+1.$
I should learn to read questions fully...
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
since the quotient ring is finite of order 4, with characteristic 2, one might wonder if we have the finite field of order 4.

the answer is no: one way to see this is that x (or rather, the coset of x, [x]) is a zero-divisor, since [x][x] = [x2] = [0].

another way to see this is that the ideal (2,x2) is not maximal, as it is contained in the proper ideal (2,x), and x is in the latter, but not the former.