Exploring Cosets of a Ring with Division Algorithm

[Stephen88]

I'm trying to list the cosets of the following ring and describe the relations that hold between these cosets.
R=Z_4[x]/((x^2+1)*Z_4[x])
I'm using the division algorithm since x^2+1 is monic in the ring Z_4[x].Now for every f that belongs to Z_4[x] by the division algorithm
f=(x^2+1)q(x)+p(x)=>the cosets are of the the form...a*x+b+I where I is an ideal generated by x^2+1.
x^2+1=0 in the quotient=>a new ring where multiplication between cosets A+I and B+I is is defined by (A+I)(B+I)=(AB)+1 where the relation x^2+1=0 exists
Is is ok?[FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT]

 

[Swlabr1]

Re: Ring and cosest

I'm trying to list the cosets of the following ring and describe the relations that hold between these cosets.
R=Z_4[x]/((x^2+1)*Z_4[x])
I'm using the division algorithm since x^2+1 is monic in the ring Z_4[x].Now for every f that belongs to Z_4[x] by the division algorithm
f=(x^2+1)q(x)+p(x)=>the cosets are of the the form...a*x+b+I where I is an ideal generated by x^2+1.
x^2+1=0 in the quotient=>a new ring where multiplication between cosets A+I and B+I is is defined by (A+I)(B+I)=(AB)+1 where the relation x^2+1=0 exists
Is is ok?

They way I would think about the elements of the quotient ring is much the same way that you would think about the elements of, say, $\mathbb{Z}/4\mathbb{Z}$. This is because they are "essentially" the same thing - you have the same division algorithm, etc.

So, every coset has a (unique!) representative of the form $ax+b$. So when you multiply $(A+I)(B+I)=AB+I$ then you do the division algorithm on $AB$ to get an element of the form $ax+b$ with $ax+b=AB\text{ mod }x^2+1$.

Does that make sense?

 

[Stephen88]

Re: Ring and cosest

So I need to apply the division algorithm on (AB)+I...ok

 

[Swlabr1]

Re: Ring and cosest

So I need to apply the division algorithm on (AB)+I...ok

Essentially, yes. For example, $(x+I)\cdot (x+1+I)=x^2+x+I=x-1+I=x+3+I$, as you know that $x^2=-1\text{ mod }I$ because $x^2+1\in I$

 

[blue_raver22]

Yes, your approach using the division algorithm is correct. The cosets of R are of the form a*x+b+I, where I is the ideal generated by x^2+1. This means that each coset can be represented by a polynomial of degree 1 or less, where the coefficients are elements of Z_4.

The relation x^2+1=0 in the quotient ring means that any polynomial of degree 2 or higher can be reduced to a polynomial of degree 1 or less by dividing it by x^2+1. This is similar to how in the real numbers, any polynomial of degree n can be reduced to a polynomial of degree n-1 by dividing it by a root of the polynomial.

In the quotient ring, multiplication between cosets is defined by (A+I)(B+I)=(AB)+I, where the relation x^2+1=0 is taken into account. This means that when multiplying two cosets, we first multiply their representative polynomials, and then reduce the result by dividing it by x^2+1.

Overall, exploring the cosets of a ring using the division algorithm allows us to understand the structure and relations within the ring, and can be a useful tool in solving problems and proving theorems.