What are the cosets of the ring R=Z_4[x]/((x^2+1)*Z_4[x])?

[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=MathJax_Main][FONT=MathJax_Main][FONT=MathJax_Main][FONT=MathJax_Math][FONT=MathJax_Math][FONT=MathJax_Main][FONT=MathJax_Main][FONT=MathJax_Math]

 

[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 $x^2+1\in I$