# Help with calculating the product of cosets of a polynomial ring.

Description of Ring and Ideal

First I'm sorry this is my first post and I don't how to entirely format math equations yet, but I'm trying to figure it out.

But I am having a difficult time understanding this specific case of a quotient ring. I know that generally speaking the product of elements of a quotient ring are

$${(a + I)(b + I) = (ab + I)}$$

where I is an ideal in the ring.

So for my specific question in the quotient ring R[x]/((x^2)-4) the problem states that each coset has the unique form $${a + bx + (x^2-4)}$$ given that a and b are real numbers. The problem then asks to express the product of two of these cosets

$${[a + bx + (x^2-4)] * [c + dx + (x^2-4)]}$$

in the same form, that is, to find the product in the form $${e + fx + (x^2-4)}$$ and write out what e and f are in terms of a,b,c, and d.

What I have so far is this:

I'm essentially taking $${(a + I)(b + I) = (ab + I)}$$ to be

$${(a + bx + (x^2-4))(c + dx + (x^2-4))}$$

and then

$${(a + bx)(c + dx) = ac + x(ad + cb) + bdx^2}$$

But from my understanding of how the quotient ring works is that since the principal ideal is $${(x^2-4)}$$ then that leads to an additional condition that we have $${x^2-4 = 0}$$ which implies $${x^2 = 4}$$. So then my equation above becomes

$${ac + x(ad + cb) + bdx^2 = ac + x(ad + cb) + 4bd}$$

and so in the form the question is asking me it would then be that for the product of the two cosets in terms of a,b,c, and d we have that

$${e + fx + (x^2-4)}$$ $${e = 4bd + ac}$$ $${f = ad + cb}$$

Is this the correct way of going about multiplying two elements of this quotient ring? Or am I entirely missing the process here?

• What do you doubt about your work? – Bill Dubuque Dec 7 '20 at 9:24
• Looks good to me. – Gerry Myerson Dec 7 '20 at 9:37
• I was both unsure of only taking the first two expressions and leaving out the (x^2-4) and having the condition of x^2 = 4. I have been having an incredibly difficult time understand abstract algebra and wrapping my head around the concepts. – Z Ham Dec 7 '20 at 19:08

You can't do $$x^2=4$$, only “sort of”. What you do know is that $$x^2-4\in(x^2-4)=I$$
and therefore $$x^2+I=4+I$$.
So when you multiply $$(a+bx+I)(c+dx+I)=ac+(ad+bc)x+bdx^2+I$$ you can use $$bdx^2+I=4bd+I$$ and you end up with $$(a+bx+I)(c+dx+I)=(ac+4bd)+(ad+bc)x+I$$