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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?

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  • $\begingroup$ What do you doubt about your work? $\endgroup$ – Bill Dubuque Dec 7 '20 at 9:24
  • $\begingroup$ Looks good to me. $\endgroup$ – Gerry Myerson Dec 7 '20 at 9:37
  • $\begingroup$ 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. $\endgroup$ – Z Ham Dec 7 '20 at 19:08
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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 $$ (a+bx+I)(c+dx+I)=(ac+4bd)+(ad+bc)x+I $$

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  • $\begingroup$ Thank you for elaborating on that concept, I wasn't sure why it was the way it was. $\endgroup$ – Z Ham Dec 7 '20 at 19:09

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