Suppose we have a ring R and we form a polynomial ring in R, R[x]. In class my professor has explained that forming the quotient ring of R with some polynomial f(x), is like adding the zero of that polynomial to the ring. What is really going on in the background with this intuition?
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Asked and answered here. – Bill Dubuque Feb 13 '14 at 04:51
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Not sure if this is what you're after, but take a simple example: say $R={\Bbb Z}$ and $f(x)=x^2-2$, and let $\alpha=\sqrt2$ be one of its roots. In $R[\sqrt2]$ we can simplify by using the fact that $\alpha^2=2$, say $$(1+2\alpha)(3+4\alpha)=3+10\alpha+8\alpha^2=19+10\alpha\ .$$ In $R[x]/\langle f(x)\rangle$ we simplify by removing multiples of $f(x)$, for example, $$\eqalign{(1+2x)(3+4x) &=3+10x+8x^2\cr &=19+10x+8(x^2-2)\cr &\equiv19+10x\pmod{f(x)}\ .\cr}$$ Hope this sheds some light on why these two processes are essentially the same.
David
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