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Let $I$ be a proper ideal of $R$. Prove that $R[X]/(X)$ is isomorphic to $R$.

I have no idea how to do this.

Brian Fitzpatrick
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user130355
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  • Did you mean $, I = (X),?\ \ $ – Bill Dubuque Mar 12 '14 at 01:07
  • You can say so if that makes sense. I have no idea why it involves I.. – user130355 Mar 12 '14 at 01:13
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    if you want some sort of intuition, think about it like this: What does an element of R[x] look like? Well, $p(x) \in R[x]$ is a polynomial with coefficients in R. So consider the element $r_0*x^n \in R[x]$. What is this quotient ring? Well it's the reduction of all multiples of x, in R[x], to 0. So the element $r_0 x^n mod x \equiv r_0 0 = 0$. So all nonconstant terms, i.e. all terms containing an x, are reduced to 0. So all you're left with is the constant term, and that will equal R once you've considered all elements of R[x]. – malxmusician212 Mar 12 '14 at 01:16

1 Answers1

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Hint $\ $ Consider the evaluation map $\,f(x)\mapsto f(0).\,$ Apply the First Isomorphism Theorem.

Bill Dubuque
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