# How to prove $A[x]/\langle x\rangle\cong A$ for a ring $A$?

I am working on this question and I was wondering if I was on the right track.

It states: "Let $A$ be a ring. Prove that $A[x]/\langle x\rangle$ is isomorphic to $A$.

So am I on the right track in saying that I must check to see if the sum of $A[x]/\langle x\rangle$ goes to the sum of $A$, and the product of $A[x]/\langle x\rangle$ goes to the product of $A$?

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What does "goes to" mean? – Chris Eagle Dec 13 '12 at 17:51
@RedPotatoe,You wrote "Prove that A[x]/ is isomorphic to A"...something seems to be missing, or else the claim's false. – DonAntonio Dec 13 '12 at 17:51
Are you asking about the definition of ring isomorphism? A ring isomorphism is a bijective ring homomorphism. – Alex B. Dec 13 '12 at 17:53
I believe <x> is the principal ideal generated by the element x. Is that correct? So A[x]/(principal ideal generated by x) is isomorphic to A. – RedPotatoe Dec 13 '12 at 17:58
Well, technically you should say that $x$ is an indeterminate; i.e. a generator of the free $A$-algebra on one variable. Otherwise if $x\in A$ then the result is not true. – Jason Polak Dec 13 '12 at 18:01

Define $f: A[x] \to A$ as $p(x) \mapsto p(0)$. The kernel of this map is $\langle x \rangle$. Hence by the first isomorphism theorem $A[x]/ \langle x \rangle \cong A$.
It remains to be verified that $f$ is indeed a ring homomorphism that is, that the "product goes to the product" and the "sum goes to the sum" as you say, so you are on the right track. You should also convince yourself that $\mathrm{ker}f = \langle x \rangle$ and $\mathrm{im}f = A$.