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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?2012-12-13
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    @RedPotatoe,You wrote "Prove that A[x]/ is isomorphic to A"...something seems to be missing, or else the claim's false.2012-12-13
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    Are you asking about the definition of ring isomorphism? A ring isomorphism is a _bijective_ ring homomorphism.2012-12-13
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    I believe is the principal ideal generated by the element x. Is that correct? So A[x]/(principal ideal generated by x) is isomorphic to A.2012-12-13
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    Yes, that is your question.2012-12-13
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    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.2012-12-13
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    @Julian Kuelshammer. Thank you Julian. In order to prove that, I must show that in fact I do have an isomorphism. In order to do that I must show that the sum is mapped to the sum and the product is mapped to the product. Right?2012-12-13
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    How about proving something like $\langle x \rangle = \ker (\phi:p(x)\mapsto p_0)$?2012-12-13
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    @RedPotatoe Yes, and you must prove that the function you define is bijective.2012-12-13
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    @Neal Thanks Neal!2012-12-13

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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 $\mathrm{im}f = A$.