Showing the following rings are isomorphic

Is it true that $\frac{\mathbb{Z}[x]}{(3,x^6+1)}\cong{\frac{\mathbb{Z_{3}[x]}}{(x^6+1)}}$? I believe that it is but am not sure how to justify it. The idea I want to use is this: $\frac{\mathbb{Z}[x]}{(3)}\cong{\mathbb{Z_{3}[x]}}$ and then use $\frac{\frac{\mathbb{Z}[x]}{(3)}}{(x^6+1)}\cong{\frac{\mathbb{Z_{3}[x]}}{(x^6+1)}}$. But now I need to show that $\frac{\mathbb{Z}[x]}{(3,x^6+1)}\cong{\frac{\frac{\mathbb{Z}[x]}{(3)}}{(x^6+1)}}$ and I'm not sure how to do this.

I would also like to know if this result can be generalized; i.e. can we replace $\mathbb{Z}$ by any integral domain $R$ and $(3, x^6+1)$ by any finitely generated ideal? If we can't, what sort of most general result can we have; Does $R$ need to be a UFD or does it need to be a PID etc.

If my guess about the rings being isomorphic is false, I'd like to see a reason for that if possible.

• I think the general fact is $R/(I+J)\cong (R/I)/((I+J)/I)$. I believe it holds for any rings and ideals. – tomasz Jun 22 '14 at 14:20
• Thanks. That works as well. – UserB1234 Jun 22 '14 at 14:30

Consider the composite surjection $$\mathbb{Z}[x] \twoheadrightarrow \mathbb{Z}_3[x] \twoheadrightarrow \mathbb{Z}_3[x]/(x^6+1).$$ What is the kernel of this map? Clearly $(3)$ is in the kernel of the composite, since it is the kernel of the first map. The kernel of the second map is $(x^6+1)$, and elements of $\mathbb{Z}[x]$ whose image is in $(x^6+1) \subset \mathbb{Z}_3[x]$ is just the ideal $(x^6+1)\subset \mathbb{Z}[x]$. Thus the kernel of the composite map is the smallest ideal containing $(3)$ and $(x^6+1)$, or $(3,x^6+1)$.