0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think the general fact is $R/(I+J)\cong (R/I)/((I+J)/I)$. I believe it holds for any rings and ideals. $\endgroup$ – tomasz Jun 22 '14 at 14:20
  • $\begingroup$ Thanks. That works as well. $\endgroup$ – UserB1234 Jun 22 '14 at 14:30
3
$\begingroup$

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 $(3,x^6+1)$.

$\endgroup$
  • $\begingroup$ Thank you. I should have caught that. $\endgroup$ – UserB1234 Jun 22 '14 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.