Let $I_1, I_2$ be ideals of $R$ — associative ring with unit. Find an example where $R/ I_1$ and $ R/I_2$ isomorphic as rings, but not isomorphic as modules. Can you check my solution?

I have an example: let $R = \mathbb{Z_2}[x]$ and it is obvious that $\mathbb{Z_2}[x]/x^2\cong \mathbb{Z_2}[x]/(x^2+1)$ as rings. I want to show that they are not isomorphic as modules. Homomorphism of modules requires such equality: $af(x) = f(ax)$ where $a\in R$.

But if we take $x\in R$ and $x\in R/I_1 = R/x^2$ and any $f\in Hom(R/I_1, R/I_2)$ we can see that $xf(x) = f(x^2) = 0$. This equation will be true iff $f(x) = 0$, because $R/(x^2+1)$ is a field. But this $f$ don't set isomorphism!

Now I see that my example is incorrect, so, can you give me a hint how can I find an example, please?


I would rather say that if $f$ is an isomorphism, we have $P$ in $R/I_1$ such that $f(P)=1$. Then, the following equalities hold in $R/I_2$ : $1 = x^2 = x^2f(P) = f(x^2P) = f(0) = 0$. Your argument isn't correct because $R/(x^2+1)$ is not a field, indeed $(x^2+1)=(x+1)^2$ in $\mathbb Z_2[x]$ so $I_2$ is not a maximal ideal.

  • $\begingroup$ I see :( I didn't noticed that it is not a field. $\endgroup$ – ErlGrey Apr 23 at 15:44
  • $\begingroup$ Can I take $I_1 = x-1$ and $I_2 = x$? Then we can see that $f(1) = f(x) = xf(1)$ ($[x] = [1]$, because $x-1\in I_1$). Then we can conclude that $f(1) =0$ and $1\in ker f$ which contradicts $f$ being isomorphism $\endgroup$ – ErlGrey Apr 23 at 15:57
  • $\begingroup$ I tried to make your solution work, is it not ok ? $\endgroup$ – elidiot Apr 23 at 15:58
  • $\begingroup$ Yes, of course! Thank you so much for your solution) I was just wondering if I can choose ideals that will be work and get concrete example $\endgroup$ – ErlGrey Apr 23 at 16:01
  • $\begingroup$ Your other solution is fine ! Your welcome $\endgroup$ – elidiot Apr 23 at 16:09

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.