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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?

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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.

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  • $\begingroup$ 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

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