# Isomorphism between quotient rings and modules

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.
• 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 – ErlGrey Apr 23 at 15:57