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.