Prove/disprove: $ I_1 \cong I_2 \iff R/I_1 \cong R/I_2 $

Question

Let $I_1,I_2$ be two ideals in a ring $R$. I thought that the following result is true: $$ I_1 \cong I_2 \iff R/I_1 \cong R/I_2 $$

i.e. If the two ideals are isomorphic (since they are sub-rings as well) then their quotient rings are also isomorphic.

My attempt:

Claim: If $\psi: I_1 \to I_2$ is an isomorphism, then $\phi : R/I_1 \to R/I_2 $ s.t. $r+I_1 \mapsto r+I_2$ is an isomorphism.

I am not able to prove that this is an isomorphism. I can't even claim that this is a well defined function. Since, if $r_1 +I_1 =r_2 +I_1 \Rightarrow r_1-r_2 \in I_1 \Rightarrow \psi (r_1-r_2) \in I_2 $. But, I don't know how to proceed after this as $\psi (r_1)$ may not be defined (in case if $r_1 \notin I_1$.)

Answer 1

No, this is not the case.

Consider, for instance, $R = \Bbb Z[x_1, x_2, x_3, \ldots]$, and the two isomorphic ideals $I_1 = (x_1, x_2, x_3, \ldots)$ and $I_2 = (x_2, x_4, x_8, \ldots)$. Then $R/I_1\cong \Bbb Z$, while $R/I_2\cong R$.

Answer 2

$\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$ with ideals $\{0\} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$ and $\{0\} \times \{0\} \times \mathbb{Z}^\mathbb{N}$.

Answer 3

Consider $R=\mathbb{Q}[x_i\ : \ i\in \mathbb{N}]$, $I_1= (x_1, x_2, \dots)$ and $I_2=(x_2, x_3, \dots)$. Those are isomorphic via $x_i \mapsto x_{i+1}$. However, $I_1$ is maximal, wheras $I_2$ is not. Thus the quotients are not isomorphic (the first quotient will be a field, and the second not).