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

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 $$r_1 \notin I_1$$.)

• @EdwardH. $2\Bbb Z$ is not isomorhic to $3\Bbb Z$ (as non-unital commutative rings). – Arthur Oct 23 '19 at 7:04
• Right. Nvm I see. – Edward H Oct 23 '19 at 7:05
• @Arthur $2 \mathbb{Z}$ is not isomorphic to $3\mathbb{Z}$, this was the problem I started with. I solved that problem, but I also thought about this problem. – MUH Oct 23 '19 at 7:07

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 $$R/I_2\cong R$$.
$$\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$$ with ideals $$\{0\} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$$ and $$\{0\} \times \{0\} \times \mathbb{Z}^\mathbb{N}$$.
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).