Do isomorphic ideals yield isomorphic quotient rings?

Question

I am interested in a proof or disproof of the following statement:

Conjecture: If $A$ and $B$ are isomorphic ideals of a ring $R$ then the quotient ring $R/A$ is isomorphic to $R/B$. $$A\cong B\implies R/A\cong R/B$$

I start by assuming the existence of an isomorphism $\phi$ from $A$ to $B$. Then I define the mapping $$\Psi:r+A\mapsto \phi(r)+B.$$

1. $\Psi$ is well-defined.

Suppose $r+A=s+A$ for some $s,r\in R$.

$$ r-s\in A\implies \phi(r)-\phi(s)=\phi(r-s)\in B.$$

Thus $\phi(r)+B=\phi(s)+B$.

2. $\Psi$ is a homomorphism.

   • $$\Psi(r+s+A)=\phi(r+s)+B$$ $$=\phi(r)+\phi(s)+B=\Psi(r+A)+\Psi(s+A).$$
   • $$\Psi(rs+A)=\phi(rs)+B$$ $$\phi(r)\phi(s)+B=\Psi(r+A)\Psi(s+A).$$

3. $\Psi$ is one-to-one.

If $\Psi(r+A)=\Psi(s+A)$ then $\phi(r)+B=\phi(s)+B$ which implies $\phi(r-s)\in B$.

How can I get $r-s\in A$?

4. $\Psi$ is onto.

For any $t+B\in R/B$ choose $\phi^{-1}(t)+A\in R/A$. $\quad\square$

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Questions: Is this work valid? Do I have sufficient conditions?

Answer 1

Why not take $(3)$ and $(10)$ in $\mathbb Z$ which are isomorphic and consider the quotients. One quotient is a PID, the other is not.

Answer 2

If isomorphism of ideals means isomorphism as $R$-modules, then here's a very simple counter-example:

Take $R=\mathbb{Z}$. Note that $2\mathbb{Z} \cong \mathbb{Z}$ as $\mathbb{Z}$-modules but $\mathbb{Z}_2 \not\cong 0$.

Here's another counter-example which shows that two ideals can be isomorphic as rings, but their corresponding quotient rings can fail to be isomorphic.

Take $\bar{R}=\bigoplus_{i=1}^{\infty} R$ with component-wise addition and multiplication in $R$ where $R$ is supposed not to be the zero ring.

Now take $I = \bar{R}$ and take $J=\{(0,a_1,a_2,\cdots): a_n\in R\}$ which is obtained by shifting components to the right by one step. Notice that $J$ is indeed an ideal of $\bar{R}$.

Then $I \cong J$ as rings with component-wise addition and multiplication that is induced from $R$. Define $\varphi: I \to J$ given by $(a_1,a_2,\cdots) \mapsto(0,a_1,a_2,\cdots)$. It can easily be verified that $$\varphi(a_1+a_1',a_2+a_2',\cdots)=\varphi(a_1,a_2,\cdots)+\varphi(a_1',a_2',\cdots)$$ $$\varphi(a_1\cdot a_1',a_2\cdot a_2',\cdots)=\varphi(a_1,a_2,\cdots)\cdot\varphi(a_1',a_2',\cdots)$$

You can see that $\ker{\varphi}=(0,0,\cdots)=\bar{0}$ and you can easily verify that $\psi: J \to I$ given by $(0,a_1,a_2,\cdots) \mapsto (a_1,a_2,\cdots)$ is a right inverse for $\varphi$. Hence, $\varphi$ is surjective too and it's a ring isomorphism.

but $\bar{R} / I = 0$ while $\bar{R}/J \not\cong 0$. Done.

Answer 3

Here's another answer, and let me be clear: I don't think it's all that great. I think stressed out's answer has the actual good counterexample. I am going to cheat a lot by reducing the problem down to abelian groups, where it is plainly false. In this answer, I am assuming:

• rings need not have identity, and
• the isomorphism between the ideals was to be a ring iso.

The point will be that rings with trivial multiplication make your conjecture false, and there are tons of counterexamples. If you do not like this and you want a counterexample with a more "honest" ring and not so much cheating, then you have other answers in this thread. However, this answers a question stressed out and I had on the side about the existence of finite counterexamples.

Given any abelian group $K$, write $K^T$ for the ring with $K$ as the underlying abelian group and multiplication $xy=0$ for all $x, y \in K$ (upper $T$ for "trivial"). The following are easy to prove:

Claim: If $A$ is any subgroup of $K$, then $A^T$ is an ideal of $K^T$.

Proof: Subgroups must contain $0$.

Claim: Every isomorphism $\phi: K \to L$ of abelian groups is also a ring isomorphism $\phi: K^T \to L^T$.

Proof: Well sure, since $\phi(xy)=\phi(0)=0$ and $\phi(x)\phi(y)=0$.

This shows that you can reduce your question down to abelian groups: if we have an abelian group $K$ with isomorphic subgroups $A$ and $B$, must it be true that $K/A$ and $K/B$ are isomorphic? If not, then $K^T$ is a ring with ideals $A^T$ and $B^T$ giving a counterexample to your original question.

So, here's an abelian groups counterexample. Take $K =\mathbb{Z}_2 \times \mathbb{Z}_4$ with subgroups $A= \mathbb{Z}_2 \times \{0\}$ and $B=\{0\} \times \{0,2\}$. Clearly $A$ and $B$ are subgroups and must be isomorphic since they are each groups of order $2$. However, $K/A \cong \mathbb{Z}_4$ yet $K/B \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. As the former is cyclic and the latter is not, the quotients cannot be isomorphic.

Answer 4

Let $R={\mathbb Z}\times 2{\mathbb Z}$, $I=2{\mathbb Z}\times\{0\}$ and $J=\{0\}\times 2{\mathbb Z}$. It is obvious that $I\simeq 2{\mathbb Z}\simeq J$.

Define $\varphi:R\to {\mathbb Z}_2\times 2{\mathbb Z}$ by $\varphi((a,2b))=([a]_2,2b)$, then we have $$ \begin{array}{cl} \varphi((a,2b)+(c,2d))&=\varphi((a+c,2(b+d)))=([a+c]_2,2(b+d)) \\ &=([a]_2+[c]_2,2b+2d)=([a]_2,2b)+([c]_2,2d) \\ &=\varphi((a,2b))+\varphi((c,2d)) \end{array}, $$ $$ \begin{array}{cl} \varphi((a,2b)(c,2d))&=\varphi((ac,4bd))=\varphi((ac,2(2bd))) \\ &=([ac]_2,2(2bd))=([a]_2[c]_2,(2b)(2d)) \\ &=([a]_2,2b)([c]_2,2d)\\ &=\varphi((a,2b))\varphi((c,2d)) \end{array}. $$ Thus, $\varphi$ is a ring epimorphism. Let $(x,2y)\in {\rm Ker}(\varphi)$, then $(0,0)=\varphi(x,2y)=([x]_2,2y)$ and so $x\in 2{\mathbb Z}, y=0$. We can conclude that ${\rm Ker}(\varphi)=2{\mathbb Z}\times\{0\}=I$. Therefore, $R/I\simeq {\mathbb Z}_2\times 2{\mathbb Z}$.

Define $\psi:R\to {\mathbb Z}$ by $\psi((a,2b))=a$, then we have $$ \begin{array}{cl} \psi((a,2b)+(c,2d))&=\psi((a+c,2(b+d))) \\ &=a+c \\ &=\psi((a,2b))+\psi((c,2d)) \end{array}, $$ $$ \begin{array}{cl} \psi((a,2b)(c,2d))&=\psi((ac,4bd)) \\ &=ac \\ &=\psi((a,2b))\psi((c,2d)) \end{array}. $$ Thus, $\psi$ is a ring epimorphism. Let $(x,2y)\in {\rm Ker}(\psi)$, then $0=\psi(x,2y)=x$ and so $x=0$, $2y\in 2{\mathbb Z}$. We can conclude that ${\rm Ker}(\psi)=\{0\}\times 2{\mathbb Z}=J$. Therefore, we have $R/J\simeq {\mathbb Z}$.

Hence, $R/I\simeq {\mathbb Z}_2\times 2{\mathbb Z}\not\simeq {\mathbb Z}\simeq R/J$.