# Is it true that if $(b)\subset (c) \subset R$, then $(c)/(b)\simeq R/(a)$ as $R$-modules, where $a\cdot c = b$?

Is it true that if $$(b)\subset (c) \subset R$$, $$b\neq 0$$, then $$(c)/(b)\simeq R/(a)$$, where $$a\cdot c = b$$?

$$R$$ is ring and $$(b)$$, $$(c)$$ are its ideals, $$(c)/(b)$$ is ideal of $$R/(b)$$ by third theorem of isomorphism for rings.

We can construct homomorphism of $$R$$-modules $$\varphi: (c)/(b) \rightarrow R/(a)$$, such as $$\varphi(x\cdot c+(b)) = x+(a)$$. It's correct construction, because if $$x\cdot c + (b) = y\cdot c+ (b)$$, then $$(x-y)\cdot c \in (b)$$, then $$(x-y) \in (a)$$. I think it's also easy to prove that it is isomorphism.

• Questions should be present in the body as well. I have added that for you. Also, it would be a good idea to mention what the objects are. From the context, I can figure out that $R$ is probably a ring, and $(b)$, $(c)$ are ideals. But what is $(b)/(c)$? Is that a quotient of $R$-modules? Is the isomorphism one of $R$-modules? Commented Oct 17, 2021 at 22:05
• I mean quotient ring. Can't we say about quotient ring $(b)/(c)$, if $(c)\subset (b)\subset R$, $(c)$ and $(b)$ are ideals of $R$? I thought it's the same object as is in third theorem of isomorphism for rings (if $I\subset J\subset R$ and $I$, $J$ are ideals of $R$, then $J/I$ is ideal of $R/I$). Commented Oct 17, 2021 at 22:13
• How is $(b)/(c)$ a ring? I agree that $(b)/(c)$ is an ideal of the ring $R/(c)$. (Of course, weird things can happen with different conventions. For example, if you consider rings without identity. Then, an ideal could also be considered as a subring...) Commented Oct 17, 2021 at 22:15
• Yes, I mean ring without identity, so ideal is a ring without identity. Commented Oct 17, 2021 at 22:18
• I have updated my answer, to include the case of $R$-modules. Commented Oct 17, 2021 at 23:02

Consider $$R=\Bbb Z$$, $$b=4, c=2$$, then we have $$a=2$$. But $$2\Bbb Z/4\Bbb Z$$ is not isomorphic to $$\Bbb Z/2\Bbb Z$$ as rings without identity.

• Are they isomorphic as $R$-modules? Commented Oct 17, 2021 at 22:38
• @JuljaMuvv: Yes, since the group structure of both is the same. (There is only one group with $2$ elements and being a $\Bbb Z$-module only talks about the group structure.) Commented Oct 17, 2021 at 22:40
• So is the state in my question is true, if isomorphism is of $R$-modules? Commented Oct 17, 2021 at 22:42
• @JuljaMuvv: Definitely not in the exact form it's stated. See my answer for some counterexamples where one quotient becomes $0$ whereas the other does not. Commented Oct 17, 2021 at 22:50
• Is it true, if we put restriction that $(b) \neq (0)$, $c\neq 0$? Commented Oct 17, 2021 at 22:58

Your map is not well-defined for the reason you state. More precisely if $$(x - y) c \in (b)$$, then you cannot simply conclude that $$x - y \in (a)$$.

For the moment, let us consider a more general situation: Let $$I, J \subset R$$ be ideals. Then, one defines the colon ideal $$I : J$$ as $$I : J := \{r \in R \mid rJ \subset I\}.$$

From your choice of $$a$$, it is clear that $$a \in (b : c)$$ or $$(a) \subset (b : c)$$. But later on, you seem to assume that $$(a) = (b : c)$$.
But this may not be the case. As a simple counterexample, we can take $$R = \Bbb Z$$, $$b = c = 0$$, and $$a = 43$$. (This is also a counterexample to your proof.)

As a less trivial example, we may take $$R = \Bbb Z/6 \times \Bbb Z$$, $$b = (0, 0)$$, $$c = (3, 0)$$, $$a = (2, 0)$$.

This can be avoided in nice situations, though. More precisely, we can say the following:

Theorem. Let $$R$$ be an integral domain, and let $$a, b, c \in R$$ be such that $$a \cdot c = b$$ and $$c \neq 0$$. Then, $$(b : c) = (a)$$.

Proof. $$(a) \subset (b : c)$$ is clear. Conversely, let $$x$$ be such that $$xc \in (b)$$. Then, $$xc = by$$ for some $$y \in R$$. We can write $$b = ac$$ to conclude that $$xc = acy$$. Since $$c \neq 0$$, we may cancel it to conclude that $$x = ay \in (a)$$. $$\qquad \square$$

But instead of restricting ourselves to integral domains, we can modify the question to ask:
$$\text{Is } (c)/(b) \simeq R/(b : c)?$$

However, your question is still not true (even in the case of domains). As a counterexample, you may take $$R = \Bbb Z$$, $$b = 4$$, and $$c = 2$$. In this case, the question is equivalent to asking $$2\Bbb Z/4\Bbb Z \simeq \Bbb Z/2\Bbb Z.$$ However, the above is not true. Indeed, the product of any two elements in the left ring is equal to $$0$$, which is not true for the right ring.

Added. What goes wrong? For the updated question, your map $$\varphi$$ is indeed well-defined.
However, it is not a homomorphism of rings. (It fails to be multiplicative, which is not surprising since you wouldn't expect $$(xc)(yc) = (xy)c$$.)

However, it is additive and $$R$$-linear. Thus, we may treat $$\varphi$$ as a map of $$R$$-modules. It is straightforward to note that $$\varphi$$ is injective since $$(x - y) \in (b : c) \Leftrightarrow (x - y) \cdot c \in (b).$$ Lastly, we check that it is onto. Let $$r \in R$$ be arbitrary. Then, we have $$\varphi(rc + (b)) = r + (a).$$ Thus, we have an isomorphism of $$R$$-modules.