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.

  • $\begingroup$ 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? $\endgroup$ Commented Oct 17, 2021 at 22:05
  • $\begingroup$ 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 $R/I$). $\endgroup$
    – Julja Muvv
    Commented Oct 17, 2021 at 22:13
  • $\begingroup$ 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...) $\endgroup$ Commented Oct 17, 2021 at 22:15
  • $\begingroup$ Yes, I mean ring without identity, so ideal is a ring without identity. $\endgroup$
    – Julja Muvv
    Commented Oct 17, 2021 at 22:18
  • $\begingroup$ I have updated my answer, to include the case of $R$-modules. $\endgroup$ Commented Oct 17, 2021 at 23:02

2 Answers 2


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.

  • $\begingroup$ Are they isomorphic as $R$-modules? $\endgroup$
    – Julja Muvv
    Commented Oct 17, 2021 at 22:38
  • $\begingroup$ @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.) $\endgroup$ Commented Oct 17, 2021 at 22:40
  • $\begingroup$ So is the state in my question is true, if isomorphism is of $R$-modules? $\endgroup$
    – Julja Muvv
    Commented Oct 17, 2021 at 22:42
  • $\begingroup$ @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. $\endgroup$ Commented Oct 17, 2021 at 22:50
  • $\begingroup$ Is it true, if we put restriction that $(b) \neq (0)$, $c\neq 0$? $\endgroup$
    – Julja Muvv
    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 $$I : J := \{r \in R \mid rJ \subset I\}.$$

Back to your question:

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.


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