# A little problem about quotient modules

• I
steenis
Let ##M## be a (right) R-module, and ##A## and ##B## two submodules of ##M##.

If ##A = B ##, then we know that ##\frac{M}{A} = \frac{M}{B}##.

But is the converse also true:

If ##\frac{M}{A} = \frac{M}{B}##, can we conclude that ##A = B ## ?

I doubt it, but I cannot find the answer. Maybe someone can help me with a proof or a counterexample ?

(If possible, also a reference.)

## Answers and Replies

Mentor
2021 Award
Did you mean ##M/A = M/B## or just ##M/A \cong M/B\,?## If it is the equality, then ##0_{M/A} = 0_{M/B}## or ##A \subseteq B## and ##B\subseteq A##. If it is the isomorphism, then we could take a direct sum of equal copies and factor out different positions.

I have no example for ##M/A \cong M/B## and ##A \ncong B\,##, yet.

Last edited:
steenis
It is the equality I am most interested in. I came across the equality in a proof. However, I am interested in both cases to be complete. Can you please expand a little, because do not quite follow you.

Mentor
2021 Award
I thought about what ##M/A = M/B## should mean. An equality of cosets means, ##\{\,m_\iota A\,\}= \{\,m_\iota B\,\}## of a set of sets, and this can only be equal, if they are pairwise equal, which leaves only one possibility for zero. At least I couldn't think of another way to interpret equality, the others would be isomorphic copies.

The example with the direct sum is a bit cheating, as ##(0,\mathbb{Z})\neq (\mathbb{Z},0)## but they both factor to ##(\mathbb{Z}\oplus \mathbb{Z})/\mathbb{Z} = \mathbb{Z}##. But if I think about it now, it's probably an isomorphism, too.

In the general case, I think we cannot have finitely generated modules. And we will need some different structures in ##A## and ##B## which are both forgotten in the factor modules. I thought whether a tensor product will help, which would exclude flat modules. I'm uncertain here, since a module doesn't provide much structure.

steenis
I think pairwise equal in this way:##\frac{M}{A} = \frac{M}{B}## means that for each ##m' \in M## there is a ##m \in M## such that ##m' + A = m + B##.

In particular there is a ##m \in M## such that ##A = m + B##, can I conclude that ##m## must be zero ? Otherwise a (sub)module equals a not necessary (sub) module ?

Mentor
2021 Award
I think pairwise equal in this way:##\frac{M}{A} = \frac{M}{B}## means that for each ##m' \in M## there is a ##m \in M## such that ##m' + A = m + B##.

In particular there is a ##m \in M## such that ##A = m + B##, can I conclude that ##m## must be zero ? Otherwise a (sub)module equals a not necessary (sub) module ?
If ##m' \in A##, then ##A=m+B## and thus ##m+0=m \in A## and thus ##B=(m'-m)+A \subseteq A## and vice versa.

steenis
This was almost my very first thought: ##A = m + B##, so ##0 = m - n##, witn ##n \in B## and so ##m = n \in B## and therefore ##A = B##. I did not write it down because I thought this cannot be correct. It helps if you say it is correct, but I remain doubtful.

Mentor
2021 Award
We can do it elementwise. Say we have ##A=m_0+B## for a certain ##m_0## corresponding to ##0\in A##. But now we can write ##0 \in B## as ##0=m_A-m_0## for a certain ##m_A \in A## which means, that ##m_0\in A##. Now any ##b\in B## can be written ##a(b) = m_0+b\,\text{ resp. }b=a(b)-m_0\in A##.

steenis
I understand that.
I am not completely convinced, but for now I have no arguments, maybe it is just a feeling.
Thank you very much for your help.

Can you say something about this case: ##\frac{M}{A} \cong \frac{M}{B}##

Mentor
2021 Award
This seems to be more difficult, since equality is quite strong. And as said before, the usual suspects won't work. There must be some kind of difference between ##A## and ##B## such that ##A \ncong B## but both be forgotten in ##M/A##. I think I'll try a proof instead of a counterexample.

steenis
I will think it over and look around. If I found something, I will post it here. Thanks very much for your help.

Mentor
2021 Award
I will think it over and look around. If I found something, I will post it here. Thanks very much for your help.
Maybe @mathwonk (ping) will have a good example.

Homework Helper
congratulations, this is quite an intertesting question! just off the top of my head, how about an infinite dimensional vector space M and two finite dimensional subspaces A, B of different finite dimensions. then the quotient spaces should be isomorphic infinite dimensional spaces.

i.e. in dimension terms, aleph null - 1 = aleph null - 2.

even easier, try modding out the direct product of Z/2 and Z/4 by two different non isomorphic submodules, but getting in both cases, the quotient Z/2.

@fresh_42 (ping).

indeed a weaker form of this question which is also not true is whether the fact that A+X ≈ B+X (direct sum) forces A and B to be isomorphic. google something like "stably isomorphic".

here are some notes by an excellent expositor:

• 