A little problem about quotient modules

[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.)

 

[fresh_42]

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.

 

[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.

 

[fresh_42]

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 ?

 

[fresh_42]

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.

 

[fresh_42]

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: $\frac{M}{A} \cong \frac{M}{B}$

 

[fresh_42]

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.

 

[fresh_42]

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.

 

[mathwonk]

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:

http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/stablyfree.pdf