A little problem about quotient modules

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  • #1
steenis
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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.)
 

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  • #2
fresh_42
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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.
 
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  • #3
steenis
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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.
 
  • #4
fresh_42
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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.
 
  • #5
steenis
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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 ?
 
  • #6
fresh_42
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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.
 
  • #7
steenis
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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.
 
  • #8
fresh_42
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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##.
 
  • #9
steenis
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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}##
 
  • #10
fresh_42
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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.
 
  • #11
steenis
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I will think it over and look around. If I found something, I will post it here. Thanks very much for your help.
 
  • #12
fresh_42
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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.
 
  • #13
mathwonk
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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
 
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