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Let $M, N, M'$ and $N'$ be R-modules. If $M'$ and $N'$ are submodules of both $M$ and $N$ then is it true that \begin{equation} \frac{M}{M'} \oplus \frac{N}{N'} \cong \frac{M \oplus N}{M' \oplus N'} \cong \frac{M \oplus N}{N' \oplus M'} \cong \frac{M}{N'} \oplus \frac{N}{M'} ? \end{equation} I think the first and the last equality is true and I could find an isomorphisms for them, but the middle equality is what confuses me. I feel that it should be true since $M' \oplus N' \cong N' \oplus M'$ but if I try to define an isomorphism as something like \begin{equation} \psi : (m + M', n + N') \rightarrow (m + N', n + M') \end{equation} there is a difficulty because $m_1 + M' = m_2 + M'$ doesn't really implies $m_1 + N' = m_2 + N'$. So at the end I'm not sure what a conclusion is. Could someone give an answer with explaination to resolve is paradox please, thank you.

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Your claim is NOT true.

$M = \mathbb{Z}, N = \mathbb{2Z}, M' = \mathbb{8Z}, N' = \mathbb{4Z}.$ All are considered modules over $\mathbb{Z}.$ Then $\frac{M}{M'} \oplus \frac{N}{N'} \cong \mathbb{Z}_8 \oplus \mathbb{Z}_2.$ On the other hand $\frac{M}{N'} \oplus \frac{N}{M'} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4.$ So this two are not isomorphic.

The mistake was to assume that since the submodules are isomorphic, so the quotients are isomorphic. This is not true in general. For example, take $M = \mathbb{Z}, M_1 = \mathbb{2Z}, M_2 = \mathbb{3Z}$ (as $\mathbb{Z}$ modules). Then $M_1 \cong M_2$, but $M/M_1 \ncong M/M_2.$ The main difficulty here is that the isomorphism between $M_1$ and $M_2$ is not induced by an isomorphism of $M$(why?). But if the isomorphism of $M_1$ and $M_2$ is induced by an isomorphism of $M$, then we will have $M/M_1 \cong M/M_2.$ Here is the actual statement: Let $A$ be a commutative ring with identity, $M$ is a $A$ module, and let $M_1, M_2$ be two submodules of $M.$ Suppose that $f: M_1 \rightarrow M_2$ is an $A$ module isomorphism. Assume that $f$ can be lifted to an $A$ module isomorphism $g$ of $M$, i.e. $g : M \rightarrow M$ is an $A$ module isomorphism, $g(M_1) = M_2$ and $g|M_1 = f.$ Then $M/M_1 \cong M/M_2.$ To see this, define a map $\phi : M \rightarrow M/M_2$ by $m \mapsto g(m) + M_2.$ Then show that $\phi$ is an $A$ module surjective homomorphism and Ker$\phi = M_1.$ (I think this is true over non-commutative ring also.)

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  • $\begingroup$ Thank you. But could you also explain where did my argument above fail? Because after all $M' \oplus N' \cong N' \oplus M'$ so shouldn't the I get the same thing if I quotient out by two things that are isomorphic? $\endgroup$
    – user113988
    Dec 21, 2014 at 3:23
  • $\begingroup$ @user113988: I've edited my answer to include a clarification of your question. I hope this will help. $\endgroup$
    – Krish
    Dec 21, 2014 at 4:02
  • $\begingroup$ Thank you Krish, that was really helpful. $\endgroup$
    – user113988
    Dec 21, 2014 at 5:09
  • $\begingroup$ While a bit late how does this stand in contrast to this? crazyproject.wordpress.com/2011/04/07/… they seem to be in opposition. $\endgroup$ Feb 6, 2016 at 8:54
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    $\begingroup$ @ZelosMalum: There is no contradiction in this. In the question, the first isomorphism is true and the last one also. But the middle one is not true, in general (I have given a clarification in the answer). In the link (the one you have given), it was about the first and last isomorphisms. $\endgroup$
    – Krish
    Feb 26, 2016 at 6:42

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