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I don't know why there is no investigation of the cancellability of quotients in category of modules. What I mean by cancellability of quotients in category of modules is the following :

Let $R$ be a ring (for simplicity, commutative) and $M$ be an $R$-module. We say that $M$ is numerator-cancellable if for every $R$-modules $M_1$ and $M_2$ and for every monomorphisms $f_1: M_1\to M$, $f_2: M_2\to M$, if the quotient modules $M/f_1(M_1)$ and $M/f_2(M_2)$ are isomorphic, then $M_1$ and $M_2$ are isomorphic.

For example, if $K$ is a commutative field, then we can prove easily that every $K$-module (ie. K-vector space) is numerator-cancellable. We can see also that any simple $R$-module is likewise numerator-cancellable. I think that is also true (under some restricted hypothesis) for the semi-simple $R$-modules.

Is there any reference concerning the problem ?

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  • $\begingroup$ If $K$ is a field, $M$ is, say, of infinite countable dimension, $M_1$ is a subspace of infinite dimension and codimension, and $M_2=0$, then $M/M_1$ and $M/M_2$ are isomorphic while $M_1$ and $M_2$ are not. So I don't understand your claim about fields. $\endgroup$
    – YCor
    Commented Sep 23, 2018 at 17:58
  • $\begingroup$ Sorry, I mean that is true for finite dimensional $K$-vector spaces. $\endgroup$
    – Rajkarov
    Commented Sep 23, 2018 at 18:02
  • $\begingroup$ One of the important cases would be, when does a finitely generated free module satisfy your property. For example, if $P\oplus R=F$, $F$ a free module, one can ask whether $P$ is free (otherwise $F$ fails your property). There is a wealth if information about this so called stably free modules in the literature. $\endgroup$
    – Mohan
    Commented Sep 23, 2018 at 21:39

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