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Let $M$ be an $R$-module, and $N'\subseteq N\subseteq M$ be two submodules. Is there a natural way to consider the module $M/N$ as a submodule of $M/N'$?

I saw sometimes in the proof, that for example $R$ is a ring, $p$ is an element. We can get an exact sequence (e.g., $\mathbb Z$ and a prime number $p$) $$0\to R/(p)\to R/(p)^n\to R/(p)^{n-1}\to0.$$ But the first map is not very natural for me.

user26857
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CO2
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    These two things are unconnected. The maps you're looking at in the second bit are almost certainly multiplication by $p^{n-1}$ and reduction modulo $p^{n-1}$, respectively. Try it with $R=\Bbb Z$, for instance. – KReiser Jan 17 '22 at 09:29
  • @KReiser Thank you that is very clear now. So in general, my first claim makes no sense? – CO2 Jan 17 '22 at 09:33
  • Not an answer, but surely relevant https://math.stackexchange.com/questions/364123/correspondence-between-submodules-and-quotient-modules – ancient mathematician Jan 17 '22 at 09:36
  • The simplest counterexample would be a module $M$ with a unique composition series $M\geqslant N\geqslant 0$ with $M/N\not\simeq N$. (Which doesn't seem to happen in the well-behaved modules I know, alas.) – ancient mathematician Jan 17 '22 at 09:39
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    ${0}\subseteq 2\mathbb{Z}\subseteq\mathbb{Z}$ as $\mathbb{Z}$-modules. However $\mathbb{Z}/2\mathbb{Z}$ does not in include in $\mathbb{Z}/{0}$. – tkf Jan 17 '22 at 10:34
  • @ancientmathematician I think $0 \to \mathbb R \to \mathbb R[x] / (x^2 + 1)(x - 1) \to \mathbb R[x] / (x^2 + 1) \to 0$, considered as a sequence of $\mathbb R[x]$-modules, should work. All three modules have different $\mathbb R$-dimension, so they cannot be isomorphic as modules. – red_trumpet Jan 17 '22 at 14:07

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