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Let $L \xrightarrow{f} M \xrightarrow{g} N$ be a short exact sequence of $R$-modules, and assume that there is a chain of submodules $0 = M_0 < M_1 < \dotsb < M_n = M$ in which the quotient $M_i / M_{i-1}$ is simple for each $i$. Setting $L_i = f^{-1}(M_i)$ and $N_i = g(M_i)$. Show that there is short exact sequence $L_i / L_{i-1} \to M_i / M_{i-1} \to N_i / N_{i-1}$ for each $i$.

Can I approach the prove using the canonical homomorphisms, and can I have a chain, say $0 = L_0 < L_1 < \dotsb < L_n = L$ from $f^{-1}$ and $0 = N_0 < N_1 < \dotsb < N_n = N$ from $g$.

I am stuck because everywhere I found short exactness, it is like this $0 \rightarrow L \xrightarrow{f} M \xrightarrow{g} N \rightarrow 0$. I get confused. Please help.

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    $\begingroup$ math.stackexchange.com/questions/629735 $\endgroup$ Commented Jan 9, 2014 at 15:04
  • $\begingroup$ If you're told that such a sequence as yours is short exact, you can simply add the initial and final 0 and the appropriate arrows; these are implied. $\endgroup$
    – Nick
    Commented Jan 9, 2014 at 15:07
  • $\begingroup$ @MartinBrandenburg I have seen it but there were no comment on it. I need some ideas or hints. $\endgroup$ Commented Jan 9, 2014 at 15:10

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You don’t need the assumption that the quotients are simple.

By constructions of $L_i$ and $N_i$, you have exact sequences $$ L_i \rightarrow M_i \rightarrow N_i \rightarrow 0 \,, $$ obtained by restricting $f$ and $g$. This fits in a commutative diagram: $$ \require{AMScd} \begin{CD} 0 @>>> L_{i-1} @>>> L_i @>>> L_i / L_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> M_{i-1} @>>> M_i @>>> M_i / M_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> N_{i-1} @>>> N_i @>>> N_i / N_{i-1} @>>> 0 \\ @. @VVV @VVV @VVV @. \\ {} @. 0 @. 0 @. 0 @. {} \end{CD} $$

The third column is obtained from the the first two and the exact rows are the definitions of a quotient of a module by a submodule.

The rows and the first two columns are exact and we want to show that the third column is exact too. This looks like the five lemma and can be proved by diagram-chasing. I encourage you to try to prove this alone.

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  • $\begingroup$ @Jermey Thanks a lot I will try it. $\endgroup$ Commented Jan 9, 2014 at 18:04

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