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While studying Module Theory, I've been stuck on this exercise; any help would be very welcome:

Let $0 \longrightarrow M_1 \longrightarrow M_2 \longrightarrow ... \longrightarrow M_{n-1} \longrightarrow M_n \longrightarrow 0$ be an exact sequence of $R$-modules such that $lh(M_i)$, the lenght of $M_i$, is finite for all $i \in \{1,...,n\}$. Show that

$\sum^n_{i=1}(-1)^ilh(M_i)=0.$

I feel that one could argument by induction (on $n$), but I don't see how.

Eric Wofsey
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Ranopano
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  • Do you know the result in the case of short exact sequences? – Eric Wofsey Nov 02 '20 at 18:57
  • Hi, Eric. Actually no, I do not – Ranopano Nov 02 '20 at 18:58
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    The general case is easy to deduce from the case of short exact sequences; see https://math.stackexchange.com/questions/2959644/length-of-modules-in-arbitrary-exact-sequences for instance (that's for the case $n=4$, but the argument easily generalizes). – Eric Wofsey Nov 02 '20 at 18:59
  • Induction ist IMHO a good idea to start with ... After the verification for $n=1$ and assuming for a certain $k$ such that $1 \le k \le n$ that you have $\sum^k_{i=1}(-1)^ilh(M_i)=0$, what would you know about $\sum^{k+1}{i=1}(-1)^ilh(M_i)$ except that $\sum^{k+1}{i=1}(-1)^ilh(M_i) = (-1)^{k+1}lh(M_{k+1}) + \sum^k_{i=1}(-1)^ilh(M_i) = (-1)^{k+1}lh(M_{k+1})$ because $\sum^k_{i=1}(-1)^ilh(M_i)=0$? Can you prove that $(-1)^{k+1}lh(M_{k+1}) = 0$ which means that $lh(M_{k+1}) = 0$? This is rather a comment than an answer. That's why I am going to delete my answer below... – Noureddine Ouertani Nov 02 '20 at 19:13

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