# A question on exact sequences and composition series

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.

• Do you know the result in the case of short exact sequences? – Eric Wofsey Nov 2 '20 at 18:57
• Hi, Eric. Actually no, I do not – Ranopano Nov 2 '20 at 18:58
• The general case is easy to deduce from the case of short exact sequences; see math.stackexchange.com/questions/2959644/… for instance (that's for the case $n=4$, but the argument easily generalizes). – Eric Wofsey Nov 2 '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 2 '20 at 19:13
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