length of modules in arbitrary exact sequences

Question

Let $R$ be a commutative, noetherian ring. Given the exact sequence of $R$-modules of finite length $ 0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0 $. Is there an equation, connecting the lengths of the modules like in the case of short exact sequences?

Answer 1

For a short exact sequence of finite length modules $0\to A\to B\to C\to 0$, it holds that $l(B)=l(A)+l(C)$.

Split your sequence into two short exact ones: \begin{gather} 0\to M_0\to M_1\to K\to 0\\ 0\to K\to M_2\to M_3\to 0 \end{gather} Then $l(K)=l(M_1)-l(M_0)$ and $l(K)=l(M_2)-l(M_3)$, so we get that $$ l(M_0)-l(M_1)+l(M_2)-l(M_3)=0 $$ which generalizes the relation for short exact sequences.

This goes on by easy induction.

Answer 2

In these situations, the alternating sum of the lengths is zero. Here, $$l(M_0)-l(M_1)+l(M_2)-l(M_3)=0.$$ One can split this into two exact sequences $$0\to M_0\to M_1\to N\to 0$$ and $$0\to N\to M_2\to M_3\to 0$$ with the same $N$, also of finite length, and $$l(N)=l(M_1)-l(M_0)=l(M_2)-l(M_3)$$ etc.