Give definitions of the following terms:
- a finite length (left) module, (ii) a composition series for a module, and (iii) the length of a module,
Let \(l(M)\) denote the length of a module \(M\). Prove that if \[\begin{align*} 0\rightarrow M_1\rightarrow M_2\rightarrow\dots\rightarrow M_n\rightarrow 0 .\end{align*}\]
is an exact sequence of modules of finite length, then \[\begin{align*} \sum_{i=1}^n(-1)^i l(M_i)=0.\end{align*}\]
Solution
1. A module \(M\) has finite length if it admits a finite composition series.A composition series is a finite chain\[ 0=M_0\subsetneq M_1\subsetneq\cdots\subsetneq M_r=M \]of submodules such that every quotient \(M_i/M_{i-1}\) is simple.The length \(l(M)\) is the number \(r\) of simple factors in a composition series.
By the Jordan–Hölder theorem, any two composition series of a finite-length module have the same number of factors, so \(l(M)\) is well-defined.
2. If\[ 0\longrightarrow A\longrightarrow B\xrightarrow{\pi} C\longrightarrow0 \]is a short exact sequence of finite-length modules, then\[ l(B)=l(A)+l(C). \]
Identify \(A\) with its image in \(B\). Choose composition series \[ 0=A_0\subsetneq A_1\subsetneq\cdots\subsetneq A_r=A \] and \[ 0=C_0\subsetneq C_1\subsetneq\cdots\subsetneq C_s=C. \] Then \[ 0=A_0\subsetneq\cdots\subsetneq A_r=A =\pi^{-1}(C_0) \subsetneq\pi^{-1}(C_1) \subsetneq\cdots\subsetneq\pi^{-1}(C_s)=B \] is a composition series for \(B\). Indeed, \[ \pi^{-1}(C_j)/\pi^{-1}(C_{j-1})\cong C_j/C_{j-1} \] for each \(j\), while the factors below \(A\) are the factors of the chosen composition series of \(A\). Hence the series has \(r+s\) factors.
3. For the exact sequence\[ 0\to M_1\xrightarrow{d_1}M_2\xrightarrow{d_2}\cdots\xrightarrow{d_{n-1}}M_n\to0, \]set\[ I_i=\operatorname{im}(d_i)\quad(1\le i\le n-1), \qquad I_0=I_n=0. \]Then for every \(1\le i\le n\) there is a short exact sequence\[ 0\longrightarrow I_{i-1}\longrightarrow M_i\longrightarrow I_i\longrightarrow0. \]
Exactness gives \[ \ker(d_i)=\operatorname{im}(d_{i-1})=I_{i-1} \] for \(1<i<n\). At the two ends, injectivity of \(M_1\to M_2\) gives \(I_0=0=\ker d_1\), while exactness at \(M_n\) gives \(I_{n-1}=M_n\) and hence the final short exact sequence \[ 0\to I_{n-1}\to M_n\to0. \] The first isomorphism theorem gives \(M_i/I_{i-1}\cong I_i\).
4. Therefore\[ l(M_i)=l(I_{i-1})+l(I_i) \qquad(1\le i\le n). \]
Apply 2 to the short exact sequences in 3.
5. The alternating sum of the lengths vanishes:\[ \sum_{i=1}^n(-1)^i l(M_i)=0. \]
Using 4, \[ \begin{aligned} \sum_{i=1}^n(-1)^i l(M_i) &=\sum_{i=1}^n(-1)^i l(I_{i-1}) +\sum_{i=1}^n(-1)^i l(I_i)\\ &=-\sum_{j=0}^{n-1}(-1)^j l(I_j) +\sum_{j=1}^{n}(-1)^j l(I_j). \end{aligned} \] All interior terms cancel, and the two endpoint terms vanish because \(I_0=I_n=0\) and \(l(0)=0\).