- For a division ring \(D\), let \(V_{i}\) be a finite dimensional vector space over \(D\) for \(i \in\{1, \ldots, k\}\). Suppose the sequence \[ 0 \longrightarrow V_{1} \longrightarrow V_{2} \longrightarrow \cdots V_{k} \longrightarrow 0 \] is exact. Prove that \(\sum_{i=1}^{k}(-1)^{i} \operatorname{dim}_{D} V_{i}=0\).
Solution
Let \[ d_i:V_i\longrightarrow V_{i+1} \] be the maps in the exact sequence, with \(V_0=V_{k+1}=0\) and the evident zero maps at the ends.
1. For every \(1\le i\le k\),\[ \dim_D V_i=\dim_D\ker d_i+\dim_D\operatorname{im}d_i. \]
This is rank-nullity over the division ring \(D\). Equivalently, choose a basis of \(\ker d_i\), extend it to a basis of \(V_i\), and observe that the images of the added basis vectors form a basis of \(\operatorname{im}d_i\).
2. Exactness gives\[ \ker d_i=\operatorname{im}d_{i-1} \qquad(1\le i\le k). \]
This is exactly the definition of exactness at \(V_i\).
3. Hence\[ \dim_D V_i =\dim_D\operatorname{im}d_{i-1} +\dim_D\operatorname{im}d_i. \]
Substitute 2 into 1.
4. The alternating sum telescopes to zero:\[ \sum_{i=1}^k(-1)^i\dim_DV_i=0. \]
Using 3, \[ \begin{aligned} \sum_{i=1}^k(-1)^i\dim_DV_i &=\sum_{i=1}^k(-1)^i\dim_D\operatorname{im}d_{i-1} +\sum_{i=1}^k(-1)^i\dim_D\operatorname{im}d_i\\ &=-\sum_{j=0}^{k-1}(-1)^j\dim_D\operatorname{im}d_j +\sum_{j=1}^{k}(-1)^j\dim_D\operatorname{im}d_j. \end{aligned} \] All interior terms cancel. The remaining boundary terms vanish because \[ \operatorname{im}d_0=0, \qquad \operatorname{im}d_k=0. \] Therefore the sum is zero.