
Lemma 10.101.6. In Situation 10.101.1, suppose the complex is isomorphic to a direct sum of trivial complexes. Then we have

1. the maps $\varphi _ i$ have rank $r_ i = n_ i - n_{i + 1} + \ldots + (-1)^{e-i-1} n_{e-1} + (-1)^{e-i} n_ e$,

2. for all $i$, $1 \leq i \leq e - 1$ we have $\text{rank}(\varphi _{i + 1}) + \text{rank}(\varphi _ i) = n_ i$,

3. each $I(\varphi _ i) = R$.

Proof. We may assume the complex is the direct sum of trivial complexes. Then for each $i$ we can split the standard basis elements of $R^{n_ i}$ into those that map to a basis element of $R^{n_{i-1}}$ and those that are mapped to zero (and these are mapped onto by basis elements of $R^{n_{i + 1}}$ if $i > 0$). Using descending induction starting with $i = e$ it is easy to prove that there are $r_{i + 1}$-basis elements of $R^{n_ i}$ which are mapped to zero and $r_ i$ which are mapped to basis elements of $R^{n_{i-1}}$. From this the result follows. $\square$

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