The Stacks project

Lemma 10.102.6. In Situation 10.102.1, 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. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.102: What makes a complex exact?

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00MW. Beware of the difference between the letter 'O' and the digit '0'.