The Stacks project

Lemma 10.102.2. Suppose $R$ is a ring. Let

\[ \ldots \xrightarrow {\varphi _{i + 1}} R^{n_ i} \xrightarrow {\varphi _ i} R^{n_{i-1}} \xrightarrow {\varphi _{i-1}} \ldots \]

be a complex of finite free $R$-modules. Suppose that for some $i$ some matrix coefficient of the map $\varphi _ i$ is invertible. Then the displayed complex is isomorphic to the direct sum of a complex

\[ \ldots \to R^{n_{i + 2}} \xrightarrow {\varphi _{i + 2}} R^{n_{i + 1}} \to R^{n_ i - 1} \to R^{n_{i - 1} - 1} \to R^{n_{i - 2}} \xrightarrow {\varphi _{i - 2}} R^{n_{i - 3}} \to \ldots \]

and the complex $\ldots \to 0 \to R \to R \to 0 \to \ldots $ where the map $R \to R$ is the identity map.

Proof. The assumption means, after a change of basis of $R^{n_ i}$ and $R^{n_{i-1}}$ that the first basis vector of $R^{n_ i}$ is mapped via $\varphi _ i$ to the first basis vector of $R^{n_{i-1}}$. Let $e_ j$ denote the $j$th basis vector of $R^{n_ i}$ and $f_ k$ the $k$th basis vector of $R^{n_{i-1}}$. Write $\varphi _ i(e_ j) = \sum a_{jk} f_ k$. So $a_{1k} = 0$ unless $k = 1$ and $a_{11} = 1$. Change basis on $R^{n_ i}$ again by setting $e'_ j = e_ j - a_{j1} e_1$ for $j > 1$. After this change of coordinates we have $a_{j1} = 0$ for $j > 1$. Note the image of $R^{n_{i + 1}} \to R^{n_ i}$ is contained in the subspace spanned by $e_ j$, $j > 1$. Note also that $R^{n_{i-1}} \to R^{n_{i-2}}$ has to annihilate $f_1$ since it is in the image. These conditions and the shape of the matrix $(a_{jk})$ for $\varphi _ i$ imply the lemma. $\square$

Comments (2)

Comment #2978 by Dario WeiƟmann on

The inequalities should go the other way.

There are also:

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

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