Lemma 15.28.8 (0629)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.28: The Koszul complex
Lemma 15.28.8 (cite)

Lemma 15.28.8. Let $R$ be a ring. Let $f_1, \ldots , f_ r$ be a sequence of elements of $R$. The complex $K_\bullet (f_1, \ldots , f_ r)$ is isomorphic to the cone of the map of complexes

\[ f_ r : K_\bullet (f_1, \ldots , f_{r - 1}) \longrightarrow K_\bullet (f_1, \ldots , f_{r - 1}). \]

Proof. Special case of Lemma 15.28.7. $\square$

Comments (2)

Comment #2447 by Raymond Cheng on March 09, 2017 at 02:42

Should be $f_r \colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$ instead of $f_n\colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$ .

Comment #2489 by Johan on April 13, 2017 at 22:57

Thanks, fixed here.

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