Let $X$ be a complex variety; one can also assume it is smooth if this helps. $\mathcal{E}$ is a locally free sheaf of rank $r$ on $X$, and $s \in H^0(X, \mathcal{E})$. Then one has a Koszul complex by contracting $s$:

$$0 \to \wedge^r \mathcal{E}^\vee \to \wedge^{r-1} \mathcal{E}^\vee \to \cdots \to \wedge^2 \mathcal{E}^\vee \to \mathcal{E}^\vee \to 1 \to 0.$$

Then it is claimed (c.f. Fulton: Intersection Theory, P431, B.3.4) this Koszul complex is exact on $X - Z(s)$. Can anyone give an explanation for this claim?


1 Answer 1


This works for every locally ringed space $X$. Since exactness can be tested stalkwise, we may assume that $\mathcal{E}$ is a free module over a (local) ring $R$. One of the sections $s_i$ generates $\mathcal{E}$. But then we can use the usual Koszul resolution (Weibel, An introduction to homological algebra, Corollary 4.5.5):

For a regular sequence $s_1,\dotsc,s_r$ in a commutative ring $R$, the sequence

$0 \to \wedge^r R^r \to \dotsc \to \wedge^2 R^r \to R^r \xrightarrow{s} R \to R/(s_1,\dotsc,s_r) \to 0$

is exact.

In your case, $s_i \in R^*$ for some $i$, hence the sequence is regular, and $R/(s_1,\dotsc,s_r)=0$. Alternatively, in this case, one can write down an explicit chain contraction.

  • 1
    $\begingroup$ Thanks for that necessary edit.+ $\endgroup$
    – Mikasa
    Apr 8, 2013 at 10:25
  • 2
    $\begingroup$ @Martin Brandenburg Thank you for your answer, I can follow your argument expect one place : why $s_i \in R^*$ for $some$ $i$ implies the sequence is a regular sequence? $\endgroup$
    – Li Yutong
    Apr 8, 2013 at 23:52

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