Let $X$ be a smooth, complex projective variety of dimension $n$. Let $F$ be a vector bundle over $X$ with rank$F=r \leq n$. Let $s$ be a regular global section of $F$ and $Z$ be the scheme of zeroes of $s$ [Here by regular I mean codim$Z$=r]. Then we have the corresponding exact Koszul complex given by :


$0 \to \wedge^rF^\vee \to \wedge^{r-1}F^\vee \to \dots \to F^\vee \to \mathcal{O}\_X \to \mathcal{O}\_Z\to 0$


Do we have an analogous sequence if rank$F=r > n$? There is a description in Fulton's Intersection theory book Page $431$, but I'm not sure if that is irrespective of any relation between $r$ and $n$.


More precisely, what I want to ask is that for $r>n$ if one has a regular global section $s$ [then $Z = \emptyset$] of $F$, then can one use the first complex mentioned in Fulton's book to conclude that there is a sequence


$0 \to \wedge^rF^\vee \to \wedge^{r-1}F^\vee \to \dots \to F^\vee \to \mathcal{O}\_X \to 0$


which is exact on $X-Z=X$?


Any suggestion is appreciated.



Koszul complex for bundles of rank higher than the dimension
Yes, this complex is exact. To prove exactness at point $x \in X$, note that the question is local, so one may assume that $F \cong \mathcal{O}^{\oplus r}$ is a trivial bundle and $s = (s\_1,\dots,s\_r)$ is a sequence of functions. Note further that the Koszul complex can be written as a tensor product
$$
\bigotimes\_{i=1}^r(\mathcal{O} \stackrel{s\_i}\to \mathcal{O}).
$$
Since $s(x) \ne 0$, we have $s\_i(x) \ne 0$ for some $i$, hence one of the factors above is acyclic at $x$, and therefore so is the tensor product.


