Elementary transformation of vector bundles Let $E$ be a vector bundle of rank $r$ on a nonsingular algebraic surface $X$. Assume that there are $r$ sections $s_1, s_2, \cdots, s_r$ of $E$ such that the zeros of the wedge of them is a curve $C$. These sections can give us a morphism from the trivial vector bundle of rank $r$ to $E$. How to show the cokernal of the morphism is a line bundle on $C$?
Thanks! 
 A: As Francesco Polizzi and Jorgen Rennemo remark, the cokernel is not always a line bundle on $C$.
For example, suppose that the vector bundle is $\mathcal E := \mathcal O(D) \oplus \mathcal O(D)$ for some curve $D$ on the surface.
Note that by construction there is a section $s:\mathcal O\to \mathcal O(D)$ whose
zero locus is exactly $D$.  Now take $s_1 = s \oplus 0$ and $s_2 = 0 \oplus s$.
Then $\wedge^2 \mathcal E = \mathcal O(D) \otimes \mathcal O(D) = \mathcal O(2D)$,
and $s_1\wedge s_2 = s^{\otimes 2}$, whose zero locus is $2D$.  So in this case the curve $C$ of your question is non-reduced (it is equal to $2 D$).  On the other hand,
the cokernel of $S$ is equal to $\mathcal O(D)/\mathcal O \oplus \mathcal O(D)/\mathcal O$,
which is a rank two bundle on $D$, rather than a line bundle on $2 D$.
For another example, consider $\mathcal E := \mathcal O \oplus \mathcal O$ on $\mathbf A^2 =$
Spec $k[x,y]$, and let $s_1 = (x,0)$ and $s_2 = (0,y)$.
Then $\wedge^2 \mathcal E = \mathcal O$, and $s_1 \wedge s_2 = xy$ as a section of
$\mathcal O$.  The zero locus of $s_1\wedge s_2$ is the curve $C$ given by $x y = 0$,
and the cokernel of the map $\mathcal O^2 \to \mathcal O^2$ is a line bundle away
from the singularity of $C$, but has a fibre of dimension two at the singular point where 
$x= 0$ and $y = 0$ cross.
So one needs some extra assumptions (which are perhaps implicit in your question),
e.g. that the curve $C$ cut out by the wedge of the $s_i$  is smooth.  (In my first
example $C$ is non-reduced, hence non-smooth, while in my second example it is reduced
but singular.)
Assuming that $C$ is smooth, one argues as follows:
Let $x \in X$, and work locally at $x$, so we let $A$ be the local ring of $X$ at $x$.
The vector bundle $\mathcal E$ is locally trivial, so the section $s_i$ is just a map
$A \to A^r$ given by $1 \mapsto (a_{i,1},\ldots, a_{i,r}).$
Taken together, these $s_i$ give a map $s: A^r \to A^r$ whose matrix is $(a_{i,j})$.
Our assumption that $C$ is a curve (rather than all of $X$) translates to the assumption that det $(a_{i,j})$ is non-zero, and the fact that $C$ is smooth at the point $x$
(if $x$ lies in $C$) translates to the assumption that det $(a_{i,j})$ lies in
$\mathfrak m_x$, but not $\mathfrak m_x^2$.
Now if some $a_{i,j}$ is a unit in $A$, then the map $s_i:A \to A^r$ splits;
a splitting is given by projection onto $j$th factor of $A^r$, followed by
multiplication by $a_{i,j}^{-1}$.  Thus in this case we may quotient out 
both the source and target of $s$ by one copy of $A$ (namely by the $i$th summand
in the source and its image under $S$ in the target) and so replace $r$ by
$r-1$.   Continuing in this way, we reduce to the case in which all the $a_{i,j}$
lie in $\mathfrak m_x$.  But then the determinant of $(a_{i,j})$ lies in $\mathfrak m_x^r$,
hence our smoothness assumption shows that $r = 1$, and so the cokernel is 
indeed free of rank one over the curve $C$.
A: Looking at the problem locally, you can assume $X = \textrm{Spec}(R)$, where $R$ is a Noetherian regular local ring. Therefore you have a short exact sequence
$0 \to R^r  \stackrel{M}{\to} R^r \to I \to 0$,
where the matrix $M$ is given by your sections $s_1, \ldots, s_r$. It follows that the cokernel $I$ is a torsion module supported where the determinat of $M$ vanishes, namely over the locus $s_1 \wedge \ldots \wedge s_r=0$, that is over your curve $C$.
Under some genericity assumptions the rank of $M$ will be exactly $r-1$ at every point of $C$, so the cokernel is actually a line bundle on $C$. 
In general, it may happen that the cokernel is not a line bundle, see the answers of Jorgen Rennemo and Emerton. 
