Let $X$ be a scheme over a field $k$ and $L$ be an invertible sheaf on it. Let $D$ be the scheme over dual numbers over $k$ with parameter $t$, i.e. $Spec(k[t]/(t^2)$.


Let $X':=X \times\_k D$ and $i:X \rightarrow X\times D$ the natural map.


One defines a first order deformation of $L$ over $D$ as a line bundle $L'$ on $X'$, such that $i^\*L'$ is isomorphic to $L$.


One knows that the iso classes of deformations of $L$ correspond to $Ext^1\_{\mathcal O\_X}(\mathcal O\_X, \mathcal O\_X)$.


One map is clear, I think: if you have a deformation $L'$, then consider the exact sequence on $D$:


$0 \rightarrow k \rightarrow \mathcal O\_D \rightarrow k \rightarrow 0$, pull it back to $X'=X\times\_k D$, tensor with $L'$ and push down to $X$. Please correct me if this is not the right way.


But how do I get explicitly a deformation in the sense of the above definition out of an exact sequence on $X$


$0\rightarrow L \rightarrow M \rightarrow L \rightarrow 0$?


In the books I read there are only hints I really don't understand, so I would be very glad about an answer which carefully constructs the deformation data out of this sequence.



Deformation of Line bundles over dual numbers
One important point is that $X$ and $X'$ have the same topological space, since $i:X'\rightarrow X$ is a nilpotent immersion. In particular $M$ is already a sheaf on $|X|=|X'|$ (of $\mathcal{O}\_X$-modules). Since $\mathcal{O}\_{X'}=\mathcal{O}\_X\oplus \mathcal{O}\_X[\varepsilon]$, to give $M$ the structure of a sheaf of $\mathcal{O}\_{X'}$-modules you only need to define the action of $\varepsilon$, as Mike Skirvin points out in his comment.


So you define multiplication by $\varepsilon$ by the composition $M\stackrel{g}{\to} L\stackrel{f}{\to} M$ of the two maps given by the extension (notice that this is $\mathcal{O}\_X$-linear, and its square is zero). This makes $M$ a sheaf of $\mathcal{O}\_{X'}$-modules, and moreover using the local criterion of flatness you can check that it is flat over the base $D$.


Now since $|X|=|X'|$ and $\mathcal{O}\_X=\mathcal{O}\_{X'}/(\varepsilon\cdot\mathcal{O}\_{X'})$, you have $i^\*(M)=M\otimes\_{\mathcal{O}\_X'}\mathcal{O}\_X=M/(\varepsilon \cdot M)=M/f(L)=L$, since by definition of the action of $\varepsilon$, $\varepsilon \cdot M=f(L)$.


