Structure sequence for a Curve on a Surface This is a very basic question of Algebraic Geometry.
Let X be a smooth algebraic surface (over $\mathbb{C}$) and $C$ be a curve (smooth irreducible) on $X$. Can somebody very kindly explain me what is the meaning of the so called ''structure sequence''
$$ 0 \longrightarrow\mathcal{O}_X(-C)\longrightarrow \mathcal{O}_X \longrightarrow \mathcal{O}_C\longrightarrow 0 $$
Also, what is meant by ''tensoring this sequence with $\mathcal{O}_X(C)$'' ?
Thanks a lot.
 A: This is the exact sequence arising from the ideal sheaf of $C$ in $X$. 
In other words: $C$ is a closed subscheme of $X$, so let us denote by $i:C\to X$ the closed immersion. When one writes $\mathcal O_X\to\mathcal O_C$, one actually means $i^\sharp:\mathcal O_X\to i_\ast\mathcal O_C$, but the abuse of notation is justified in the sense that $i_\ast\mathcal O_C$ is zero outside of $C$. Now, this arrow is surjective by definition of closed immersion. This explains the right part of the sequence. The left part is just: take the kernel of this surjective map $\mathcal O_X\to\mathcal O_C$. This kernel is by definition the ideal sheaf $\mathcal I_{C/X}$ of $C$ in $X$, and it coincides with $\mathcal O_X(-C)$ (in this case, $C$ is a divisor). 
When you tensor an exact sequence, it just means that you apply a functor. In this case, the functor $$\mathcal O_X(C)\otimes_{\mathcal O_X}-$$ applied to your short exact sequence gives a new exact sequence on $X$: $$0\to \mathcal O_X\to \mathcal O_X(C)\to \mathcal O_X(C)\otimes_{\mathcal O_X} \mathcal O_C\to 0.$$
(It stays exact since you tensor with a locally free sheaf)
