This question arised when I was trying to use [this answer](https://mathoverflow.net/questions/54311/determinant-and-exact-sequences-of-sheaves/54321#54321) to understand Reid's ["Young Person's guide to Canonical Singularities"](http://math.unice.fr/~sb/SpringSchool2009/YPG.pdf). In particular page 352 when computing the blow-up $Y\rightarrow A^2/\mu\_3$, the affine plane quotient the cyclic group of order 3, arises to the conclusion that the exceptional divisor is $E\sim P^1$, (no problems there) and $\mathcal{O}\_E(-E)\sim \mathcal{O}(3)$ (problems here).


Given a variety $Y$ and an effective Cartier divisor $D$ on it, there seems to be a pretty standard exact sequence:


$$0 \longrightarrow \mathcal{O}\_Y \longrightarrow \mathcal{O}\_Y(D) \longrightarrow\mathcal{O}\_D(D)\longrightarrow 0 $$
As far as I understand, if $U$ is an open set in $S$ and $D\cap U = div(g)\_U$ (for $D$ a hypersurface, if you want, and extend by linearity), then


$$ \mathcal{O}\_Y(D)(U)= \{g \in \mathcal{O}\_Y(U) \vert div(g)\geq D \}$$


or equivalently $g/f$ is regular. The first map must be something like $g\rightarrow gf$ maybe with some order. A good answer to my question would include:


1. Is this correct?
2. What is $\mathcal{O}\_D(D)$?
3. What is the second map?
4. What does $\mathcal{O}\_D(-D)$ mean
5. Why $\mathcal{O}\_E(-E) \sim \mathcal{O}(3)$? I understood the RHS is generated by polynomials of degree 3?


I am aware this is a simple question and probably everyone knows why, but I could not find a proper answer for it.



Elementary short exact sequence of sheaves
Expanding the comment of Donu Arapura, let $X$ be a variety and $Y\subset X$ a subvariety.
Then, you have a short exact sequence of sheaves
$$
0\to\mathcal I\_Y\to\mathcal O\_X\to\mathcal O\_X/\mathcal I\_Y\to 0,
$$
where $\mathcal I\_Y$ is the ideal sheaf of $Y$. By definition, $\mathcal O\_X/\mathcal I\_Y=\mathcal O\_Y$ is the structure sheaf of $Y$.


If $\mathcal F$ is any invertible sheaf, then tensoring by $\mathcal F$ leaves the sequence exact, so that you have a short exact sequence
$$
0\to\mathcal I\_Y\otimes\mathcal F\to\mathcal F\to\mathcal O\_Y\otimes\mathcal F\to 0
$$
and $\mathcal O\_Y\otimes\mathcal F$ is just the restriction of $\mathcal F$ to $Y$.


Now, suppose that your $Y=D$ is a (Cartier) divisor, and $\mathcal F=\mathcal O\_X(D)$ is its associated (invertible) sheaf of sections (meromorphic functions with poles allowed along $D$). In this case, $\mathcal I\_D=\mathcal O\_X(-D)$ and the above-mentioned short exact sequence becomes
$$
0\to\mathcal O\_X\to\mathcal O\_X(D)\to\mathcal O\_D(D)\to 0,
$$
and $\mathcal O\_D(D)$ is nothing but the restriction $\mathcal O\_X(D)\otimes\mathcal O\_D$ of the invertible sheaf $\mathcal O\_X(D)$ to the hypersurface $D$.


You can argue dually for $\mathcal O\_D(-D)$, which is thus just the restriction to $D$ of the invertible sheaf $\mathcal O\_X(-D)$.


So your "second map", is just the restriction map.


For your last question, an heuristic explanation is the following: blow-up a smooth point on a surface to obtain a new surface $\widetilde X$, and call the exceptional divisor $E$. Then $\mathcal O\_{\widetilde X}(-E)$ restricted to $E$, which is precisely $\mathcal O\_E(-E)$, can be easily shown to be isomorphic to the (anti)tautological line bundle $\mathcal O(1)$ over $\mathbb P^1\simeq E$ (you can find that on every introductory book in algebraic geometry). Now, you are blowing up a singular point which is an isolated quotient singularity of order three, thus in some sense you are "counting three times" your point, so that $\mathcal O\_E(-E)$ now becomes isomorphic to $\mathcal O(3)$.


