I am reading Artin's notes "Lipman's Proof of Resolution of Singularities for Surfaces" from the book "Arithmetic Geometry". I am very confused by the proof of Lemma $6.5.$ (I am formulating it below in a little bit different way than it appears in the text)



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> Lemma 6.5: Let $(A,\mathfrak m, k)$ be a normal complete excellent ring of dimension $2$ that defines a rational double point (rational Gorenstein singularity). Denote by $X$ the blow-up of the unique closed point in Spec $A$. Assume that the exceptional divisor $E$ is equal to $2C$, where $C$ is a line in $\mathbb P^2\_k$. And let $X' \to X$ be a sequence of blow-ups in closed points $p\_1, \dots, p\_n$, s.t. $X'$ is regular at evert point of the strict transform $C'$, then $\Sigma\_{i} [k(p\_i):k]=3$. 
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The key step is to compute $\deg\_C \mathcal O\_X(-C)|\_C$ (Note that $\mathcal O\_X(-C)$ isn't locally free since $C$ isn't a Cartier divisor, but it is always reflexive, in particular torsion-free. Hence, $\mathcal O\_C(-C)$ is always an invertible sheaf). Artin claims that it is equal to $-1$, but I don't understand his argument. 



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> In our case, since $2C$ is isomorphic to a double line in $\mathbb P^2$, the degree is the same as for such a line, i.e., $[-C,C]=-1$.
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How could one put this into a rigorous argument? It is not clear how to relate $\deg\_C \mathcal O\_C(-C)$ with this immersion since $\mathcal O\_C(-C)^{\otimes 2} \neq \mathcal O\_C(-E)$.


P.S. By a rational singularity I mean that for any normal modification $f:X \to Spec A$ we have $H^1(X,\mathcal O\_X)=0$. If $A$ is also Gorenstein, it is called rational double point. The latter condition is equivalent to $\dim\_k \mathfrak m/\mathfrak m^2 \leq 3$.


P.S.2. In the formulation of Lemma $E$ should be equal to a double line $2C$ with respect to the natural immersion $X \to \mathbb P^2\_A$ defined by the sheaf $\mathcal O\_X(-E)$.


UPD: Jason Starr mentioned in the comments that if $A$ is defined over a field $k$, then $A\cong k[[x,y,z]]/(F(x,y,z)-G(x,y,z))$, where $F$ is homogeneous quadratic polynomial and $G$ is of degree at least $3$. We can do almost the same without assuming that $A$ is defined over a field. Namely, since $A$ is a rational double point $\dim\_k \mathfrak m^n/\mathfrak m^{n+1}=2n+1$. Then we have $3$ generators for $\mathfrak m$ and there is precisely one relation in degree $2$ between them in $gr\_{\mathfrak m} A$. Let this relation be $F(x,y,z)=G(x,y,z)$, where $F,G\in k[T\_1,T\_2,T\_3]$ are polynomials of degree $2$ and $3$ respectively ($F$ is also homogeneous). Since $E\cong Proj(gr\_{\mathfrak m} A)$ we conclude that $E\cong V(F) \subset \mathbb P^2\_k$. Taking into account that $E=2C$ we can actually choose (after a suitable linear change of coordinates) $F(T\_1,T\_2,T\_3)=T\_1^2$.


But I still don't understand what is the connection between $\deg \mathcal O\_C(-C)$ and the intersection number $[-C,C]$ inside $\mathbb P^2\_k$.



Resolution of Gorenstein rational singularities on a surface
The explanation given in the text indeed seems to be too terse. What follows is taken from an insert I have in my copy of that book (since the argument I came up with when I read the article many years ago did not fit in the margin as other clarifications did). I hope it is helpful, and that I haven't made some blunder.


We'll use the notation in Artin's article (which I won't explain here). The idea is to give a description of $O\_C(-C)$ that is intrinsic to the infinitesimal thickening $C \hookrightarrow Z$, without direct reference to the ambient $X\_1$. That will allow us to make a switch to transfer the degree calculation to a more convenient choice of $X\_1$! 


By definition the $O\_C$-module $O\_C(-C)$ on $C \simeq \mathbf{P}^1\_k$ is
$$(O\_{X\_1}(-C) \otimes\_{O\_{X\_1}} O\_C)/(O\_C\mbox{-}{\rm{torsion}}) = (O\_{X\_1}(-C) \otimes\_{O\_{X\_1}} (O\_{X\_1}/O\_{X\_1}(-C)))/(O\_C\mbox{-}{\rm{torsion}})$$
that is torsion-free with generic rank 1 by design, so it is invertible. Moreover, we claim that the canonical surjection of $O\_{X\_1}$-modules
$$q:O\_{X\_1}(-C) \twoheadrightarrow O\_C(-C)$$
has kernel exactly $O\_{X\_1}(-2C)$. The quotient $O\_{X\_1}(-C)/O\_{X\_1}(-2C)$ is easily checked to be torsion-free as an $O\_C$-module and has generic rank 1 as such, so it is invertible as such. Thus, since a surjection between invertible sheaves is an isomorphism, to identify $\ker q$ as claimed it suffices to show that the $O\_{X\_1}$-linear restriction $q: O\_{X\_1}(-2C) \to O\_C(-C)$ vanishes. But the target
of this restriction is a torsion-free $O\_C$-module, so it suffices to check the vanishing near the generic point of $C$, hence over the regular locus $X\_1^{\rm{reg}}$, where it is clear. The upshot is that we have
$$O\_C(-C) \simeq O\_{X\_1}(-C)/O\_{X\_1}(-2C)$$
as $O\_C$-modules.


Let $I$ denote the coherent ideal sheaf of $C$ in $Z$, so this is a square-zero ideal by definition of $Z$ (as the ideal sheaf $O\_{X\_1}(-C)$ inside $O\_{X\_1}$ has square contained in $O\_{X\_1}(-2C)$) and hence $I$ is naturally an $O\_C$-module. As such, we have 
$$I := {\rm{image}}(O\_{X\_1}(-C) \to O\_Z = O\_{X\_1}/O\_{X\_1}(-2C)) =
O\_{X\_1}(-C)/O\_{X\_1}(-2C) \simeq O\_C(-C),$$
the final isomorphism being what we established above.
Thus, we have an *intrinsic* description of the $O\_C$-module $O\_C(-C)$ as the coherent ideal sheaf of $C$ inside the square-zero thickening $Z$ of $C$. The conclusion is that $O\_C(-C)$ as an $O\_C$-module depends
only in the data of the infinitesimal closed immersion $C \hookrightarrow Z$ and not on the ambient $X\_1$.


But we can identify $Z$ scheme-theoretically as a doubled-line in $\mathbf{P}^2\_k$ with $C = Z\_{\rm{red}}$ a straight line in this projective plane. Hence, to compute the degree of the $O\_C$-module $O\_C(-C)$ it suffices to do the calculation in an ambient $\mathbf{P}^2\_k$. Now we can run the calculations in reverse, with $C$ inside $\mathbf{P}^2\_k$ having ideal sheaf $O\_{\mathbf{P}^2}(-1)$:
$$O\_C(-C) \simeq O\_{\mathbf{P}^2}(-C)/O\_{\mathbf{P}^2}(-2C) \simeq
O\_{\mathbf{P}^2}(-1)/O\_{\mathbf{P}^2}(-2).$$
Since $C = \mathbf{P}^1\_k$ has structure sheaf $O\_C$ with Euler characteristic equal to 1, so ${\rm{deg}}\_C(L) = \chi(L) - 1$ for any line bundle $L$ on $C$, to show ${\rm{deg}}\_C(O\_C(-C)) = -1$ is the same as showing $\chi(O\_C(-C))=0$. From the above displayed expression for $O\_C(-C)$ in terms of $O\_{\mathbf{P}^2}(-r)$'s, it is the same to show
that $\chi(O\_{\mathbf{P}^2}(-1))=\chi(O\_{\mathbf{P}^2}(-2))$. Both of
these latter Euler characteristics are equal to 0.


