Resolution of Gorenstein rational singularities on a surface

Question

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)

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 $\Sigma_{i} [k(p_i):k]=3$.

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.

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., $[-C,C]=-1$.

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 $\mathbb P^2_k$.

Answer 1

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.