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 $\mathbb P^2_k$.

  • 1
    $\begingroup$ By definition, a "double point" is a local ring whose completion $\widehat{A}$ is isomorphic to a hypersurface singularity, $k[[X_0,X_1,X_2]]/\langle F(X_0,X_1,X_2)+g\rangle$, with $F$ a nonzero homogeneous polynomial of degree $2$ and $g$ an element of $\langle X_0,X_1,X_2\rangle^3$. Thus, you can do an embedded blowing up of this hypersurface inside the blowing up at the origin of the regular ring. The exceptional divisor of the ambient blowing up, resp. the embedded resolution, is $\text{Proj}\ k[X_0,X_1,X_2]\cong \mathbb{P}^2$, resp. the zero scheme of $F$ in $\mathbb{P}^2$. $\endgroup$ Commented Nov 24, 2017 at 11:20
  • $\begingroup$ . . . If $F$ is a rank $3$ quadric, then the embedded resolution is already smooth, and $A$ has an "ordinary double point". Artin's hypothesis is that $F$ equals the square of a linear polynomial, $F=\lambda H(X_0,X_1,X_2)^2$ for some nonzero $\lambda$. For the associated line $C=\text{Zero}(H)$ in the exceptional surface $E\cong \mathbb{P}^2$, the intersection $E\cap X$ equals $2C$ as Weil divisors on $X$. Also $\mathcal{O}_C(-\underline{E})$ equals $1$ since $\mathcal{O}_E(-\underline{E})$ equals $\mathbb{O}_{\mathbb{P}^2}(1)$. $\endgroup$ Commented Nov 24, 2017 at 11:33
  • $\begingroup$ @JasonStarr Sorry, I don't quite get your argument, you showed that $\deg \mathcal O_C(-E)=1$ and $E\cap X=2C$. Basically, this is by the definition. But how do you relate it with $\deg \mathcal O_C(-C)$? My question can be rephrased as "why does $[-C,C]$ computed in $X$ agree with one computed in $\mathbb P^2_k$?". $\endgroup$
    – gdb
    Commented Nov 24, 2017 at 17:52
  • $\begingroup$ @JasonStarr Also, I don't assume that $A$ is defined over a field. Probably, it is not so important, even without this assumption we know that 𝔪m is generated by $3$ elements and there is exactly $1$ relation in $\mathfrak m^2/\mathfrak m^3$. In any case, I will be glad to understand the proof at least for algebras over fields. $\endgroup$
    – gdb
    Commented Nov 24, 2017 at 23:25
  • $\begingroup$ I do not currently have access to "Arithmetic Geometry", so I cannot address what Artin wrote. $\endgroup$ Commented Nov 25, 2017 at 2:30

1 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.


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