# Artin's “On Isolated Rational Singularities of Surfaces”

My question refers to M. Artin's paper "On Isolated Rational Singularities of Surfaces"; more precisely the proof of Thm 4 on page 133. Here the relevant excerpt:

The Setting: Let $$\bar{V}=Spec(A)$$ where $$A$$ is a local, normal $$2$$-dimensional ring with algebraically closed residue field $$k=A/\frak{m}$$.

Let $$\pi:V \to \bar{V}$$ be a birational proper map with $$V$$ regular, i.e. it "resolves" the singularity $$s$$ of $$\bar{V}$$, where the point $$s$$ corresponds to the maximal ideal $$\frak{m}$$ $$\subset A$$.

Denote by $$Z= \sum_i X_i$$ the fundamental cycle; here the definition:

$$Z$$ is defined as the unique smallest cycle satisfying property

One of the intermediate steps in the proof is to show that for the ideal $$I_Z \mathcal{O}_V$$ that determines the fundamental cycle (that is a closed subscheme of $$V$$) we have

$$\mathfrak{m} \cdot \mathcal{O}_{\text{V}}= I_Z$$

Artin reduces the problem to verification of the surjectivity of

(*) $$H^0(Z, \mathcal{O}_{(n+1)Z}) \to H^0(Z, \mathcal{O}_{nZ})$$

for each $$n$$.

Lemma 5 proves it.

Problem/Question: Then it is claimed that "moreover, because of (*), it follows that the canonical map $$A/\mathfrak{m}$$ $$\to H^0(nZ, \mathcal{O}_{nZ})$$ is surjective.

Why? I don't understand it. How the argument works?

What I tried: My first approach was to argue by induction but this gives an obstacle that I can't solve.

Denote $$K_n:= ker(\mathcal{O}_{(n+1)Z} \to \mathcal{O}_{nZ})$$. One can show that $$K_n \cong I_{nZ} \otimes_V \mathcal{O}_Z$$ and one obtain the diagram

$$\require{AMScd} \begin{CD} \mathfrak{m}^n/\mathfrak{m}^{n+1} = \mathfrak{m}^n \otimes A/\mathfrak{m} @>{} >> A/\mathfrak{m}^{n+1} @>{} >> A/\mathfrak{m}^n \\ @VaVV @VbVV @VcVV \\ H^0(V, K_n) @>{} >> H^0(Z, \mathcal{O}_{(n+1)Z}) @>{}>> H^0(Z, \mathcal{O}_{nZ}); \end{CD}$$

By induction hypothesis we may assume that $$c$$ is surjective. Question/Problem: Why is $$a$$ surjective? (we need surjectivity of $$a$$ to conclude that $$b$$ is surjective).

• I don't know whether this will answer the question, but you write $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ while Artin's homomorphism goes in the opposite direction. – inkspot Feb 8 at 11:24
• I think that it was just a typo in the paper. That is we consider indeed $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ induced canonically by $nZ \subset V \to \bar{V}=Spec(A)$ on the ring side ($A \to H^0(nZ,O_{nZ})$ factorizes through $A/{\mathfrak m}^n$) – KarlPeter Feb 9 at 3:00
• I think I have (almost) the answer I was looking for: The problem is to show that $a$ surjective and by induction hypothesis $c_n: A/m^k \to H^0(Z,O_{kZ})$ is surjective for all $k < n+1$. Assume $A/m \to H^0(Z,O_{Z})$ is surjective (that's a seriuos problem: see below problem 2, but let at first assume we know it). We tensor it with $m^n$ and obtain the surjection $m^n/m^{n+1} \to m^n \otimes H^0(V, K_n)$. The goal would be to show $m^n \otimes H^0(Z,O_{Z})= H^0(V, K_n)$. – KarlPeter Feb 13 at 1:33
• That is that $H^0$ and $\otimes$ commute. I'm facing two problems now: first one: to show that $m O_V$ is invertible $O_V$-module. (then it would flat and $H^0$ and $\otimes$ commute and we are done). The second problem is the initial induction step: how to show that $A/m \to H^0(Z,O_{Z})$ is surjection? – KarlPeter Feb 13 at 1:34