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My question deals with Michael Artin's paper "On isolated rational singularities of surfaces"; more precisely the proof of Theorem 4 on page 133. Here the relevant excerpt:

enter image description here

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:

enter image description here

$Z$ 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).

Additionally let me loose some words on inkspot's observation about the map $ H^0(nZ,O_{nZ}) \to A/{\mathfrak m}^n $: I think that it is a typo and the map should be $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ because considering the composition $nZ \subset V \to \bar{V}=Spec(A)$ induces naturally $A \to H^0(nZ,O_{nZ})$ which factorize through $\mathfrak m^n$, thus I think the only map which can be extracted can only go in the opposite direction of that one in Artin's paper.

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    $\begingroup$ 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. $\endgroup$
    – inkspot
    Commented Feb 8, 2020 at 11:24
  • $\begingroup$ 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 $A/{\mathfrak m}^n$) $\endgroup$
    – user267839
    Commented Feb 9, 2020 at 3:00
  • $\begingroup$ 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 $m^n \otimes H^0(Z,O_{Z})= H^0(V, K_n)$. $\endgroup$
    – user267839
    Commented Feb 13, 2020 at 1:33
  • $\begingroup$ 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? $\endgroup$
    – user267839
    Commented Feb 13, 2020 at 1:34

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