My question refers to M. Artin's paper "On isolated rational singularities of surfaces"; more precisely the proof of Theorem 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).

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.

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:

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).

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.

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).

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.

My question refers to M. Artin's paper "On isolated rational singularities of surfaces"; more precisely the proof of Theorem 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).

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.

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).

Additionally let lose some of 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.

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).

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.