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I have few questions about the proof of Thm 7.5 from P. Wagreich's paper "Elliptic Singularities of Surfaces" (page 452):

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Denote by $\mathcal{N}$ the kernel of $A \to H^0(O_Z)$ (sorry, I haven't found the latex command for such winged "N" from the excerpt :(

  1. QUESTION:The author says that $\mathcal{N}$ is primary to the maximal ideal $\mathcal{M}$ of $A$ "since $H^0(O_Z)$ is Artin ring".

I don't understand this argument. $\mathcal{N}$ primary to $\mathcal{M}$ means by definition that $\sqrt{\mathcal{N}}=\mathcal{M}$. Every kernel is prime ideal and so this would imply $\mathcal{N}= \mathcal{M}$. This seems absurd, following futher steps in the proof. Or do I miss something?

next problem:

The proof shows that for a $f \in I_x$ the power $f^n \in \mathcal{N}O_{\mathcal{X}} (mod \text{ } I^{n+1})$.

Rmk: It was implicitely used Lemma 7.4 form the paper which implies that for $n$ big enough $H^1(I^{n+1})=0$ so $H^0(I^n) \to H^0(I^n/I^{n+1})$ is indeed surjective.

  1. QUESTION:I don't know how to perform the induction step to show that for $f \in I_x$ we have $f^n \in \mathcal{N}O_{\mathcal{X}} (mod \text{ } I^{n+i+1})$ under assumption that $f^n \in \mathcal{N}O_{\mathcal{X}} (mod \text{ } I^{n+i})$ already holds.
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  • $\begingroup$ "every kernel is prime ideal": why??? $H^0(\mathcal{O}_Z)$ is certainly not a domain. $\endgroup$
    – abx
    Commented Oct 28, 2019 at 7:13
  • $\begingroup$ @abx: ah yes, you are right, sorry for dumb statement. I think the argument for QUESTION 1 is that we consider for $\phi: A \to H^0(O_Z)$ the ideal $\widehat{\mathcal{M}}:= \phi(\mathcal{M}) \cdot H^0(O_Z)$ generated by image of $\mathcal{M}$ and find a $n >0$ with $\widehat{\mathcal{M}}^n=0$ (such $n$ exists because $H^0(O_Z)$ Artin), And this means exactly that the kernel is primary to $\mathcal{M}$. That is it,right? $\endgroup$
    – user267839
    Commented Oct 28, 2019 at 22:01

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