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Let $M$ be a complex manifold with its sheaf $\mathcal{O}_M$ of holomorphic functions.

Fix a point $z\in M$ and denote by $\mathcal{O}_z$ the stalk of $\mathcal{O}_M$ at $z.$

Cosider ideals $\mathfrak{m}_z$ and $\overline{\mathfrak{m}_z}$ of functions vanishing at $z$ in $\mathcal{O}_M(M)$ and $\mathcal{O}_z$ respectively. I.e.

$$\mathfrak{m}_z=\lbrace f\in \mathcal{O}_M(M): f(z)=0\rbrace\hspace{5pt}\text{and}\hspace{5pt} \overline{\mathfrak{m}_z}=\lbrace f\in \mathcal{O}_z: f(z)=0\rbrace$$

Question. Does $\mathfrak{m}_z/\mathfrak{m}_z^2\cong\overline{\mathfrak{m}_z}/\overline{\mathfrak{m}_z}^2 $?

My guess is that it should not hold, cause holomorphic functions are too rigid. I wonder if counterexample can be find in $M$ being open subset of $\mathbb{C}.$

Update after Ben's comment.

If $M$ is a compact manifold then $O_M(M)=\mathbb{C},$ so $\mathfrak{m}_z=0$ and hence $\mathfrak{m}_z/\mathfrak{m}_z^2=0.$

So if $\mathfrak{m}_z/\mathfrak{m}_z^2\not\cong\overline{\mathfrak{m}_z}/\overline{\mathfrak{m}_z}^2 $, then all we need to do is compute $\overline{\mathfrak{m}_z}/\overline{\mathfrak{m}_z}^2.$

Footnote

Recently I asked in this question if more general result holds, but it occurred that is does not.

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    $\begingroup$ It's the same problem as in the MSE question: if $\mathfrak{m}_z$ means the sheaf of ideals in $\mathcal{O}_M$, then $\mathfrak{m}_z/\mathfrak{m}_z^2$ is supported at $z$ and there it is isomorphic to $\overline{\mathfrak{m}}_z/\overline{\mathfrak{m}}_z^2$, but if (as stressed in the MSE thread) $\mathfrak{m}_z$ means a hypothetical ideal in the ring of global sections $\mathcal{O}_M(M)$, then the question is not well-posed. (If $M$ is compact, $\mathcal{O}_M(M) = \mathbb{C}$.) $\endgroup$
    – Ben
    Commented Aug 9, 2017 at 11:39
  • $\begingroup$ @Ben Sorry, I ment that $\mathfrak{m}_z=\lbrace f\in \mathcal{O}_M(M): f(z)=0\rbrace.$ That $\mathfrak{m}_z$ is an ideal in $O_M(M).$ So for compact $M$ we have that $\mathfrak{m}_z/\mathfrak{m}_z^2=0.$ Hence the only thing left is to compute $\overline{\mathfrak{m}_z}/\overline{\mathfrak{m}_z}^2.$ $\endgroup$
    – Fallen Apart
    Commented Aug 9, 2017 at 12:10
  • $\begingroup$ Well, $\overline{\mathfrak{m}}_z/\overline{\mathfrak{m}}_z^2\cong\mathbb{C}^{\dim(M)}$...?! $\endgroup$
    – Ben
    Commented Aug 9, 2017 at 13:30

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