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Consider a complex vector space $V \cong {\mathbb C}^n$. Consider the ring of germs of holomorphic functions ${\mathcal O}_0 (V)$ at $0\in V$. We know that this ring is Noetherian.

Now consider a finite group $G$ acting linearly on $V$ and consider the subring of $G$-invariant germs, denoted by $${\mathcal O}_0 (V)^G.$$ Consider another $G$-representation $W$ and the set of germs of $G$-equivariant holomorphic maps $f: V \to W$ at $0\in V$, denoted by ${\rm Map}_0 (V, W)^G$. This is a module over ${\mathcal O}_0 (V)^G$.

Question: is ${\rm Map}_0(V, W)^G$ finitely generated over ${\mathcal O}_0(V)^G$? Can we take generators to be polynomial maps?

Background: we know that for the ring of invariant polynomial maps and the module of equivariant polynomial maps, the answer is yes (see the Mathoverflow post) It is easy to see that the version for formal power series is also true.

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  • $\begingroup$ Is the ring $\mathcal{O}_0(V)$ a finite type $\mathbf{C}$-algebra? I'm a bit rusty. $\endgroup$
    – David Benjamin Lim
    Commented Mar 27, 2023 at 13:02
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    $\begingroup$ The local ring $\mathcal{O}_0(V)^G$ is an algebra over $\mathbb{C}[V^G]$, the ring $\mathbb{C}[V]$ is a finite module over $\mathbb{C}[V^G]$, and thus the module $\mathcal{O}_0(V) = \mathbb{C}[V]\otimes_{\mathbb{C}[V]}\mathcal{O}_0(V)^G$ is a finite module over $\mathcal{O}_0(V)^G$. Since $\text{Poly}_0(V,W)$ is a finite (free) module over $\mathbb{C}[V]$, also $\text{Map}_0(V,W)=\text{Poly}_0(V,W)\otimes_{\mathbb{C}[V]}\mathcal{O}_0(V)$ is a finite (free) module over $\mathcal{O}_0(V)$, and thus also a finite module over $\mathcal{O}_0(V)^G$. $\endgroup$
    – Jason Starr
    Commented Mar 27, 2023 at 14:39
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    $\begingroup$ . . . Since $\mathcal{O}_0(V)^G=\mathcal{O}_0(V/G)$ is Noetherian, also the $\mathcal{O}_0(V)^G$ submodule $\text{Map}_0(V,W)^G$ of the finitely generated $\mathcal{O}_0(V)^G$ module $\text{Map}_0(V,W)$ is a finitely generated $\mathcal{O}_0(V)^G$-module. (I forgot to add that step to the previous comment.) $\endgroup$
    – Jason Starr
    Commented Mar 27, 2023 at 22:48

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