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Mar 27, 2023 at 22:48 comment added Jason Starr . . . 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.)
Mar 27, 2023 at 14:39 comment added Jason Starr 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 $\mathcal{O}_0(V)^G$.
Mar 27, 2023 at 13:02 comment added David Benjamin Lim Is the ring $\mathcal{O}_0(V)$ a finite type $\mathbf{C}$-algebra? I'm a bit rusty.
Mar 26, 2023 at 22:06 history asked UVIR CC BY-SA 4.0