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From the Geometric Invariant Theory book [Mumford - Fogarty - Kirwan], we have the following theorem

([MFK,Theorem 1.1) Let $X$ be an affine scheme over a field $k$, let $G$ be a reductive algebraic group over $k$, let $\sigma:G\times X \to X$ be an action of $G$ on X. Then, a universal categorical quotient $(Y,\Phi)$ of $X$ by $G$ exists.

To fix ideas let's suppose $X:=\operatorname{Spec} A$ for a finitely generated $k$-algebra $A$.

Since $G$ is reductive, we have that $A^G$ is a finitely generated $k$-algebra.

I don't see where the finite generation of $A^G$ is used in the proof of the theorem? For instance we can drop the assumption that $G$ is reductive but then we still need to require that : $A^G$ is finitely generated $k$-algebra in order to conclude that $\operatorname{Spec} A^G$ is the categorical quotient. Where this assumption is used in the proof of the theorem above?

Thank you for your help.

KReiser
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Conjecture
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    Well, which category are you working in? If you're working with finite type schemes over $k$, then you need $A^G$ to be finitely generated in order for the quotient $\operatorname{Spec} A^G$ to belong to your category and therefore be the categorical quotient. (There may be other reasons, but this is the thing that immediately jumps out.) – KReiser Feb 08 '22 at 09:12
  • Thank you @KReiser. I was indeed thinking about the category of finite-type schemes. – Conjecture Feb 09 '22 at 10:33
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    Great, I'll write that as an answer to close the problem then. – KReiser Feb 09 '22 at 21:33

1 Answers1

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As noted in the comments, if one is working with (locally) finite type schemes over $k$, then one needs $A^G$ to be finitely generated in order for the quotient $\operatorname{Spec} A^G$ to belong to the category of (locally) finite type schemes over $k$ and therefore be the categorical quotient.

KReiser
  • 74,746