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Does one need to work over an algebraic closed field in ordre to construct GIT quotients à la Mumford? If yes, why?

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    $\begingroup$ Have you looked in Mumford's book? $\endgroup$
    – Angelo
    Nov 8, 2012 at 15:29
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    $\begingroup$ $X = {\rm Proj}\ R$, $X//G := {\rm Proj}\ R^G$ $\endgroup$ Nov 8, 2012 at 20:16

1 Answer 1

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Hey MBeasy,

you can construct the quotients over an arbitrary "not necessarily algebraically closed field", as Mumford states at the beginning of chapter 1.

Greetings

Daniel

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