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Let $k$ be a field, $A$ a finitely generated $k$-algebra, put $A^{G}:=\{a \in A \mid g(a)=a ~ \mbox{for all}~g \in G\}$, where $G$ is a finite group of automorphisms of $A$. If

(1) the order of $G$ is not divisible by the characteristic of $k$,

then $A^{G}$ is a finitely generated $k$-algebra.

I saw this statement in I. R. Safarevich's Basic Algebraic Geometry. But I don't know where (1) is used. In in Atiyah Macdonald's Commutative Algebra (exercise 5 of Chapter 7), Condition (1) is omitted.

I wonder what the correct statement is.

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Dear Sang Cheol Lee: I edited your question. I hope the changes I made are OK. I've already added your interesting question to my Google list about Atiyah-MacDonald. –  Pierre-Yves Gaillard Sep 26 '11 at 1:28
Thanks. Pierre-Yves Gaillard –  Sang Cheol Lee Sep 26 '11 at 1:40

2 Answers 2

up vote 3 down vote accepted

The condition on the characteristic is not necessary (nor is the condition that $k$ be a field, though it should be noetherian); see for instance Corollary 1.19 of http://people.fas.harvard.edu/~amathew/chgraded.pdf

Essentially, the point is that $A$ will be integral over $A^G$, and $A^G$ is "sandwiched" between $k, A^G, A$. From these facts, the result is not too complicated to prove.

The result that I only know under conditions on the characteristic is the strengthening to the case of $G$ a reductive algebraic group, acting (algebraically) on a finitely generated $k$-algebra: then $k$ is required to be of characteristic zero. The reason is that the proof uses semisimplicity of the category of $G$-representations, which is only true in characteristic zero.

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Indeed the result is true with no restrictions on the characteristic: see $\S 14.6$, Theorem 339 of my commutative algebra notes. As I mention there, the result was first proved by Hilbert in characteristic zero and then in 1928 by Emmy Noether in arbitrary characteristic.

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