# The fixed subalgebra of a finitely generated algebra

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.

• Thanks. Pierre-Yves Gaillard – Sang Cheol Lee Sep 26 '11 at 1:40

## 3 Answers

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.

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.

As pointed out above, we don't need $$k$$ to be a field; nor do we need $$|G|$$ to be invertible in $$k$$. Let $$A=k[a_1,\ldots, a_m]$$, and notice $$a_i$$ satisfies the polynomial $$\prod_{g\in G}(X-g(a_i))\in A^G[X]$$. Taking all of these coefficients, as $$i$$ varies from 1 to $$m$$, gives us $$m|G|$$ elements which generate a $$k$$-algebra $$Y$$. Clearly $$Y\subset A^G$$ and $$Y$$ is finitely generated as a $$k$$-algebra, thus by Hilbert's Basis Theorem, $$Y$$ is Noetherian. We have $$k\subset Y\subset A^G\subset A$$.

Now, $$A$$ is a finitely generated integral $$Y$$-algebra, because all the $$a_i$$ are integral over $$Y$$, and is thus a finitely generated module over $$Y$$. (This is a general fact that can be easily proven by induction on the number of generators needed to display $$A$$ as a $$Y$$-algebra). Since $$Y$$ is Noetherian this implies $$A^G$$ is also a finitely generated $$Y$$-module.

Finally, $$A^G$$ is a finitely generated module over a finitely generated $$k$$-algebra and is thus itself a finitely generated $$k$$-algebra, as desired.