Let $G$ be a finite group acting on a topological space $X$. (In my applications, $X$ is the classifying space of a compact Lie group.) Let $H^\ast(-)=H^\ast(-;\mathbb{Z}/2\mathbb{Z})$ denote mod 2 singular cohomology.

By the work of Quillen ("The Spectrum of an Equivariant Cohomology Ring: I"), if $H^\ast(X)$ is a finitely generated $\mathbb{Z}/2\mathbb{Z}$-module, then the equivariant cohomology $$H^\ast_G(X) = H^\ast(EG\times_G X)$$ is a finitely generated as a $\mathbb{Z}/2\mathbb{Z}$-algebra.

My question is, if we only assume that $H^\ast(X)$ is finitely generated as an algebra, does the conclusion still hold, ie is the equivariant cohomology $H^\ast_G(X)$ necessarily finitely generated as an algebra?

More background

The standard tool for computing equivariant cohomology is the Leray-Serre spectral sequence of the fibration $X\to EG\times_G X \to BG$, which has $$H^\ast(BG,H^\ast(X)) \Longrightarrow H^\ast(EG\times_G X)$$ as algebras.

Now if $G$ acts trivially on $H^\ast(X)$ then the $E_2$-page can be identified with $$H^\ast(BG)\otimes H^\ast(X)$$ as algebras. By a classical result of Evens, the mod 2 cohomology algebra of a finite group is finitely generated. Since the tensor product of finitely generated algebras is again finitely generated, so is this $E_2$-page, and hence so is the equivariant cohomology algebra. (Please correct me if I'm doing something wrong!see the comments made by Algori and Ralph below.)

However, in general the coefficients in the $E_2$-page are twisted by the action of $G$ on $H^\ast(X)$ (which in my case happens to be non-trivial), and so I can't see how the argument would go.

At the other extreme, if $G$ acts freely on $X$ then we are basically asking if there is a finite covering space $X\to Y$ such that $H^\ast(X)$ is finitely generated as an algebra but $H^\ast(Y)$ is not. Such a beast would give a counter-example.

(I'm also rather sure that if the answer to my question was yes, Quillen would have told us so!)

More background

The standard tool for computing equivariant cohomology is the Leray-Serre spectral sequence of the fibration $X\to EG\times_G X \to BG$, which has$$H^\ast(BG,H^\ast(X)) \Longrightarrow H^\ast(EG\times_G X)$$as algebras.

Now if $G$ acts trivially on $H^\ast(X)$ then the $E_2$-page can be identified with $$H^\ast(BG)\otimes H^\ast(X)$$as algebras. By a classical result of Evens, the mod 2 cohomology algebra of a finite group is finitely generated. Since the tensor product of finitely generated algebras is again finitely generated, so is this $E_2$-page, and hence so is the equivariant cohomology algebra. (Please correct me if I'm doing something wrong!)

However, in general the coefficients in the $E_2$-page are twisted by the action of $G$ on $H^\ast(X)$ (which in my case happens to be non-trivial), and so I can't see how the argument would go.

At the other extreme, if $G$ acts freely on $X$ then we are basically asking if there is a finite covering space $X\to Y$ such that $H^\ast(X)$ is finitely generated as an algebra but $H^\ast(Y)$ is not. Such a beast would give a counter-example.

(I'm also rather sure that if the answer to my question was yes, Quillen would have told us so!)

Let $G$ be a finite group acting on a topological space $X$. (In my applications, $X$ is the classifying space of a compact Lie group.) Let $H^\ast(-)=H^\ast(-;\mathbb{Z}/2\mathbb{Z})$ denote mod 2 singular cohomology.

By the work of Quillen ("The Spectrum of an Equivariant Cohomology Ring: I"), if $H^\ast(X)$ is a finitely generated $\mathbb{Z}/2\mathbb{Z}$-module, then the equivariant cohomology $$H^\ast_G(X) = H^\ast(EG\times_G X)$$ is a finitely generated as a $\mathbb{Z}/2\mathbb{Z}$-algebra.

My question is, if we only assume that $H^\ast(X)$ is finitely generated as an algebra, does the conclusion still hold, ie is the equivariant cohomology $H^\ast_G(X)$ necessarily finitely generated as an algebra?