Consider a commutative group $G$ of finite type, a subgroup of finite index $H\subseteq G$, a noetherian commutative ring $A$, and a $G$-graded $A$-algebra $R=\bigoplus_{g\in G}R_g$ with no zero-divisors, and denote by $R_H=\bigoplus_{g\in H}R_g$ the degree restriction of $R$ to $H$.

It is well-known that if $R$ is of finite type over $A$, then so is $R_H$. So, we can ask whether the following statement is true:

(+) If $R_H$ is of finite type over $A$, then so is $R$.

Using the fact that $R$ is integral over $R_H$, one can show that (+) holds in the following two cases:

  • $R_H$ is integrally closed and the field of fractions of $R$ is a separable extension of the field of fractions of $R_H$;

  • $R_H$ is a japanese ring (see EGA 0$_{\rm IV}$.23).

In particular, (+) holds if $A$ is universally japanese, e.g. excellent, e.g. of finite type over a field.

My question is now as follows:

Is there an example where (+) does not hold?


In case $G$ is finite, this cannot happen.

(This might extend to the general case of finitely generated groups, as Fred told me when we talked about this in my office :-) )

First, let me show that $R$ is of finite type over $R_0$ in case $R_0$ is noetherian and $G$ is finite. For that, it suffices to show that every $R_g$, $g \in G$ is finitely generated as an $R_0$-module.

Now fix some $g \in G$. In case $R_g = \{ 0 \}$, we are done. Otherwise, let $\alpha \in R_g \setminus \{ 0 \}$. As $G$ is a finite group, there exists some $n > 0$ such that $n g = 0$. As $R$ is integral, $\beta := \alpha^{n-1} \neq 0$ is a non-trivial element of $R_{g^{-1}}$.

Now the map $\varphi : R_g \to R_0$, $x \mapsto \beta x$ is an injective $R_0$-module homomorphism. The image, $\beta R_g$, is therefore isomorphic to $R_g$ as an $R_0$-module. As $R_0$ is noetherian, every $R_0$-submodule of $R_0$ is finitely generated, whence $\beta R_g$ and thus $R_g$ is a finitely generated $R_0$-module.

This shows that $R$ is of finite type over $R_0$.

Now let me go back to the original problem. As $R_0$ is a subring of both $R$ and $R_{(H)}$, it is of finite type over the noetherian ring $A$. Therefore, $R_0$ is noetherian as well. So in case $G$ is finite, the above shows that $R$ is of finite type over $R_0$, and thus also over $A$.

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    $\begingroup$ Doesn't this solve the general case as well? (which I assume is what you're saying). It's really about the grading induced by $G/H$, which is a finite group. (I.e., you should lump $R_H$ into one ring, $R_0$, and grade by the cosets in $G/H$). $\endgroup$ – Mike Roth Apr 3 '12 at 16:39
  • $\begingroup$ Thanks, @Felix! The general case follows as @Mike suggests. In particular, $G$ need not be of finite type. $\endgroup$ – Fred Rohrer Apr 4 '12 at 11:38
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    $\begingroup$ The hypothesis that $R$ is a domain is crucial. Take $R=k[x_2,x_3,x_5,x_7,\dots]/I$ where $I$ contains every binomial $x_ix_j-x_2^{(i+j)/2}$ for $i,j \geq 3$ odd. Then $G=\mathbb{Z}/2\mathbb{Z}$ acts $R$ by $x_j \mapsto (-1)^j x_j$; the invariants are $R^G \cong k[x_2]$, but $R$ is not finitely generated (has generators in every odd degree $\geq 3$). $R$ has plenty of zerodivisors, e.g., $x_5(x_5-x_2x_3)=0$. $\endgroup$ – John Voight Dec 10 '13 at 21:49

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