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: 



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> (+) If $R\_H$ is of finite type over $A$, then so is $R$.
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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:



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



Finiteness conditions and Veronese subrings
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 :-) )




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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 $R\_0$.




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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 $A$.


