## Finiteness conditions and Veronese subrings

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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?

-

## 1 Answer

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 $A$.

-

1
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 $G/H$). – Mike Roth Apr 3 2012 at 16:39
Thanks, @Felix! The general case follows as @Mike suggests. In particular, $G$ need not be of finite type. – Fred Rohrer Apr 4 2012 at 11:38
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9388204216957092, "perplexity_flag": "head"}