Finiteness conditions and Veronese subrings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:14:14Z http://mathoverflow.net/feeds/question/92683 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92683/finiteness-conditions-and-veronese-subrings Finiteness conditions and Veronese subrings Fred Rohrer 2012-03-30T13:33:11Z 2012-04-03T14:05:02Z <p>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 $H$.</p> <p>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: </p> <blockquote> <p>(+) If $R_H$ is of finite type over $A$, then so is $R$.</p> </blockquote> <p>Using the fact that $R$ is integral over $R_H$, one can show that (+) holds in the following two cases:</p> <ul> <li><p>$R_H$ is integrally closed and the field of fractions of $R$ is a separable extension of the field of fractions of $R_H$;</p></li> <li><p>$R_H$ is a japanese ring (see EGA 0$_{\rm IV}$.23).</p></li> </ul> <p>In particular, (+) holds if $A$ is universally japanese, e.g. excellent, e.g. of finite type over a field.</p> <p>My question is now as follows:</p> <blockquote> <p>Is there an example where (+) does not hold?</p> </blockquote> http://mathoverflow.net/questions/92683/finiteness-conditions-and-veronese-subrings/93001#93001 Answer by felix for Finiteness conditions and Veronese subrings felix 2012-04-03T14:05:02Z 2012-04-03T14:05:02Z <p>In case $G$ is finite, this cannot happen.</p> <p>(This might extend to the general case of finitely generated groups, as Fred told me when we talked about this in my office :-) )</p> <hr> <p>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.</p> <p>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 $R_{g^{-1}}$.</p> <p>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.</p> <p>This shows that $R$ is of finite type over $R_0$.</p> <hr> <p>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 $A$.</p>