1
$\begingroup$

Let $S$ be a graded ring i.e. $S_d\cdot S_e \subset S_{d+e}$ where $S_d$ is the degree $d$ part of $S$. Assume $S_{<0}$ vanishes. For simplicity, you may think of $S$ as a graded subring of $\mathbb C[T_0,\cdots, T_N]$, the polynomial ring of multi-variables. Indeed, this question is from algebraic geometry and what I want to study is $\mathrm{Proj}(S) \setminus Zero(S_{>0})$. In my mind, $S$ is the ring $\mathbb C[T]^G$ of $G$-invariant polynomials for a reductive algebraic group $G$.

Question: (1) Can we find an integer $m>0$ such that $S_{\ge m}$ can be generated by finitely many $s_1,\dots,s_n$ of the same degree $m$? If not, can we add some additional conditions to make it become true?

(2) In the case $S$ is a graded subring of $\mathbb C [T_0,\dots, T_N]$, can we find degree-$m$ elements $s_0,\dots,s_n$ as above that generate $S_{\ge m}$ such that $$ Zero(s_0,\dots,s_n)=Zero(S_{>0})\subset \mathbb P^{N}$$

(3) In general, how to find an integer $m$ as above? how about the smallest $m$?

EDIT: My purpose is to give a more concrete description of projective GIT quotient $\pi: X_{ss} \to X_{ss}// G$ where $X_{ss}$ is the set of semi-stable points of a projective variety $X$. As we know the set $X_{unstable}$ of unstable points is just the complement of $X_{ss}$, and $X_{unstable} = X - X_{ss}$ is equal to $Zero( \mathbb C[X^*]^G_{>0})$ where $X^*$ is the affine cone of $X$. If (some modification of) (2) is true, then $X_{unstable} = Zero(s_0, \dots, s_N)$. And the quotient will be given explicitly by: $$ X_{ss} \equiv X-X_{unstable} \to \mathbb P^N, \ \ \ \ x\mapsto [s_0(x),\dots, s_N(s)]$$

An example of this explicit map is as follows. Let $\mathbb C^*$ acts on $\mathbb C^3$ by $(x,y,z)\mapsto (tx,t^{-1}y,z)$, and $X=\mathbb P^2$. Then $\mathbb C[x,y,z]^G=\mathbb C[xy,z]$. In this example, one can choose $m=2$: $$ \mathbb C[x,y,z]^G_{\ge 2} = \mathbb C[xy,z^2] $$ The quotient map then will be $$ X_{ss} = \mathbb P^2 - \{[1,0,0],[0,1,0]\} \ni [x,y,z]\mapsto[xy,z^2] \in \mathbb P^1 =X_{ss} / / G $$

$\endgroup$
  • $\begingroup$ Under your assumptions, $S$ is not even necessarily a finitely generated algebra over $\mathbb C$. For example, let $S = \mathbb C[xy, xy^2, xy^3, \ldots] \subseteq \mathbb C[x,y]$. Then $S_{\geq m}$ is not a finitely generated ideal for any $m > 0$. You might want to put some finiteness conditions on $S$. $\endgroup$ – R. van Dobben de Bruyn Jun 27 '17 at 2:30
  • $\begingroup$ Thanks. In fact, I want to consider the case $S=\mathbb C[X]^G$. $\endgroup$ – Hang Jun 27 '17 at 2:35
  • $\begingroup$ The answer to (1) is no for trivial reasons: Take, e.g., $S=\mathbb C[u_2,u_3]$ with $\deg u_i=i$. Then $S_1=0$ and $S_m\ne0$ for all $m>1$. Therefore, the ideal $I$ generated by $S_m$ will have $I_{m+1}=0$. In particular $I\ne S_{\ge m}$. $\endgroup$ – Friedrich Knop Jun 27 '17 at 6:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.