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 $$ X_{ss} = \mathbb P^2 - \{[1,0,0],[0,1,0]\} \ni [x,y,z]\mapsto[xy,z^2] \in \mathbb P^1 =X_{ss} / / G $$

nofor 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