I wanted to prove the following equivalence. Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i \gg 0$ if and only if $\text{Supp}_R M \subset \mathcal{V}(R^{\geq 1})$. Here $\text{Supp}_R M = \{ \mathcal{P} \in \text{Spec}(R) : M_{\mathcal{P}}\neq 0 \}$ is the support of $M$, there are other definitions but this is the easiest.

For the context of the exercise see corollary 4.25 of the following notes https://arxiv.org/pdf/1107.4815.pdf.

The implication $\Rightarrow$ is easy: take a prime ideal $\mathcal{P}$ and suppose $R^{\geq 1}\not\subset \mathcal{P}$, then this means there exists an element $r \in R \setminus \mathcal{P}$ of positive degree, thus $r^km=0$ for any $m \in M$ and $k \in \mathbb{N}$ such that $k|r|$ is big enough. This implies that $M_{\mathcal{P}}=0$, thus $\mathcal{P}$ cannot be in the support of $M$ and we deduce the inclusion $\text{Supp}_R M \subset \mathcal{V}(R^{\geq 1})$.

The other implication is the difficult part. I tried a direct approach: using the fact that $R^{\geq 1}$ is in the support of $M$ I get that there exists $m \in M$ such that for any $r\in{R^0}$ $rm \neq 0$ but from this I cannot deduce that $M^i=0$ for $i$ big enough. I tried other approaches but he fact is that the starting assumption gives information on how $R$ interact on $M$ while I want to prove $M^i=0$ for $i\gg0$ which is a property of the additive structure of the module. So I suppose that I cannot conclude using only the definitions but I have to use some result like Nakayama lemma.


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