# Support of a graded module of a ring concentrated in non-negative dimensions

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.