Support of a tail of a graded module.

Question

Suppose that $R$ is a non-negatively, graded commutative ring. I have been trying to decide if the following is true for a graded $R$-module $M$ (not necessarily finite over $R$): $$\text{Supp}_R M=\text{Supp}_R M_{\geq n}.$$ The right to left containment is obvious but I am failing to see if the other containment is true in general. Also, if we define $$\text{Graded-Supp}_R M:=\{\mathfrak{p}\in \text{Supp}_RM: \mathfrak{p}\text{ is homogeneous and } \mathfrak{p}\not\supseteq R_+\},$$ does equality above hold replacing support with graded support (again I'm interested in the non-finitely generated case)? Any thoughts on this would be appreciated.

EDIT: As the counterexample below shows, the first equality does not hold. So thanks for clarifying that one!

Answer 1

Counterexample: View $R = \mathbb{Z}$ as a graded ring concentrated in degree zero and take $M = \mathbb{Z}[X]/(pX)$ with $\deg(x)=1$. As $\mathbb{Z}$-module: $$M =\mathbb{Z}\oplus \mathbb{Z}/p \oplus \mathbb{Z}/p \oplus \cdots$$ Since $M_0$ is torsion-free, $\text{Supp}_R M=\{(q)\mid q \text{ prime}\}$ while $\text{Supp}_R M_{\geq 1} = \{ (p)\}$.

Taking $M= \mathbb{Z} \oplus \mathbb{Z}/p$ (degrees 0, 1) also gives a counterexample in the finitely generated case.

Answer 2

I think I may have figured out the second equality. Let me know if I've overlooked anything or if there is a better way to see this!

First, we note that $$\text{Graded-Supp}_RM=\text{Graded-Supp}_RM_{\geq n} \cup \text{Graded-Supp}_R (M/M_{\geq n}).$$ I claim that $\text{Graded-Supp}_R (M/M_{\geq n})=\emptyset.$ Observe that $\sqrt{\text{ann}_R (M/M_{\geq n})}\supseteq R_+$ (the irrelevant maximal ideal). Suppose that there exists $\mathfrak{p}\in \text{Graded-Supp}_R (M/M_{\geq n}),$ then $\mathfrak{p}\neq R_+$ and so there exists $r\in R_+$ with $r\notin \mathfrak{p}$. Thus, $r^t(M/M_{\geq n})=0$ and hence, for any $\overline{m}\in M/M_{\geq n}$ we have that $r^t\overline{m}=0$ and so $\overline{m}/1=0$ in $(M/M_{\geq n})_\mathfrak{p}$ (since $r^t\notin \mathfrak{p}$). Thus, we've reached a contradiction and this proves the claim.