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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!

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  • $\begingroup$ I suppose $M_{\ge n}=\oplus_{i\ge n}M_i$ but you should make it clear in the question. $\endgroup$
    – user26857
    Commented Mar 17, 2016 at 17:54

2 Answers 2

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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.

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  • $\begingroup$ Thanks! I've also edited the definition for graded support so that the primes can't contain the irrelevant maximal ideal. So this example shows the equality fails for the usual support but not the graded support. Still that helps, thanks again! $\endgroup$
    – mwmjp
    Commented Mar 18, 2016 at 12:57
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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.

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    $\begingroup$ I think $R_+ \subseteq \sqrt{\text{ann}_R (M/M_{\geq n})}$ (there could be zero-divisors in degree 0) but that suffices to make your proof work. $\endgroup$ Commented Mar 18, 2016 at 13:15
  • $\begingroup$ You're completely right, thanks for pointing that out! $\endgroup$
    – mwmjp
    Commented Mar 18, 2016 at 13:31

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