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Let $M$ be a graded module over an $\mathbb{N}$-graded ring $S$ and $S_+$ be the ideal of positive degree elements. Is it true that $M=0$ iff the homogeneous localization $M_{(\mathfrak p)}=0$ for all homogeneous prime ideals $\mathfrak p$ of $S$ not containing $S_+$?

The proof in the non graded case uses the fact that a nonzero element has a proper annihilator ideal. To follow the same proof I'd have to conclude that the annihilator does not contain all of $S_+$, which I'm not sure about, and then argue that the annihilator is contained in some prime graded ideal which does not contain all of $S_+$. We can assume $S$ is finitely generated by $S_1$ as an $S_0$-algebra if this helps.

EDIT: As shown by user26857 this is not true. I'd like to follow up by asking: if all homogeneous localizations at homogeneous primes $M_{(\mathfrak p)}=0$ is $M$ necessarily $0$?

user26857
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Seth
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  • @user26857 good thinking, I guess a $0$ is the only homogeneous prime ideal of $S$ not containing $S_+$. Is the statement true if we leave off all the stuff about $S_+$? Using the standard argument and Zorn's lemma it would be sufficient to prove that an ideal maximal with respect to proper homogeneous ideals is prime. I'm not sure if this is true or not. – Seth Oct 11 '14 at 01:08
  • Seth: I don't understand your comment but leaving all the stuff about $S_+$. For a counter example, all you need is a module whose support is $S_+$. For instance, take $S/S_+$ provided that $S_0$ is a field. You are probably asking being finite length instead of being zero. – Youngsu Oct 11 '14 at 05:31
  • @Seth Do you mean if $M=0$ if all its homogeneous localizations are $0$? – user26857 Oct 11 '14 at 07:32
  • @user26857 I mean, if all homogeneous localizations at homogeneous primes $M_p=0$ then is $M=0$? I don't think your example is a counter example to that statement because $M_{S_+}$ is not zero. – Seth Oct 11 '14 at 11:42
  • @Seth My example concerns your original question (and now I have posted it as an answer to not mix up the things). About your second question the answer is obviously positive: take a homogeneous non-zero element of M and note that ann(x) is a homogeneous ideal. (If you add this to the body of the question I can edit my answer and post a full explanation.) – user26857 Oct 11 '14 at 13:40
  • @user26857 Hey, thanks for your continued help. I know that the annihilator of a nonzero homogeneous element is a proper homogeneous ideal, but what I don't know is whether or not it is necessarily contained in a prime homogeneous ideal or not. – Seth Oct 11 '14 at 14:06
  • This shouldn't be that hard: it is contained in a maximal ideal, say $m$, and being homogeneous it is contained in $m^h$ which is a prime ideal. – user26857 Oct 11 '14 at 14:13
  • @user26857 So take $m$ maximal, not necessarily homogeneous? What is $m^h$? – Seth Oct 11 '14 at 14:15
  • @Seth $m^h$ is the ideal generated by the homogeneous elements of $m$, so $m^h\subseteq m$. – user26857 Oct 11 '14 at 14:17
  • Take a look at http://math.stackexchange.com/questions/385292/homogeneous-ideals-are-contained-in-homogeneous-prime-ideals/967919#967919 – user26857 Oct 11 '14 at 14:27
  • @user26857 Thanks, that answer was exactly what I was looking for. – Seth Oct 11 '14 at 14:38

1 Answers1

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No, this isn't true.

If $S=K[X]$ and $M=S/(X^2)$ then $M_{(0)}=0$ and $(0)$ is the only homogeneous prime of $S$ different from $S_+$. In general, I can't see any reason to exclude $S_+$ from such a question.

Edit. The answer to your edit is positive: take $x$ a homogeneous non-zero element of $M$ and note that $\operatorname{Ann}(x)$ is a homogeneous ideal, and therefore this is contained in a homogeneous prime ideal.

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