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Jacob Lurie gave a very simple example for primary decomposition:

Let $R=\mathbb{Z}$, let $M=\mathbb{Z}\oplus \mathbb{Z}/p$. Then $0=\mathbb{Z}\cap \mathbb{Z}/p$. Here $\mathbb{Z}$ is $p$-coprimary, $\mathbb{Z}/p$ is $0$-coprimary.

Further he stated that due to uniqueness of minimal primary decomposition, the zero-coprimary part has to be the $p$-torsion, since obviously $0$ is the minimal ideal.

However, when I recall during the proof we showed the $p$-coprimary part of $M$ is the kernel of $$M\rightarrow M_{p}$$I ran into trouble. If I localize at $0$ I would expecting to have $\mathbb{Z}\oplus \mathbb{Z}/p\rightarrow \mathbb{Z}$.

However I do not know how to write down the localization explicitly. To me it seems $\mathbb{Z}_{0}=\mathbb{Q}$, since every element $n$ has an inverse $1/n$; and $\mathbb{Z}/p$'s localization at $0$ is nothing but $\mathbb{Z}_{p}$. So I feel I must be confused with something really fundamental. Maybe Jacob Lurie use $M_{p}$ to mean $M\otimes R/p$? Here the $M_{p}$ notation is from the support of a module,and if I am not mistaken I think it means the localization of $M$ at $p$.

I suspect reason for this discrepency maybe because we treat $\mathbb{Z}$ and $\mathbb{Z}/p$ as a $\mathbb{Z}$-module, not a ring itself; but still how can we localize $\mathbb{Z}/p$ at $0$ to get $0$? I did googled and found the second example in the wikipedia article. But it give no reason other than $\mathbb{Z}/p$ is already a local ring itself, which does not make sense since a field can be localized at $0$ to get the field back.


Finally I realized this is a conceptual mistake. Thanks YACP for pointing it out.

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Where did Lurie give this example? – Martin Jan 22 '13 at 9:33
From Matthew's course notes in his blog. See, page 75, lecture 17. – Bombyx mori Jan 22 '13 at 9:39
That's what I suspected. But let us take the most simple case $\mathbb{Z}/2\mathbb{Z}$, localize at $1$ gives us elements $1/1,0/1$, and they are equal if and only if $1(1*1-0*1)=0$. But this does not make sense; so $\mathbb{Z}/2\mathbb{Z}$'s localization at $0$ cannot be 0. – Bombyx mori Jan 22 '13 at 17:50
I am not joking, you are localizing at $0$ as a set, I am localizing at $0$ as the prime ideal. So $S=R-p=1$. – Bombyx mori Jan 22 '13 at 22:39
We localize at a multiplicative closed subset $S$. In this case $0$ is the prime ideal, so $1$ is the multiplicative closed subset. Now to localize $\mathbb{Z}/2\mathbb{Z}$ it is the same as $\mathbb{Z}2/\mathbb{Z}\otimes \mathbb{Z}_{0}$. But $\mathbb{Z}_{0}=\mathbb{Q}$. So $\mathbb{Z}/2\mathbb{Z}\otimes \mathbb{Q}=0$. I see. – Bombyx mori Jan 22 '13 at 22:57
up vote 1 down vote accepted

If I understand correctly you have $M = \Bbb{Z}/p\Bbb{Z}$ considered as a $\Bbb{Z}$ - module. Now when you localise at $0$ you are localising at the multiplicative set $S = \Bbb{Z} - \{0\}$. Since $p \in \Bbb{Z}$ and $M$ is finitely - generated as a $\Bbb{Z}$ - module the fact that $S^{-1}M = 0$ will now follow from the following more general fact:

Atiyah - Macdonald Problem 3.1: Let $S$ be a multiplicatively closed subset of a ring $A$ and $M$ a finitely generated $A$ - module. Prove that $S^{-1}M = 0$ iff there exists $s \in S$ such that $sM = 0$.

Proof: One direction of this problem is a triviality. We prove the other direction, namely that if $M$ is a finitely generated $A$ - module and $S$ a multiplicative set, then $S^{-1}M = 0$ implies that there exists $s \in S$ such that $sM= 0$. If $0 \in S$ there is nothing to prove, so we assume that $0 \notin S$

If $S^{-1}M=0$ this implies that $\operatorname{Ann}(S^{-1}M) = S^{-1}A$. By proposition $3.14$ of Atiyah - Macdonald, this means that $S^{-1}(\operatorname{Ann}(M)) = S^{-1}A$. Now $\operatorname{Ann}(M)$ cannot be the zero ideal of $A$ for if it is, this would mean that $S^{-1}(A) = 0$ meaning that $0 \in S$. This contradicts our assumption that $0 \notin S$. Therefore $\operatorname{Ann}(M)$ is an ideal of $A$ that strictly contains the zero ideal.

Now because localisation distributes over quotients, we have the following ${S}^{-1}A$ module isomorphism, namely

$$\overline{S}^{-1}(A/\operatorname{Ann}(M)) \cong S^{-1}A/S^{-1}(\operatorname{Ann}(M))$$

where $\overline{S}$ is the image of $S$ in the quotient ring $A/\operatorname{Ann}(M)$. But then as noted before $S^{-1}A = S^{-1}(\operatorname{Ann}(M))$ which means that $\operatorname{Ann}(M) \subset S$ so that in particular $S \cap \operatorname{Ann}(M)$ is not empty and does not only contain zero. It follows that there exists $s \in S$ such that $sM = 0$ which completes the proof.

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This is overkill; the second part of your proof can be simplified. Also the problem is basically solved; so I am not looking for an answer at this point. But thanks nevertheless. – Bombyx mori Jan 23 '13 at 4:34
@user32240 The second part I added for the sake of completeness. But the fact that $(\Bbb{Z}/p)_0 = 0$ follows immediately from the definition of what it means for an element in the localisation to be zero. – user38268 Jan 23 '13 at 4:36
@BenjiaLim: I was thinking about $Z/pZ$ as a commuative ring, so there is the confusion. Otherwise it is immediate. – Bombyx mori Jan 23 '13 at 4:40

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