Localization at $0$? 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. 
Update:
Finally I realized this is a conceptual mistake. Thanks YACP for pointing it out. 
 A: 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.
