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Let $R$ be a ring, $P\subset R$ a prime ideal, and $M$ a finitely generated $R$-module. Suppose that $M_P$ (the localization of the module $M$ at the prime $P$) is the zero module over the ring $R_P$ (the localization of $R$ at $P$). Is it true that

  1. for any prime ideal $Q$ in $R$ such that $P\subset Q$, $M_Q=0$;
  2. $M/PM$ is the zero $R/P$-module, i.e., $M=PM$?

My thoughts about the first question.

Let $S=R\setminus P$. The module $M_P$ consists of all fractions of the form $m/s$ (by fraction I mean the corresponding equivalence class). If $M_P$ is the zero module, then it consists of all fractions of the form $0/s$ with $s\in S$. Another way of saying this is the following: given any element $m/s'$ in $M_P$, there exists an element $\widetilde s\in S$ such that $\widetilde s ms=0$.

Let's try to show that $M_Q$ is the zero module (which consists of all elements of the form $0/t$) whenever $P\subset Q$. To this end we need to show that given any $\overline m /t'$ in $M_Q$ (with $\overline m \in M $ and $t' \in \bar S:=R\setminus Q$) there is a $\widetilde t\in \bar S$ such that $\widetilde t \overline m t=0$. But since $P\subset Q$, it follows that $\bar S \subset S$. So we can't take $\widetilde t=\widetilde s$ (it might be the case that $\widetilde s\in S\setminus \bar S$.

So in general the statement fails? If yes, what if we require that $M/PM$ be the zero module instead of requiring that $M_P$ be zero? (This would certainly imply $M_P=0$.)

Now for the second question. I know that for any $R$-module $M$, TFAE:

  • $M=0$
  • $M_P=0$ for all primes $P$
  • $M_m=0$ for all maximals $m$

So if $M_P=0$ then $M=0$ and hence $M/PM=0$ and $M=PM$. So this statement is true. Is it correct?

Also, what I don't understand is where it is used (in my "proofs" of both the first and the second statements) that $M/PM$ is an $R/P$-module (we may consider it as an $R$-module) and that $M_P$ is an $R_P$-module (again we may consider it as an $R$-module)?

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Both are false:

  1. Take any local integral domain $R$ (a discrete valuation domain, for instance), and $M$ a non-zero torsion $R$-module. Let $K$ be the fraction field of $R$, $\mathfrak m$ the maximal ideal of $R$. Then $M_{\mathfrak m}=M\ne 0$, but tje localisation of $M$ at the $(0)$ prime ideal, i.e. $\;M\otimes_R K=0$.
  2. The same example shows we do not have $M=(0)M$.

For this second assertion, you misunderstood the equivalence: it asserts that $$\color{red}(\forall\mathfrak p\in\operatorname{Spec}R,\;M_{\mathfrak p}= 0\color{red})\implies M=0,$$ and not that$$\forall\mathfrak p\in\operatorname{Spec}R,\;\color{red}(M_{\mathfrak p}= 0\implies M=0\color{red}).$$

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