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
- for any prime ideal $Q$ in $R$ such that $P\subset Q$, $M_Q=0$;
- $M/PM$ 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:
- $M_P=0$ 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)?