Localization of a non-zero module is non-zero?

Question

Hy everybody, this is my first question on stackexchange after many many readings :)

I have the following problem:

Let $A$ be a ring (commutative with unit) .

Suppose $M\neq0$ is a non-zero $A$-module , and $p$ is a prime ideal of A such that the localization of $M$ at $p$ is nonzero , i.e $M_p\neq0$.

Now let $f\in A\backslash p$ , I want to show that the localization of $M$ by the set $S_f := \{1,f,f^2,...\}$ is non-zero.

Here is my idea:

1) $p$ is prime and $f\notin p$ , so we have $p\cap S_f=\emptyset$, this imply that p extend to a prime ideal in $A_f$.

I denote the extension of $p$ with $q:=A_f \cdot p $

2)Now I consider the localization of $M_f$ at the prime ideal $q$...

I think there may be a relation btw $(M_f)_q$ and $M_p$ , i tried to construct an explicit ring-homomorphism but it didn't we work, how should I proceed?

EDIT: I forgot the keypoint of my idea : I want to prove that $(M_f)_q$ is non zero because I know that beeing zero is a local property...so if $q$ is prime and $(M_f)_q \neq 0$ then this imply $M_f \neq 0$ right?

So I'm trying to show that $M_p \neq0$ somehow imply that $(M_f)_q \neq0$

Answer 1

I think you're working too hard here. Suppose the localization $M_{f} = 0$. Then for every $m \in M$, there exists some $n \in \mathbb{N}$ such that $f^{n} \cdot m = 0$. Since $f \in A \setminus p$, $f^{n} \in A \setminus p$ for all $n \in \mathbb{N}$ by the primality of $p$, which shows that $M_{p} = 0$.

Answer 2

As $p$ is prime, then $A\setminus p$ is multiplicatively closed, so that $S_f\subseteq A\setminus p$. This means that a localisation of module $M$ by $A\setminus p$ is a localisation of the localisation of $M$ by $S_f$. So if $M_f=0$ then $M_p=0$.

Answer 3

The idea of your proof is good, but there's a danger of getting bogged down in information about $q$, which is irrelevant. I'd proceed by using the universal property of localizations. The canonical homomorphism $h:A\to A_p$ sends all elements $f^n$ of $S_f$ to invertible elements (since $f^n\notin p$), so $h$ factors through the canonical homomorphism $j:A\to{S_f}^{-1}A$, say $h=k\circ j$, where $k:{S_f}^{-1}A\to A_p$. As a result, $M_p=M\otimes_A A_p=(M\otimes_A {S_f}^{-1}A)\otimes_{{S_f}^{-1}A}A_p$. (To be pedantic, each equality sign here means canonical isomorphism, and $A_p$, ${S_f}^{-1}A$, and $A_p$ are algebras over $A$, $A$, and ${S_f}^{-1}A$ via the homomorphisms $h$, $j$, and $k$, respectively.) Since $M_p\neq0$, it follows that $M\otimes_A {S_f}^{-1}A\neq0$, as required.

(To connect with your notation, the homomorphism that I called $k$ is the localization at the ideal that you called $q$.)