If $M_p$ is a localization, why doesn't $M_p=0$ imply $M=0$?

Question

Let $R$ be a unital commutative ring and let $p\subseteq R$ be a prime ideal. We define localization (regarding to $p$) as $$M_p:=\{{m\over t}:m\in M, t\notin p\}$$ (with the apropriate realtion on the elements of $M_p$).

I don't understand why doesn't $M_p=0\Rightarrow M=0$?

(I have a counter example but I don't see where is the problem with my "proof" of the previous claim):

Suppose $M_p=0$. $p$ is prime so $1\notin p$. Let $m\in M$. $m"="{m\over 1}\in M_p=\{0\}$. Thus $m=0$. Where is my mistake?

Answer 1

The fact that $\frac m1=0$ doesn't imply $m=0$. It only means that there exists some $t\in R\setminus\mathfrak p$ such that $tm=0$.

Answer 2

Your mistake lies in $\color{red}{"="}$. There is, indeed, an $R$-linear map $\phi : M \rightarrow M_p$ that sends $m\in M$ to $\frac{m}{1} \in M_p$, but it may fail to be injective, so $\phi(m)=0$ doesn't imply $m=0$.