prove that $ \phi _\mathfrak{m}: M_\mathfrak m \rightarrow N_ \mathfrak m $ is injective then $\phi : M \rightarrow N$ is injective.

Question

I want to prove that $ \phi _\mathfrak{m}: M_\mathfrak m \rightarrow N_ \mathfrak m $ is injective then $ \phi: M \rightarrow N$ is injective, where $M_ \mathfrak m$ is localization at maximal ideal $\mathfrak m$.
If $\phi (m_1)= \phi(m_2) $ then $\frac{\phi(m_1)}{1}=\frac{\phi(m_2)}{1} $. As $\phi _\mathfrak m$ is injective this gives $\frac{m_1}{1}=\frac{m_2}{1} $ which gives $m_1=m_2 $.
Is this argument correct? It seems surprisingly elementary and short.

Answer 1

It is not true in general. For example set $R:=\mathbb{Z}$, $m:=5\mathbb{Z}$, and $M=N:=\mathbb{Z}_4\cong \mathbb{Z}/4\mathbb{Z}$, and let $\phi: M\rightarrow N$ by $\phi(x)=2x$. Clearly $\phi$ is a non injective $R$-module homomorphism, but $M_m=N_m=0$ since $4\in Ann_\mathbb{Z}(M_m=N_m)\setminus 5\mathbb{Z}$. Hence, $ \phi_m: M_m\rightarrow N_m $ is injective.

Note the following theorem:

Theorem: Let $M$ and $N$ be an $R$-modules and $f: M \rightarrow N$ an $R$-homomorphism. The following conditions are equivalent. (i) $f$ is injective (surjective). (ii) $f_P: M_p\rightarrow N_p$ is injective (surjective) for all prime ideals $P$. (iii) $f_m: M_m\rightarrow N_m$ is injective (surjective) for all maximal ideals $m$.