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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.

user26857
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mathemather
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    $\frac {m_1}1=\frac{m_2}1\implies \exists t\in A\setminus \mathfrak m $ such that $t (m_1-m_2)=0$. I don't think we can conclude $m_1=m_2$ unless $t $ is a unit. – cqfd Sep 02 '19 at 07:19
  • True but $t$ is unit by construction right? – mathemather Sep 02 '19 at 10:27
  • Why? If $A$ is a local ring, it's true. Otherwise, I can't see how. – cqfd Sep 02 '19 at 11:22
  • @ThomasShelby $\mathfrak m$ is a maximal ideal so any element outside is unit. – mathemather Sep 02 '19 at 15:07
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    @mathemather If $t \in A \setminus \mathfrak{m}$, then $t/1$ is a unit in $A_\mathfrak{m}$, but $t$ need not be a unit in $A$. For instance, $5 \in \mathbb{Z} \setminus 2 \mathbb{Z}$, but is $5$ a unit in $\mathbb{Z}$? – Viktor Vaughn Sep 02 '19 at 15:11
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    @mathemather If $A$ is a commutative ring with $1$ and $a \in A$ is a non-unit, then there exists a maximal ideal $M$ of $A$ such that $a \in M$. Convince yourself that this is different from what you wrote above. Also, please read this. – cqfd Sep 02 '19 at 15:52

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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 $m$.

E.R
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