Let $ \phi : M_1 \to M_2 $ be an $A$-module homomorphism. Prove that $\phi$ is injective iff $\forall \mathfrak{m} \in \text{Max}(A)$, $\phi_{\mathfrak{m}}$ is injective.
I am aiming to show that $\text{ker}(\phi_{\mathfrak{m}}) \subset (\text{ker}\phi)_{\mathfrak{m}}$, as a step to prove this problem.
I have the following:
\begin{align} \frac{m}{s} \in \text{ker}(\phi_\mathfrak{m}) &\implies \frac{m}{s} \in \text{ker}(S_\mathfrak{m}^{-1}\phi) \\ &\implies S_\mathfrak{m}^{-1}\phi\left(\frac{m}{s}\right) = 0 \\ &\implies \frac{\phi(m)}{s} = 0 \\ &\implies \exists s_0 \in A - \mathfrak{m}, s_0 \cdot (1 \cdot \phi(m) - s \cdot 0) = 0 \\ &\implies \exists s_0 \in A - \mathfrak{m}, s_0 \cdot\phi(m) = 0 \\ &\implies \exists s_0 \in A - \mathfrak{m}, \phi( s_0 \cdot m) = 0 \\ &\implies \exists s_0 \in A - \mathfrak{m}, s_0 \cdot \phi( m) \in \text{ker}(\phi) \end{align} and that $s_0 \notin \mathfrak{m} \implies s_0 \neq 0$.
I am struggling to finish this with $\implies \frac{m}{s} \in \left(\text{ker}(\phi)\right)_\mathfrak{m}$. Essentially, why does $s_0 \cdot \phi (m) \in \text{ker}(\phi) \implies \phi(m) \in \text{ker}(\phi)$ when $s_0 \neq 0$. Couldn't $s_0$ simply be an annihilator of $\phi(m)$?