If $I\subseteq J\subseteq A$ have same image in localization by all maximal ideals, then $I=J$

If $I\subseteq J\subseteq A$ have same image in localization by all
maximal ideals, then $I=J$

I will state my question first:
Suppose $I\subseteq J\subseteq A$ are two ideals in a commutative ring
$A$. Furthermore, assume that for every maximal ideal $\mathfrak{m}$ of
$A$, the image of $I$ and $J$ under the canonical map $A\to
A_{\mathfrak{m}}$ is the same. How can I prove that $I=J$ ?
The aforementioned canonical map $A\to A_{\mathfrak{m}}$ is $a\mapsto a/1$.
My attempts: I think we might need to use the following fact. If $f: M\to
N$ is $A$-module homomorphism, then the following statements are
equivalent:
1) $f$ is injective.
2) The induced map $f_{\mathfrak{m}}: M_{\mathfrak{m}}\to
N_{\mathfrak{m}}$ is injective for every maximal ideal $\mathfrak{m}$ of
$A$.
Now, if we let $M=I$ and $N=J$ (as ideals of $A$ are naturally
$A$-modules), then the inclusion map $i: I\to J$ is injective. So, the
induced maps $i_{\mathfrak{m}}: I_{\mathfrak{m}}\to I_{\mathfrak{m}}$ are
also injective, for each maximal ideal $\mathfrak{m}$ of $A$. Now I would
like to use the fact that $I$ and $J$ have same extensions in
$A_{\mathfrak{m}}$. Can this approach be made to work?
Thanks for your time.