Let $f:A\rightarrow B$ be a ring homomorphism. Let $\alpha$ be an ideal of $A$ . Denote by $\alpha^{e}={f(\alpha)}B$ the extension of $\alpha.$ In general, we know if $\alpha^{e} \subseteq \beta^{e}$, it is not true that $\alpha\subseteq \beta$. My question is
If $f$ is injective, $\alpha^{e} \subseteq \beta^{e}$, then is it true that $\alpha\subseteq \beta$?
I want to prove as follows. $\alpha^{e} \subseteq \beta^{e}\Rightarrow f(\alpha)B\subseteq f(\beta)B\stackrel{?}\Rightarrow f(\alpha)\subseteq f(\beta)\stackrel{f \text{ is injective}}\Longrightarrow \alpha \subseteq \beta$.
So my question reduced to "if $f(\alpha)B\subseteq f(\beta)B$ then is it true $f(\alpha)\subseteq f(\beta)$?" It looks natural, but how to prove it ?
The answer to the highlighted question is no:
In the inclusion $f:\mathbb Z\hookrightarrow \mathbb Q$ we have $(2\mathbb Z)^e=(3\mathbb Z)^e=\mathbb Q$ but $2\mathbb Z \not\subset 3\mathbb Z$