This SE answer claims [see the edit] (in the beginning paragraph) that if $A$, $B$ are rings with $B$ being Noetherian as an $A$-module, then $B$ is Noetherian as a ring because
any chain of $B$-ideals is a chain of $A$-submodules of $B$.
However, I am not able to see this. I guess what I need to show is that any $B$-ideal $\mathfrak b$ is also an $A$-submodule of $B$. But I am unable to prove closure-under-scalar-multiplication for $\mathfrak b$.
Can you help?
Edit:
I realised that the linked answer was also working with the hypothesis that $A$ is a subring of $B$.
So, I ask if this is true for arbitrary rings $A$, $B$.