The following is a question I have asked [here](https://math.stackexchange.com/questions/1381867/when-mb-neq-b-m-is-a-maximal-ideal-of-a-a-subseteq-b) without receiving any comments, therefore I post it here:


Let $A \subseteq B$ be commutative rings, $m$ a maximal ideal of $A$.
When $mB \neq B$?


This is true when $A \subseteq B$ is faithfully flat.
(If I am not wrong, this is also true when $A \subseteq B$ is integral).
Any other ideas are welcome.


Please notice: A similar (but not identical, I think) question is: <https://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306>, since the property I am talking about is slightly more general then $IB \cap A =I$, for every ideal $I$ of $A$.
Indeed, let $m$ be a maximal ideal of $A$. Then, in particular, $mB \cap A=m$ and if $mB=B$ we would get $A= B \cap A = mB \cap A=m$, a contradiction to the maximality of $m$. So, "$IB \cap A =I$ implies $mB \neq B$".
(An exercise in Atiyah-MacDonald, which was mentioned in the second answer of <https://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306> shows that for a flat extension those two properties are equivalent).



When $mB \neq B$? $m$ is a maximal ideal of $A$, $A \subseteq B$
Here are a few comments that may help.


1. As abx explains, a good way to think about it is in terms whether or not $m$ lies in image of the map on spectra.
2. As you surmised, either condition of faithful flatness or integrality are sufficient,
3. but not necessary! For example, take $A=k[x,y]$ and the affine blow up $B=k[x,y,z]/(y-zt)$. I'll let you convince yourself that it is surjective on spectra but neither flat nor integral (think about the fibres).


