When $mB \neq B$? $m$ is a maximal ideal of $A$, $A \subseteq B$ The following is a question I have asked here 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).
 A: Here are a few  comments that may help.


*

*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.

*As you surmised, either condition of faithful flatness or integrality are sufficient,

*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).
A: This is a bit of a non-answer, but it doesn't fit in a comment, so I put it here. You write "If I am not wrong, this is also true when $A\subseteq B$ is integral".
I think you are mistaken here. Let $B = \mathbb{R}$ (which as a field is clearly an integral domain) and $A = \mathbb{Z} \subseteq \mathbb{R}$. This is a maximal ideal of $A = \mathbb{Z}$ (but of course it is not an ideal of $B=\mathbb{R}$ because $\mathbb{R}$ as a field doesn't have non-trivial ideals).
Let $m = 2\mathbb{Z}\subseteq A$. Then we have $\frac{1}{2}\in \mathbb{R}=B$ and therefore $\frac{1}{2}\cdot 2 = 1\in Bm = mB$ therefore $mB = B$.
