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: http://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 http://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 shows that for a flat extension those two properties are equivalent).