What can be said about an ideal contracted to a maximal ideal?

Question

Let $A \subseteq B$ be commutative (finitely generated) $k$-algebras, $k$ is a field of characteristic zero. Let $J$ be an ideal of $B$ and assume that its contraction $I:=J \cap A$ is a maximal ideal of $A$.

(1) Is it true that $J$ is a maximal ideal of $B$ or at least a prime ideal of $B$? Probably not?

(2) What additional conditions are required in order to guarantee that $J$ is a prime ideal in $B$? For example, integrality of $A \subseteq B$.

(3) What if we assume that $J$ is a radical ideal?

Relevant questions: i and ii.

Any hints and coments are welcome!

Answer 1

Hint: Consider $k \subset k[t]$ and some proper ideal $J \subset k[t]$. Then $J$ cannot contain any non-zero element of $k$ so that the intersection with $k$ is the zero ideal and hence a maximal ideal.

Edit: For an example that is not a field, consider $k[t] \subset k[t,s]$ and the (non-prime) ideal $J = (t,s^2) \subset k[t,s]$. Then $J \cap k[t] = (t) \subset k[t]$ is maximal.