Lying Over for Algebraic Ring Extensions

Question

Let $B$ be a finitely generated algebraic $A$-algebra (but not necessarily integral). Is it true that for any prime in $A$ we can find a prime in $B$ which contracts to $A$? What if we also allow the map $A\to B$ to be inclusion?

Answer 1

No. Consider $k[x]\subset k[x,x^{-1}]$.