Let $k \subseteq B \subseteq A$ be $k$-algebras where $k$ is algebraically closed, and $A$ is a finitely generated $B$-module. Then any $k$-algebra homomorphism $B \rightarrow k$ can be lifted to a $k$-algebra homomorphism $A \rightarrow k$.
I am looking for a proof/hint for this statement (not a homework exercise).
Note that if $B \subset A$ with $A$ a finitely generated $B$ module then $A$ is an integral extension of $B$. Now you can use the lying over and incomparability properties for integral extensions to show that for any maximal ideal $\mathfrak{m} \subset B$ there is a maximal ideal $\mathfrak{m}' \subset A$ s.t. $\mathfrak{m} = \mathfrak{m}' \cap B$. By the second isomorphism theorem, any map into a field must factor through a quotient by a maximal ideal. Now use the finitely generated and algebraically closed properties to get you the result.