I assume each algebra here is commutative with a unit.
Let $\nu : A \hookrightarrow B$ be a finite embedding of algebras (that is $B$ is finitely-generated as an $A$-module). I want to show the following:
That the map $\nu^{-1} : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is onto and has finite fibers. That is, every prime ideal in $A$ has a preimage which is a prime ideal in $B$, and there are only finitely many such ideals in $B$.
An ideal $P \lhd B$ is maximal if and only if $\nu^{-1}(P) \lhd A$ is maximal.
My attempts:
My thoughts on 1) where to extend the ideal $P \lhd A$ to $B$ via $P \mapsto B\nu(P)$ and then take $\nu^{-1}(B \nu(P))$ and show it is equal to $P$, however, there is no $\nu^{-1}(B)$ since $\nu$ is not surjective.
We note that $B\nu(P)$ is not necessarily a prime ideal. Therefore we must take the maximal ideal $I$ in $B$ containing $B\nu(P)$ and such that $I \cap \nu(A) = \nu(P)$. We need to prove that $I$ is prime ideal and that $\nu^{-1}(I) = P$.
My thoughts on 2) where to use the characteriziation that $P \lhd B$ is maximal if and only if $B/P$ is a field. However, lack of surjectivity still causing me troubles. By using a Proposition that say that if $\beta$ is algebraic over a field $F$ then $F[\beta]$ is a field, using the fact that $B$ is finite over $A$ I can prove that $ A/\nu^{-1}(P) \mbox{ is a field } \Rightarrow B/P \mbox{ is a field } $ since the quotient $B/P$ is finite over $A/\nu^{-1}(P)$ and has a finite number of generators, all are algebraic. Now I need to prove the other direction. Any thoughts?