Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.
I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.
I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.
We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct?
For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n < \infty$.
I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{ for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct?
However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$).