TITLE: Homomorphism extensions over finite ring extensions
QUESTION [8 upvotes]: 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 $\nu$).

REPLY [4 votes]: Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\sigma : A \to \Omega$, where $\Omega$ is an algebraically closed field. Prove that there is at least one, but finitely many ring homomorphisms $\overline{\sigma}: B \to \Omega$ that extend $\sigma$.

Existence. We don't need to assume that the ring extension is finite. It is enough to be integral (see exercise 2, Chapter 5, from Atiyah and MacDonald). 
Let $\mathfrak p=\ker\sigma$. This is a prime ideal of $A$ and there is a prime ideal $P$ of $B$ lying over $\mathfrak p$, that is, $P\cap A=\mathfrak p$. So we have an integral extension of integral domains $A/\mathfrak p\subset B/P$. Since $A/\mathfrak p\to\Omega$ is injective and every nonzero element of $A/\mathfrak p$ is invertible in $\Omega$ we can extend this to a homomorphism $Q(A/\mathfrak p)\to\Omega$. But $Q(A/\mathfrak p)\subset Q(B/P)$ is an algebraic field extension and then we can extend $Q(A/\mathfrak p)\to\Omega $ to $Q(B/P)\to\Omega$. The extension of $\sigma$ we are looking for is $B\to B/P\to Q(B/P)\to\Omega$.
Finiteness. Now only two observations are enough to show that there are only finitely many extensions of $\sigma$ when $A\subset B$ is finite. The first is the following: if $A\subset B$ is finite, then there are only finitely many prime ideals of $B$ lying over $\mathfrak p$. The second comes from field theory and says that for a finite field extension $K\subset L$ and a homomorphism $\sigma:K\to\Omega$, where $\Omega$ is an algebraic closed field, there are only finitely many extensions of $\sigma$ to $L$.