TITLE: Extending a homomorphism of a ring to an algebraically closed field
QUESTION [14 upvotes]: Let $A$ be a subring of a ring $B$.
Suppose $B$ is integral over $A$.
Let $\Omega$ be an algebraically closed field.
Then every homomorphism $\psi\colon A \rightarrow \Omega$ can be extended to a homomorphism 
$\phi\colon B \rightarrow \Omega$?

REPLY [5 votes]: As a complement to BenjaLim's perfect answer, let me give an example of an integral extension $A\subset B$ and  of a morphism $\psi:A\to \Omega$ with infinitely many extensions $B\to \Omega$ .  
Take $A=\mathbb F_p$ and fix an algebraic closure $B=\Omega=\bar {\mathbb F_p}$ of $\mathbb F_p$.
The inclusion $\psi: \mathbb F_p\to \Omega$ extends to $\phi: B=\Omega\to \Omega:x\to x^p$, the Frobenius automorphism .
But there are many other extensions, the simplest being the infinitely many powers of that Frobenius automorphism, namely the $\phi^n: B=\Omega\to \Omega:x\to x^{p^n}$ where $n\in \mathbb Z$ is arbitrary.