Extending a homomorphism of a ring to an algebraically closed field 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 
$\phi\colon B \rightarrow \Omega$?
 A: 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.   
A: Hint: Using Zorn's lemma, there is a maximal ring $M\subset B$ to which $\psi$ can be extended. Assume $b\in B\setminus M$. What can you conclude?
A: We know that $\mathfrak{p} = \ker \psi$ is a prime ideal of $A$ since the image of $\psi$ is an subring of $\Omega$, in particular an integral domain. The lying-over theorem implies that there exists a prime ideal $\mathfrak{q}$ of $B$ lying over $\mathfrak{p}$ since $B$ is integral over $A$.
Now consider the compositum $ A \to B \to B/\mathfrak{q}$. The kernel of this compositum is precisely $\mathfrak{p}$ and so we have an injective homomorphism
$$f : A/\mathfrak{p} \to B/\mathfrak{q}$$
such that $B/\mathfrak{q}$ is integral over $A/\mathfrak{p}$. Taking fraction fields, we have that $\textrm{Frac}(B/\mathfrak{q})$ is an algebraic extension of $\textrm{Frac}(A/\mathfrak{p})$. The following steps will now complete the problem:


*

*The universal property of quotients gives us a unique ring homomorphism $\overline{\psi} : A/\mathfrak{p} \to \Omega$ that is injective.

*The universal property of the fraction field now gives a unique ring homomorphism $\Psi: \textrm{Frac}(A/\mathfrak{p} ) \to \Omega$ that extends $\psi$. 

*Use Theorem 1. of Keith Conrad's notes here to give us a homomorphism $\Phi : \textrm{Frac}(B/\mathfrak{q}) \to \Omega$.

*The desired homomorphism from $B$ to $\Omega$ is obtained by doing $$B \to B/\mathfrak{q} \to \textrm{Frac}(B/\mathfrak{q}) \stackrel{\Phi}{\longrightarrow} \Omega. $$
We can put this all in two diagrams:


