Example of a ring extension $A\subseteq B$ and $F:A\to \Omega$ such that $f$ cannot be extended to $B$. Exercise 5.2 in Atiyah-Macdonald asks to show the following:
"Let $A$ be a subring of a ring $B$ such that $B$ is integral over $A$, and let $f: A \to \Omega$ be a homomorphism of $A$ into an algebraically closed field $\Omega$. Show that $f$ can be extended to a homomorphism of $B$ into $\Omega.$"
My question:  what are some notable counterexamples of this property, which occur when we take a ring $B$ which fails to be integral over a subring $A$?
Although I don't doubt that this property fails in general (i.e. for "most" rings), the only example I can come up with seems seriously contrived and unnatural:  let $B$ be the complex numbers, let $A$ be $\mathbb{Z},$ let $\Omega$ be some algebraically closed field included in $\mathbb{C}$ (say the complex algebraic numbers) with $f$ the inclusion map. 
 A: The issue is that some elements in $B$ satisfy equations  over $A$ and those equations transform into incompatible equations by the map $A\to \Omega$. Here is a typical example
$$A = \mathbb{C}[x] \to \mathbb{C}[x,y]/(x y -1)=B$$
and take the map $A \to \Omega= \mathbb{C}$, $x\mapsto 0$. The equation $xy-1$ satisfied by $y$ over $\mathbb{C}[x]$ is transformed to an impossible equation $0\cdot \bar y -1 =0$.
One issue : some elements of $A$ getting mapped to $0$ in $\Omega$. For instance if the map $A \to \Omega$ is injective and the extension $A\to B$ is finitely generated then we can always extend it to a map $B\to \Omega$ (not necessarily injective). There is a more general result due to Chevalley on extensions of maps to an algebraically close field for finitely generated extensions, should be in the same book.
Another issue ( @Slade: thanks!) is that the extension $A\to B$ may not be finitely generated. Example $A= \mathbb{C} \to \mathbb{C}(x)=B$, cannot extend the map $A \to \mathbb{C}$ to $B$. 
