Can you find a domain where $ax+by=1$ has a solution for all $a$ and $b$ relatively prime, but which is not a PID? In Intro Number Theory a key lemma is that if $a$ and $b$ are relatively prime integers, then there exist integers $x$ and $y$ such that $ax+by=1$.  In a more advanced course instead you would use the theorem that the integers are a PID, i.e. that all ideals are principal.  Then the old lemma can be used to prove that "any ideal generated by two elements is actually principal."  Induction then says that any finitely generated ideal is principal.  But, what if all finitely generated ideals are principal but there are some ideals that aren't finitely generated?  Can that happen?

If I'm not mistaken, the integral domain of holomorphic functions on a connected open set $U 
\subset \mathbb{C}$ works. It is a theorem (in Chapter 15 of Rudin's Real and Complex Analysis, and essentially a corollary of the Weierstrass factorization theorem), that every finitely generated ideal in this domain is principal. This implies that if $a,b$ have no common factor, they generate the unit ideal.  However, for instance, the ideal of holomorphic functions in the unit disk that vanish on all but finitely many of ${1-\frac{1}{n}}$ is nonprincipal.
