Non-existence ofintegral basis of integral closure in a finite extension of Frac(A), A Dedekind. Let $A$ be a Dedekind domain, $K:=\text{Frac}(A)$ and $L/K$ finite so that the integral closure $B$ of $A$ in $L$ is Dedekind.  If $A$ is a PID, for example, then there exists an integral basis : $B$  is free over $A$, and this basis gives a basis of $L/K$.
In particular, if $A=\mathbb{Z}$ or $\mathbb{Z}_p$ (or even the ring of integers of a finite extension of $\mathbb{Q}_p$), and $L/K$ finite then an integral basis exists.  
I'm looking for a non-example ; $A$ Dedekind, $L/K$ finite for which there does not exist an integral basis.  I suspect that taking $A$ to be the ring of algebraic integers of a finite extension of $\mathbb{Q}$ which is a UFD but not PID might do the trick.  
Highly pathological examples (e.g. outside the realm of number fields) are very much welcome !
 A: Watch out: just because $A$ is a PID does not make $B$ a free $A$-module. You need to know that $B$ is finitely generated over $A$ to conclude $B$ has an $A$-basis when $A$ is a PID.  If $L/K$ were separable then using discriminants you can stuff $B$ inside a finitely generated $A$-module so $B$ is finite free if $A$ is a PID.  
If we drop the separability condition on $L/K$ in this corollary, then the whole discriminant 
argument breaks down and in fact $B$ need not be a finitely generated $A$-module, even if $A$ is a PID! For an example, see Exercise 11 on p. 205 of Borevich and Shafarevich's "Number Theory". 
I wrote up a course handout with examples in number fields where the top ring of integers is not a free module over the bottom ring of integers: look at 
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf.  Note that the bottom ring having class number greater than 1 is a necessary condition for such a phenomenon to occur but it's not sufficient. An example illustrating that is at the end of the handout.
