Counterexample to the finiteness of integral closure of a Dedekind domain. Let A be a Dedekind domain, K its field of fractions, L/K a finite extension,
B the integral closure of A in L.
By the Krull-Akizuki theorem, B is noetherian, hence B is a Dedekind domain.
In the following two cases, B is an A-module of finite type.
1) L/K is separable.
2) A is an algebra of finite type over a field.
I'm trying to find an example where B is not an A-module of finite type.
After spending some time, I gave up.
It seems to be a very hard problem.

Look at $k(x)$ with $k$ a field of characteristic $p>0$. The extension $k((x))/k(x)$ is transcendental so let $\alpha\in k[[x]]$ be transcendental and let $\beta =\alpha^p$. Then $L:=k(x,\alpha )$ is a purely inseparable extension of $K:=k(x,\beta)$. (Classic.)  The DVR $B=k[[x]]\cap L$ is the integral closure of the DVR $A=k[[x]]\cap K$ and $x$ is a prime element of $A$. Now each $n\in\mathbb{N}$ the element $\beta$ can be written as $$\beta=f_n^p + x^{pn}y_n , f_n\in k[x], y_n\in k[[x]].$$ (Look at $\alpha =c_0+c_1x+c_2x^2+c_3x^3+\ldots$ so that $\beta =c_0^p+c_1^px^p+c_2^px^{2p}+c_3^px^{3p}+\ldots$.)
Then $y_n\in A$ and you get $$y_n^{1/p} =x^{-n}(-f_n+\alpha )\;\; (Eq).$$ Now if $B$ is a finite $A$-module you can find $d\in A\backslash \{0\}$ such that $$dB\subseteq A+A\alpha   +A\alpha^2 +\ldots +A\alpha^{p-1} :=R.$$ Then $d y_n^{1/p}\in R$, which using (Eq) gives that $d$ is divisible by $x^n$ for every $n$. (Note that $1, \alpha ,\ldots ,\alpha^{p-1}$ are linearly independent  over $K$.)
