Is the integral closure of an integral domain a dedekind domain?

Question

I am reading through Milne's number theory notes, and he shows that the ring of integers of a field is a dedekind domain. He also shows that the integral closure of a dedekind domain A in a finite separable extension of Frac(A) is a dedekind domain. But what if we only know that A is an integral domain? Can we still conclude that its integral closure is a dedekind domain? Any help is appreciated.

Answer 1

Dedekind domains must be one-dimensional, so here is an easy counter-example: the integral closure of $A=\mathbb C[x,y]\subset\mathbb C(x,y)$ is itself, which is $2$-dimensional.