Q:

A Noetherian ring with Krull dimension one which is not a Dedekind domain

Can someone give me, with proof, an example of a Noetherian ring which has Krull dimension one but is not a Dedekind domain?  
I think it would also be instructive to see other "near misses."  

A:

The ring $\mathbb{Z}[2i]$ is an example.  It satisfies all the properties of a Dedekind domain except that it is not integrally closed. ( To see that it satisfies these properties, note that it is integral over $\mathbb{Z}$, so the Krull dimension is one, as integral extensions preserve dimension.  It is also clearly noetherian.) 

A:

The ring $R_1 = \mathbb{Z} \times \mathbb{Z}$ gives a counterexample to your claim (it is not a domain).   However $\operatorname{Spec}(R_1)$ is a Dedekind scheme, so this is a somewhat cheap counterexample.   The counterexample $R_2 = \mathbb{Z}[t]/(t^2)$ is more serious.