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."
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."
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.)
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.
If by "other near misses" you mean other rings that satisfy two of the three conditions for being a Dedekind domain, another such ring is k[x, y], which is Noetherian (by the Hilbert basis theorem), integrally closed (it is a UFD, and UFDs are integrally closed) but not every prime ideal is maximal (for example (x)).
Consider $f(x,y)$ = $x^3-y^2$ $\in$ $\mathbb{C}[x,y]$. Then the quotient ring $\mathbb{C}[x,y]/(f)$ is a Noetherian ring with Krull dimension 1 but is not a Dedekind Domain. In general for an algebraically closed field $\bar k$, coordinate ring of the affine plane curve $Z_{f}(\bar k)$ is integrally closed in its field of fractions if and only if the curve $Z_{f}(\bar k)$ is nonsingular.