Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 1 / b

The ring $\mathbb{Z}[T]$ is Noetherian and integrally closed, but it is not a Dedekind domain because it has Krull dimension two. Concretely, the nonzero prime ideal $(2)$ is properly contained in the prime ideal $(2,T)$ and is therefore not maximal.

The element $T$ belongs to the fraction field of $R=\mathbb{C}[T^2,T^3]$ and is integral over $R$, since it satisfies the monic polynomial $X^2-T^2$. But $T\notin R$. Thus $R$ is not integrally closed domain and hence is not a Dedekind domain.

The ring $\mathbb{Z}[{\zeta }_3]$ is the full ring of integers of the quadratic number field $\mathbb{Q}(\sqrt{-3})$. Every ring of integers of a number field is a Dedekind domain, so this ring is Dedekind.