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TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.