= Solution
{c}
Take
$$
\boxed{D=\mathbb Z[\sqrt{-5}].}
$$
We 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 $d\equiv1\pmod4$, and $\mathbb Z[\sqrt d]$ otherwise. Since $-5\equiv3\pmod4$, $D$ 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 <ideal>
$$
I=(2,1+\sqrt{-5}).
$$
Reduction modulo $2$ with $\sqrt{-5}\mapsto1$ 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 matrix
$$
\begin{pmatrix}a&-5b\\b&a\end{pmatrix}
$$
in the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, so
$$
2=[D:I]=a^2+5b^2.
$$
There are no integer solutions: $b\ne0$ 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 \b[$D$ is a Dedekind domain that is not a principal ideal domain].
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