= Solution
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 $x_1,\ldots,x_m$. 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), take
$$
R=\mathbb Z[\sqrt{-5}].
$$
It is an <integral> domain and has additive <group> $\mathbb Z^2$. The <ideal>
$$
I=(2,1+\sqrt{-5})
$$
has 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 index
$$
a^2+5b^2=2,
$$
which 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.
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