= Solution
\b[The printed domain assertion is false.] For example, take $K=\mathbb Q$ and $I=(4)$. Then $\mathcal O_K/I=\mathbb Z/4\mathbb Z$ has the nonzero <zero divisor> $2+(4)$, so it is not an <integral domain> and cannot be a <principal ideal domain>. The valid conclusion is that $\mathcal O_K/I$ is a <principal ideal ring>. This error is present in the original PDF, not just its transcription.
To prove the corrected conclusion, factor
$$
I=\prod_{j=1}^sP_j^{n_j}
$$
into powers of distinct nonzero <prime ideals>. These powers are pairwise <comaximal ideals>. Indeed, a <maximal ideal> containing two such powers would contain both underlying <prime ideals>, forcing them to be equal. The <Chinese remainder theorem> gives
$$
R/I\simeq\prod_{j=1}^s R/P_j^{n_j}.
$$
Every factor is a <principal ideal ring> by the preceding part. Every <ideal> $H$ in a finite <direct product of rings> is a product $H_1\times\cdots\times H_s$ of <ideals> in the factors: multiplication by the coordinate <idempotent elements> separates the coordinates of every element of $H$. Write $H_j=(h_j)$. Then
$$
H_1\times\cdots\times H_s=((h_1,\ldots,h_s)),
$$
so $R/I$ is a <principal ideal ring>. This proves the <principal ideal property of quotients of a Dedekind domain> in the present setting. In fact, this finite <quotient ring> is an <integral domain> precisely when $I$ is a <prime ideal>, in which case it is a <field>.
For the final deduction, choose any $0\ne a\in I$ and apply the result to the nonzero proper <ideal> $(a)$. The <ideal> $I/(a)$ of $R/(a)$ is generated by one residue class $b+(a)$, which can be represented by $b\in I$. For each $x\in I$, some $r\in R$ satisfies $x-rb\in(a)$. Hence
$$
\boxed{I=(a,b)}.
$$
This proves <two generators for an ideal in a Dedekind domain>; moreover, its first generator can be any prescribed nonzero element of $I$.
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