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Let $D$ be a Dedekind domain. The zero ideal is generated by $0$, and $D$ is generated by $1$, so consider a nonzero proper ideal $I$. Choose $0\ne a\in I$. We use these consequences of the unique factorization of ideals in a Dedekind domain: every nonzero ideal has finitely many nonzero prime-ideal valuations in a Dedekind domain; containment reverses the valuations as in part (b); andThe last formula follows from the containment rule: $J+K$ is the smallest ideal containing both $J$ and $K$, and the product with the indicated minimum exponents is exactly that ideal.

Let $T$ be the finite set of prime ideals dividing $(a)$. For each $P\in T$, set $r_P=v_P(I)$. Since $(a)\subseteq I$, we have $r_P\le v_P((a))$. The containment rule also gives $I⊈P^{r_P+1}$, so chooseDistinct nonzero prime ideals of $D$ are distinct maximal ideals and hence comaximal ideals; their positive powers are also comaximal. Indeed, if $u+v=1$ with $u\in P$, $v\in Q$, for positive integers $m,n$ expanding $(u+v{)}^{m+n-1}$ shows $1\in P^m+Q^n$, since each term contains either at least $m$ factors of $u$ or at least $n$ factors of $v$. We use the Chinese remainder theorem in its ideal form: for pairwise comaximal ideals $J_1,\dots ,J_s$, the map $D\to {\prod }_{j}D/J_j$ is surjective. Apply it to the powers $P^{r_P+1}$ to choose $e_P\in D$ withThen $b={\sum }_{P\in T}{e}_P{b}_P$ belongs to $I$ and satisfies $b\equiv b_P(\mathrm{mod}{P}^{r_P+1})$ for every $P\in T$. Thus $b\notin P^{r_P+1}$, whereas $b\in I\subseteq P^{r_P}$, proving $v_P((b))=r_P$.

For $P\in T$, therefore,For $P\notin T$, $v_P((a))=0$, and $(a)\subseteq I$ forces $v_P(I)=0$ as well. Hence $(a,b)$ and $I$ have the same valuation at every nonzero prime ideal. Uniqueness of their factorizations givesThis proves two generators for an ideal in a Dedekind domain; in fact the first nonzero generator $a\in I$ can be chosen in advance.