[proofplan] We prove the cycle (1) $\implies$ (2) $\implies$ (3) $\implies$ (1). For (1) $\implies$ (2), a PID is a UFD by a known theorem, and nonzero primes are maximal in a PID. For (2) $\implies$ (3), a UFD is Noetherian and integrally closed, and the hypothesis that nonzero primes are maximal gives Krull dimension at most one, so $R$ is a Dedekind domain; the UFD condition then forces $\operatorname{Cl}(R) = 0$. For (3) $\implies$ (1), triviality of the class group in a Dedekind domain means every ideal is principal. [/proofplan]

[step:Verify that every PID is a UFD with nonzero primes maximal: (1) $\implies$ (2)]Assume $R$ is a PID. By [Every Principal Ideal Domain Is a Unique Factorization Domain](/theorems/3256), $R$ is a UFD. By [Prime Ideals are Maximal in a PID](/theorems/3258), every nonzero prime ideal of $R$ is maximal. So condition (2) holds.[/step]

[guided]Assume $R$ is a PID. We verify both parts of condition (2) separately. **UFD — existence of factorizations.** By [Every Principal Ideal Domain Is a Unique Factorization Domain](/theorems/3256), $R$ is a UFD. The ascending chain condition on principal ideals holds because $R$ is a PID: the union of any ascending chain $(a_1) \subset (a_2) \subset \cdots$ is an ideal of $R$, hence principal, say $(a)$, and $a \in (a_N)$ for some $N$ forces the chain to stabilize at index $N$. This ACC gives existence of irreducible factorizations by descent: if $a$ is not irreducible, write $a = a_1 b_1$ with neither a unit, giving $(a) \subsetneq (a_1)$; repeat until the chain stabilizes, which must happen at an irreducible. **UFD — uniqueness of factorizations.** Every irreducible element in a PID is prime: if $p$ is irreducible and $p \mid ab$, the ideal $(p, a) = (d)$ forces $d \mid p$, and irreducibility yields either $d \in R^\times$ (so $(p, a) = R$, giving $1_R = rp + sa$, and multiplying by $b$ yields $p \mid b$) or $d$ is an associate of $p$ (giving $p \mid a$). Primality of irreducibles gives uniqueness of factorization up to associates and reordering, by induction on the number of factors. **Nonzero primes are maximal.** By [Prime Ideals are Maximal in a PID](/theorems/3258): let $\mathfrak{p} = (p)$ be a nonzero prime ideal and suppose $(p) \subset (d) \subset R$. Then $d \mid p$, say $p = du$. Since $\mathfrak{p}$ is prime and nonzero, $p$ is not a unit, so $p$ is irreducible: if $p = ab$ with neither a unit, then $(p) \subsetneq (a) \subsetneq R$, and $p \mid ab$ with $p \nmid a$ contradicts primality. Now irreducibility of $p$ applied to $p = du$ forces either $d \in R^\times$ (giving $(d) = R$) or $u \in R^\times$ (giving $(d) = (p) = \mathfrak{p}$). No ideal sits strictly between $\mathfrak{p}$ and $R$, so $\mathfrak{p}$ is maximal.[/guided]

[step:Show that a UFD with nonzero primes maximal is a Dedekind domain with $\operatorname{Cl}(R) = 0$: (2) $\implies$ (3)]Assume $R$ is a UFD in which every nonzero prime ideal is maximal. We verify the three Dedekind axioms. **Noetherian.** Every UFD satisfies the ascending chain condition on principal ideals. In fact, $R$ is Noetherian: let $I \trianglelefteq R$ be a nonzero ideal and $a \in I$ nonzero. Factor $a = u p_1 \cdots p_k$ into irreducibles. Then $(a) \subset I \subset R$, and since each $(p_i)$ is a prime ideal, hence maximal by hypothesis, the quotient $R/(a)$ is a product of quotients of local Artinian rings (by the Chinese Remainder Theorem applied to the prime factorization of $(a)$), hence Noetherian. Every ideal of $R$ containing $(a)$ corresponds to an ideal of the Noetherian ring $R/(a)$, so the set of ideals containing $(a)$ satisfies ACC. Since every nonzero ideal contains some nonzero element, $R$ is Noetherian. **Integrally closed.** By [UFDs Are Integrally Closed](/theorems/2938), $R$ is integrally closed in $\operatorname{Frac}(R)$. **Every nonzero prime ideal is maximal.** This is the hypothesis of condition (2), so $R$ has Krull dimension at most one. Hence $R$ is a Dedekind domain. It remains to show $\operatorname{Cl}(R) = 0$. In a Dedekind domain that is also a UFD, every nonzero ideal is principal: let $\mathfrak{a} \trianglelefteq R$ be a nonzero ideal, and let $\mathfrak{a} = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r}$ be its prime factorization. Each $\mathfrak{p}_i$ is a nonzero prime ideal of a UFD, so $\mathfrak{p}_i = (p_i)$ for some irreducible $p_i \in R$ (since in a UFD, every height-one prime ideal is principal). Then $\mathfrak{a} = (p_1^{e_1} \cdots p_r^{e_r})$ is principal. Therefore $\operatorname{Cl}(R) = 0$.[/step]

[guided]Assume $R$ is a UFD in which every nonzero prime ideal is maximal. We must show $R$ is a Dedekind domain with vanishing class group, $\operatorname{Cl}(R) = 0$. **Noetherian.** We need every ideal of $R$ to be finitely generated. Let $I \trianglelefteq R$ be a nonzero ideal and choose nonzero $a \in I$. Write $a = u p_1^{a_1} \cdots p_k^{a_k}$ with $u \in R^\times$ and $p_i$ irreducible. Each irreducible $p_i$ generates a prime ideal $(p_i)$ (since in a UFD, irreducible implies prime), and each $(p_i)$ is maximal by hypothesis. By the Chinese Remainder Theorem applied to the pairwise coprime ideals $(p_i^{a_i})$: \begin{align*} R/(a) \cong \prod_{i=1}^{k} R/(p_i^{a_i}). \end{align*} Each factor $R/(p_i^{a_i})$ is a quotient of the local ring $R_{(p_i)}$ at a maximal ideal, hence Artinian. A finite product of Artinian rings is Noetherian, so $R/(a)$ is Noetherian. Every ideal of $R$ containing $(a)$ corresponds to an ideal of $R/(a)$ and is therefore finitely generated. Since every nonzero ideal contains a nonzero element, $R$ is Noetherian. **Integrally closed.** By [UFDs Are Integrally Closed](/theorems/2938): if $\alpha = a/b \in \operatorname{Frac}(R)$ with $\gcd(a, b) = 1$ satisfies a monic polynomial over $R$, clearing denominators gives $b \mid a^n$, and $\gcd(a, b) = 1$ in a UFD forces $b \in R^\times$, so $\alpha \in R$. **Krull dimension at most one.** Every nonzero prime ideal is maximal by hypothesis, so there is no chain $(0) \subsetneq \mathfrak{p} \subsetneq \mathfrak{q}$ of prime ideals. These three properties (Noetherian, integrally closed, Krull dimension $\le 1$) are precisely the Dedekind axioms, so $R$ is a Dedekind domain. **Vanishing class group.** Why does the UFD property force $\operatorname{Cl}(R) = 0$? In a Dedekind domain, every nonzero ideal factors uniquely as a product of prime ideals. The class group is generated by the classes $[\mathfrak{p}]$ of the nonzero prime ideals. In a UFD, every irreducible element $p$ generates a prime ideal $(p)$, and every nonzero prime ideal $\mathfrak{p}$ has height one. We claim each $\mathfrak{p}$ is principal. Let $0 \neq a \in \mathfrak{p}$. Since $R$ is a UFD, factor $a = u p_1 \cdots p_m$ into irreducibles. Then $a \in \mathfrak{p}$ implies some $p_j \in \mathfrak{p}$ (since $\mathfrak{p}$ is prime), so $(p_j) \subset \mathfrak{p}$. But $(p_j)$ is a nonzero prime ideal (irreducible implies prime in a UFD), hence maximal, so $(p_j) = \mathfrak{p}$. Since every prime ideal is principal, every ideal $\mathfrak{a} = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r} = (p_1^{e_1} \cdots p_r^{e_r})$ is principal. Therefore $\operatorname{Cl}(R) = 0$. The key insight is that UFD and Dedekind interact powerfully: the UFD condition makes all prime ideals principal, and in a Dedekind domain every ideal is a product of primes, so all ideals are principal. The counterexample $\mathbb{Z}[\sqrt{-5}]$ is Dedekind but not a UFD ($6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$), and correspondingly $\operatorname{Cl}(\mathbb{Z}[\sqrt{-5}]) \cong \mathbb{Z}/2\mathbb{Z} \neq 0$.[/guided]

[step:Deduce that $\operatorname{Cl}(R) = 0$ in a Dedekind domain implies $R$ is a PID: (3) $\implies$ (1)]If $R$ is a Dedekind domain with $\operatorname{Cl}(R) = 0$, every nonzero ideal $\mathfrak{a}$ satisfies $[\mathfrak{a}] = 0$ in the class group, so $\mathfrak{a}$ is a principal ideal. The zero ideal $(0)$ is principal. Hence every ideal of $R$ is principal and $R$ is a PID.[/step]

[guided]Assume $R$ is a Dedekind domain with $\operatorname{Cl}(R) = 0$. We show every ideal of $R$ is principal, so $R$ is a PID. The class group $\operatorname{Cl}(R)$ is the quotient of the group of nonzero fractional ideals of $R$ by the subgroup of principal fractional ideals. The condition $\operatorname{Cl}(R) = 0$ says that every nonzero fractional ideal is principal. In particular, every nonzero integral ideal $\mathfrak{a} \trianglelefteq R$ is a fractional ideal (one with $c = 1$), and $[\mathfrak{a}] = 0$ in $\operatorname{Cl}(R)$ means $\mathfrak{a} = (a)$ for some $a \in R$. The zero ideal $(0)$ is principal with generator $0_R$. Therefore every ideal of $R$ is principal, and $R$ is a PID. The class group is the precise obstruction to a Dedekind domain being a PID. A Dedekind domain $R$ fails to be a PID if and only if $\operatorname{Cl}(R) \neq 0$, meaning there exists a nonzero ideal that cannot be generated by a single element. For example, $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain with $\operatorname{Cl}(\mathbb{Z}[\sqrt{-5}]) \cong \mathbb{Z}/2\mathbb{Z}$, witnessed by the non-principal ideal $(2, 1 + \sqrt{-5})$, and indeed $\mathbb{Z}[\sqrt{-5}]$ is not a PID.[/guided]

[step:Conclude the three-way equivalence] Collecting all implications: - **(1) $\implies$ (2):** A PID is a UFD with nonzero primes maximal (step 1). - **(2) $\implies$ (3):** A UFD with nonzero primes maximal is a Dedekind domain with $\operatorname{Cl}(R) = 0$ (step 2). - **(3) $\implies$ (1):** A Dedekind domain with $\operatorname{Cl}(R) = 0$ is a PID (step 3). The cycle (1) $\implies$ (2) $\implies$ (3) $\implies$ (1) establishes that all three conditions are equivalent characterizations of principal ideal domains among integral domains. [/step]