[proofplan] We prove the three-way equivalence (1) $\mathcal{O}_L$ is a PID, (2) $\mathcal{O}_L$ is a UFD, (3) $\mathrm{Cl}_L$ is trivial. The implications (1) $\Leftrightarrow$ (3) are tautological from the definition of the class group $\mathrm{Cl}_L = I_L / P_L$: the class group is trivial precisely when every nonzero fractional ideal is principal. The implication (1) $\Rightarrow$ (2) is a standard theorem of commutative algebra (every PID is a UFD). The substantive direction is (2) $\Rightarrow$ (1), proved for Dedekind domains by showing that in a UFD every prime ideal of $\mathcal{O}_L$ is principal, and then appealing to [Ideal Group Is Free Abelian](/theorems/1590) to conclude that every (fractional) ideal is principal. [/proofplan]

[step:State the structure to be proved and recall the class group definition]Recall the class group: \begin{align*} \mathrm{Cl}_L := I_L / P_L, \end{align*} where $I_L$ is the group of nonzero fractional ideals under multiplication (by [Ideal Group Is Free Abelian](/theorems/1590)) and $P_L \leq I_L$ is the subgroup of **principal** fractional ideals, i.e., those of the form $\langle x \rangle := x\mathcal{O}_L$ for some $x \in L^\times$. We show the three conditions are equivalent by proving the cycle \begin{align*} (1) \Longrightarrow (2) \Longrightarrow (1), \qquad (1) \Longleftrightarrow (3). \end{align*}[/step]

[guided]We structure the proof as: prove (1) $\Leftrightarrow$ (3) first (easy, essentially definitions), then prove (1) $\Rightarrow$ (2) (classical, every PID is UFD), then prove (2) $\Rightarrow$ (1) (the substantive direction, using the Dedekind structure of $\mathcal{O}_L$). Recall the class group construction: - $I_L$: the group of nonzero fractional ideals of $\mathcal{O}_L$. By [Ideal Group Is Free Abelian](/theorems/1590), this is a free abelian group on the nonzero prime ideals. - $P_L \leq I_L$: the subgroup of principal fractional ideals, $\{x\mathcal{O}_L : x \in L^\times\}$. That $P_L$ is a subgroup is immediate: $x\mathcal{O}_L \cdot y\mathcal{O}_L = (xy)\mathcal{O}_L$, $(x\mathcal{O}_L)^{-1} = x^{-1}\mathcal{O}_L$, $\mathcal{O}_L = 1 \cdot \mathcal{O}_L$. - $\mathrm{Cl}_L := I_L / P_L$: the quotient abelian group. An element $[\mathfrak{q}] \in \mathrm{Cl}_L$ is zero (identity element) iff $\mathfrak{q} \in P_L$, iff $\mathfrak{q}$ is principal. Hence $\mathrm{Cl}_L$ is trivial iff every $\mathfrak{q} \in I_L$ is principal. This is the content of the equivalence $(1) \Leftrightarrow (3)$ that we will unpack in the next step.[/guided]

[step:Prove (1) $\Leftrightarrow$ (3) — $\mathcal{O}_L$ PID iff every (fractional) ideal is principal iff class group of order one]**(1) $\Rightarrow$ (3).** Suppose $\mathcal{O}_L$ is a PID, i.e., every integral ideal is principal. Let $\mathfrak{q} \in I_L$. By Step 2 of the proof of [Ideal Group Is Free Abelian](/theorems/1590) (or by definition of fractional ideal), $\mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1}$ with $\mathfrak{a}, \mathfrak{b}$ integral. Since $\mathcal{O}_L$ is a PID, $\mathfrak{a} = \langle \alpha \rangle$ and $\mathfrak{b} = \langle \beta \rangle$ for some $\alpha \in \mathcal{O}_L$, $\beta \in \mathcal{O}_L \setminus \{0\}$. Then \begin{align*} \mathfrak{q} = \langle \alpha \rangle \langle \beta \rangle^{-1} = \langle \alpha/\beta \rangle \end{align*} is principal (as a fractional ideal, generated by $\alpha/\beta \in L^\times$). Hence $I_L = P_L$ and $\mathrm{Cl}_L = I_L / P_L$ is trivial. **(3) $\Rightarrow$ (1).** Suppose $\mathrm{Cl}_L$ is trivial, i.e., $I_L = P_L$: every nonzero fractional ideal is principal. An integral ideal $\mathfrak{a} \unlhd \mathcal{O}_L$ (if nonzero) is in particular a fractional ideal, so $\mathfrak{a} = \langle x \rangle$ for some $x \in L^\times$. Since $\mathfrak{a} \subseteq \mathcal{O}_L$ and $x = x \cdot 1 \in x\mathcal{O}_L = \mathfrak{a} \subseteq \mathcal{O}_L$, we have $x \in \mathcal{O}_L$. Thus $\mathfrak{a}$ is generated by an element of $\mathcal{O}_L$. The zero ideal $\langle 0 \rangle$ is principal by definition. Hence every ideal of $\mathcal{O}_L$ is principal: $\mathcal{O}_L$ is a PID.[/step]

[guided]**The direction $\mathrm{Cl}_L$ of order one $\Rightarrow$ PID.** The nontriviality to check is that "every nonzero *fractional* ideal is principal" implies "every *integral* ideal is principal (by an element of $\mathcal{O}_L$)". Given an integral ideal $\mathfrak{a} \subseteq \mathcal{O}_L$ (nonzero case), $\mathfrak{a}$ is a fractional ideal, hence of the form $\mathfrak{a} = x\mathcal{O}_L$ for some $x \in L^\times$. We claim $x \in \mathcal{O}_L$. Indeed $x = x \cdot 1 \in x\mathcal{O}_L = \mathfrak{a} \subseteq \mathcal{O}_L$. So $\mathfrak{a} = \langle x \rangle_{\mathcal{O}_L}$ with $x \in \mathcal{O}_L$, i.e., principal in the usual PID sense. The zero ideal $\{0\} = \langle 0 \rangle$ is principal by default. **The direction PID $\Rightarrow$ $\mathrm{Cl}_L$ of order one.** Every element of $I_L$ writes as $\mathfrak{a}\mathfrak{b}^{-1}$ with $\mathfrak{a}, \mathfrak{b}$ integral. Principality of $\mathfrak{a}, \mathfrak{b}$ gives $\mathfrak{a} = \alpha\mathcal{O}_L$, $\mathfrak{b} = \beta\mathcal{O}_L$, and then $\mathfrak{a}\mathfrak{b}^{-1} = (\alpha/\beta)\mathcal{O}_L$ — a principal fractional ideal. So $I_L = P_L$, $\mathrm{Cl}_L = 0$. **Structural remark.** The equivalence (1) $\Leftrightarrow$ (3) is essentially a reformulation of the definition of $\mathrm{Cl}_L$. The class group is *designed* to measure how far $\mathcal{O}_L$ is from being a PID; triviality of $\mathrm{Cl}_L$ is the statement "no obstruction to principality", i.e., every ideal is principal.[/guided]

[step:Recall (1) $\Rightarrow$ (2): every PID is a UFD]Every PID is a UFD. This is a standard theorem of commutative algebra; a brief proof: - **Irreducibles exist in any factorization.** In a PID, every nonzero non-unit $x$ lies in some maximal (hence prime, hence nonzero) ideal $\langle p \rangle$; $p$ is then irreducible. - **Existence of factorizations.** In a Noetherian integral domain (and a PID is Noetherian), if $x$ is a nonzero non-unit, iterated factorization $x = y_1 z_1$ with $y_1, z_1$ non-units must terminate: the chain $\langle x \rangle \subsetneq \langle y_1 \rangle \subsetneq \langle y_1/y_2 \rangle \subsetneq \cdots$ (where at each step we factor off a non-unit divisor) is ascending, hence stabilizes by the ACC. - **Uniqueness.** In a PID, irreducibles are prime (irreducible $p$ implies $\langle p \rangle$ is maximal, hence prime; $\langle p \rangle$ prime iff $p$ prime in integral domains). Prime elements give uniqueness by the usual cancellation argument (Euclid's lemma). Hence $\mathcal{O}_L$ being a PID implies it is a UFD.[/step]

[guided]This implication is not specific to $\mathcal{O}_L$ — it is the general ring-theoretic fact "PID $\Rightarrow$ UFD". We outline the argument, as it is classical. **Existence of factorizations (uses Noetherianity).** A PID is Noetherian: every ideal is generated by a single element, and in particular the ACC holds. Let $x \in R$ be a nonzero non-unit. Suppose for contradiction $x$ has no factorization into irreducibles; then in particular $x$ is not itself irreducible, so $x = y_1 z_1$ with $y_1, z_1$ non-units; at least one of them, say $y_1$, has no factorization into irreducibles. Iterating, $y_1 = y_2 z_2$ with $y_2$ non-unit and no factorization, and so on. This produces a strictly ascending chain $\langle x \rangle \subsetneq \langle y_1 \rangle \subsetneq \langle y_2 \rangle \subsetneq \cdots$ (strict because each $z_i$ is a non-unit, preventing $\langle y_i \rangle = \langle y_{i+1} \rangle$), violating the ACC. Hence factorization exists. **Uniqueness (uses "irreducible $\Leftrightarrow$ prime" in a PID).** In a PID, every irreducible $p$ generates a prime ideal: $\langle p \rangle$ is maximal (nonzero and any $\langle q \rangle \supsetneq \langle p \rangle$ forces $p = qr$ with $r$ a non-unit, so $q$ a unit, so $\langle q \rangle = R$), hence prime, hence $p$ is a prime element. Given two factorizations $p_1 \cdots p_r = q_1 \cdots q_s$ into irreducibles, $p_1$ is prime and divides the right-hand side, so divides some $q_i$; since $q_i$ is irreducible and $p_1$ is not a unit, $p_1$ and $q_i$ differ by a unit. Cancelling and inducting on $r$ gives uniqueness up to reordering and unit factors. Hence (1) $\Rightarrow$ (2).[/guided]

[step:Prove (2) $\Rightarrow$ (1): in $\mathcal{O}_L$, UFD forces every prime ideal to be principal]This is the substantive step. Assume $\mathcal{O}_L$ is a UFD; we show every nonzero prime ideal $\mathfrak{p} \unlhd \mathcal{O}_L$ is principal. Combined with [Ideal Group Is Free Abelian](/theorems/1590) (which expresses every fractional ideal as a product of integer powers of primes) and the subgroup property of $P_L$ (closed under products and inverses), this will show every nonzero fractional ideal is principal, hence (3), hence (1). Fix a nonzero prime ideal $\mathfrak{p} \unlhd \mathcal{O}_L$. Pick any nonzero $x \in \mathfrak{p}$ (possible since $\mathfrak{p} \neq (0)$). [claim:A nonzero element of $\mathcal{O}_L$ has a finite factorization into irreducibles] Every nonzero $x \in \mathcal{O}_L$ has a factorization $x = u \cdot \pi_1 \cdots \pi_k$ with $u \in \mathcal{O}_L^\times$ a unit and $\pi_1, \ldots, \pi_k \in \mathcal{O}_L$ irreducible (including the empty product case $k = 0$ where $x = u$ is a unit). [/claim] [proof] By hypothesis $\mathcal{O}_L$ is a UFD, which is defined as: every nonzero non-unit has a factorization into irreducibles, unique up to reordering and unit factors. If $x$ is a unit, take $k = 0$, $u = x$. If $x$ is a nonzero non-unit, UFD gives the factorization directly. Irreducibles multiplied by a unit remain irreducible, so writing $x = \pi_1 \cdots \pi_k$ and setting $u = 1$ produces the desired form. (A more hands-on proof independent of the UFD assumption: since $|N(x)|$ is a positive integer, we can argue by strong induction on $|N(x)|$, using multiplicativity of the norm and [Units via Norm](/theorems/1578) to bound the irreducible factors. But we will use the UFD hypothesis here since we assume it.) [/proof] By the claim, the fixed $x \in \mathfrak{p}$, $x \neq 0$, factors as $x = u \pi_1 \cdots \pi_k$. Since $\mathfrak{p}$ is prime and $x \in \mathfrak{p}$, by iterated application of the defining property of prime ideals (if $yz \in \mathfrak{p}$ then $y \in \mathfrak{p}$ or $z \in \mathfrak{p}$), some factor $u, \pi_1, \ldots, \pi_k$ lies in $\mathfrak{p}$. The unit $u$ cannot lie in $\mathfrak{p}$ (otherwise $\mathfrak{p}$ would contain a unit, hence equal $\mathcal{O}_L$, contradicting $\mathfrak{p}$ being a proper prime ideal). So some $\pi_i \in \mathfrak{p}$. Relabel so $\pi_i = \pi$. The principal ideal $\langle \pi \rangle \subseteq \mathfrak{p}$. [claim:In a UFD, an irreducible generates a prime ideal] Let $R$ be a UFD and $\pi \in R$ irreducible. Then $\langle \pi \rangle$ is a prime ideal. [/claim] [proof] First, $\langle \pi \rangle$ is proper: $\pi$ is not a unit (by definition of irreducible), so $1 \notin \langle \pi \rangle$, and $\langle \pi \rangle \neq R$. Suppose $ab \in \langle \pi \rangle$, i.e., $\pi \mid ab$, for some $a, b \in R$. We must show $a \in \langle \pi \rangle$ or $b \in \langle \pi \rangle$. Without loss of generality $a, b$ are nonzero: if $a = 0$, then $a = 0 \cdot \pi \in \langle \pi \rangle$ and the implication holds; similarly if $b = 0$. So assume $a, b \neq 0$. By the UFD hypothesis, $a = u_a \alpha_1 \cdots \alpha_m$, $b = u_b \beta_1 \cdots \beta_n$ with $u_a, u_b$ units and $\alpha_i, \beta_j$ irreducible. Then \begin{align*} ab = (u_a u_b) \alpha_1 \cdots \alpha_m \beta_1 \cdots \beta_n. \end{align*} The hypothesis $\pi \mid ab$ means $ab = \pi \cdot c$ for some $c \in R$. If $c = 0$, then $ab = 0$ forces $a = 0$ or $b = 0$ (integral domain), handled above. Otherwise $c$ factors as $c = u_c \gamma_1 \cdots \gamma_l$, so \begin{align*} (u_a u_b) \alpha_1 \cdots \alpha_m \beta_1 \cdots \beta_n = u_c \pi \gamma_1 \cdots \gamma_l. \end{align*} By uniqueness of factorization in a UFD (up to reordering and units), $\pi$ must be associate to some $\alpha_i$ or $\beta_j$. If $\pi \sim \alpha_i$, then $\pi \mid \alpha_i \mid a$, so $a \in \langle \pi \rangle$. If $\pi \sim \beta_j$, then $b \in \langle \pi \rangle$. Either way, the required disjunction holds. Hence $\langle \pi \rangle$ is prime. [/proof] Applying this claim with our $\pi \in \mathfrak{p}$: the ideal $\langle \pi \rangle$ is a nonzero prime ideal, and $\langle \pi \rangle \subseteq \mathfrak{p}$. By [Ring of Integers is a Dedekind Domain](/theorems/1582), every nonzero prime ideal in $\mathcal{O}_L$ is maximal. So $\langle \pi \rangle$ is a maximal ideal. From $\langle \pi \rangle \subseteq \mathfrak{p}$ and $\langle \pi \rangle$ maximal among proper ideals, together with $\mathfrak{p}$ proper, we get $\langle \pi \rangle = \mathfrak{p}$. Hence $\mathfrak{p}$ is principal.[/step]

[guided]**Goal.** Assuming $\mathcal{O}_L$ is a UFD, show every nonzero prime ideal $\mathfrak{p}$ is principal. **Strategy.** Find an irreducible element $\pi \in \mathfrak{p}$, show $\langle \pi \rangle$ is prime, and use the Dedekind property (nonzero primes are maximal) to conclude $\langle \pi \rangle = \mathfrak{p}$. **Finding $\pi$.** Pick any nonzero $x \in \mathfrak{p}$. In a UFD, $x$ factors as $x = u \pi_1 \cdots \pi_k$ with $u$ a unit and $\pi_i$ irreducible. Since $\mathfrak{p}$ is prime and $x \in \mathfrak{p}$, primality forces some factor of $x$ into $\mathfrak{p}$. The unit $u$ cannot lie in a proper ideal. So some $\pi_i \in \mathfrak{p}$; relabel it as $\pi$. **Showing $\langle \pi \rangle$ is prime.** In a UFD, irreducible $\Leftrightarrow$ prime (the key property of UFDs that distinguishes them from general integral domains). Explicitly, if $\pi \mid ab$, factor $a, b$ into irreducibles, write $ab$ as a product, and use the unique-factorization property to match $\pi$ to a factor of $a$ or $b$. **Upgrading to equality.** We have $\langle \pi \rangle \subseteq \mathfrak{p}$, both nonzero. Both are prime ideals (the left by irreducible-to-prime in UFDs; the right by hypothesis). By $\mathcal{O}_L$ being a Dedekind domain ([Ring of Integers is a Dedekind Domain](/theorems/1582)), nonzero primes are maximal. Hence $\langle \pi \rangle$ is maximal. A maximal ideal contained in a proper ideal ($\mathfrak{p}$ is proper — prime ideals are proper by definition) must equal it. So $\langle \pi \rangle = \mathfrak{p}$: $\mathfrak{p}$ is principal, generated by $\pi$. **Why the Dedekind dimension-1 property is essential.** In a general UFD like $k[x, y]$, a nonzero prime may not be maximal: $\langle x \rangle \subsetneq \langle x, y \rangle \subsetneq k[x, y]$. There, the argument above gives $\langle \pi \rangle$ prime but not necessarily $= \mathfrak{p}$. For $\mathcal{O}_L$, the one-dimensionality ensures the containment upgrades to equality. This is why the theorem is specific to Dedekind domains.[/guided]

[step:Extend principality from prime ideals to all fractional ideals, closing (2) $\Rightarrow$ (1)]From Step 4, every nonzero prime ideal of $\mathcal{O}_L$ is principal. We show every nonzero fractional ideal is principal. By [Ideal Group Is Free Abelian](/theorems/1590), every $\mathfrak{q} \in I_L$ factors as \begin{align*} \mathfrak{q} = \prod_i \mathfrak{p}_i^{a_i} \end{align*} with $\mathfrak{p}_i$ distinct nonzero prime ideals and $a_i \in \mathbb{Z}$. By Step 4, each $\mathfrak{p}_i = \langle \pi_i \rangle$ for some $\pi_i \in \mathcal{O}_L$. In the group $I_L$, \begin{align*} \mathfrak{q} = \prod_i \langle \pi_i \rangle^{a_i} = \prod_i \langle \pi_i^{a_i} \rangle = \langle \prod_i \pi_i^{a_i} \rangle, \end{align*} using that (i) $\langle \pi \rangle^k = \langle \pi^k \rangle$ for $k \in \mathbb{Z}$, understanding $\langle \pi \rangle^{-1} = \langle \pi^{-1} \rangle$ in $L^\times$, and (ii) $\langle \alpha \rangle \langle \beta \rangle = \langle \alpha\beta \rangle$ for $\alpha, \beta \in L^\times$. Hence $\mathfrak{q}$ is principal in $I_L$, generated by $x := \prod_i \pi_i^{a_i} \in L^\times$. Thus $P_L = I_L$ and $\mathrm{Cl}_L = 0$. By the equivalence (1) $\Leftrightarrow$ (3) of Step 2, $\mathcal{O}_L$ is a PID. This closes (2) $\Rightarrow$ (1). Combining Steps 2–5: (1) $\Leftrightarrow$ (3) (Step 2), (1) $\Rightarrow$ (2) (Step 3), (2) $\Rightarrow$ (1) (Steps 4–5). Hence all three conditions are equivalent.[/step]

[guided]**From prime-to-principal to general-to-principal.** Step 4 does the hard work — every nonzero prime ideal is principal. The rest is a bookkeeping argument using the group structure of $I_L$. **The extension.** [Ideal Group Is Free Abelian](/theorems/1590) gives every $\mathfrak{q} \in I_L$ as a finite product $\prod_i \mathfrak{p}_i^{a_i}$ with $a_i \in \mathbb{Z}$. Rewrite each $\mathfrak{p}_i$ as $\langle \pi_i \rangle$ using Step 4. Then \begin{align*} \mathfrak{q} = \prod_i \langle \pi_i \rangle^{a_i}. \end{align*} **Computing in $I_L$.** The principal fractional ideals $P_L$ form a subgroup of $I_L$ isomorphic to $L^\times / \mathcal{O}_L^\times$ via $x \mapsto \langle x \rangle$. Products and integer powers of principal ideals stay in $P_L$: \begin{align*} \langle \alpha \rangle \langle \beta \rangle = \langle \alpha\beta \rangle, \qquad \langle \alpha \rangle^{k} = \langle \alpha^k \rangle \text{ for } k \in \mathbb{Z}, \end{align*} where $\alpha^k$ is computed in $L^\times$ (allowing negative exponents). Hence $\prod_i \langle \pi_i \rangle^{a_i} = \langle \prod_i \pi_i^{a_i} \rangle$, a principal fractional ideal. **Conclusion.** Every $\mathfrak{q} \in I_L$ is principal. So $I_L = P_L$, hence $\mathrm{Cl}_L = 0$. By the tautological equivalence (1) $\Leftrightarrow$ (3), $\mathcal{O}_L$ is a PID. **The tautology revisited.** The three conditions fit together as follows: \begin{align*} (1) \text{ PID} \;\stackrel{\text{def}}{\Longleftrightarrow}\; \text{every integral ideal principal} \;\Longleftrightarrow\; \text{every fractional ideal principal} \;\stackrel{\text{def}}{\Longleftrightarrow}\; (3) \mathrm{Cl}_L = 0. \end{align*} The only nontrivial equivalence is (2) $\Leftrightarrow$ (1), which splits into: - (1) $\Rightarrow$ (2): general "PID $\Rightarrow$ UFD" (Step 3). - (2) $\Rightarrow$ (1): Dedekind-specific argument that in a UFD, every prime ideal is principal, hence every ideal is principal (Steps 4–5). **Where Dedekind structure enters.** The (2) $\Rightarrow$ (1) direction uses two properties specific to Dedekind domains: - Nonzero primes are maximal (used to go from $\langle \pi \rangle \subseteq \mathfrak{p}$ to $\langle \pi \rangle = \mathfrak{p}$). - $I_L$ is free abelian on primes, so principality at primes extends to principality at all fractional ideals. In a general UFD neither holds. For instance $\mathbb{Z}[x]$ is a UFD but not a PID: the ideal $\langle 2, x \rangle$ is not principal. **Historical remark.** This equivalence was the starting point for Dedekind's theory of ideals: $\mathbb{Z}[\sqrt{-5}]$ is *not* a UFD at the element level ($6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$), and correspondingly $\mathrm{Cl}_{\mathbb{Q}(\sqrt{-5})} = \mathbb{Z}/2\mathbb{Z}$ is nontrivial. Passing to *ideals* restores unique factorization; the class group measures exactly the obstruction to unique factorization at the element level.[/guided]