[proofplan] The claim has two parts. First, nonzero fractional ideals form a group under multiplication: closure is automatic, associativity is inherited from the ring structure, the identity is $\mathcal{O}_L$, and invertibility is [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586). Second, the group is free abelian with basis the set of nonzero prime ideals: every nonzero fractional ideal $\mathfrak{q}$ can be written as a quotient of integral ideals $\mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1}$, and applying [Unique Factorization of Ideals](/theorems/1589) to each integral factor yields a unique expression $\mathfrak{q} = \prod_i \mathfrak{p}_i^{a_i}$ with $a_i \in \mathbb{Z}$. [/proofplan]

[step:Verify that nonzero fractional ideals form a group under multiplication]Let $I_L$ denote the set of nonzero fractional ideals of $\mathcal{O}_L$, together with the operation of ideal multiplication. We check the four group axioms. 1. **Closure.** For fractional ideals $\mathfrak{q}_1, \mathfrak{q}_2 \in I_L$, the product $\mathfrak{q}_1 \mathfrak{q}_2$ is the $\mathcal{O}_L$-submodule of $L$ generated by products $x_1 x_2$ with $x_i \in \mathfrak{q}_i$. Because each $\mathfrak{q}_i$ is a finitely generated $\mathcal{O}_L$-submodule of $L$ (the defining property of a fractional ideal), so is their product. Since $\mathfrak{q}_1, \mathfrak{q}_2 \neq 0$, there exist nonzero $x_i \in \mathfrak{q}_i$, and $x_1 x_2 \neq 0$ lies in $\mathfrak{q}_1 \mathfrak{q}_2$, so the product is nonzero. 2. **Associativity.** Inherited from $L$: for $\mathfrak{q}_1, \mathfrak{q}_2, \mathfrak{q}_3 \in I_L$, \begin{align*} (\mathfrak{q}_1 \mathfrak{q}_2) \mathfrak{q}_3 = \mathfrak{q}_1 (\mathfrak{q}_2 \mathfrak{q}_3) \end{align*} holds elementwise in $L$. 3. **Identity.** $\mathcal{O}_L$ is the identity: $\mathcal{O}_L \cdot \mathfrak{q} = \mathfrak{q}$ for every $\mathfrak{q} \in I_L$ since $1 \in \mathcal{O}_L$ gives $\mathfrak{q} \subseteq \mathcal{O}_L \mathfrak{q}$, and $\mathfrak{q}$ being an $\mathcal{O}_L$-module gives $\mathcal{O}_L \mathfrak{q} \subseteq \mathfrak{q}$. 4. **Inverses.** By [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586), for every $\mathfrak{q} \in I_L$ there exists a fractional ideal $\mathfrak{q}^{-1}$ (namely $\{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}$) with $\mathfrak{q} \mathfrak{q}^{-1} = \mathcal{O}_L$. Moreover, multiplication is commutative: $\mathfrak{q}_1 \mathfrak{q}_2 = \mathfrak{q}_2 \mathfrak{q}_1$, since $L$ is commutative. Hence $(I_L, \cdot)$ is an abelian group.[/step]

[guided]Recall the definition: a **fractional ideal** of $\mathcal{O}_L$ is a nonzero $\mathcal{O}_L$-submodule $\mathfrak{q} \subseteq L$ such that there exists $d \in \mathcal{O}_L \setminus \{0\}$ with $d\mathfrak{q} \subseteq \mathcal{O}_L$. Equivalently, $\mathfrak{q}$ is a finitely generated $\mathcal{O}_L$-submodule of $L$ (using Noetherianity of $\mathcal{O}_L$, since $d\mathfrak{q} \subseteq \mathcal{O}_L$ is finitely generated and multiplying by $d^{-1}$ preserves finite generation). We verify the group axioms on $I_L$. **Closure.** Given $\mathfrak{q}_1, \mathfrak{q}_2 \in I_L$, their product $\mathfrak{q}_1 \mathfrak{q}_2$ is defined as the $\mathcal{O}_L$-submodule generated by $\{x_1 x_2 : x_i \in \mathfrak{q}_i\}$; concretely, this is $\{\sum_k x_{1,k} x_{2,k} : x_{i,k} \in \mathfrak{q}_i, \text{ finite sum}\}$. - *Fractional-ness.* If $d_i \mathfrak{q}_i \subseteq \mathcal{O}_L$ for $d_i \in \mathcal{O}_L \setminus \{0\}$, then $(d_1 d_2)(\mathfrak{q}_1 \mathfrak{q}_2) \subseteq \mathcal{O}_L$, so $\mathfrak{q}_1 \mathfrak{q}_2$ is a fractional ideal. - *Nonzero.* Pick $0 \neq x_1 \in \mathfrak{q}_1$, $0 \neq x_2 \in \mathfrak{q}_2$. Since $L$ is a field, $x_1 x_2 \neq 0$, so $x_1 x_2 \in \mathfrak{q}_1 \mathfrak{q}_2$ shows the product is nonzero. Thus $\mathfrak{q}_1 \mathfrak{q}_2 \in I_L$. **Associativity and commutativity.** Both hold elementwise: for $x_i \in \mathfrak{q}_i$, \begin{align*} (x_1 x_2) x_3 = x_1 (x_2 x_3), \qquad x_1 x_2 = x_2 x_1, \end{align*} in the commutative ring $L$. Passing to generating sets gives the corresponding identities for ideals. **Identity $\mathcal{O}_L$.** $\mathcal{O}_L \in I_L$ (it is the "unit" fractional ideal — $d = 1$ works). For $\mathfrak{q} \in I_L$: - $\mathfrak{q} \subseteq \mathcal{O}_L \mathfrak{q}$: take $1 \cdot x$ for $x \in \mathfrak{q}$. - $\mathcal{O}_L \mathfrak{q} \subseteq \mathfrak{q}$: $\mathfrak{q}$ is an $\mathcal{O}_L$-submodule, so closed under multiplication by elements of $\mathcal{O}_L$. Thus $\mathcal{O}_L \mathfrak{q} = \mathfrak{q}$. **Inverses.** This is the substantive axiom. [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586) provides, for each $\mathfrak{q} \in I_L$, a fractional ideal $\mathfrak{q}^{-1} = \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}$ with $\mathfrak{q}\mathfrak{q}^{-1} = \mathcal{O}_L$. All axioms of an abelian group are verified, so $(I_L, \cdot)$ is an abelian group with identity $\mathcal{O}_L$.[/guided]

[step:Represent each nonzero fractional ideal as a quotient of nonzero integral ideals]Let $\mathfrak{q} \in I_L$. We show there exist nonzero integral ideals $\mathfrak{a}, \mathfrak{b} \unlhd \mathcal{O}_L$ with \begin{align*} \mathfrak{q} = \mathfrak{a} \mathfrak{b}^{-1}. \end{align*} By the definition of fractional ideal, there exists $d \in \mathcal{O}_L \setminus \{0\}$ with $d\mathfrak{q} \subseteq \mathcal{O}_L$. Set $\mathfrak{b} := \langle d \rangle = d\mathcal{O}_L$, a nonzero principal integral ideal, and set $\mathfrak{a} := d\mathfrak{q}$, which is a nonzero integral ideal (as $\mathfrak{a} \subseteq \mathcal{O}_L$ and $\mathfrak{a} \neq 0$ since $\mathfrak{q} \neq 0$ and $d \neq 0$). Then \begin{align*} \mathfrak{a} \mathfrak{b}^{-1} = (d\mathfrak{q})(d\mathcal{O}_L)^{-1} = (d\mathfrak{q}) \cdot (d^{-1} \mathcal{O}_L) = \mathfrak{q}, \end{align*} using $(d\mathcal{O}_L)^{-1} = d^{-1}\mathcal{O}_L$ (inverses of principal fractional ideals are principal with inverse generator).[/step]

[guided]Every nonzero fractional ideal comes within a bounded scalar multiple of an integral ideal. We make this precise by exhibiting a "denominator" and a "numerator". **The denominator $d$.** Being a fractional ideal, $\mathfrak{q}$ is finitely generated as an $\mathcal{O}_L$-module, say by $x_1, \ldots, x_m \in L$. Each $x_i = \alpha_i/\beta_i$ with $\alpha_i, \beta_i \in \mathcal{O}_L$, $\beta_i \neq 0$ (using $L = \operatorname{Frac}(\mathcal{O}_L)$ from [Fraction Field of the Ring of Integers](/theorems/1571)). Taking $d := \beta_1 \beta_2 \cdots \beta_m \in \mathcal{O}_L \setminus \{0\}$, we have $d x_i \in \mathcal{O}_L$ for every $i$, hence $d\mathfrak{q} \subseteq \mathcal{O}_L$. Alternatively, this "common denominator" property is *built into* the definition of a fractional ideal: one standard definition of fractional ideal is "an $\mathcal{O}_L$-submodule of $L$ admitting a nonzero $d$ with $d\mathfrak{q} \subseteq \mathcal{O}_L$". **The numerator $\mathfrak{a}$.** Set $\mathfrak{a} := d \mathfrak{q}$. Then $\mathfrak{a}$ is an $\mathcal{O}_L$-submodule of $L$ contained in $\mathcal{O}_L$, i.e., an integral ideal. It is nonzero since $\mathfrak{q} \neq 0$ and $d \neq 0$ and $L$ is a field. **The denominator as an ideal $\mathfrak{b}$.** Set $\mathfrak{b} := d\mathcal{O}_L = \langle d \rangle$, the principal integral ideal generated by $d$. Its inverse as a fractional ideal is $d^{-1}\mathcal{O}_L$ (one verifies $(d\mathcal{O}_L)(d^{-1}\mathcal{O}_L) = d \cdot d^{-1} \cdot \mathcal{O}_L \cdot \mathcal{O}_L = \mathcal{O}_L$, and inverses in an abelian group are unique). **Recovery of $\mathfrak{q}$.** Combining, \begin{align*} \mathfrak{a}\mathfrak{b}^{-1} = (d\mathfrak{q})(d^{-1}\mathcal{O}_L) = (dd^{-1}) (\mathfrak{q} \mathcal{O}_L) = 1 \cdot \mathfrak{q} = \mathfrak{q}, \end{align*} using commutativity and the identity property of $\mathcal{O}_L$. So $\mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1}$, a quotient of two integral ideals.[/guided]

[step:Factor the numerator and denominator into prime ideals and combine]By Step 2, $\mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1}$ with $\mathfrak{a}, \mathfrak{b}$ nonzero integral ideals. Apply [Unique Factorization of Ideals](/theorems/1589) separately to $\mathfrak{a}$ and $\mathfrak{b}$: \begin{align*} \mathfrak{a} = \prod_{i=1}^{t} \mathfrak{p}_i^{u_i}, \qquad \mathfrak{b} = \prod_{i=1}^{t} \mathfrak{p}_i^{v_i}, \end{align*} where $\mathfrak{p}_1, \ldots, \mathfrak{p}_t$ enumerate the distinct prime ideals appearing in either factorization (padded with zero exponents where a prime is absent), and $u_i, v_i \geq 0$. Multiplying the factorization for $\mathfrak{a}$ by the inverse of that for $\mathfrak{b}$ in the group $I_L$, \begin{align*} \mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1} = \left(\prod_i \mathfrak{p}_i^{u_i}\right) \left(\prod_i \mathfrak{p}_i^{v_i}\right)^{-1} = \prod_i \mathfrak{p}_i^{u_i - v_i} = \prod_i \mathfrak{p}_i^{a_i} \end{align*} with $a_i := u_i - v_i \in \mathbb{Z}$. Dropping primes for which $a_i = 0$ gives \begin{align*} \mathfrak{q} = \mathfrak{p}_{i_1}^{a_{i_1}} \cdots \mathfrak{p}_{i_r}^{a_{i_r}} \end{align*} with distinct primes and $a_{i_k} \in \mathbb{Z} \setminus \{0\}$. This is the claimed representation.[/step]

[guided]**Factoring the parts.** By [Unique Factorization of Ideals](/theorems/1589), every nonzero integral ideal factors uniquely as a product of prime ideals. Apply this separately to $\mathfrak{a}$ and $\mathfrak{b}$. Let $S = \{\mathfrak{p}_1, \ldots, \mathfrak{p}_t\}$ be the finite set of primes appearing in the factorization of $\mathfrak{a}$ *or* $\mathfrak{b}$ (we take the union of primes, padding with zero exponents if needed). Then uniquely \begin{align*} \mathfrak{a} = \mathfrak{p}_1^{u_1} \cdots \mathfrak{p}_t^{u_t}, \qquad \mathfrak{b} = \mathfrak{p}_1^{v_1} \cdots \mathfrak{p}_t^{v_t}, \qquad u_i, v_i \in \mathbb{Z}_{\geq 0}. \end{align*} **Inverting the denominator.** In the group $I_L$, inverses distribute over products (since the group is abelian): \begin{align*} \mathfrak{b}^{-1} = \left(\prod_{i=1}^t \mathfrak{p}_i^{v_i}\right)^{-1} = \prod_{i=1}^t \mathfrak{p}_i^{-v_i}. \end{align*} This uses that $\mathfrak{p}_i^{-1} = (\mathfrak{p}_i^{-1})$ is well-defined in $I_L$ (by Step 1), and $(\mathfrak{p}_i^{v_i})^{-1} = \mathfrak{p}_i^{-v_i}$ in any abelian group. **Combining.** In the abelian group $I_L$, \begin{align*} \mathfrak{q} = \mathfrak{a}\mathfrak{b}^{-1} = \prod_{i=1}^t \mathfrak{p}_i^{u_i} \cdot \prod_{i=1}^t \mathfrak{p}_i^{-v_i} = \prod_{i=1}^t \mathfrak{p}_i^{u_i - v_i}. \end{align*} Setting $a_i := u_i - v_i \in \mathbb{Z}$, this is $\mathfrak{q} = \prod_i \mathfrak{p}_i^{a_i}$. **Sign of $a_i$.** Note $a_i$ can be negative: this happens precisely when the prime $\mathfrak{p}_i$ divides the "denominator" $\mathfrak{b}$ to higher order than the "numerator" $\mathfrak{a}$. An $a_i > 0$ means $\mathfrak{p}_i$ appears in the integral part; $a_i < 0$ means it appears in the fractional part; $a_i = 0$ means $\mathfrak{p}_i$ has no net contribution, and we may drop it from the representation. **Integral ideal iff all exponents nonnegative.** If $\mathfrak{q} \subseteq \mathcal{O}_L$, then in particular $\mathfrak{q}$ is itself an integral ideal; applying unique factorization to $\mathfrak{q}$ directly yields nonnegative exponents. Conversely, a product $\prod \mathfrak{p}_i^{a_i}$ with all $a_i \geq 0$ is integral, being a product of integral ideals. We collect this remark since the theorem statement claims "$\mathfrak{q}$ integral iff $a_i \geq 0$ for all $i$".[/guided]

[step:Prove uniqueness of the exponents $(a_i)$]We show that the integer exponents $(a_i)$ in the representation $\mathfrak{q} = \prod_i \mathfrak{p}_i^{a_i}$ (with distinct primes and $a_i \in \mathbb{Z}$) are unique. Suppose \begin{align*} \mathfrak{q} = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_t^{a_t} = \mathfrak{p}_1^{b_1} \cdots \mathfrak{p}_t^{b_t} \end{align*} (by taking the union of the primes appearing in each side, padding with zero exponents, we may assume the same set of primes). Rearranging using the group structure of $I_L$, \begin{align*} \mathfrak{p}_1^{a_1 - b_1} \cdots \mathfrak{p}_t^{a_t - b_t} = \mathcal{O}_L. \end{align*} Split into positive and negative exponents: let $P = \{i : a_i - b_i > 0\}$ and $N = \{i : a_i - b_i < 0\}$. Set $c_i = a_i - b_i$ for $i \in P$ and $d_i = b_i - a_i$ for $i \in N$ (so $c_i, d_i > 0$). Multiplying by $\prod_{i \in N} \mathfrak{p}_i^{d_i}$: \begin{align*} \prod_{i \in P} \mathfrak{p}_i^{c_i} = \prod_{i \in N} \mathfrak{p}_i^{d_i}. \end{align*} Both sides are products of *nonnegative* powers of prime ideals, i.e., integral ideals. By [Unique Factorization of Ideals](/theorems/1589), the multisets of prime factors agree. But $P$ and $N$ are disjoint sets of distinct primes, so the only way for two products of primes to have matching multisets with the factors pulled from disjoint sets is for both sides to be the empty product. Hence $P = N = \varnothing$, which means $a_i = b_i$ for all $i$.[/step]

[guided]**Goal.** Given two integer-exponent representations of the same fractional ideal, show the exponents match. **Setup.** Write $\mathfrak{q} = \prod_i \mathfrak{p}_i^{a_i} = \prod_i \mathfrak{p}_i^{b_i}$; after taking the union of primes in the two factorizations and padding absent primes with exponent zero, we assume the same indexing set. **Reducing to the identity.** In the group $I_L$ (Step 1), \begin{align*} \mathcal{O}_L = \mathfrak{q} \cdot \mathfrak{q}^{-1} = \prod_i \mathfrak{p}_i^{a_i} \cdot \prod_i \mathfrak{p}_i^{-b_i} = \prod_i \mathfrak{p}_i^{a_i - b_i}. \end{align*} Setting $e_i := a_i - b_i \in \mathbb{Z}$, we must show \begin{align*} \prod_i \mathfrak{p}_i^{e_i} = \mathcal{O}_L \implies e_i = 0 \text{ for all } i. \end{align*} **Separating signs.** Split the index set: $P := \{i : e_i > 0\}$, $N := \{i : e_i < 0\}$. With $c_i := e_i$ for $i \in P$, $d_i := -e_i$ for $i \in N$ (both positive integers): \begin{align*} \prod_{i \in P} \mathfrak{p}_i^{c_i} \cdot \prod_{i \in N} \mathfrak{p}_i^{-d_i} = \mathcal{O}_L. \end{align*} Multiplying by $\prod_{i \in N} \mathfrak{p}_i^{d_i}$ (an integral ideal): \begin{align*} \prod_{i \in P} \mathfrak{p}_i^{c_i} = \prod_{i \in N} \mathfrak{p}_i^{d_i}. \end{align*} Both sides are integral ideals (products of nonnegative prime powers), and they are equal. **Applying unique factorization.** [Unique Factorization of Ideals](/theorems/1589) tells us: an integral ideal has a unique multiset of prime factors. The left-hand side has $\mathfrak{p}_i$ (for $i \in P$) appearing $c_i$ times; the right-hand side has $\mathfrak{p}_i$ (for $i \in N$) appearing $d_i$ times. Since $P \cap N = \varnothing$ (an index cannot simultaneously have $e_i > 0$ and $e_i < 0$) and all the $\mathfrak{p}_i$ are distinct, the multisets agree only if both are empty: $P = N = \varnothing$ and all $c_i, d_i = 0$. Hence $e_i = 0$ for all $i$, i.e., $a_i = b_i$ for all $i$, proving uniqueness. **Remark on the integral-iff-nonnegative characterization.** If $\mathfrak{q}$ is integral, it has a unique prime factorization with nonnegative exponents by [Unique Factorization of Ideals](/theorems/1589); by the uniqueness just proved, this is the *same* factorization as any integer-exponent representation, so all exponents are $\geq 0$. Conversely, all exponents $\geq 0$ gives $\mathfrak{q} = \prod \mathfrak{p}_i^{a_i}$ as a product of integral ideals, which is integral.[/guided]

[step:Assemble the group-theoretic conclusion: $I_L$ is free abelian on the prime ideals]Steps 1–4 together establish the theorem: - $I_L$ is an abelian group (Step 1). - Every $\mathfrak{q} \in I_L$ is a product $\prod_i \mathfrak{p}_i^{a_i}$ with integer exponents (Steps 2–3). - The exponents $(a_i)$ are unique (Step 4). This says: the map \begin{align*} \bigoplus_{\mathfrak{p} \text{ prime}} \mathbb{Z} &\to I_L \\ (a_\mathfrak{p})_{\mathfrak{p}} &\mapsto \prod_{\mathfrak{p}} \mathfrak{p}^{a_\mathfrak{p}} \end{align*} is an isomorphism of abelian groups. Hence $I_L$ is a free abelian group with basis the set of nonzero prime ideals. The integral ideals correspond to the subset $\bigoplus \mathbb{Z}_{\geq 0}$, i.e., tuples with all nonnegative components.[/step]

[guided]We collect the three ingredients. 1. **$I_L$ is an abelian group under ideal multiplication** (Step 1) — closure, associativity, commutativity, identity $\mathcal{O}_L$, inverses from [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586). 2. **Every $\mathfrak{q} \in I_L$ is expressible as $\prod \mathfrak{p}_i^{a_i}$ with $a_i \in \mathbb{Z}$** (Steps 2–3) — write $\mathfrak{q}$ as a quotient of integral ideals, factor each using [Unique Factorization of Ideals](/theorems/1589), combine with signed exponents. 3. **The exponent sequence $(a_i)$ is unique** (Step 4) — follows from uniqueness of prime factorization of integral ideals. Phrased group-theoretically: there is a map \begin{align*} \Phi: \bigoplus_{\mathfrak{p} \text{ prime}} \mathbb{Z} &\to I_L \\ (a_\mathfrak{p}) &\mapsto \prod_{\mathfrak{p}} \mathfrak{p}^{a_\mathfrak{p}}, \end{align*} where the direct sum is over the (possibly infinite) set of nonzero prime ideals of $\mathcal{O}_L$, and the support of any element is finite by definition of direct sum. The map $\Phi$ is: - A homomorphism: $\Phi((a_\mathfrak{p}) + (b_\mathfrak{p})) = \prod \mathfrak{p}^{a_\mathfrak{p} + b_\mathfrak{p}} = (\prod \mathfrak{p}^{a_\mathfrak{p}})(\prod \mathfrak{p}^{b_\mathfrak{p}}) = \Phi((a_\mathfrak{p})) \Phi((b_\mathfrak{p}))$, using commutativity and the laws of exponents in an abelian group. - Surjective: Steps 2–3 show every $\mathfrak{q}$ is in the image. - Injective: Step 4 shows the exponent sequence is unique, i.e., $\Phi((a_\mathfrak{p})) = \mathcal{O}_L$ forces $a_\mathfrak{p} = 0$ for all $\mathfrak{p}$. So $\Phi$ is an isomorphism of abelian groups. **Restatement.** $I_L$ is the free abelian group on the set of nonzero prime ideals of $\mathcal{O}_L$. **Filtration by integral ideals.** Integral ideals correspond under $\Phi$ to tuples in $\bigoplus \mathbb{Z}_{\geq 0}$, a submonoid of $\bigoplus \mathbb{Z}$. This recovers [Unique Factorization of Ideals](/theorems/1589) as the statement that $\Phi$ restricts to a bijection on the "positive cone". **Consequence — valuations.** Given a fixed prime $\mathfrak{p}$, the map $v_\mathfrak{p}: I_L \to \mathbb{Z}$, $\mathfrak{q} \mapsto a_\mathfrak{p}$ (the exponent of $\mathfrak{p}$), is a group homomorphism. It is the **$\mathfrak{p}$-adic valuation**. Its restriction to nonzero elements $L^\times \to \mathbb{Z}$ (via $x \mapsto \langle x \rangle$, then reading off the $\mathfrak{p}$-exponent) is a discrete valuation on $L$. The collection of these for all $\mathfrak{p}$ determines a fractional ideal completely, which is the "finite-adele" reformulation of the theorem.[/guided]