[proofplan] We first reduce invertibility of a general nonzero fractional ideal to invertibility of an integral ideal: any fractional $\mathfrak{q}$ has the form $\frac{1}{c}\mathfrak{a}$ for some integral $\mathfrak{a}$ and $c \in \mathcal{O}_L \setminus \{0\}$, and $\mathfrak{q}$ is invertible iff $\mathfrak{a}$ is. The core is then a Noetherian induction on integral ideals. Suppose there were an integral ideal not invertible; by Noetherianity choose $\mathfrak{a}$ maximal with this defect. Let $\mathfrak{b} = \{x \in L : x\mathfrak{a} \subseteq \mathcal{O}_L\}$. Using [Stability Implies Integrality](/theorems/1585), $\mathfrak{b}$ strictly contains $\mathcal{O}_L$, so $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$. A second invocation of Theorem 1585 gives $\mathfrak{a} \neq \mathfrak{a}\mathfrak{b}$. If $\mathfrak{a}\mathfrak{b} = \mathcal{O}_L$ we are done; otherwise $\mathfrak{a}\mathfrak{b}$ is a strictly larger proper integral ideal, which by maximality of $\mathfrak{a}$ must be invertible, and its inverse multiplied by $\mathfrak{b}$ inverts $\mathfrak{a}$ — contradiction. Finally, once invertibility is known, the formula for the inverse follows from a chain of two inclusions forced to be equalities by $\mathcal{O}_L = \mathfrak{q} \mathfrak{q}^{-1}$. [/proofplan] [step:Reduce to the case of integral ideals] Let $\mathfrak{q}$ be a nonzero fractional ideal of $\mathcal{O}_L$. By definition of a fractional ideal, there exists $c \in \mathcal{O}_L \setminus \{0\}$ with $c\mathfrak{q} \subseteq \mathcal{O}_L$; the set \begin{align*} \mathfrak{a} &:= c \mathfrak{q} \subseteq \mathcal{O}_L \end{align*} is therefore an integral ideal, and it is nonzero because $\mathfrak{q} \neq 0$ and $c \neq 0$ (and $L$ is a field). Hence $\mathfrak{q} = c^{-1} \mathfrak{a}$ inside $L$. If $\mathfrak{a}$ is invertible with inverse $\mathfrak{a}^{-1}$, then $\mathfrak{q} \cdot (c \mathfrak{a}^{-1}) = c^{-1} \mathfrak{a} \cdot c \mathfrak{a}^{-1} = \mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$, so $\mathfrak{q}$ is invertible with inverse $c\mathfrak{a}^{-1}$ (a fractional ideal of $\mathcal{O}_L$, since $\mathfrak{a}^{-1}$ is and scaling by $c \in L^\times$ preserves the fractional-ideal property). Hence it suffices to prove that every **nonzero integral ideal** $\mathfrak{a} \unlhd \mathcal{O}_L$ is invertible. [guided] A nonzero fractional ideal $\mathfrak{q}$ of $\mathcal{O}_L$ is, by definition, a finitely generated nonzero $\mathcal{O}_L$-submodule of $L$. Equivalently, it is a nonzero $\mathcal{O}_L$-submodule $\mathfrak{q} \subseteq L$ for which there exists a common denominator: some $c \in \mathcal{O}_L \setminus \{0\}$ with $c\mathfrak{q} \subseteq \mathcal{O}_L$. **Reduction strategy.** Given such $\mathfrak{q}$, we can scale it into $\mathcal{O}_L$ by the common denominator $c$: \begin{align*} \mathfrak{a} := c \mathfrak{q} = \{c\xi : \xi \in \mathfrak{q}\} \subseteq \mathcal{O}_L. \end{align*} This $\mathfrak{a}$ is an $\mathcal{O}_L$-submodule of $\mathcal{O}_L$, i.e., an integral ideal. It is nonzero: if $c\mathfrak{q} = \{0\}$ then every $\xi \in \mathfrak{q}$ satisfies $c\xi = 0$ in the field $L$, forcing $\xi = 0$ (since $c \neq 0$), contradicting $\mathfrak{q} \neq \{0\}$. **Transferring invertibility.** We have $\mathfrak{q} = c^{-1} \mathfrak{a}$ in the fractional-ideal algebra. Invertibility transfers easily: if $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$, then \begin{align*} \mathfrak{q} \cdot (c\mathfrak{a}^{-1}) &= (c^{-1} \mathfrak{a}) \cdot (c \mathfrak{a}^{-1}) = c^{-1} c \cdot \mathfrak{a} \mathfrak{a}^{-1} = \mathcal{O}_L. \end{align*} The set $c \mathfrak{a}^{-1} \subseteq L$ is a nonzero $\mathcal{O}_L$-submodule (scalar multiplication by $c$ is an $\mathcal{O}_L$-module automorphism of $L$, preserving ideal structure), hence a fractional ideal, serving as the inverse of $\mathfrak{q}$. In the other direction, if $\mathfrak{q}$ is invertible with inverse $\mathfrak{q}^{-1}$, then $\mathfrak{a} \cdot (c^{-1} \mathfrak{q}^{-1}) = (c\mathfrak{q}) \cdot (c^{-1} \mathfrak{q}^{-1}) = \mathfrak{q}\mathfrak{q}^{-1} = \mathcal{O}_L$, so $\mathfrak{a}$ is invertible. Thus invertibility of fractional ideals is equivalent to invertibility of their integral scalings. **Reduced problem.** It suffices to prove: every nonzero integral ideal $\mathfrak{a} \unlhd \mathcal{O}_L$ is invertible. [/guided] [/step] [step:Set up the Noetherian induction — suppose $\mathfrak{a}$ is maximal among non-invertible integral ideals] Suppose for contradiction there exists a nonzero integral ideal that is not invertible. Let \begin{align*} \mathcal{N} &:= \{\mathfrak{a} \unlhd \mathcal{O}_L : \mathfrak{a} \neq 0, \mathfrak{a} \text{ is not invertible}\}. \end{align*} By hypothesis $\mathcal{N} \neq \varnothing$. By [Ring of Integers is a Dedekind Domain](/theorems/1582), $\mathcal{O}_L$ is Noetherian — every nonempty set of ideals has a maximal element under inclusion. So $\mathcal{N}$ has a maximal element $\mathfrak{a}$. Note $\mathfrak{a} \neq \mathcal{O}_L$: the unit ideal $\mathcal{O}_L$ is invertible (its own inverse), hence $\mathcal{O}_L \notin \mathcal{N}$. So $\mathfrak{a}$ is a nonzero proper integral ideal that is not invertible, and every nonzero integral ideal strictly larger than $\mathfrak{a}$ is either invertible or equals $\mathcal{O}_L$ (itself invertible). [guided] This is the setup for a Noetherian induction (or equivalently, induction on the partial order of ideals). The strategy: - Assume the conclusion fails: there is a non-invertible nonzero integral ideal. - Use Noetherianity to pick one that is maximal with this property. - Derive a contradiction. **Why Noetherianity supplies a maximal counterexample.** A ring is Noetherian iff every nonempty family of ideals has a maximal element under inclusion. (Equivalently, the ascending chain condition: any increasing sequence of ideals stabilizes.) For $\mathcal{O}_L$ Noetherian, see [Ring of Integers is a Dedekind Domain](/theorems/1582). Apply this to the family $\mathcal{N}$ of nonzero non-invertible ideals, assumed nonempty. A maximal element $\mathfrak{a}$ exists. **Properties of $\mathfrak{a}$.** 1. $\mathfrak{a}$ is a nonzero integral ideal (by $\mathfrak{a} \in \mathcal{N}$). 2. $\mathfrak{a}$ is not invertible (by $\mathfrak{a} \in \mathcal{N}$). 3. $\mathfrak{a} \neq \mathcal{O}_L$: because $\mathcal{O}_L \cdot \mathcal{O}_L = \mathcal{O}_L$, so $\mathcal{O}_L$ is invertible, hence $\mathcal{O}_L \notin \mathcal{N}$. 4. Any nonzero integral ideal $\mathfrak{a}' \supsetneq \mathfrak{a}$ is *not* in $\mathcal{N}$ (by maximality of $\mathfrak{a}$ in $\mathcal{N}$). Since $\mathfrak{a}'$ is nonzero, $\mathfrak{a}' \notin \mathcal{N}$ means $\mathfrak{a}'$ is invertible. In the following steps, we construct an ideal $\mathfrak{a}\mathfrak{b}$ strictly larger than $\mathfrak{a}$, verify it is a nonzero proper integral ideal, and then use its invertibility (from item 4) to deduce invertibility of $\mathfrak{a}$ — contradicting item 2. [/guided] [/step] [step:Construct the "candidate inverse" $\mathfrak{b}$ and show $\mathcal{O}_L \subsetneq \mathfrak{b}$ via Stability Implies Integrality] Define \begin{align*} \mathfrak{b} &:= \{x \in L : x\mathfrak{a} \subseteq \mathcal{O}_L\}. \end{align*} This is a nonzero $\mathcal{O}_L$-submodule of $L$ (it is closed under $\mathcal{O}_L$-linear combinations: if $x_1, x_2 \in \mathfrak{b}$ and $c_1, c_2 \in \mathcal{O}_L$, then $(c_1 x_1 + c_2 x_2)\mathfrak{a} \subseteq c_1 \mathcal{O}_L + c_2 \mathcal{O}_L \subseteq \mathcal{O}_L$). It is a fractional ideal: any nonzero $\alpha \in \mathfrak{a}$ satisfies $\alpha \mathfrak{b} \subseteq \mathcal{O}_L$, so $\alpha$ is a common denominator. **$\mathcal{O}_L \subseteq \mathfrak{b}$.** For any $c \in \mathcal{O}_L$, we have $c\mathfrak{a} \subseteq \mathfrak{a} \subseteq \mathcal{O}_L$, so $c \in \mathfrak{b}$. **$\mathcal{O}_L \subsetneq \mathfrak{b}$.** Apply part (2) of [Stability Implies Integrality](/theorems/1585): since $\mathfrak{a}$ is a proper ideal (by Step 2), the set $\{y \in L : y\mathfrak{a} \subseteq \mathcal{O}_L\} = \mathfrak{b}$ strictly contains $\mathcal{O}_L$. So there exists $y \in \mathfrak{b} \setminus \mathcal{O}_L$. [guided] Having fixed our maximal non-invertible $\mathfrak{a}$, we construct the "would-be inverse" $\mathfrak{b}$ as the largest possible set of $L$-elements that multiply $\mathfrak{a}$ back into $\mathcal{O}_L$. The first key observation is that $\mathfrak{b}$ is strictly bigger than $\mathcal{O}_L$ — this is exactly what [Stability Implies Integrality](/theorems/1585) gives us. **Definition of $\mathfrak{b}$.** \begin{align*} \mathfrak{b} &= \{x \in L : x\mathfrak{a} \subseteq \mathcal{O}_L\}. \end{align*} Intuitively, $\mathfrak{b}$ consists of the "$\mathcal{O}_L$-rationals" of $\mathfrak{a}^{-1}$: if $\mathfrak{a}$ were invertible with inverse $\mathfrak{a}^{-1}$, then $\mathfrak{a}^{-1}$ would consist precisely of elements sending $\mathfrak{a}$ into $\mathcal{O}_L$ (more precisely, sending $\mathfrak{a}$ onto $\mathcal{O}_L$). So $\mathfrak{b}$ is a natural "maximal candidate inverse". **Is $\mathfrak{b}$ a fractional ideal?** *$\mathcal{O}_L$-submodule of $L$:* if $x_1, x_2 \in \mathfrak{b}$ and $c_1, c_2 \in \mathcal{O}_L$, then for any $a \in \mathfrak{a}$, \begin{align*} (c_1 x_1 + c_2 x_2) \cdot a &= c_1 (x_1 a) + c_2 (x_2 a) \in c_1 \mathcal{O}_L + c_2 \mathcal{O}_L \subseteq \mathcal{O}_L. \end{align*} So $c_1 x_1 + c_2 x_2 \in \mathfrak{b}$. *Common denominator:* pick any nonzero $\alpha \in \mathfrak{a}$ (possible since $\mathfrak{a} \neq \{0\}$). For $x \in \mathfrak{b}$, $x\alpha \in x\mathfrak{a} \subseteq \mathcal{O}_L$, so $\alpha \cdot x \in \mathcal{O}_L$ for every $x \in \mathfrak{b}$, i.e., $\alpha \mathfrak{b} \subseteq \mathcal{O}_L$. Hence $\alpha$ is a common denominator for $\mathfrak{b}$, verifying $\mathfrak{b}$ is a fractional ideal. *Nonzero:* $\mathcal{O}_L \subseteq \mathfrak{b}$ (see below), so $1 \in \mathfrak{b}$ and $\mathfrak{b} \neq \{0\}$. **$\mathcal{O}_L \subseteq \mathfrak{b}$.** For any $c \in \mathcal{O}_L$, multiplication by $c$ sends $\mathfrak{a}$ into $\mathfrak{a}$ (since $\mathfrak{a}$ is an ideal), hence in particular into $\mathcal{O}_L$. So $c \in \mathfrak{b}$. **$\mathcal{O}_L \subsetneq \mathfrak{b}$ — the crucial strict containment.** This is where we invoke [Stability Implies Integrality](/theorems/1585), part (2). That theorem states: > If $\mathfrak{a}$ is a proper ideal, then $\{y \in L : y\mathfrak{a} \subseteq \mathcal{O}_L\}$ strictly contains $\mathcal{O}_L$. **Hypothesis verification.** $\mathfrak{a}$ is a nonzero proper ideal (from Step 2, $\mathfrak{a} \neq \mathcal{O}_L$ and $\mathfrak{a} \neq 0$). Theorem 1585(2) applies. **Conclusion.** $\{y \in L : y\mathfrak{a} \subseteq \mathcal{O}_L\} \supsetneq \mathcal{O}_L$. But this set is exactly $\mathfrak{b}$. So $\mathfrak{b} \supsetneq \mathcal{O}_L$. In particular, there exists $y_0 \in \mathfrak{b}$ with $y_0 \notin \mathcal{O}_L$. This "extra" element will drive the strict containment $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b}$ in the next step. [/guided] [/step] [step:Derive $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$, again using Stability Implies Integrality] Consider $\mathfrak{a}\mathfrak{b}$ — the fractional ideal generated by products of elements of $\mathfrak{a}$ and $\mathfrak{b}$. **$\mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$** by definition of $\mathfrak{b}$: for any $\xi \in \mathfrak{a}\mathfrak{b}$, $\xi = \sum_i a_i x_i$ with $a_i \in \mathfrak{a}$, $x_i \in \mathfrak{b}$, and $x_i \mathfrak{a} \subseteq \mathcal{O}_L$ gives $x_i a_i \in \mathcal{O}_L$; summing, $\xi \in \mathcal{O}_L$. **$\mathfrak{a} \subseteq \mathfrak{a}\mathfrak{b}$** because $1 \in \mathcal{O}_L \subseteq \mathfrak{b}$, so $\mathfrak{a} = \mathfrak{a} \cdot 1 \subseteq \mathfrak{a}\mathfrak{b}$. **$\mathfrak{a} \neq \mathfrak{a}\mathfrak{b}$.** Suppose for contradiction $\mathfrak{a}\mathfrak{b} = \mathfrak{a}$. Then for every $x \in \mathfrak{b}$, $x\mathfrak{a} \subseteq \mathfrak{a}\mathfrak{b} = \mathfrak{a}$; i.e., $x$ satisfies the stability condition $x\mathfrak{a} \subseteq \mathfrak{a}$. By [Stability Implies Integrality](/theorems/1585), part (1), $x \in \mathcal{O}_L$. Hence $\mathfrak{b} \subseteq \mathcal{O}_L$, contradicting Step 3 ($\mathfrak{b} \supsetneq \mathcal{O}_L$). Hence $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$. [guided] We have $\mathfrak{a}$ and $\mathfrak{b}$, with $\mathcal{O}_L \subsetneq \mathfrak{b}$. We now examine $\mathfrak{a}\mathfrak{b}$. **Upper bound: $\mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$.** Any $\xi \in \mathfrak{a}\mathfrak{b}$ is a *finite* sum $\xi = \sum_i a_i x_i$ with $a_i \in \mathfrak{a}$, $x_i \in \mathfrak{b}$. For each $i$, since $x_i \in \mathfrak{b}$ and $a_i \in \mathfrak{a}$, we have $x_i a_i \in \mathcal{O}_L$ (definition of $\mathfrak{b}$). Summing, $\xi \in \mathcal{O}_L$. **Lower bound: $\mathfrak{a} \subseteq \mathfrak{a}\mathfrak{b}$.** We have $1 \in \mathcal{O}_L \subseteq \mathfrak{b}$. For any $a \in \mathfrak{a}$, $a = a \cdot 1 \in \mathfrak{a} \cdot \mathfrak{b}$. **Strict lower bound: $\mathfrak{a} \neq \mathfrak{a}\mathfrak{b}$.** Assume for contradiction $\mathfrak{a}\mathfrak{b} = \mathfrak{a}$. Then for every $x \in \mathfrak{b}$ and every $a \in \mathfrak{a}$, \begin{align*} xa \in x\mathfrak{a} \subseteq \mathfrak{a}\mathfrak{b} = \mathfrak{a}, \end{align*} so $x\mathfrak{a} \subseteq \mathfrak{a}$ — that is, $x$ satisfies the stability condition of [Stability Implies Integrality](/theorems/1585), part (1). **Hypothesis verification.** Theorem 1585(1) requires $\mathfrak{a}$ to be a nonzero ideal of $\mathcal{O}_L$, which holds by Step 2. **Conclusion of Theorem 1585(1).** $x \in \mathcal{O}_L$. Since $x \in \mathfrak{b}$ was arbitrary, $\mathfrak{b} \subseteq \mathcal{O}_L$ — contradicting Step 3's conclusion $\mathfrak{b} \supsetneq \mathcal{O}_L$. Hence $\mathfrak{a} \neq \mathfrak{a}\mathfrak{b}$, and combined with $\mathfrak{a} \subseteq \mathfrak{a}\mathfrak{b}$, \begin{align*} \mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L. \end{align*} **Structural remark.** The key interplay is that Theorem 1585's two parts act as a "pincer": - Part (2) ensures $\mathfrak{b}$ is *strictly bigger* than $\mathcal{O}_L$, giving some element that when multiplied into $\mathfrak{a}$ cannot stay inside $\mathfrak{a}$ (it might leave $\mathfrak{a}$ but still land in $\mathcal{O}_L$). - Part (1) forbids the scenario where $\mathfrak{b}$ — via multiplication — stays inside $\mathfrak{a}$; this is the contrapositive of "elements stabilizing $\mathfrak{a}$ are in $\mathcal{O}_L$." These two facts together force $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$. [/guided] [/step] [step:Invoke maximality of $\mathfrak{a}$ to deduce $\mathfrak{a}\mathfrak{b} = \mathcal{O}_L$ or extract a contradiction] We have $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subseteq \mathcal{O}_L$. Two cases: **Case A: $\mathfrak{a}\mathfrak{b} = \mathcal{O}_L$.** Then by definition, $\mathfrak{b}$ is a multiplicative inverse of $\mathfrak{a}$ in the monoid of fractional ideals. So $\mathfrak{a}$ is invertible, contradicting $\mathfrak{a} \in \mathcal{N}$. **Case B: $\mathfrak{a}\mathfrak{b} \subsetneq \mathcal{O}_L$.** Then $\mathfrak{a}\mathfrak{b}$ is a nonzero proper integral ideal strictly containing $\mathfrak{a}$ (by the strict containment from Step 4). By maximality of $\mathfrak{a}$ in $\mathcal{N}$, the ideal $\mathfrak{a}\mathfrak{b}$ is not in $\mathcal{N}$; since $\mathfrak{a}\mathfrak{b}$ is nonzero, this means $\mathfrak{a}\mathfrak{b}$ is invertible. Let $\mathfrak{c}$ be a fractional-ideal inverse: $(\mathfrak{a}\mathfrak{b})\mathfrak{c} = \mathcal{O}_L$. Then $\mathfrak{a} \cdot (\mathfrak{b}\mathfrak{c}) = \mathfrak{a}\mathfrak{b}\mathfrak{c} = \mathcal{O}_L$, so $\mathfrak{b}\mathfrak{c}$ is a fractional-ideal inverse of $\mathfrak{a}$. Hence $\mathfrak{a}$ is invertible, again contradicting $\mathfrak{a} \in \mathcal{N}$. In either case we obtain a contradiction. Therefore $\mathcal{N} = \varnothing$ — every nonzero integral ideal of $\mathcal{O}_L$ is invertible. Combined with Step 1's reduction, every nonzero fractional ideal is invertible. [guided] We have constructed $\mathfrak{a}\mathfrak{b}$ strictly between $\mathfrak{a}$ and $\mathcal{O}_L$. We now split into the two possible positions. **Case A: $\mathfrak{a}\mathfrak{b} = \mathcal{O}_L$.** This is the "easy" case: it says $\mathfrak{b}$ is an inverse of $\mathfrak{a}$ in the monoid of fractional ideals. But $\mathfrak{a}$ is *not* invertible (by its membership in $\mathcal{N}$), contradicting this. Case A yields a contradiction. **Case B: $\mathfrak{a}\mathfrak{b} \subsetneq \mathcal{O}_L$, i.e., $\mathfrak{a}\mathfrak{b}$ is a nonzero proper integral ideal.** By the strict inclusion $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b}$ from Step 4, the ideal $\mathfrak{a}\mathfrak{b}$ is strictly larger than $\mathfrak{a}$ in the poset of integral ideals. *Applying maximality of $\mathfrak{a}$ in $\mathcal{N}$:* $\mathfrak{a}\mathfrak{b} \notin \mathcal{N}$. Since $\mathfrak{a}\mathfrak{b}$ is nonzero (containing $\mathfrak{a}$, a nonzero ideal), the only way to not be in $\mathcal{N}$ is to be invertible. So $\mathfrak{a}\mathfrak{b}$ is invertible: there exists a fractional ideal $\mathfrak{c}$ with \begin{align*} (\mathfrak{a}\mathfrak{b}) \cdot \mathfrak{c} = \mathcal{O}_L. \end{align*} *Extracting an inverse for $\mathfrak{a}$:* multiplication of fractional ideals is associative (and commutative), so \begin{align*} \mathfrak{a} \cdot (\mathfrak{b}\mathfrak{c}) = (\mathfrak{a}\mathfrak{b}) \cdot \mathfrak{c} = \mathcal{O}_L. \end{align*} The product $\mathfrak{b}\mathfrak{c}$ is a fractional ideal (the product of two fractional ideals in the fractional-ideal monoid), serving as an inverse of $\mathfrak{a}$. So $\mathfrak{a}$ is invertible — but $\mathfrak{a} \in \mathcal{N}$ says $\mathfrak{a}$ is not invertible. Contradiction. **Conclusion.** Both cases yield a contradiction, so the assumption $\mathcal{N} \neq \varnothing$ must fail. Hence $\mathcal{N} = \varnothing$: every nonzero integral ideal of $\mathcal{O}_L$ is invertible. By Step 1, every nonzero fractional ideal is invertible. **Structural remark.** Although the final argument uses a contradiction, the intuition is constructive: for any nonzero integral ideal $\mathfrak{a}$, iterating the "multiply by $\mathfrak{b}$" operation produces a strictly ascending chain of ideals $\mathfrak{a} \subsetneq \mathfrak{a}\mathfrak{b} \subsetneq \mathfrak{a}\mathfrak{b}^2 \subsetneq \cdots$ (when $\mathfrak{a}$ is not yet invertible), which must terminate by Noetherianity — and the terminal state is $\mathcal{O}_L$, producing the inverse. [/guided] [/step] [step:Establish the formula $\mathfrak{q}^{-1} = \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}$] Let $\mathfrak{q}$ be a nonzero fractional ideal; by the preceding steps, $\mathfrak{q}$ is invertible. Denote by $\mathfrak{q}^{-1}$ its inverse (unique in the fractional-ideal monoid, since inverses in a commutative monoid are unique when they exist). Define \begin{align*} \mathfrak{c} &:= \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}, \end{align*} the set we have called $\mathfrak{b}$ above, applied now to $\mathfrak{q}$. We claim $\mathfrak{q}^{-1} = \mathfrak{c}$. **$\mathfrak{q}^{-1} \subseteq \mathfrak{c}$.** For any $x \in \mathfrak{q}^{-1}$ and $\xi \in \mathfrak{q}$, the element $x\xi$ lies in $\mathfrak{q}^{-1} \mathfrak{q} = \mathcal{O}_L$. Hence $x\mathfrak{q} \subseteq \mathcal{O}_L$, i.e., $x \in \mathfrak{c}$. **Sandwiching via multiplication by $\mathfrak{q}$.** Multiply the inclusion $\mathfrak{q}^{-1} \subseteq \mathfrak{c}$ by $\mathfrak{q}$ from the left: \begin{align*} \mathcal{O}_L = \mathfrak{q}\mathfrak{q}^{-1} \subseteq \mathfrak{q}\mathfrak{c}. \end{align*} But by definition of $\mathfrak{c}$, $\mathfrak{c}\mathfrak{q} \subseteq \mathcal{O}_L$, hence (by commutativity of the fractional-ideal monoid) $\mathfrak{q}\mathfrak{c} \subseteq \mathcal{O}_L$. Combining: \begin{align*} \mathcal{O}_L \subseteq \mathfrak{q}\mathfrak{c} \subseteq \mathcal{O}_L, \qquad \text{hence} \qquad \mathfrak{q}\mathfrak{c} = \mathcal{O}_L. \end{align*} **Uniqueness forces $\mathfrak{c} = \mathfrak{q}^{-1}$.** The equality $\mathfrak{q}\mathfrak{c} = \mathcal{O}_L$ says $\mathfrak{c}$ is a fractional-ideal inverse of $\mathfrak{q}$. Inverses are unique in a commutative monoid (if $\mathfrak{q}\mathfrak{c}_1 = \mathfrak{q}\mathfrak{c}_2 = \mathcal{O}_L$, multiplying by $\mathfrak{q}^{-1}$ gives $\mathfrak{c}_1 = \mathfrak{c}_2$). Therefore \begin{align*} \mathfrak{c} &= \mathfrak{q}^{-1}, \end{align*} i.e., $\mathfrak{q}^{-1} = \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}$ as claimed. [guided] We know $\mathfrak{q}$ is invertible; we now identify its inverse explicitly. The natural candidate is \begin{align*} \mathfrak{c} = \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\} \end{align*} — the largest set of $L$-elements that multiply $\mathfrak{q}$ into $\mathcal{O}_L$. We show $\mathfrak{c}$ is exactly the inverse. **First inclusion: $\mathfrak{q}^{-1} \subseteq \mathfrak{c}$.** The abstract inverse $\mathfrak{q}^{-1}$ satisfies $\mathfrak{q}\mathfrak{q}^{-1} = \mathcal{O}_L$. This means: for any $x \in \mathfrak{q}^{-1}$ and $\xi \in \mathfrak{q}$, the product $x\xi \in \mathfrak{q}\mathfrak{q}^{-1} = \mathcal{O}_L$. So $x\mathfrak{q} \subseteq \mathcal{O}_L$, i.e., $x \in \mathfrak{c}$. This gives $\mathfrak{q}^{-1} \subseteq \mathfrak{c}$. **Second inclusion: $\mathfrak{c} \subseteq \mathfrak{q}^{-1}$, i.e., $\mathfrak{c}$ is the inverse.** We show $\mathfrak{q}\mathfrak{c} = \mathcal{O}_L$, then invoke uniqueness of inverses. *$\mathfrak{q}\mathfrak{c} \supseteq \mathcal{O}_L$:* Since $\mathfrak{q}^{-1} \subseteq \mathfrak{c}$, multiplying both sides by $\mathfrak{q}$: \begin{align*} \mathcal{O}_L = \mathfrak{q}\mathfrak{q}^{-1} \subseteq \mathfrak{q}\mathfrak{c}. \end{align*} *$\mathfrak{q}\mathfrak{c} \subseteq \mathcal{O}_L$:* By definition of $\mathfrak{c}$, every $x \in \mathfrak{c}$ satisfies $x\mathfrak{q} \subseteq \mathcal{O}_L$. A general element of $\mathfrak{q}\mathfrak{c}$ has the form $\sum_i \xi_i x_i$ with $\xi_i \in \mathfrak{q}$, $x_i \in \mathfrak{c}$; each term $\xi_i x_i = x_i \xi_i \in x_i \mathfrak{q} \subseteq \mathcal{O}_L$, so the sum lies in $\mathcal{O}_L$. *Combining:* $\mathcal{O}_L \subseteq \mathfrak{q}\mathfrak{c} \subseteq \mathcal{O}_L$, giving $\mathfrak{q}\mathfrak{c} = \mathcal{O}_L$. **Uniqueness of inverses.** In the commutative monoid of nonzero fractional ideals (with multiplication and identity $\mathcal{O}_L$), if $\mathfrak{q}\mathfrak{c}_1 = \mathfrak{q}\mathfrak{c}_2 = \mathcal{O}_L$, then multiplying by $\mathfrak{q}^{-1}$ (which exists — we just showed it does): $\mathfrak{c}_1 = \mathfrak{q}^{-1}\mathfrak{q}\mathfrak{c}_1 = \mathfrak{q}^{-1}\mathcal{O}_L = \mathfrak{q}^{-1}$, and similarly $\mathfrak{c}_2 = \mathfrak{q}^{-1}$. So inverses are unique. Therefore $\mathfrak{c} = \mathfrak{q}^{-1}$. This is exactly the formula in the theorem statement: \begin{align*} \mathfrak{q}^{-1} &= \{x \in L : x\mathfrak{q} \subseteq \mathcal{O}_L\}. \end{align*} **Interpretation.** The formula says the inverse of $\mathfrak{q}$ is the "largest denominator set" — the set of all elements that can cancel a copy of $\mathfrak{q}$ into $\mathcal{O}_L$. Both inclusions $\supseteq$ (abstract inverse is certainly such a denominator) and $\subseteq$ (such a denominator, multiplied by $\mathfrak{q}$, lands exactly in $\mathcal{O}_L$) must hold. [/guided] [/step]